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

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

Crossrefs

Cf. A000142, A064237, A092983.

Sequence in context: A281191 A324496 A137454 * A338428 A297382 A258257

Adjacent sequences: A092981 A092982 A092983 * A092985 A092986 A092987

Keyword

nonn

Author

Amarnath Murthy, Mar 28 2004

Extensions

More terms from Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 29 2004

More terms from David Wasserman, Sep 27 2006

Typo in description corrected by Antti Karttunen, Aug 22 2017

Status

approved