Tag Archives: maths

The Hill to Power

by in Base Behavior, Digit-sums, Mathematics, Narcissistic numbers, Powers and tagged , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

89 is special because it’s a prime number, divisible by only itself and 1. It’s also a sum of powers in a special way: 89 = 8^1 + 9^2. In base ten, no other two-digit number is equal to its own ascending power-sum like that. But the same pattern appears in these three-digit numbers, as the powers climb with the digits:

135 = 1^1 + 3^2 + 5^3 = 1 + 9 + 125 = 135
175 = 1^1 + 7^2 + 5^3 = 1 + 49 + 125 = 175
518 = 5^1 + 1^2 + 8^3 = 5 + 1 + 512 = 518
598 = 5^1 + 9^2 + 8^3 = 5 + 81 + 512 = 598

And in these four-digit numbers:

1306 = 1^1 + 3^2 + 0^3 + 6^4 = 1 + 9 + 0 + 1296 = 1306
1676 = 1^1 + 6^2 + 7^3 + 6^4 = 1 + 36 + 343 + 1296 = 1676
2427 = 2^1 + 4^2 + 2^3 + 7^4 = 2 + 16 + 8 + 2401 = 2427

The pattern doesn’t apply to any five-digit number in base-10 and six-digit numbers supply only this near miss:

263248 + 1 = 2^1 + 6^2 + 3^3 + 2^4 + 4^5 + 8^6 = 2 + 36 + 27 + 16 + 1024 + 262144 = 263249

But the pattern re-appears among seven-digit numbers:

2646798 = 2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7 = 2 + 36 + 64 + 1296 + 16807 + 531441 + 2097152 = 2646798

Now try some base behaviour. Some power-sums in base-10 are power-sums in another base:

175 = 1^1 + 7^2 + 5^3 = 1 + 49 + 125 = 175
175 = 6D[b=27] = 6^1 + 13^2 = 6 + 169 = 175

1306 = 1^1 + 3^2 + 0^3 + 6^4 = 1 + 9 + 0 + 1296 = 1306
1306 = A[36][b=127] = 10^1 + 36^2 = 10 + 1296 = 1306

Here is an incomplete list of double-base power-sums:

83 = 1103[b=4] = 1^1 + 1^2 + 0^3 + 3^4 = 1 + 1 + 0 + 81 = 83
83 = 29[b=37] = 2^1 + 9^2 = 2 + 81 = 83

126 = 105[b=11] = 1^1 + 0^2 + 5^3 = 1 + 0 + 125 = 126
126 = 5B[b=23] = 5^1 + 11^2 = 5 + 121 = 126

175 = 1^1 + 7^2 + 5^3 = 1 + 49 + 125 = 175
175 = 6D[b=27] = 6^1 + 13^2 = 6 + 169 = 175

259 = 2014[b=5] = 2^1 + 0^2 + 1^3 + 4^4 = 2 + 0 + 1 + 256 = 259
259 = 3G[b=81] = 3^1 + 16^2 = 3 + 256 = 259

266 = 176[b=13] = 1^1 + 7^2 + 6^3 = 1 + 49 + 216 = 266
266 = AG[b=25] = 10^1 + 16^2 = 10 + 256 = 266

578 = 288[b=15] = 2^1 + 8^2 + 8^3 = 2 + 64 + 512 = 578
578 = 2[24][b=277] = 2^1 + 24^2 = 2 + 576 = 578

580 = 488[b=11] = 4^1 + 8^2 + 8^3 = 4 + 64 + 512 = 580
580 = 4[24][b=139] = 4^1 + 24^2 = 4 + 576 = 580

731 = 209[b=19] = 2^1 + 0^2 + 9^3 = 2 + 0 + 729 = 731
731 = 2[27][b=352] = 2^1 + 27^2 = 2 + 729 = 731

735 = 609[b=11] = 6^1 + 0^2 + 9^3 = 6 + 0 + 729 = 735
735 = 6[27][b=118] = 6^1 + 27^2 = 6 + 729 = 735

1306 = 1^1 + 3^2 + 0^3 + 6^4 = 1 + 9 + 0 + 1296 = 1306
1306 = A[36][b=127] = 10^1 + 36^2 = 10 + 1296 = 1306

