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A240620
a(n) is the smallest k such that in the prime power factorization of k! at least the first n positive exponents are even.
19
6, 6, 10, 20, 48, 54, 338, 816, 816, 816, 816, 816, 37514, 37514, 37514, 37514, 268836, 268836, 591360, 855368, 1475128, 1475128, 1475128, 1475128, 1475128, 1475128, 127632241, 472077979, 472077979, 472077979, 472077979, 472077979, 472077979, 16072818386
OFFSET
1,1
COMMENTS
The sequence is nondecreasing and, by Berend's theorem, a(n) --> infinity as n goes to infinity.
The distinct terms 6, 10, 20, 48, 54, 338, 816, 37514, 268836, ... repeat 2, 1, 1, 1, 1, 1, 5, 4, 2, ... times.
REFERENCES
P. Erdős, P. L. Graham, Old and new problems and results in combinatorial number theory, L'Enseignement Mathématique, Imprimerie Kunding, Geneva, 1980.
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..46 (first 36 terms from Hiroaki Yamanouchi)
D. Berend, Parity of exponents in the factorization of n!, J. Number Theory, 64 (1997), 13-19.
Y.-G. Chen, On the parity of exponents in the standard factorization of n!, J. Number Theory, 100 (2003), 326-331.
EXAMPLE
Prime power factorizations of k! for k = 2, 3, 4, 5, 6 are 2, 2*3, 2^3*3, 2^3*3*5, 2^4*3^2*5. Thus the least k having at least 1 first even exponent is 6, and 6 is also the least k having at least 2 first even exponents. So a(1) = a(2) = 6.
PROG
(PARI) isokp(n, k) = {my(fk = k!, f = factor(fk)); if (#f~ < n, return (0)); if (f[n, 1] != prime(n), return (0)); for (j=1, n, if (f[j, 2] % 2, return(0)); ); return(1); }
a(n) = {my(k=1); while (! isokp(n, k), k++); k; } \\ Michel Marcus, Feb 04 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Apr 09 2014
EXTENSIONS
a(17)-a(18) corrected and a(19)-a(34) added by Hiroaki Yamanouchi, Sep 05 2014
STATUS
approved