A232098 a(n) is the largest m such that m! divides n^2; a(n) = A055881(n^2).

1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1

Offset

1,2

Comments

For all n, A055881(n) <= a(n), and probably also a(n) <= A055874(n).

Moreover, a(n) > A055881(n) if and only if A055874(n) > A055881(n), thus A055926 gives (also) all the positions where this sequence differs from A055881. Please see Comments section in A055926 for the proof.

Differs from A055874 for the first time at n=840, where a(840)=7, while A055874(840)=8. A232099 gives all the positions where such differences occur.

Links

Antti Karttunen, Table of n, a(n) for n = 1..10080

Formula

a(n) = A055881(A000290(n)) = A055881(n^2).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A065887(k) = 1.78672776922161809767... . - Amiram Eldar, Jan 01 2024

Prog

(Scheme)
(define (A232098 n) (A055881 (A000290 n)))

Crossrefs

Cf. A055881, A055874, A055926, A065887, A232096, A232099, A233267.

Sequence in context: A055881 A332202 A204917 * A055874 A195155 A178544

Adjacent sequences: A232095 A232096 A232097 * A232099 A232100 A232101

Keyword

nonn

Author

Antti Karttunen, Nov 18 2013

Status

approved