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A174795 Number of admissible sequences of order j; related to 5x+1 problem 2
1, 2, 5, 14, 56, 202, 715, 3244, 12945, 49742, 238048, 996158, 3991995, 19676358, 84649176, 347993910, 1747681160, 7656026268, 32018897274, 162900330572, 722787262419, 3060338457400, 15718332812096, 70407433197686 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
FORMULA
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.
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
EXAMPLE
The unique admissible sequence of order 1 is 5/2, 1/2, 1/2.
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.
MATHEMATICA
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]]]]
PROG
(PARI)
n=20; a=vector(n); log52=log(5)/log(2);
{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
CROSSREFS
Similar to A100982, but for the 5x+1 problem.
Sequence in context: A267559 A047136 A047042 * A243551 A110043 A006847
KEYWORD
nonn
AUTHOR
T.M.M. Laarhoven (t.laarhoven(AT)gmail.com), Mar 29 2010
STATUS
approved

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Last modified March 29 00:26 EDT 2024. Contains 371264 sequences. (Running on oeis4.)