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A241786
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Smallest k such that the number of the first even exponents in prime power factorization of (2*k)! is n, or a(n)=0 if there is no such k.
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1
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1, 6, 3, 5, 10, 24, 27, 169, 924, 3168, 720, 3208, 408, 35421, 50878, 73920, 18757, 204513, 134418, 295680, 427684, 2746710, 6867848, 14476645, 7278558, 3668406, 737564, 245340483, 1931850660, 1514239096, 3228582476, 1325085081, 16188866895, 33517640073
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OFFSET
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0,2
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COMMENTS
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Conjecture: 1) All a(n)>0; 2) a(2*n+1)>a(2*n).
Conjecture (2) is wrong because a(24) = 7278558 >= a(25) = 3668406.
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REFERENCES
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P. Erdős, P. L. Graham, Old and new problems and results in combinatorial number theory, L'Enseignement Mathematique, Imprimerie Kunding, Geneva, 1980.
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LINKS
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EXAMPLE
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a(2)=3, since (2*3)!= 2^4*3^2*5, and here the number of the first even exponents is 2.
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PROG
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(PARI) nbev(n) = {f = factor(n); nbe = 0; i = 1; while ((i <= #f~) && ((f[i, 2] % 2) == 0), i++; nbe++); nbe; }
a(n) = {k = 0; while(nbev((2*k)!) != n, k++); k; } \\ Michel Marcus, Apr 30 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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