A137411 Weak Goodstein sequence starting at 11.

11, 30, 67, 127, 217, 343, 511, 636, 775, 928, 1095, 1276, 1471, 1680, 1903, 2139, 2389, 2653, 2931, 3223, 3529, 3849, 4183, 4531, 4893, 5269, 5659, 6063, 6481, 6913, 7359, 7818, 8291, 8778, 9279, 9794, 10323, 10866, 11423, 11994, 12579, 13178

Offset: 0

Author

Nicholas Matteo (kundor(AT)kundor.org), Apr 15 2008

Keywords

• nonn
• fini

Comments

The sequence eventually goes to zero, as can be seen by noting that multiples of the highest exponent (3 in this case) only go down; in fact the 8th term, a(8) = 7*8^2 + 7*8 + 7 = 511; after which the multiple of the square term will only go down, etc.

This sequence, for 11, grows beyond the quintillions of digits before going to zero.

From Zhuorui He, Aug 07 2025: (Start)

For more info see A266201-A266202.

This sequence has A266203(11)+1 terms and a(A266203(11))=0 is the last term of this sequence. The maximum term in this sequence is a((A266203(11)-1)/2)=(A266203(11)+1)/2. 10^^8 < A266203(11) < 10^^9.

More precisely, 10^(10^(10^(10^(10^(10^(10^619.29937)))))) < A266203(11) < 10^(10^(10^(10^(10^(10^(10^619.299371)))))). (End)

Examples

			a(0) = 11 = 2^3 + 2^1 + 2^0
a(1) = 3^3 + 3^1 + 3^0 - 1 = 30
a(2) = 4^3 + 4^1 - 1 = 4^3 + 3*4^0 = 67
		

References

• K. Hrbacek and T. Jech, Introduction to Set Theory, Taylor & Francis Group, 1999, pp. 125-127.

Links

• Zhuorui He, Table of n, a(n) for n = 0..10000 (Terms 0..998 from Harvey P. Dale)

Crossrefs

Cf. A056004 (strong Goodstein sequences), A059933 (strong Goodstein sequence for 16.).

Weak Goodstein sequences: A267647: g_n(4); A267648: g_n(5); A271987: g_n(6); A271988: g_n(7); A271989: g_n(8); A271990: g_n(9); A271991: g_n(10); A137411: g_n(11); A271992: g_n(16); A265034: g_n(266); A266202: g_n(n); A266203: a(n)=k such that g_k(n)=0;

Programs

Mathematica

nxt[{n_,a_}]:={n+1,FromDigits[IntegerDigits[a,n+1],n+2]-1}; Transpose[ NestList[ nxt,{1,11},50]][[2]] (* Harvey P. Dale, Feb 09 2015 *)

PARI

a(n, m=11) = { my(wn = m); for (k=2, n+1, wn = fromdigits(digits(wn, k), k+1) - 1); wn; } \\ Zhuorui He, Aug 08 2025

Formula

To obtain a(n + 1), write a(n) in base n + 2, increase the base to n + 3 and subtract 1.

Extensions

Offset changed to 0 by Zhuorui He, Aug 07 2025