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Second step in Goodstein sequences, i.e., g(4) if g(2)=n: (first step) write g(2)=n in hereditary representation base 2, bump to base 3, then subtract 1 to produce g(3)=A056004(n), then (second step) write g(3) in hereditary representation base 3, bump to base 4, then subtract 1 to produce g(4).
18

%I #28 Apr 24 2024 11:11:54

%S 1,3,41,255,257,259,553,1023,1025,1027,1065,1279,1281,1283,

%T 50973998591214355139406377,

%U 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084095

%N Second step in Goodstein sequences, i.e., g(4) if g(2)=n: (first step) write g(2)=n in hereditary representation base 2, bump to base 3, then subtract 1 to produce g(3)=A056004(n), then (second step) write g(3) in hereditary representation base 3, bump to base 4, then subtract 1 to produce g(4).

%H Reinhard Zumkeller, <a href="/A057650/b057650.txt">Table of n, a(n) for n = 2..1000</a>

%H R. L. Goodstein, <a href="http://www.jstor.org/stable/2268019">On the Restricted Ordinal Theorem</a>, J. Symb. Logic 9, 33-41, 1944.

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

%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 a(12)=1065 since with g(2) = 12 = 2^(2+1) + 2^2, we get g(3) = 3^(3+1) + 3^3 - 1 = 107 = 3^(3+1) + 2*3^2 + 2*3 + 2 and g(4) = 4^(4+1) + 2*4^2 + 2*4 + 2 - 1 = 1065.

%e a(17) = 4^(4^4) - 1, with g(2) = 17 = 2^(2^2) + 1 and g(3) = 3^(3^3).

%e Similarly a(18) = 4^(4^4) + 1, with g(2) = 18 = 2^(2^2) + 2 and g(3) = 3^(3^3) + 2.

%o (Haskell) -- See Link

%Y Cf. A056004, A059933, A059934, A059935, A059936.

%K nonn

%O 2,2

%A _Henry Bottomley_, Oct 13 2000

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