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Goodstein sequence starting with 15.
25

%I #24 Apr 24 2024 11:12:21

%S 15,111,1283,18752,326593,6588344,150994943,3524450280,100077777775,

%T 3138578427934,106993479003783,3937376861542204,155568096352467863,

%U 6568408356994335930,295147905181357143919,14063084452070776884879,708235345355342213988445

%N Goodstein sequence starting with 15.

%C To calculate a(n+1), write a(n) in the hereditary representation base n+2, then bump the base to n+3, then subtract 1;

%C Compare to A222113: the underlying variants to define Goodstein sequences are equivalent.

%H Nicholas Matteo, <a href="/A222117/b222117.txt">Table of n, a(n) for n = 0..383</a> (first 249 terms from Reinhard Zumkeller)

%H R. L. Goodstein, <a href="https://www.jstor.org/stable/2268019">On the Restricted Ordinal Theorem</a>, The Journal of Symbolic Logic, Vol. 9, No. 2, Jun., 1944

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GoodsteinSequence.html">Goodstein Sequence</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Goodstein&#39;s_theorem">Goodstein's Theorem</a>

%H Reinhard Zumkeller, <a href="/A211378/a211378.hs.txt">Haskell programs for Goodstein sequences</a>

%e The first terms are:

%e a(0) = 2^(2+1) + 2^2 + 2^1 + 2^0 = 15;

%e a(1) = 3^(3+1) + 3^3 + 3^1 + 3^0 - 1 = 111;

%e a(2) = 4^(4+1) + 4^4 + 4^1 - 1 = 4^(4+1) + 4^4 + 3*4^0 = 1283;

%e a(3) = 5^(5+1) + 5^5 + 3*5^0 - 1 = 5^(5+1) + 5^5 + 2*5^0 = 18752;

%e a(4) = 6^(6+1) + 6^6 + 2*6^0 - 1 = 6^(6+1) + 6^6 + 1 = 326593;

%e a(5) = 7^(7+1) + 7^7 + 1 - 1 = 6588344;

%e a(6) = 8^(8+1) + 8^8 - 1 = 150994943.

%o (Haskell) -- See Link

%o (PARI) lista(nn) = {print1(a = 15, ", "); for (n=2, nn, pd = Pol(digits(a, n)); q = sum(k=0, poldegree(pd), if (c=polcoeff(pd, k), c*x^subst(Pol(digits(k, n)), x, n+1), 0)); a = subst(q, x, n+1) - 1; print1(a, ", "); ); } \\ _Michel Marcus_, Feb 24 2016

%Y Cf. A215409 (start=3), A056193 (start=4), A059933 (start=16), A211378 (start=19).

%K nonn,fini

%O 0,1

%A _Reinhard Zumkeller_, Feb 13 2013

%E Offset changed to 0 by _Nicholas Matteo_, Aug 21 2019