

A306467


Let S(n)_k be the smallest positive integer t that t!k is a multiple of n (t!k is ktuple factorial of t); then a(n) is the smallest k for which S(n)_k = n.


0



1, 1, 1, 1, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 3, 5, 1, 4, 1, 5, 3, 2, 1, 9, 4, 2, 7, 7, 1, 6, 1, 9, 3, 2, 5, 13, 1, 2, 3, 15, 1, 6, 1, 11, 10, 2, 1, 15, 6, 8, 3, 13, 1, 14, 5, 21, 3, 2, 1, 15, 1, 2, 14, 17, 5, 6, 1, 17, 3, 10, 1, 35, 1, 2, 12, 19, 7, 6, 1, 25
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,6


COMMENTS

If p is prime, a(p) = 1.
Conjecture: consecutive primes p satisfying the equation a(p+1) = 2 are consecutive elements of A005383 (primes p such that (p+1)/2 are also primes, for p > 3). The conjecture was checked for all primes < 10^4.
Conjecture: consecutive primes p satisfying the equations a(p+1) = 2 and a(p+2) = 3 are consecutive elements of A036570 (primes p such that (p+1)/2 and (p+2)/3 are also primes). The conjecture was checked for all primes < 10^4.
The first six solutions of the equation a(n) = a(n+1) are 1, 2, 3, 4, 9, 27. Is there a larger n? If such a number n exists, it is larger than 4000.


LINKS



EXAMPLE

a(8) = 3 because:
 for k = 1 is: 1!1, 2!1, 3!1 are not multiples of 8 and 4!1 is a multiple of 8, then (t = 4 = S(8)_1) <> (n = 8);
 for k = 2 is: 1!2, 2!2, 3!2 are not multiples of 8 and 4!2 is a multiple of 8, then (t = 4 = S(8)_2) <> (n = 8);
 for k = 3 is: 1!3, 2!3, 3!3, 4!3, 5!3, 6!3, 7!3 are not multiples of 8 and 8!3 is a multiple of 8, then (t = 8 = S(8)_3) = (n = 8), hence a(8) = k = 3.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



