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

1,1

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

Cf. A240537, A240606, A240619.

Sequence in context: A096474 A220439 A363324 * A344328 A168282 A122762

Adjacent sequences: A240617 A240618 A240619 * A240621 A240622 A240623

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