Power Trip

Here are the first few powers of 2:

2 = 1 * 2
4 = 2 * 2
8 = 4 * 2
16 = 8 * 2
32 = 16 * 2
64 = 32 * 2
128 = 64 * 2
256 = 128 * 2
512 = 256 * 2
1024 = 512 * 2
2048 = 1024 * 2
4096 = 2048 * 2
8192 = 4096 * 2
16384 = 8192 * 2
32768 = 16384 * 2
65536 = 32768 * 2
131072 = 65536 * 2
262144 = 131072 * 2
524288 = 262144 * 2
1048576 = 524288 * 2
2097152 = 1048576 * 2
4194304 = 2097152 * 2
8388608 = 4194304 * 2
16777216 = 8388608 * 2
33554432 = 16777216 * 2
67108864 = 33554432 * 2…

As you can see, it’s a one-way power-trip: the numbers simply get larger. But what happens if you delete the digit 0 whenever it appears in a result? For example, 512 * 2 = 1024, which becomes 124. If you apply this rule, the sequence looks like this:

2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 124
124 * 2 = 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 5794
5794 * 2 = 11588
11588 * 2 = 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 9274…

Is this a power-trip? Not quite: it’s a return trip, because the numbers can never grow beyond a certain size and the sequence falls into a loop. If the result 2n contains a zero, then zerodelete(2n) < n, so the sequence has an upper limit and a number will eventually occur twice. This happens at step 526 with 366784, which matches 366784 at step 490.

The rate at which we delete zeros can obviously be varied. Call it 1:z. The sequence above sets z = 1, so 1:z = 1:1. But what if z = 2, so that 1:z = 1:2? In other words, the procedure deletes every second zero. The first zero occurs when 1024 = 2 * 512, so 1024 is left as it is. The second zero occurs when 2 * 1024 = 2048, so 2048 becomes 248. When z = 2 and every second zero is deleted, the sequence begins like this:

1 * 2 = 2
2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 1024
1024 * 2 = 2048 → 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 50794
50794 * 2 = 101588 → 101588
101588 * 2 = 203176 → 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 92704
92704 * 2 = 185408 → 18548

This sequence also has a ceiling and repeats at step 9134 with 5458864, which matches 5458864 at step 4166. And what about the sequence in which z = 3 and every third zero is deleted? Does this have a ceiling or does the act of multiplying by 2 compensate for the slower removal of zeros?

In fact, it can’t do so. The larger 2n becomes, the more zeros it will tend to contain. If 2n is large enough to contain 3 zeros on average, the deletion of zeros will overpower multiplication by 2 and the sequence will not rise any higher. Therefore the sequence that deletes every third zero will eventually repeat, although I haven’t been able to discover the relevant number.

But this reasoning applies to any rate, 1:z, of zero-deletion. If z = 100 and every hundredth zero is deleted, numbers in the sequence will rise to the point at which 2n contains sufficient zeros on average to counteract multiplication by 2. The sequence will have a ceiling and will eventually repeat. If z = 10^100 or z = 10^(10^100) and every googolth or googolplexth zero is deleted, the same is true. For any rate, 1:z, at which zeros are deleted, the sequence n = zerodelete(2n,z) has an upper limit and will eventually repeat.

Summus

I’m interested in digit-sums and in palindromic numbers. Looking at one, I found the other. It started like this: 9^2 = 81 and 9 = 8 + 1, so digitsum(9^1) = digitsum(9^2). I wondered how long such a sequence of powers could be (excluding powers of 10). I quickly found that the digit-sum of 468 is equal to the digit-sum of its square and cube:

digsum(468) = digsum(219024) = digsum(102503232)

But I couldn’t find any longer sequence, although plenty of other numbers are similar to 468:

digsum(585) = digsum(342225) = digsum(200201625)
digsum(4680) = digsum(21902400) = digsum(102503232000)
digsum(5850) = digsum(34222500) = digsum(200201625000)
digsum(5851) = digsum(34234201) = digsum(200304310051)
digsum(5868) = digsum(34433424) = digsum(202055332032)
digsum(28845) = digsum(832034025) = digsum(24000021451125) […]
digsum(589680) = digsum(347722502400) = digsum(205045005215232000)

What about other bases? First came this sequence:

digsum(2) = digsum(11) (base = 3) (highest power = 2)

Then these:

digsum(4) = digsum(22) = digsum(121) (b=7) (highest power = 3)
digsum(8) = digsum(44) = digsum(242) = digsum(1331) (b=15) (hp=4)
digsum([16]) = digsum(88) = digsum(484) = digsum(2662) = digsum(14641) (b=31) (hp=5)