1852 = 3BC[b=23] = 3^1 + 11^2 + 12^3 = 3 + 121 + 1728 = 1852
1852 = 3[43][b=603] = 3^1 + 43^2 = 3 + 1849 = 1852

2943 = 3EE[b=29] = 3^1 + 14^2 + 14^3 = 3 + 196 + 2744 = 2943
2943 = [27][54][b=107] = 27^1 + 54^2 = 27 + 2916 = 2943

Previously pre-posted (please peruse):

• Narcissarithmetic #1
• Narcissarithmetic #2

Know Your Limaçons

by in Book Reviews, Fractals, Geometry, Mathematics and tagged , , , , , , , , , , , , , , , , , , , | Leave a comment

Front cover of The Penguin Dictionary of Curious and Interesting Geometry by David WellsThe Penguin Dictionary of Curious and Interesting Geometry, David Wells (1991)

Mathematics is an ocean in which a child can paddle and an elephant can swim. Or a whale, indeed. This book, a sequel to Wells’ excellent Penguin Dictionary of Curious and Interesting Mathematics, is suitable for both paddlers and plungers. Plumbers, even, because you can dive into some very deep mathematics here.

Far too deep for me, I have to admit, but I can wade a little way into the shallows and enjoy looking further out at what I don’t understand, because the advantage of geometry over number theory is that it can appeal to the eye even when it baffles the brain. If this book is more expensive than its prequel, that’s because it needs to be. It’s a paperback, but a large one, to accommodate the illustrations.

Fortunately, plenty of them appeal to the eye without baffling the brain, like the absurdly simple yet mindstretching Koch snowflake. Take a triangle and divide each side into thirds. Erect another triangle on each middle third. Take each new line of the shape and do the same: divide into thirds, erect another triangle on the middle third. Then repeat. And repeat. For ever.

A Koch snowflake (from Wikipedia)

A Koch snowflake (from Wikipedia)

The result is a shape with a finite area enclosed by an infinite perimeter, and it is in fact a very early example of a fractal. Early in this case means it was invented in 1907, but many of the other beautiful shapes and theorems in this book stretch back much further: through Étienne Pascal and his oddly organic limaçon (which looks like a kidney) to the ancient Greeks and beyond. Some, on the other hand, are very modern, and this book was out-of-date on the day it was printed. Despite the thousands of years devoted by mathematicians to shapes and the relationship between them, new discoveries are being made all the time. Knots have probably been tied by human beings for as long as human beings have existed, but we’ve only now started to classify them properly and even find new uses for them in biology and physics.

Which is not to say knots are not included here, because they are. But even the older geometry Wells looks at would be enough to keep amateur and recreational mathematicians happy for years, proving, re-creating, and generalizing as they work their way through variations on all manner of trigonomic, topological, and tessellatory themes.

Previously pre-posted (please peruse):

• Poulet’s Propeller — discussion of Wells’ Penguin Dictionary of Curious and Interesting Numbers (1986)

Persecution Complex

by in Geometry, Mathematics, Pursuit Curves and tagged , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

Imagine four mice sitting on the corners of a square. Each mouse begins to run towards its clockwise neighbour. What happens? This:

Four mice chasing each other

Four mice chasing each other

The mice spiral to the centre and meet, creating what are called pursuit curves. Now imagine eight mice on a square, four sitting on the corners, four sitting on the midpoints of the sides. Each mouse begins to run towards its clockwise neighbour. Now what happens? This:

Eight mice chasing each other

Eight mice chasing each other

But what happens if each of the eight mice begins to run towards its neighbour-but-one? Or its neighbour-but-two? And so on. The curves begin to get more complex:

square+midpoint+2

(Please open the following image in a new window if it fails to animate.)

square+midpoint+3

You can also make the mice run at different speeds or towards neighbours displaced by different amounts. As these variables change, so do the patterns traced by the mice:

• Continue reading Persecution Complexified

Poulet’s Propeller

by in Biology, History, Magistra Mundi, Mathematics, Prime numbers, Psychology and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

The Penguin Dictionary of Curious and Interesting Numbers (1986) is one of my favourite books. It’s a fascinating mixture of math, mathecdote and math-joke:

2·618 0333…

The square of φ, the golden ratio, and the only positive number such that √n = n-1. (pg. 45)

6

Kepler discussed the 6-fold symmetry of snowflakes, and attempted to explain it by considering the close packing of spheres in a hexagonal array. (pg. 69)

39

This appears to be the first uninteresting number, which of course makes it an especially interesting number, because it is the smallest number to have the property of being uninteresting.

