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.

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

Author

Vladimir Shevelev, Apr 09 2014

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

Cf. A240537, A240606, A240619.

Programs

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

Extensions

a(17)-a(18) corrected and a(19)-a(34) added by Hiroaki Yamanouchi, Sep 05 2014