The pattern continues (a number between square brackets represents a single digit in the base):

digsum([32]) = digsum([16][16]) = digsum(8[16]8) = digsum(4[12][12]4) = digsum(28[12]82) = digsum(15[10][10]51) (b=63) (hp=6)
digsum([64]) = digsum([32][32]) = digsum([16][32][16]) = digsum(8[24][24]8) = digsum(4[16][24][16]4) = digsum(2[10][20][20][10]2) = digsum(16[15][20][15]61) (b=127) (hp=7)
digsum([128]) = digsum([64][64]) = digsum([32][64][32]) = digsum([16][48][48][16]) = digsum(8[32][48][32]8) = digsum(4[20][40][40][20]4) = digsum(2[12][30][40][30][12]2) = digsum(17[21][35][35][21]71) (b=255) (hp=8)
digsum([256]) = digsum([128][128]) = digsum([64][128][64]) = digsum([32][96][96][32]) = digsum([16][64][96][64][16]) = digsum(8[40][80][80][40]8) = digsum(4[24][60][80][60][24]4) = digsum(2[14][42][70][70][42][14]2) = digsum(18[28][56][70][56][28]81) (b=511) (hp=9)

After this, I looked at sequences in which n(i) = n(i-1) + digitsum(n(i-1)). How long could digitsum(n(i)) be greater than or equal to digitsum(n(i-1))? In base 10, I found these sequences:

1 (digitsum=1) → 2 → 4 → 8 → 16 (sum=7) (count=4) (base=10)
9 → 18 (sum=9) → 27 (s=9) → 36 (s=9) → 45 (s=9) → 54 (s=9) → 63 (s=9) → 72 (s=9) → 81 (s=9) → 90 (s=9) → 99 (s=18) → 117 (s=9) (c=11) (b=10)
801 (s=9) → 810 (s=9) → 819 (s=18) → 837 (s=18) → 855 (s=18) → 873 (s=18) → 891 (s=18) → 909 (s=18) → 927 (s=18) → 945 (s=18) → 963 (s=18) → 981 (s=18) → 999 (s=27) → 1026 (s=9) (c=13)

Base 2 does better:

1 → 10 (s=1) → 11 (s=2) → 101 (s=2) → 111 (s=3) → 1010 (s=2) (c=5) (b=2)
16 = 10000 (s=1) → 10001 (s=2) → 10011 (s=3) → 10110 (s=3) → 11001 (s=3) → 11100 (s=3) → 11111 (s=5) → 100100 (s=2) (c=7) (b=2)
962 = 1111000010 (s=5) → 1111000111 (s=7) → 1111001110 (s=7) → 1111010101 (s=7) → 1111011100 (s=7) → 1111100011 (s=7) → 1111101010 (s=7) → 1111110001 (s=7) → 1111111000 (s=7) → 1111111111 (s=10) → 10000001001 (s=3) (c=10) (b=2)
524047 = 1111111111100001111 (s=15) → 1111111111100011110 (s=15) → 1111111111100101101 (s=15) → 1111111111100111100 (s=15) → 1111111111101001011 (s=15) → 1111111111101011010 (s=15) → 1111111111101101001(s=15) → 1111111111101111000 (s=15) → 1111111111110000111 (s=15) → 1111111111110010110 (s=15) → 1111111111110100101 (s=15) → 1111111111110110100 (s=15) → 1111111111111000011 (s=15) → 1111111111111010010 (s=15) → 1111111111111100001 (s=15) → 1111111111111110000 (s=15) → 1111111111111111111 (s=19) → 10000000000000010010 (s=3) (c=17) (b=2)

The best sequence I found in base 3 is shorter than in base 10, but there are more sequences:

1 → 2 → 11 (s=2) → 20 (s=2) → 22 (s=4) → 110 (s=2) (c=5) (b=3)
31 = 1011 (s=3) → 1021 (s=4) → 1102 (s=4) → 1120 (s=4) → 1201 (s=4) → 1212 (s=6) → 2002 (s=4) (c=6) (b=3)
54 = 2000 (s=2) → 2002 (s=4) → 2020 (s=4) → 2101 (s=4) → 2112 (s=6) → 2202 (s=6) → 2222 (s=8) → 10021(s=4) (c=7) (b=3)
432 = 121000 (s=4) → 121011 (s=6) → 121101 (s=6) → 121121 (s=8) → 121220 (s=8) → 122012 (s=8) → 122111 (s=8) → 122210 (s=8) → 200002 (s=4) (c=8) (b=3)
648 = 220000 (s=4) → 220011 (s=6) → 220101 (s=6) → 220121 (s=8) → 220220 (s=8) → 221012 (s=8) → 221111 (s=8) → 221210 (s=8) → 222002 (s=8) → 222101 (s=8) → 222200 (s=8) → 222222 (s=12) → 1000102 (s=4) (c=12) (b=3)

