A372237 a(0) = 4; to obtain a(k), write out the base-(2^k) expansion of a(k-1), bump to base 2^(k+1), then subtract 1.

4, 15, 26, 49, 96, 191, 318, 573, 1084, 2107, 4154, 8249, 16440, 32823, 65590, 131125, 262196, 524339, 1048626, 2097201, 4194352, 8388655, 16777262, 33554477, 67108908, 134217771, 268435498, 536870953, 1073741864, 2147483687, 4294967334, 8589934629, 17179869220

Offset

0,1

Comments

Applying to the proof of the usual Goodstein's theorem to the ordinal number omega^omega shows that: for no matter what initial value and no matter what increasing sequence of bases b(0), b(1), ... with b(0) >= 2, the (weak) Goodstein sequence eventually terminates with 0. Here b(k) = 2^(k+1).

Sequence terminates at a(2^(2^70+70) + 2^70 + 68) = 0.

Links

Jianing Song, Table of n, a(n) for n = 0..1000

Googology Wiki, Goodstein sequence.

Wikipedia, Goodstein's Theorem

Formula

a(k) = 2^(k+2) + 68 - k for 5 <= k <= 68. The base-(2^(k+1)) expansion of a(k) consists of two digits 2 and 68 - k.

a(k) = 2^(k+1) + 2^70 + 68 - k for 69 <= 2^70 + 68. The base-(2^(k+1)) expansion of a(k) consists of two digits 1 and 2^70 + 68 - k.

a(k) = 2^(2^70+70) + 2^70 + 68 - k for 2^70 + 69 <= k <= 2^(2^70+70) + 2^70 + 68. The base-(2^(k+1)) expansion of a(k) consists of a single digit 2^(2^70+70) + 2^70 + 68 - k.

Example

a(0) = 100_2 = 4;
a(1) = 100_4 - 1 = 15 = 33_4;
a(2) = 33_8 - 1 = 26 = 32_8;
a(3) = 32_16 - 1 = 49 = 31_16;
a(4) = 31_32 - 1 = 96 = 30_32;
a(5) = 30_64 - 1 = 191 = (2,63)_64.

Prog

(PARI) A372237_first_N_terms(N) = my(v=vector(N+1)); v[1] = 4; for(i=1, N, v[i+1] = fromdigits(digits(v[i], 2^i), 2^(i+1))-1); v

Crossrefs

Cf. A056193, A266202.

Sequence in context: A061873 A017437 A281264 * A366869 A267103 A030553

Adjacent sequences: A372234 A372235 A372236 * A372238 A372239 A372240

Keyword

nonn,easy,fini

Author

Jianing Song, Apr 23 2024

Status

approved