Get Your Ox Off

Boustrophedon (pronounced “bough-stra-FEE-dun” or “boo-stra-FEE-dun”) is an ancient Greek word literally meaning “as the ox turns (in ploughing)”, that is, moving left-right, right-left, and so on. The word is used of writing that runs down the page in the same way. To see what that means, examine two versions of the first paragraph of Clark Ashton Smith’s story “The Demon of the Flower” (1933). The first is written in the usual way, the second is written boustrophedon:

Not as the plants and flowers of Earth, growing peacefully beneath a simple sun, were the blossoms of the planet Lophai. Coiling and uncoiling in double dawns; tossing tumultuously under vast suns of jade green and balas-ruby orange; swaying and weltering in rich twilights, in aurora-curtained nights, they resembled fields of rooted servants that dance eternally to an other-worldly music.


Not as the plants and flowers of Earth, growing peacefully
.iahpoL tenalp eht fo smossolb eht erew ,nus elpmis a htaeneb
Coiling and uncoiling in double dawns; tossing tumultuously
;egnaro ybur-salab dna neerg edaj fo snus tsav rednu
swaying and weltering in rich twilights, in aurora-curtained
ecnad taht stnavres detoor fo sdleif delbmeser yeht ,sthgin
eternally to an other-worldly music.


Boustrophedon writing was once common and sometimes the left-right lines would also be mirror-reversed, like this:
CAS_text1


You could also use the term “boustrophedon” to describe the way this table of numbers is filled:

primes_table


The table begins with “1” in the top left-hand corner, then moves right for “2”, then down for “3”, then right-and-up for “4”, “5” and “6”, then right for “7”, then left-and-down for “8”, “9” and “10”, and so on. You could also say that the numbers snake through the table. I’ve marked the primes among them, because I was interested in the patterns made by the primes when the numbers were represented as blocks on a grid, like this:

primes_large


Primes are in solid white (compare the Ulam spiral). Here’s the boustrophedon prime-grid on a finer scale:

primes

(click for full image)


And what about other number-tests? Here are the even numbers marked on the grid (i.e. n mod 2 = 0):

mod2

n mod 2 = 0


And here are some more examples of a modulus test:

mod3

n mod 3 = 0


mod5

n mod 5 = 0


mod9

n mod 9 = 0


mod15

n mod 15 = 0


mod_various

n mod various = 0 (animated gif)


Next I looked at reciprocals (numbers divided into 1) marked on the grid, with the digits of a reciprocal marking the number of blank squares before a square is filled in (if the digit is “0”, the square is filled immediately). For example, in base ten 1/7 = 0.142857142857142857…, where the block “142857” repeats for ever. When represented on the grid, 1/7 has 1 blank square, then a filled square, then 4 blank squares, then a filled square, then 2 blank squares, then a filled square, and so on:

recip7_base10

1/7 in base 10


And here are some more reciprocals (click for full images):

recip9_base2

1/9 in base 2


recip13_base10

1/13 in base 10


recip27_base10

1/27 in base 10


recip41_base10

1/41 in base 10


recip63_base10

1/63 in base 10


recip82_base10

1/82 in base 10


recip101_base10

1/101 in base 10


recip104_base10

1/104 in base 10


recip124_base10

1/124 in base 10


recip143_base10

1/143 in base 10


recip175_base10

1/175 in base 10


recip604_base8

1/604 in base 8


recip_various

1/n in various bases (animated gif)


Magistra Rules the Waves

One of my favourite integer sequences has the simple formula n(i) = n(i-1) + digitsum(n(i-1)). If it’s seeded with 1, its first few terms go like this:

n(1) = 1
n(2) = n(1) + digitsum(n(1)) = 1 + digitsum(1) = 2
n(3) = 2 + digitsum(2) = 4
n(4) = 4 + digitsum(4) = 8
n(5) = 8 + digitsum(8) = 16
n(6) = 16 + digitsum(16) = 16 + 1+6 = 16 + 7 = 23
n(7) = 23 + digitsum(23) = 23 + 2+3 = 23 + 5 = 28
n(8) = 28 + digitsum(28) = 28 + 2+8 = 28 + 10 = 38

As a sequence, it looks like this:

