

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
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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), 1319.
Y.G. Chen, On the parity of exponents in the standard factorization of n!, J. Number Theory, 100 (2003), 326331.


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: A301690 A096474 A220439 * A168282 A122762 A046605
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



