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%I #11 Feb 26 2019 10:51:46
%S 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,
%T 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,
%U 17,5,6,1,17,3,10,1,35,1,2,12,19,7,6,1,25
%N Let S(n)_k be the smallest positive integer t that t!k is a multiple of n (t!k is k-tuple factorial of t); then a(n) is the smallest k for which S(n)_k = n.
%C If p is prime, a(p) = 1.
%C 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.
%C 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.
%C 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.
%H J. Sondow and E. W. Weisstein, <a href="http://mathworld.wolfram.com/SmarandacheFunction.html">MathWorld: Smarandache Function</a>
%e a(8) = 3 because:
%e - 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);
%e - 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);
%e - 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.
%Y Cf. A002034, A005383, A007922, A036570, A063917.
%K nonn
%O 1,6
%A _Lechoslaw Ratajczak_, Feb 17 2019
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