1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, 107, 115, 122, 127, 137, 148, 161, 169, 185, 199, 218, 229, 242, 250, 257, 271, 281, 292, 305, 313, 320, 325, 335, 346, 359, 376, 392, 406, 416, 427, 440, 448, 464, 478, 497, 517, 530, 538, 554, 568, 587, 607, 620, 628, 644, 658, 677, 697, 719, 736, 752, 766, 785, 805, 818, 835, 851, 865, 884, 904, 917, 934, 950, 964, 983, 1003…

Given a number at random, is there a quick way to say whether it appears in the sequence seeded with 1? Not that I know, with one exception. If the number is divisible by 3, it doesn’t appear, at least in base 10. In base 2, that rule doesn’t apply:

n(1) = 1
n(2) = 1 + digitsum(1) = 10 = 1 + 1 = 2
n(3) = 10 + digitsum(10) = 10 + 1 = 11 = 2 + 1 = 3
n(4) = 11 + digitsum(11) = 11 + 1+1 = 101 = 3 + 2 = 5
n(5) = 101 + digitsum(101) = 101 + 1+0+1 = 111 = 5 + 2 = 7
n(6) = 111 + digitsum(111) = 111 + 11 = 1010 = 7 + 3 = 10
n(7) = 1010 + digitsum(1010) = 1010 + 10 = 1100 = 10 + 2 = 12
n(8) = 1100 + digitsum(1100) = 1100 + 10 = 1110 = 12 + 2 = 14

1, 2, 3, 5, 7, 10, 12, 14, 17, 19, 22, 25, 28, 31, 36, 38, 41, 44, 47, 52, 55, 60, 64, 65, 67, 70, 73, 76, 79, 84, 87, 92, 96, 98, 101, 105, 109, 114, 118, 123, 129, 131, 134, 137, 140, 143, 148, 151, 156, 160, 162, 165, 169, 173, 178, 182, 187, 193, 196, 199, 204, 208, 211, 216, 220, 225, 229, 234, 239, 246, 252, 258, 260, 262, 265, 268, 271, 276, 279, 284, 288, 290, 293, 297, 301, 306, 310, 315, 321, 324, 327, 332, 336, 339, 344, 348, 353, 357, 362, 367, 374…

What patterns are there in these sequences? It’s easier to check when they’re represented graphically, so I converted them into patterns à la the Ulam spiral, where n is represented as a dot on a spiral of integers. This is the spiral for base 10:

ulambase10Base 10


And these are the spirals for bases 2 and 3:

ulambase2

Base 2


ulambase3

Base 3


These sequences look fairly random to me: there are no obvious patterns in the jumps from n(i) to n(i+1), i.e. in the values for digitsum(n(i)). Now try the spirals for bases 9 and 33:

ulambase9

Base 9


ulambase33

Base 33


Patterns have appeared: there is some regularity in the jumps. You can see these regularities more clearly if you represent digitsum(n(i)) as a graph, with n(i) on the x axis and digitsum(n(i)) on the y axis. If the graph starts with n(i) = 1 on the lower left and proceeds left-right, left-right up the screen, it looks like this in base 10:

base10

Base 10 (click to enlarge)


Here are bases 2 and 3:

base2

Base 2


base3

Base 3


The jumps seem fairly random. Now try bases 9, 13, 16, 17, 25, 33 and 49:

base9

Base 9


base13

Base 13


base16

Base 16


base17

Base 17


base25

Base 25


base33

Base 33


base49

Base 49


In some bases, the formula n(i) = n(i-1) + digitsum(n(i-1)) generates mild randomness. In others, it generates strong regularity, like waves rolling ashore under a steady wind. I don’t understand why, but regularity seems to occur in bases that are one more than a power of 2 and also in some bases that are primes or squares.


Elsewhere other-posted:

Mathematica Magistra Mundi
8200_idf_insignia

Breeding Bunnies

Front cover of The Golden Ratio by Mario Livio
The Golden Ratio: The Story of Phi, the Extraordinary Number of Nature, Art and Beauty, Mario Livio (Headline Review 2003)

