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A216814
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Given n and a constant C, define a sequence b(m) by the recurrence in the comments; a(n) = smallest positive integer C such that for some prime p the denominators of all b(m) are powers of p (conjectured).
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0
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2, 4, 10, 12, 84, 264, 990, 2860, 9724, 18564, 117572, 45220, 19380, 1782960, 6463230, 25092540, 58549260, 95527740
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OFFSET
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2,1
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COMMENTS
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The sequence b(m) is defined by b(1)=C, and for m>=2, b(m) = (1/(2*binomial(m+n-1,m-1))) * Sum_{k=1..m-1} binomial(m+n-1,m-k-1)*binomial(m+n-1,k-1)*b(k)*b(m-k).
For n=2..19, the corresponding primes p are 3, 5, 7, 7, 11, 13, 13, 17, 19, 19, 23, 23, 23, 29, 31, 31, 31, 37.
The terms up to a(19) have been obtained by generating the first 2000 terms of the relative sequences. - Giovanni Resta, Oct 07 2019
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LINKS
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MATHEMATICA
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ppQ[n_] := n == 1 || PrimePowerQ[n]; isOk[n_, c_, mmax_] := Block[{d, p=1, ret=True, vb = 0 Range@ mmax}, vb[[1]] = c; Do[ vb[[m]] = (1/(2 * Binomial[m+n-1, m-1]) Sum[ Binomial[ m+n-1, m-k-1] * Binomial[m+n-1, k-1] * vb[[k]]*vb[[m - k]], {k, m-1}]); If[! ppQ[d = Denominator[vb[[m]]]], ret = False; Break[]]; If[d != 1, d = FactorInteger[d][[1, 1]]; If[p == 1, p = d, If[p != d, ret = False; Break[]]]], {m, 2, mmax}]; ret]; a[n_] := Block[{c = 1}, While[! isOk[n, c, 100], c++]; c]; a/@ Range[2, 10] (* Giovanni Resta, Oct 07 2019 *)
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PROG
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(PARI) ispp(n) = (n==1) || isprimepower(n);
isokC(n, C, mmax) = {my(vb = vector(mmax)); vb[1] = C; for (m=2, mmax, vb[m] = (1/(2*binomial(m+n-1, m-1))*sum(k=1, m-1, binomial(m+n-1, m-k-1)*binomial(m+n-1, k-1)*vb[k]*vb[m-k])); if (!ispp(denominator(vb[m])), return (0)); ); return (1); }
a(n) = {my(C=1, mmax = 1000); while(!isokC(n, C, mmax), C++); C; } \\ Michel Marcus, Sep 29 2019
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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Conjectured added to name and a(10)-a(15) from Michel Marcus, Oct 06 2019
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STATUS
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approved
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