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

Author

Jianing Song, Apr 23 2024

Keywords

• nonn
• easy
• fini

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.

Examples

			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.
		

Links

• Jianing Song, Table of n, a(n) for n = 0..1000
• Googology Wiki, Goodstein sequence.
• Wikipedia, Goodstein's Theorem

Crossrefs

Cf. A056193, A266202.

Programs

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

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.