It is therefore also the first number to be simultaneously interesting and uninteresting. (pg. 120)

David Wells, who wrote the Dictionary, “had the rare distinction of being a Cambridge scholar in mathematics and failing his degree”. He must be the mathematical equivalent of the astronomer Patrick Moore: a popularizer responsible for opening many minds and inspiring many careers. He’s also written books on geometry and mathematical puzzles. But not everyone appreciates his efforts. This is a sideswipe in a review of William Hartston’s The Book of Numbers:

Thankfully, this book is more concerned with facts than mathematics. Anyone wanting to learn more about [π] or the Fibonacci sequence should turn to the Penguin Dictionary of Curious and Interesting Numbers, a volume which none but propeller-heads will find either curious or interesting. (Review in The Independent, 18th December 1997)

Continue reading: Poulet’s Propeller…

Narcissarithmetic #2

by in Base Behavior, Digit-sums, Mathematics, Narcissistic numbers, Powers and tagged , , , , , , , , , , , , , , , , , , , , , | Leave a comment

It’s easy to find patterns like these in base ten:

81 = (8 + 1)^2 = 9^2 = 81

512 = (5 + 1 + 2)^3 = 8^3 = 512
4913 = (4 + 9 + 1 + 3)^3 = 17^3 = 4913
5832 = (5 + 8 + 3 + 2)^3 = 18^3 = 5832
17576 = (1 + 7 + 5 + 7 + 6)^3 = 26^3 = 17576
19683 = (1 + 9 + 6 + 8 + 3)^3 = 27^3 = 19683

2401 = (2 + 4 + 0 + 1)^4 = 7^4 = 2401
234256 = (2 + 3 + 4 + 2 + 5 + 6)^4 = 22^4 = 234256
390625 = (3 + 9 + 0 + 6 + 2 + 5)^4 = 25^4 = 390625
614656 = (6 + 1 + 4 + 6 + 5 + 6)^4 = 28^4 = 614656
1679616 = (1 + 6 + 7 + 9 + 6 + 1 + 6)^4 = 36^4 = 1679616

17210368 = (1 + 7 + 2 + 1 + 0 + 3 + 6 + 8)^5 = 28^5 = 17210368
52521875 = (5 + 2 + 5 + 2 + 1 + 8 + 7 + 5)^5 = 35^5 = 52521875
60466176 = (6 + 0 + 4 + 6 + 6 + 1 + 7 + 6)^5 = 36^5 = 60466176
205962976 = (2 + 0 + 5 + 9 + 6 + 2 + 9 + 7 + 6)^5 = 46^5 = 205962976

1215766545905692880100000000000000000000 = (1 + 2 + 1 + 5 + 7 + 6 + 6 + 5 + 4 + 5 + 9 + 0 + 5 + 6 + 9 + 2 + 8 + 8 + 0 + 1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0)^20 = 90^20 = 1215766545905692880100000000000000000000

Patterns like this are much rarer:

914457600 = (9 x 1 x 4 x 4 x 5 x 7 x 6)^2 = 30240^2 = 914457600

3657830400 = (3 x 6 x 5 x 7 x 8 x 3 x 4)^2 = 60480^2 = 3657830400

I haven’t found a cube like that in base ten, but base six supplies them:

2212 = (2 x 2 x 1 x 2)^3 = 12^3 = 2212 (b=6) = 8^3 = 512 (b=10)
325000 = (3 x 2 x 5)^3 = 50^3 = 325000 (b=6) = 30^3 = 27000 (b=10)
411412 = (4 x 1 x 1 x 4 x 1 x 2)^3 = 52^3 = 411412 (b=6) = 32^3 = 32768 (b=10)

And base nine supplies a fourth and fifth power:

31400 = (3 x 1 x 4)^4 = 13^4 = 31400 (b=9) = 12^4 = 20736 (b=10)
11600 = (1 x 1 x 6)^5 = 6^5 = 11600 (b=9) = 6^5 = 7776 (b=10)

Then base ten is rich in patterns like these:

