A092984 - OEIS

A092984

a(n) = the least k >= 1 such that n! + k is squarefree.

1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

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OFFSET

1,4

COMMENTS

Conjecture: There exists a finite k such that a(n) < k for all n. Subsidiary sequence: Index of the first occurrence of n in this sequence. In case the conjecture is true, this sequence would be finite.

If a(n) = 2 ==> n!+1 is divisible by a square (sequence A064237). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 29 2004

LINKS

Table of n, a(n) for n=1..105.

FORMULA

a(n) = A092983(n) - n!.

EXAMPLE

a(5) = 2 = 122 - 5! = 122 - 120 (as 121 = 11^2 is not squarefree).

MATHEMATICA

Table[SelectFirst[Range@ 10, SquareFreeQ[n! + #] &], {n, 45}] (* Michael De Vlieger, Aug 23 2017 *)

PROG

(PARI) a(n)=for(i=1, n!, if(issquarefree(n!+i), return(i)))

(PARI) A092984(n) = { my(k=1); while(!issquarefree(n!+k), k++); k; }; \\ Antti Karttunen, Aug 22 2017