And what about sequences in which digitsum(n(i)) is always greater than digitsum(n(i-1))? Base 10 is disappointing:

1 → 2 → 4 → 8 → 16 (sum=7) (count=4) (base=10)
50 (s=5) → 55 (s=10) → 65 (s=11) → 76 (s=13) → 89 (s=17) → 106 (s=7) (c=5) (b=10)

Some other bases do better:

2 = 10 (s=1) → 11 (s=2) → 101 (s=2) (c=2) (b=2)
4 = 100 (s=1) → 101 (s=2) → 111 (s=3) → 1010 (s=2) (c=3) (b=2)
240 = 11110000 (s=4) → 11110100 (s=5) → 11111001 (s=6) → 11111111 (s=8) → 100000111 (s=4) (c=4) (b=2)

1 → 2 → 11 (s=2) (c=2) (b=3)
19 = 201 (s=3) → 211 (s=4) → 222 (s=6) → 1012 (s=4) (c=3) (b=3)
58999 = 2222221011 (s=15) → 2222221201 (s=16) → 2222222022 (s=18) → 2222222222 (s=20) → 10000000201 (s=4) (c=4) (b=3)

1 → 2 → 10 (s=1) (c=2) (b=4)
4 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 23 (s=5) → 100 (s=1) (c=4) (b=4)
977 = 33101 (s=8) → 33121 (s=10) → 33203 (s=11) → 33232 (s=13) → 33323 (s=14) → 100021 (s=4) (c=5) (b=4)

1 → 2 → 4 → 13 (s=4) (c=3) (b=5)
105 = 410 (s=5) → 420 (s=6) → 431 (s=8) → 444 (s=12) → 1021 (s=4) (c=4) (b=5)

1 → 2 → 4 → 12 (s=3) (c=3) (b=6)
13 = 21 (s=3) → 24 (s=6) → 34 (s=7) → 45 (s=9) → 102 (s=3) (c=4) (b=6)
396 = 1500 (s=6) → 1510 (s=7) → 1521 (s=9) → 1534 (s=13) → 1555 (s=16) → 2023 (s=7) (c=5) (b=6)

1 → 2 → 4 → 11 (s=2) (c=3) (b=7)
121 = 232 (s=7) → 242 (s=8) → 253 (s=10) → 266 (s=14) → 316 (s=10) (c=4) (b=7)
205 = 412 (s=7) → 422 (s=8) → 433 (s=10) → 446 (s=14) → 466 (s=16) → 521 (s=8) (c=5) (b=7)

1 → 2 → 4 → 10 (s=1) (c=3) (b=8)
8 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 17 (s=8) → 27 (s=9) → 40 (s=4) (c=5) (b=8)
323 = 503 (s=8) → 513 (s=9) → 524 (s=11) → 537 (s=15) → 556 (s=16) → 576 (s=18) → 620 (s=8) (c=6) (b=8)

1 → 2 → 4 → 8 → 17 (s=8) (c=4) (b=9)
6481 = 8801 (s=17) → 8820 (s=18) → 8840 (s=20) → 8862 (s=24) → 8888 (s=32) → 10034 (s=8) (c=5) (b=9)

1 → 2 → 4 → 8 → 16 (s=7) (c=4) (b=10)
50 (s=5) → 55 (s=10) → 65 (s=11) → 76 (s=13) → 89 (s=17) → 106 (s=7) (c=5) (b=10)

1 → 2 → 4 → 8 → 15 (s=6) (c=4) (b=11)
1013 = 841 (s=13) → 853 (s=16) → 868 (s=22) → 888 (s=24) → 8[10][10] (s=28) → 925 (s=16) (c=5) (b=11)

1 → 2 → 4 → 8 → 14 (s=5) (c=4) (b=12)
25 = 21 (s=3) → 24 (s=6) → 2[10] (s=12) → 3[10] (s=13) → 4[11] (s=15) → 62 (s=8) (c=5) (b=12)
1191 = 833 (s=14) → 845 (s=17) → 85[10] (s=23) → 879 (s=24) → 899 (s=26) → 8[11][11] (s=30) → 925 (s=16) (c=6) (b=12)