81 = (8^1 + 1^1) x (8 + 1) = 9 x 9 = 81

133 = (1^2 + 3^2 + 3^2) x (1 + 3 + 3) = 19 x 7 = 133
315 = (3^2 + 1^2 + 5^2) x (3 + 1 + 5) = 35 x 9 = 315
803 = (8^2 + 0^2 + 3^2) x (8 + 0 + 3) = 73 x 11 = 803
1148 = (1^2 + 1^2 + 4^2 + 8^2) x (1 + 1 + 4 + 8) = 82 x 14 = 1148
1547 = (1^2 + 5^2 + 4^2 + 7^2) x (1 + 5 + 4 + 7) = 91 x 17 = 1547
2196 = (2^2 + 1^2 + 9^2 + 6^2) x (2 + 1 + 9 + 6) = 122 x 18 = 2196

1215 = (1^3 + 2^3 + 1^3 + 5^3) x (1 + 2 + 1 + 5) = 135 x 9 = 1215
3700 = (3^3 + 7^3 + 0^3 + 0^3) x (3 + 7 + 0 + 0) = 370 x 10 = 3700
11680 = (1^3 + 1^3 + 6^3 + 8^3 + 0^3) x (1 + 1 + 6 + 8 + 0) = 730 x 16 = 11680
13608 = (1^3 + 3^3 + 6^3 + 0^3 + 8^3) x (1 + 3 + 6 + 0 + 8) = 756 x 18 = 13608
87949 = (8^3 + 7^3 + 9^3 + 4^3 + 9^3) x (8 + 7 + 9 + 4 + 9) = 2377 x 37 = 87949

182380 = (1^4 + 8^4 + 2^4 + 3^4 + 8^4 + 0^4) x (1 + 8 + 2 + 3 + 8 + 0) = 8290 x 22 = 182380
444992 = (4^4 + 4^4 + 4^4 + 9^4 + 9^4 + 2^4) x (4 + 4 + 4 + 9 + 9 + 2) = 13906 x 32 = 444992

41500 = (4^5 + 1^5 + 5^5 + 0^5 + 0^5) x (4 + 1 + 5 + 0 + 0) = 4150 x 10 = 41500
3508936 = (3^5 + 5^5 + 0^5 + 8^5 + 9^5 + 3^5 + 6^5) x (3 + 5 + 0 + 8 + 9 + 3 + 6) = 103204 x 34 = 3508936
3828816 = (3^5 + 8^5 + 2^5 + 8^5 + 8^5 + 1^5 + 6^5) x (3 + 8 + 2 + 8 + 8 + 1 + 6) = 106356 x 36 = 3828816
4801896 = (4^5 + 8^5 + 0^5 + 1^5 + 8^5 + 9^5 + 6^5) x (4 + 8 + 0 + 1 + 8 + 9 + 6) = 133386 x 36 = 4801896
5659875 = (5^5 + 6^5 + 5^5 + 9^5 + 8^5 + 7^5 + 5^5) x (5 + 6 + 5 + 9 + 8 + 7 + 5) = 125775 x 45 = 5659875

Previously pre-posted (please peruse):

• Narcissarithmetic

Narcissarithmetic

by in Base Behavior, Digit-sums, Mathematics, Powers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

Why is 438,579,088 a beautiful number? Simple: it may seem entirely arbitrary, but it’s actually self-empowered:

438,579,088 = 4^4 + 3^3 + 8^8 + 5^5 + 7^7 + 9^9 + 0^0 + 8^8 + 8^8 = 256 + 27 + 16777216 + 3125 + 823543 + 387420489 + 0 + 16777216 + 16777216 (usually 0^0 = 1, but the rule is slightly varied here)

438,579,088 is so beautiful, in fact, that it’s in love with itself as a narcissistic number, or number that can be generated by manipulation of its own digits. 89 = 8^1 + 9^2 = 8 + 81 and 135 = 1^1 + 3^2 + 5^3 = 1 + 9 + 125 are different kinds of narcissistic number. 3435 is self-empowered again:

3435 = 3^3 + 4^4 + 3^3 + 5^5 = 27 + 256 + 27 + 3125

But that’s your lot: there are no more numbers in base-10 that are equal to the sum of their self-empowered digits (apart from the trivial 0 and 1). To prove this, start by considering that there is a limit to the size of a self-empowered number. 9^9 is 387,420,489, which is nine digits long. The function autopower(999,999,999) = 387,420,489 x 9 = 3,486,784,401, which is ten digits long. But autopower(999,999,999,999) = 387,420,489 x 12 = 4,649,045,868, also ten digits long.

