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A056193
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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.
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27
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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
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,fini
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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