A good short popular guide to perhaps the most interesting, and certainly the most irrational, of all numbers: the golden ratio or phi (φ), which is approximately equal to 1·6180339887498948482… Prominent in mathematics since at least the ancient Greeks and Euclid, phi is found in many places in nature too, from pineapples and sunflowers to the flight of hawks. Livio catalogues its appearances in both maths and nature, looking closely at the Fibonacci sequence and rabbit-breeding, before going on to debunk mistaken claims that phi also appears a lot in art, music and poetry. Dalí certainly used it, but da Vinci, Debussy and Virgil almost certainly didn’t. Nor, almost certainly, did the builders of the Parthenon and pyramids. Finally, he examines what has famously been called (by the physicist Eugene Wiegner) the unreasonable effectiveness of mathematics: why is this human invention so good at describing the behaviour of the Universe? Livio quotes one of the best short answers I’ve seen:

Human logic was forced on us by the physical world and is therefore consistent with it. Mathematics derives from logic. That is why mathematics is consistent with the physical world. (ch. 9, “Is God a mathematician?”, pg. 252)

It’s not hard to recommend a book that quotes everyone from Johannes Kepler and William Blake to Lewis Carroll, Christopher Marlowe and Jef Raskin, “the creator of the Macintosh computer”, whose answer is given above. Recreational mathematicians should also find lots of ideas for further investigation, from fractal strings to the fascinating number patterns governed by Benford’s law. It isn’t just human beings who look after number one: as a leading figure, 1 turns up much more often in data from the real world, and in mathematical constructs like the Fibonacci sequence, than intuition would lead you to expect. If you’d like to learn more about that and about many other aspects of mathematics, hunt down a copy of this book.


Elsewhere other-posted:

Roses Are Golden – φ and floral homicide

Factory Records

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

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

Multi-Magic

A magic square is a square of numbers in which all rows, all columns and both diagonals add to the same number, or magic total. The simplest magic square using distinct numbers is this:

6 1 8
7 5 3
2 9 4

It’s easy to prove that the magic total of a 3×3 magic square must be three times the central number. Accordingly, if the central number is 37, the magic total must be 111. There are lots of ways to create a magic square with 37 at its heart, but this is my favourite:

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

The square is special because all the numbers are prime, or divisible by only themselves and 1 (though 1 itself is not usually defined as prime in modern mathematics). I like the 37-square even more now that I’ve discovered it can be found inside another all-prime magic square:

0619 = 0006[37] | 0097 = 00000010 | 1123 = [11][56]
1117 = [11][50] | 0613 = 0006[31] | 0109 = 0001[12]
0103 = 00000016 | 1129 = [11][62] | 0607 = 0006[25]

Magic total = 1839

The square is shown in both base-10 and base-97. If the digit-sums of the base-97 square are calculated, this is the result (e.g., the digit-sum of 6[37][b=97] = 6 + 37 = 43):

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

This makes me wonder whether the 613-square might nest in another all-prime square, and so on, perhaps ad infinitum [Update: yes, the 613-square is a nestling]. There are certainly many nested all-prime squares. Here is square-631 in base-187:

661 = 003[100] | 379 = 00000025 | 853 = 004[105]
823 = 004[075] | 631 = 003[070] | 439 = 002[065]
409 = 002[035] | 883 = 004[135] | 601 = 003[040]

Magic total = 1893

Digit-sums:

103 | 007 | 109
079 | 073 | 067
037 | 139 | 043

Magic total = 219

There are also all-prime magic squares that have two kinds of nestlings inside them: digit-sum magic squares and digit-product magic squares. The digit-product of a number is calculated by multiplying its digits (except 0): digit-product(37) = 3 x 7 = 21, digit-product(103) = 1 x 3 = 3, and so on. In base-331, this all-prime magic square yields both a digit-sum square and a digit-product square:

503 = 1[172] | 359 = 1[028] | 521 = 1[190]
479 = 1[148] | 461 = 1[130] | 443 = 1[112]
401 = 1[070] | 563 = 1[232] | 419 = 1[088]

Magic total = 1383

Digit-sums:

173 | 029 | 191
149 | 131 | 113
071 | 233 | 089

Magic total = 393

Digit-products:

172 | 028 | 190
148 | 130 | 112
070 | 232 | 088

Magic total = 390

Here are two more twin-bearing all-prime magic squares:

Square-719 in base-451:

761 = 1[310] | 557 = 1[106] | 839 = 1[388]
797 = 1[346] | 719 = 1[268] | 641 = 1[190]
599 = 1[148] | 881 = 1[430] | 677 = 1[226]

Magic total = 2157

Digit-sums:

