A174796 Number of admissible sequences of order j; related to 7x+1 problem.
1, 2, 7, 30, 143, 728, 3148, 15986, 86009, 478907, 2731365, 13131703, 72135374, 412835191, 2416852480, 14369476066, 72067537808, 409636973141, 2412844770335, 14479410843183, 87964452906330, 451313038006432
Offset: 1
Keywords
Examples
The unique admissible sequence of order 1 is 7/2, 1/2, 1/2. The two admissible sequences of order 2 are 7/2, 7/2, 1/2, 1/2, 1/2, 1/2 and 7/2, 1/2, 7/2, 1/2, 1/2, 1/2.
Programs
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Mathematica
h[n_] := Module[{L = {{1}}}, For[i = 1, i <= n, i++, K = {}; S = 0; j = 1; While[7^i >= 2^(i + j - 1), If[7^(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]]]] -
PARI
n=20; a=vector(n); log72=log(7)/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)*log72)+m-1,m)*a[k-m+1] ); print1(a[k], ", ") );} \\ Vladimir M. Zarubin, Sep 25 2015
Formula
A sequence s(k), where k=1, 2, ..., n, is *admissible* if it satisfies s(k)=7/2 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) = Sum_{m=1..k-1} (-1)^(m-1)*binomial(floor((k-m)*(log(7)/log(2)))+m-1, m)*a(k-m) for k >= 2. - Vladimir M. Zarubin, Sep 25 2015
Comments