The Metamorphosis of Narcissus by Salvador Dalí

Salvador Dalí, La Metamorfosis de Narciso (1937)

So you don’t need to check numbers above a certain size. There still seem a lot of numbers to check: 438,579,088 is a long way above 3435. However, the search is easy to shorten if you consider that checking 3-3-4-5 is equivalent to checking 3-4-3-5, just as checking 034,578,889 is equivalent to checking 438,579,088. If you self-empower a number and the result has the same digits as the original number, you’ve found what you’re looking for. The order of digits in the original number doesn’t matter, because the result has automatically sorted them for you. The function autopower(3345) produces 3435, therefore 3435 must be self-empowered.

So the rule is simple: Check only the numbers in which any digit is greater than or equal to all digits to its left. In other words, you check 12 and skip 21, check 34 and skip 43, check 567 and skip 576, 657, 675, 756 and 765. That reduces the search-time considerably: discarding numbers is computationally simpler than self-empowering them. It’s also computationally simple to vary the base in which you’re searching. Base-10 produces only two self-empowered numbers, but its neighbours base-9 and base-11 are much more fertile:

30 = 3^3 + 0^0 = 30 + 0 (b=9)
27 = 27 + 0 (b=10)

31 = 3^3 + 1^1 = 30 + 1 (b=9)
28 = 27 + 1 (b=10)

156262 = 1^1 + 5^5 + 6^6 + 2^2 + 6^6 + 2^2 = 1 + 4252 + 71000 + 4 + 71000 + 4 (b=9)
96446 = 1 + 3125 + 46656 + 4 + 46656 + 4 (b=10)

1647063 = 1^1 + 6^6 + 4^4 + 7^7 + 0^0 + 6^6 + 3^3 = 1 + 71000 + 314 + 1484617 + 0 + 71000 + 30 (b=9)
917139 = 1 + 46656 + 256 + 823543 + 0 + 46656 + 27 (b=10)

1656547 = 1^1 + 6^6 + 5^5 + 6^6 + 5^5 + 4^4 + 7^7 = 1 + 71000 + 4252 + 71000 + 4252 + 314 + 1484617 (b=9)
923362 = 1 + 46656 + 3125 + 46656 + 3125 + 256 + 823543 (b=10)

34664084 = 3^3 + 4^4 + 6^6 + 6^6 + 4^4 + 0^0 + 8^8 + 4^4 = 30 + 314 + 71000 + 71000 + 314 + 0 + 34511011 + 314 (b=9)
16871323 = 27 + 256 + 46656 + 46656 + 256 + 0 + 16777216 + 256 (b=10)

66500 = 6^6 + 6^6 + 5^5 + 0^0 + 0^0 = 32065 + 32065 + 2391 + 0 + 0 (b=11)
96437 = 46656 + 46656 + 3125 + 0 + 0 (b=10)

66501 = 6^6 + 6^6 + 5^5 + 0^0 + 1^1 = 32065 + 32065 + 2391 + 0 + 1 (b=11)
96438 = 46656 + 46656 + 3125 + 0 + 1 (b=10)

517503 = 5^5 + 1^1 + 7^7 + 5^5 + 0^0 + 3^3 = 2391 + 1 + 512816 + 2391 + 0 + 25 (b=11)
829821 = 3125 + 1 + 823543 + 3125 + 0 + 27 (b=10)

18453278 = 1^1 + 8^8 + 4^4 + 5^5 + 3^3 + 2^2 + 7^7 + 8^8 = 1 + 9519A75 + 213 + 2391 + 25 + 4 + 512816 + 9519A75 (b=11)
34381388 = 1 + 16777216 + 256 + 3125 + 27 + 4 + 823543 + 16777216 (b=10)

18453487 = 1^1 + 8^8 + 4^4 + 5^5 + 3^3 + 4^4 + 8^8 + 7^7 = 1 + 9519A75 + 213 + 2391 + 25 + 213 + 9519A75 + 512816 (b=11)
34381640 = 1 + 16777216 + 256 + 3125 + 27 + 256 + 16777216 + 823543 (b=10)

It’s easy to extend the concept of self-empowered narcisso-numbers. The prime 71 = 131 in base-7 and the prime 83 = 146 in base-7. If 131[b=7] is empowered to the digits of 146[b=7], you get 146[b=7]; and if 146[b=7] is empowered to the digits of 131[b=7], you get 131[b=7], like this:

71 = 131[b=7] → 1^1 + 3^4 + 1^6 = 1 + 81 + 1 = 83 = 146[b=7]

83 = 146[b=7] → 1^1 + 4^3 + 6^1 = 1 + 64 + 6 = 71 = 131[b=7]

But it’s not easy to find more examples. Are there other-empowering pairs like that in base-10? I don’t know.