311 | 107 | 389
347 | 269 | 191
149 | 431 | 227

Magic total = 807

Digit-products:

310 | 106 | 388
346 | 268 | 190
148 | 430 | 226

Magic total = 804

Square-853 in base-344:

883 = 2[195] | 709 = 2[021] | 967 = 2[279]
937 = 2[249] | 853 = 2[165] | 769 = 2[081]
739 = 2[051] | 997 = 2[309] | 823 = 2[135]

Magic total = 2559

Digit-sums:

197 | 023 | 281
251 | 167 | 083
053 | 311 | 137

Magic total = 501

Digit-products:

390 | 042 | 558
498 | 330 | 162
102 | 618 | 270

Magic total = 990

Proviously Post-Posted (please peruse):

More Multi-Magic

Prummer-Time Views

East, west, home’s best. And for human beings, base-10 is a kind of home. We have ten fingers and we use ten digits. Base-10 comes naturally to us: it feels like home. So it’s disappointing that there is no number in base-10 that is equal to the sum of the squares of its digits (apart from the trivial 0^2 = 0 and 1^2 = 1). Base-9 and base-11 do better:

41 = 45[b=9] = 4^2 + 5^2 = 16 + 25 = 41
50 = 55[b=9] = 5^2 + 5^2 = 25 + 25 = 50

61 = 56[b=11] = 5^2 + 6^2 = 25 + 36 = 61
72 = 66[b=11] = 6^2 + 6^2 = 36 + 36 = 72

Base-47 does better still, with fourteen 2-sumbers. And base-10 does have 3-sumbers, or numbers equal to the sum of the cubes of their digits:

153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153
370 = 3^3 + 7^3 + 0^3 = 27 + 343 + 0 = 370
371 = 3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371
407 = 4^3 + 0^3 + 7^3 = 64 + 0 + 343 = 407

But base-10 disappoints again when it comes to prumbers, or prime sumbers, or numbers that are equal to the sum of the primes whose indices are equal to the digits of the number. The index of a prime number is its position in the list of primes. Here are the first nine primes and their indices (with 0 as a pseudo-prime at position 0):

prime(0) = 0
prime(1) = 2
prime(2) = 3
prime(3) = 5
prime(4) = 7
prime(5) = 11
prime(6) = 13
prime(7) = 17
prime(8) = 19
prime(9) = 23

So the prumber, or prime-sumber, of 1 = prime(1) = 2. The prumber of 104 = prime(1) + prime(0) + prime(4) = 2 + 0 + 7 = 9. The prumber of 186 = 2 + 19 + 13 = 34. But no number in base-10 is equal to its prime sumber. Base-2 and base-3 do better:

Base-2 has 1 prumber:

2 = 10[b=2] = 2 + 0 = 2

Base-3 has 2 prumbers:

4 = 11[b=3] = 2 + 2 = 4
5 = 12[b=3] = 2 + 3 = 5

But prumbers are rare. The next record is set by base-127, with 4 prumbers:

165 = 1[38][b=127] = 2 + 163 = 165
320 = 2[66][b=127] = 3 + 317 = 320
472 = 3[91][b=127] = 5 + 467 = 472
620 = 4[112][b=127] = 7 + 613 = 620

Base-479 has 4 prumbers:

1702 = 3[265] = 5 + 1697 = 1702
2250 = 4[334] = 7 + 2243 = 2250
2800 = 5[405] = 11 + 2789 = 2800
3344 = 6[470] = 13 + 3331 = 3344

Base-637 has 4 prumbers:

1514 = 2[240] = 3 + 1511 = 1514
2244 = 3[333] = 5 + 2239 = 2244
2976 = 4[428] = 7 + 2969 = 2976
4422 = 6[600] = 13 + 4409 = 4422

Base-831 has 4 prumbers:

999 = 1[168] = 2 + 997 = 999
2914 = 3[421] = 5 + 2909 = 2914
3858 = 4[534] = 7 + 3851 = 3858
4798 = 5[643] = 11 + 4787 = 4798

Base-876 has 4 prumbers:

1053 = 1[177] = 2 + 1051 = 1053
3066 = 3[438] = 5 + 3061 = 3066
4064 = 4[560] = 7 + 4057 = 4064
6042 = 6[786] = 13 + 6029 = 6042

Previously pre-posted (please peruse):

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