1 → 2 → 4 → 8 → 13 (s=4) (c=4) (b=13)
781 = 481 (s=13) → 491 (s=14) → 4[10]2 (s=16) → 4[11]5 (s=20) → 4[12][12] (s=28) → 521 (s=8) (c=5) (b=13)
19621 = 8[12]14 (s=25) → 8[12]33 (s=26) → 8[12]53 (s=28) → 8[12]75 (s=32) → 8[12]9[11] (s=40) → 8[12][12][12] (s=44) → 9034 (s=16) (c=6) (b=13)

1 → 2 → 4 → 8 → 12 (s=3) (c=4) (b=14)
72 = 52 (s=7) → 59 (s=14) → 69 (s=15) → 7[10] (s=17) → 8[13] (s=21) → [10]6 (s=16) (c=5) (b=14)
1275 = 671 (s=14) → 681 (s=15) → 692 (s=17) → 6[10]5 (s=21) → 6[11][12] (s=29) → 6[13][13] (s=32) → 723 (s=12) (c=6) (b=14)
19026 = 6[13]10 (s=20) → 6[13]26 (s=27) → 6[13]45 (s=28) → 6[13]65 (s=30) → 6[13]87 (s=34) → 6[13][10][13] (s=42) → 6[13][13][13] (s=45) → 7032 (s=12) (c=7) (b=14)

1 → 2 → 4 → 8 → 11 (s=2) (c=4) (b=15)
603 = 2[10]3 (s=15) → 2[11]3 (s=16) → 2[12]4 (s=18) → 2[13]7 (s=22) → 2[14][14] (s=30) → 31[14] (s=18) (c=5) (b=15)
1023 = 483 (s=15) → 493 (s=16) → 4[10]4 (s=18) → 4[11]7 (s=22) → 4[12][14] (s=30) → 4[14][14] (s=32) → 521 (s=8) (c=6) (b=15)
1891 = 861 (s=15) → 871 (s=16) → 882 (s=18) → 895 (s=22) → 8[10][12] (s=30) → 8[12][12] (s=32) → 8[14][14] (s=36) → 925 (s=16) (c=7) (b=15)

1 → 2 → 4 → 8 → 10 (s=1) (c=4) (b=16)
16 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 17 (s=8) → 1[15] (s=16) → 2[15] (s=17) → 40 (s=4) (c=6) (b=16)
1396 = 574 (s=16) → 584 (s=17) → 595 (s=19) → 5[10]8 (s=23) → 5[11][15] (s=31) → 5[13][14] (s=32) → 5[15][14] (s=34) → 620 (s=8) (c=7) (b=16)
2131 = 853 (s=16) → 863 (s=17) → 874 (s=19) → 887 (s=23) → 89[14] (s=31) → 8[11][13] (s=32) → 8[13][13] (s=34) → 8[15][15] (s=38) → 925 (s=16) (c=8) (b=16)

1 → 2 → 4 → 8 → [16] (s=16) → 1[15] (s=16) (c=5) (b=17)

1 → 2 → 4 → 8 → [16] (s=16) → 1[14] (s=15) (c=5) (b=18)
5330 = [16]82 (s=26) → [16]9[10] (s=35) → [16][11]9 (s=36) → [16][13]9 (s=38) → [16][15][11] (s=42) → [16][17][17] (s=50) → [17]2[13] (s=32) (c=6) (b=18)

1 → 2 → 4 → 8 → [16] (s=16) → 1[13] (s=14) (c=5) (b=19)
116339 = [16][18]52 (s=41) → [16][18]75 (s=46) → [16][18]9[13] (s=56) → [16][18][12][12] (s=58) → [16][18][15][13] (s=62) → [16][18][18][18] (s=70) → [17]03[12] (s=32) (c=6) (b=19)

1 → 2 → 4 → 8 → [16] (s=16) → 1[12] (s=13) (c=5) (b=20)
100 = 50 (s=5) → 55 (s=10) → 5[15] (s=20) → 6[15] (s=21) → 7[16] (s=23) → 8[19] (s=27) → [10]6 (s=16) (c=6) (b=20)
135665 = [16][19]35 (s=43) → [16][19]58 (s=48) → [16][19]7[16] (s=58) → [16][19][10][14] (s=59) → [16][19][13][13] (s=61) → [16][19][16][14] (s=65) → [16][19][19][19] (s=73) → [17]03[12] (s=32) (c=7) (b=20)