In Perms Of

by in Base Behavior, Mathematics, Prime numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

13 is a prime number, divisible only by itself and 1. Perm 13 and you get 31, which is also a prime number. The same is true of 17, 37 and 79. There are only two possible permutations – 2 x 1 – of a two-digit number, so base-10 is terminally permal for two-digit primes:

13, 31
17, 71
37, 73
79, 97

What about three-digit primes? Now there are six possible permutations: 3 x 2 x 1. But base-10 is not terminally permal for three-digit primes. This is the best it does:

149, 419, 491, 941
179, 197, 719, 971
379, 397, 739, 937

Fortunately, we aren’t restricted to base-10. Take a step up and you’ll find that base-11 is terminally permal for three-digit primes (139 in base-11 = 1 x 11^2 + 3 x 11 + 9 = 163 in base-10):

139, 193, 319, 391, 913, 931 (6 primes) (base=11)

163, 223, 383, 463, 1103, 1123 (base=10)

Four-digit primes have twenty-four possible permutations – 4 x 3 x 2 x 1 – and base-10 again falls short:

1237, 1327, 1723, 2137, 2371, 
2713, 2731, 3217, 3271, 7213,
7321 (11 primes)

1279, 1297, 2179, 2719, 2791,
2917, 2971, 7129, 7219, 9127,
9721

For four-digit primes, the most permal base I’ve discovered so far is base-13 (where B represents [11]):

134B, 13B4, 14B3, 1B34, 1B43,
314B, 31B4, 34B1, 3B14, 413B,
41B3, 431B, 43B1, 4B13, 4B31,
B134, B143, B314, B413 (19 primes) (base=13)

2767, 2851, 3019, 4099, 4111,
6823, 6907, 7411, 8467, 9007,
9103, 9319, 9439, 10663, 10687,
24379, 24391, 24691, 24859 (base=10)

Is there a base in which all permutations of some four-digit number are prime? I think so, but I haven’t found it yet. Is there always some base, b, in which all permutations of some d-digit number are prime? Is there an infinity of bases in which all permutations of some d-digit number are prime? Easy to ask, difficult to answer. For me, anyway.

Factory Records

by in Base Behavior, Mathematics, Palindromic numbers, Prime numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

The factors of n are those numbers that divide n without remainder. So the factors of 6 are 1, 2, 3 and 6. If the function s(n) is defined as “the sum of the factors of n, excluding n”, then s(6) = 1 + 2 + 3 = 6. This makes 6 a perfect number: its factors re-create it. 28 is another perfect number. The factors of 28 are 1, 2, 4, 7, 14 and 28, so s(28) = 1 + 2 + 4 + 7 + 14 = 28. Other perfect numbers are 496 and 8128. And they’re perfect in any base.

Amicable numbers are amicable in any base too. The factors of an amicable number sum to a second number whose factors sum to the first number. So s(220) = 284, s(284) = 220. That pair may have been known to Pythagoras (c.570-c.495 BC), but s(1184) = 1210, s(1210) = 1184 was discovered by an Italian schoolboy called Nicolò Paganini in 1866. There are also sociable chains, in which s(n), s(s(n)), s(s(s(n))) create a chain of numbers that leads back to n, like this:

12496 → 14288 → 15472 → 14536 → 14264 → 12496 (c=5)

Or this:

14316 → 19116 → 31704 → 47616 → 83328 → 177792 → 295488 → 629072 → 589786 → 294896 → 358336 → 418904 → 366556 → 274924 → 275444 → 243760 → 376736 → 381028 → 285778 → 152990 → 122410 → 97946 → 48976 → 45946 → 22976 → 22744 → 19916 → 17716 → 14316 (c=28)

