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A241786
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.
1
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
OFFSET
0,2
COMMENTS
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.
a(35) > 10^11; a(36) = 8036409193. - Hiroaki Yamanouchi, Sep 29 2014
REFERENCES
P. Erdős, P. L. Graham, Old and new problems and results in combinatorial number theory, L'Enseignement Mathematique, Imprimerie Kunding, Geneva, 1980.
LINKS
Giovanni Resta, Table of n, a(n) for n = 0..44 (terms a(0)-a(34) and a(36) from Hiroaki Yamanouchi)
D. Berend, Parity of exponents in the factorization of n!, J. Number Theory, 64 (1997), 13-19.
EXAMPLE
a(2)=3, since (2*3)!= 2^4*3^2*5, and here the number of the first even exponents is 2.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Apr 28 2014
EXTENSIONS
More terms from Peter J. C. Moses, May 06 2014
a(21)-a(33) from Hiroaki Yamanouchi, Sep 29 2014
STATUS
approved