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A056193
Goodstein sequence starting with 4: to calculate a(n+1), write a(n) in the hereditary representation in base n+2, then bump the base to n+3, then subtract 1.
28
4, 26, 41, 60, 83, 109, 139, 173, 211, 253, 299, 348, 401, 458, 519, 584, 653, 726, 803, 884, 969, 1058, 1151, 1222, 1295, 1370, 1447, 1526, 1607, 1690, 1775, 1862, 1951, 2042, 2135, 2230, 2327, 2426, 2527, 2630, 2735, 2842, 2951, 3062, 3175, 3290, 3407
OFFSET
0,1
COMMENTS
Goodstein's theorem shows that such a sequence converges to zero for any starting value [e.g. if a(0)=1 then a(1)=0; if a(0)=2 then a(3)=0; and if a(0)=3 then a(5)=0]. With a(0)=4 we have a(3*2^(3*2^27 + 27) - 3)=0, which is well beyond the 10^(10^8)-th term.
The second half of such sequences is declining and the previous quarter is stable.
The resulting sequence 0,1,3,5,3*2^402653211 - 3, ... (see Comments in A056041) grows too rapidly to have its own entry.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 (final 2 terms from Nicholas Matteo)
R. L. Goodstein, On the Restricted Ordinal Theorem, The Journal of Symbolic Logic, Vol. 9, No. 2 (1944), 33-41.
Eric Weisstein's World of Mathematics, Goodstein Sequence.
EXAMPLE
a(0) = 4 = 2^2,
a(1) = 3^3 - 1 = 26 = 2*3^2 + 2*3 + 2,
a(2) = 2*4^2 + 2*4 + 2 - 1 = 41 = 2*4^2 + 2*4 + 1,
a(3) = 2*5^2 + 2*5 + 1 - 1 = 60 = 2*5^2 + 2*5,
a(4) = 2*6^2 + 2*6 - 1 = 83 = 2*6^2 + 6 + 5,
a(5) = 2*7^2 + 7 + 5 - 1 = 109 etc.
PROG
(Haskell) See Zumkeller link
(PARI) lista(nn) = {print1(a = 4, ", "); 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 22 2016
KEYWORD
nonn,fini
AUTHOR
Henry Bottomley, Aug 02 2000
EXTENSIONS
Edited by N. J. A. Sloane, Mar 06 2006
Offset changed to 0 by Nicholas Matteo, Sep 04 2019
STATUS
approved