Those sociable chains were discovered (and christened) in 1918 by the Belgian mathematician Paul Poulet (1887-1946). Other factor-sum patterns are dependant on the base they’re expressed in. For example, s(333) = 161. So both n and s(n) are palindromes in base-10. Here are more examples — the numbers in brackets are the prime factors of n and s(n):

333 (3^2, 37) → 161 (7, 23)
646 (2, 17, 19) → 434 (2, 7, 31)
656 (2^4, 41) → 646 (2, 17, 19)
979 (11, 89) → 101 (prime)
1001 (7, 11, 13) → 343 (7^3)
3553 (11, 17, 19) → 767 (13, 59)
10801 (7, 1543) → 1551 (3, 11, 47)
11111 (41, 271) → 313 (prime)
18581 (17, 1093) → 1111 (11, 101)
31713 (3, 11, 31^2) → 15951 (3, 13, 409)
34943 (83, 421) → 505 (5, 101)
48484 (2^2, 17, 23, 31) → 48284 (2^2, 12071)
57375 (3^3, 5^3, 17) → 54945 (3^3, 5, 11, 37)
95259 (3, 113, 281) → 33333 (3, 41, 271)
99099 (3^2, 7, 11^2, 13) → 94549 (7, 13, 1039)
158851 (7, 11, 2063) → 39293 (prime)
262262 (2, 7, 11, 13, 131) → 269962 (2, 7, 11, 1753)
569965 (5, 11, 43, 241) → 196691 (11, 17881)
1173711 (3, 7, 11, 5081) → 777777 (3, 7^2, 11, 13, 37)

Note how s(656) = 646 and s(646) = 434. There’s an even longer sequence in base-495:

33 → 55 → 77 → 99 → [17][17] → [19][19] → [21][21] → [43][43] → [45][45] → [111][111] → [193][193] → [195][195] → [477][477] (b=495) (c=13)
1488 (2^4, 3, 31) → 2480 (2^4, 5, 31) → 3472 (2^4, 7, 31) → 4464 (2^4, 3^2, 31) → 8432 (2^4, 17, 31) → 9424 (2^4, 19, 31) → 10416 (2^4, 3, 7, 31) → 21328 (2^4, 31, 43) → 22320 (2^4, 3^2, 5, 31) → 55056 (2^4, 3, 31, 37) → 95728 (2^4, 31, 193) → 96720 (2^4, 3, 5, 13, 31) → 236592 (2^4, 3^2, 31, 53)

I also tried looking for n whose s(n) mirrors n. But they’re hard to find in base-10. The first example is this:

498906 (2, 3^3, 9239) → 609894 (2, 3^2, 31, 1093)

498906 mirrors 609894, because the digits of each run in reverse to the digits of the other. Base-9 does better for mirror-sums, clocking up four in the same range of integers:

42 → 24 (base=9)
38 (2, 19) → 22 (2, 11)
402 → 204 (base=9)
326 (2, 163) → 166 (2, 83)
4002 → 2004 (base=9)
2918 (2, 1459) → 1462 (2, 17, 43)
5544 → 4455 (base=9)
4090 (2, 5, 409) → 3290 (2, 5, 7, 47)

Base-11 does better still, clocking up eight in the same range:

42 → 24 (base=11)
46 (2, 23) → 26 (2, 13)
2927 → 7292 (base=11)
3780 (2^2, 3^3, 5, 7) → 9660 (2^2, 3, 5, 7, 23)
4002 → 2004 (base=11)
5326 (2, 2663) → 2666 (2, 31, 43)
13772 → 27731 (base=11)
19560 (2^3, 3, 5, 163) → 39480 (2^3, 3, 5, 7, 47)
4[10]7[10]9 → 9[10]7[10]4 (base=11)
72840 (2^3, 3, 5, 607) → 146040 (2^3, 3, 5, 1217)
6929[10] → [10]9296 (base=11)
100176 (2^4, 3, 2087) → 158736 (2^4, 3, 3307)
171623 → 326171 (base=11)
265620 (2^2, 3, 5, 19, 233) → 520620 (2^2, 3, 5, 8677)
263702 → 207362 (base=11)
414790 (2, 5, 41479) → 331850 (2, 5^2, 6637)

Note that 42 mirrors its factor-sum in both base-9 and base-11. But s(42) = 24 in infinitely many bases, because when 42 = 2 x prime, s(42) = 1 + 2 + prime. So (prime-1) / 2 will give the base in which 24 = s(42). For example, 2 x 11 = 22 and 22 = 42 in base (11-1) / 2 or base-5. So s(42) = 1 + 2 + 11 = 14 = 2 x 5 + 4 = 24[b=5]. There are infinitely many primes, so infinitely many bases in which s(42) = 24.

