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%I #24 Oct 11 2015 04:45:09
%S 1,2,5,14,56,202,715,3244,12945,49742,238048,996158,3991995,19676358,
%T 84649176,347993910,1747681160,7656026268,32018897274,162900330572,
%U 722787262419,3060338457400,15718332812096,70407433197686
%N Number of admissible sequences of order j; related to 5x+1 problem
%F A sequence s(k), where k=1, 2, ..., n, is *admissible* if it satisfies s(k)=5/2 exactly j times, s(k)=1/2 exactly n-j times, s(1)*s(2)*...*s(n) < 1 but s(1)*s(2)*...*s(m) > 1 for all 1 < m < n.
%F a(1)=1 and a(k+1) = Sum_{m=1..k}(-1)^(m-1)*binomial(floor((k-m+1)*(log(5)/log(2)))+m-1, m)*a(k-m+1) for k >= 1. - _Vladimir M. Zarubin_, Sep 25 2015
%e The unique admissible sequence of order 1 is 5/2, 1/2, 1/2.
%e The two admissible sequences of order 2 are 5/2, 5/2, 1/2, 1/2, 1/2 and 5/2, 1/2, 5/2, 1/2, 1/2.
%t h[n_] := Module[{L = {{1}}}, For[i = 1, i <= n, i++, K = {}; S = 0; j = 1; While[5^i >= 2^(i + j - 1), If[5^(i - 1) >= 2^(i + j - 2), S = S + L[[i, j]]]; AppendTo[K, S]; j = j + 1]; AppendTo[L, K]; ]; Return[Map[Last, Drop[L, 1]]]]
%o (PARI)
%o n=20; a=vector(n); log52=log(5)/log(2);
%o {a[1]=1; for ( k=1, n-1, a[k+1]=sum( m=1,k,(-1)^(m-1)*binomial( floor( (k-m+1)*log52)+m-1,m)*a[k-m+1] );print1(a[k], ", ") );} \\ _Vladimir M. Zarubin_, Sep 25 2015
%Y Similar to A100982, but for the 5x+1 problem.
%K nonn
%O 1,2
%A T.M.M. Laarhoven (t.laarhoven(AT)gmail.com), Mar 29 2010
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