A232094 a(n) = A060130(A000217(n)); number of nonzero digits in factorial base representation (A007623) of 0+1+2+...+n.

0, 1, 2, 1, 2, 3, 3, 2, 2, 4, 3, 2, 2, 3, 4, 1, 3, 5, 4, 4, 3, 5, 3, 3, 3, 4, 5, 2, 4, 4, 5, 3, 2, 5, 4, 3, 3, 4, 4, 3, 3, 5, 6, 5, 4, 5, 3, 3, 3, 4, 5, 3, 5, 6, 5, 3, 4, 6, 5, 4, 4, 5, 6, 3, 5, 6, 4, 4, 4, 5, 5, 4, 4, 5, 5, 4, 4, 4, 6, 5, 2, 6, 5, 3, 4, 4, 5

Offset: 0

Author

Antti Karttunen, Nov 18 2013

Keywords

• nonn
• base

Comments

The next 1 after a(1), a(3) and a(15) occurs at n=224, as A000217(224) = 25200 = 5 * 7!.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000217, A007623, A060130, A226061, A230410, A232095.

Programs

Mathematica

a[n_] := Module[{k = n*(n+1)/2, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Count[s, ?(# > 0 &)]]; Array[a, 100, 0] (* _Amiram Eldar, Feb 07 2024 *)

Scheme

(define (A232094 n) (A060130 (A000217 n)))

Formula

a(n) = A060130(A000217(n)).

a(n) = A230410(A226061(n+1)). [Not a practical way to compute this sequence. Please see comments at A230410.]