Base-10 does better for mirror-sums when s(n) is re-defined to include n itself. So s(69) = 1 + 3 + 23 + 69 = 96. Here are the first examples of all-factor mirror-sums in base-10:

69 (3, 23) → 96 (2^5, 3)
276 (2^2, 3, 23) → 672 (2^5, 3, 7)
639 (3^2, 71) → 936 (2^3, 3^2, 13)
2556 (2^2, 3^2, 71) → 6552 (2^3, 3^2, 7, 13)

In the same range, base-9 now produces one mirror-sum, 13 → 31 = 12 (2^2, 3) → 28 (2^2, 7). Base-11 produces no mirror-sums in the same range. Base behaviour is eccentric, but that’s what makes it interesting.

More Multi-Magic

by in Digit-sums, Magic squares, Mathematics and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

The answer, I’m glad to say, is yes. The question is: Can a prime magic-square nest inside a second prime magic-square that nests inside a third prime magic-square? I asked this in Multi-Magic, where I described how a magic square is a square of numbers where all rows, all columns and both diagonals add to the same number, or magic total. This magic square consists entirely of prime numbers, or numbers divisible only by themselves and 1:

43 | 01 | 67
61 | 37 | 13
07 | 73 | 31

Base = 10, magic total = 111

It nests inside this prime magic-square, whose digit-sums in base-97 re-create it:

0619  =  [06][37] | 0097  =  [01][00] | 1123  =  [11][56]
1117  =  [11][50] | 0613  =  [06][31] | 0109  =  [01][12]
0103  =  [01][06] | 1129  =  [11][62] | 0607  =  [06][25]

Base = 97, magic total = 1839

And that prime magic-square nests inside this one:

2803  =  [1][0618] | 2281  =  [1][0096] | 3307  =  [1][1122]
3301  =  [1][1116] | 2797  =  [1][0612] | 2293  =  [1][0108]
2287  =  [1][0102] | 3313  =  [1][1128] | 2791  =  [1][0606]

Base = 2185, magic total = 8391

I don’t know whether that prime magic-square nests inside a fourth square, but a 3-nest is good for 3×3 magic squares. On the other hand, this famous 3×3 magic square is easy to nest inside an infinite series of other magic squares:

6 | 1 | 8
7 | 5 | 3
2 | 9 | 4

Base = 10, magic total = 15

It’s created by the digit-sums of this square in base-9 (“14 = 15” means that the number 14 is represented as “15” in base-9):

14 = 15 → 6 | 09 = 10 → 1 | 16 = 17 → 8
15 = 16 → 7 | 13 = 14 → 5 | 11 = 12 → 3
10 = 11 → 2 | 17 = 18 → 9 | 12 = 13 → 4

Base = 9, magic total = 39

And that square in base-9 is created by the digit-sums of this square in base-17:

30 = 1[13] → 14 | 25 = 00018 → 09 | 32 = 1[15] → 16
31 = 1[14] → 15 | 29 = 1[12] → 13 | 27 = 1[10] → 11
26 = 00019 → 10 | 33 = 1[16] → 17 | 28 = 1[11] → 12

Base = 17, magic total = 87

And so on:

62 = 1[29] → 30 | 57 = 1[24] → 25 | 64 = 1[31] → 32
63 = 1[30] → 31 | 61 = 1[28] → 29 | 59 = 1[26] → 27
58 = 1[25] → 26 | 65 = 1[32] → 33 | 60 = 1[27] → 28

Base = 33, magic total = 183

126 = 1[61] → 62 | 121 = 1[56] → 57 | 128 = 1[63] → 64
127 = 1[62] → 63 | 125 = 1[60] → 61 | 123 = 1[58] → 59
122 = 1[57] → 58 | 129 = 1[64] → 65 | 124 = 1[59] → 60

Base = 65, magic total = 375

Previously Pre-Posted (please peruse):

• Multi-Magic