A063917 -id:A063917 - cp's OEIS frontend

A306467 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.

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

Offset: 1

Lechoslaw Ratajczak, Feb 17 2019

nonn

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.

			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.

J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function

Cf. A002034, A005383, A007922, A036570, A063917.

A322445 Smallest positive integer m such that n divides A297707(m).

Original entry on oeis.org

1, 2, 3, 4, 5, 3, 7, 4, 3, 5, 11, 4, 13, 7, 5, 4, 17, 3, 19, 5, 7, 11, 23, 4, 5, 13, 6, 7, 29, 5, 31, 4, 11, 17, 7, 5, 37, 19, 13, 5, 41, 7, 43, 11, 5, 23, 47, 4, 7, 5, 17, 13, 53, 6, 11, 7, 19, 29, 59, 5, 61, 31, 7, 4, 13

Offset: 1

Lechoslaw Ratajczak, Dec 08 2018

nonn

If p is prime, a(p) = p.
The first three integers n for which a(n!) is not a prime number are: 1 (a(1!) = 1), 4 (a(4!) = 4), 10 (a(10!) = 8). Is there a larger n? If such a number n exists, it is greater than 2000.
The smallest integer n satisfying the equation a(n) = a(n+1) is 2400 (a(2400) = a(2401) = 7). Is there a larger n? If such a number n exists, it is greater than 3000.

			a(12) = 4 because 12 is not divisible by A297707(1) = 1, A297707(2) = 2*1, A297707(3) = 3*2*1*3*1, and is divisible by A297707(4) = 4*3*2*1*4*2*4*1.

J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function

Cf. A002034, A007922, A063917.

f[n_] := n^(n - 1) *  Product[k^DivisorSigma[0, n - k], {k, n - 1}]; a[n_] := Module[{k = 1}, While[! Divisible[f[k], n], k++]; k]; Array[a, 60] (* Amiram Eldar, Dec 08 2018 *)

f(n) = (n^(n-1))*prod(k=1, n-1, k^numdiv(n-k)); \\ A297707
a(n) = {my(k=1); while (f(k) % n, k++); k;} \\ Michel Marcus, Dec 09 2018

cp's OEIS Frontend

A306467 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.

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