A260736 a(0) = 0; for n >= 1, a(n) = A000035(n) + a(A257684(n)); in the factorial representation of n the number of digits with maximal possible value allowed in its location.

0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 1, 2, 2, 3, 1, 2, 1, 2, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 0

Offset: 0

Author

Antti Karttunen, Aug 27 2015

Keywords

• nonn
• base

Comments

In the factorial representation of n, given as {d_k, ..., d_3, d_2, d_1}, the maximal allowed digit for any position j is j. This sequence gives the number of digits in the whole representation [A007623(n)] that attain that maximum allowed value.

Examples

			For n=19, which has factorial representation "301", the digits at position 1 and 3, namely "1" and "3" are equal to their one-based position index, in other words, the maximal digits allowed in those positions (while "0" at position 2 is not), thus a(19) = 2.
		

Links

• Index entries for sequences related to factorial base representation.

Crossrefs

Cf. A000035, A007623, A257684.

Cf. also A257511.

Programs

Mathematica

a[n_] := Module[{k = n, m = 2, c = 0, r}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r == m - 1, c++]; m++]; c]; Array[a, 100, 0] (* Amiram Eldar, Jan 23 2024 *)

Python

from sympy import factorial as f
def a007623(n, p=2): return n if n
0 else '0' for i in x)[::-1]
    return 0 if n==1 else sum(int(y[i])*f(i + 1) for i in range(len(y)))
def a(n): return 0 if n==0 else n%2 + a(a257684(n))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 20 2017

Formula

a(0) = 0; for n >= 1, a(n) = A000035(n) + a(A257684(n)).

Other identities. For all n >= 1:

a(n!-1) = n-1. [n!-1 also gives the first position where n-1 occurs.]