A276008 Substitute ones for all nonzero digits in factorial base representation of n: a(n) = A059590(A275727(n)).

0, 1, 2, 3, 2, 3, 6, 7, 8, 9, 8, 9, 6, 7, 8, 9, 8, 9, 6, 7, 8, 9, 8, 9, 24, 25, 26, 27, 26, 27, 30, 31, 32, 33, 32, 33, 30, 31, 32, 33, 32, 33, 30, 31, 32, 33, 32, 33, 24, 25, 26, 27, 26, 27, 30, 31, 32, 33, 32, 33, 30, 31, 32, 33, 32, 33, 30, 31, 32, 33, 32, 33, 24, 25, 26, 27, 26, 27, 30, 31, 32, 33, 32, 33, 30, 31, 32, 33, 32, 33, 30, 31, 32, 33, 32, 33, 24

Offset: 0

Author

Antti Karttunen, Aug 18 2016

Keywords

• nonn
• base

Examples

			For n=23 whose factorial base representation is "321", when we replace each nonzero digit with 1, we get "111", the factorial base representation of 9, thus a(23) = 9.
From n=37 ("1201") we get "1101", thus a(37) = 31 as A007623(31) = 1101.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..40319
• Index entries for sequences related to factorial base representation.

Crossrefs

Cf. A059590, A275727.

Cf. also A276009.

Programs

Mathematica

a[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; s = Min[#, 1]& /@ s; Total[s*Range[Length[s]]!]]; Array[a, 100, 0] (* Amiram Eldar, Feb 14 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 sum([int(y[i])*f(i + 1) for i in range(len(y))])
print([a(n) for n in range(201)]) # Indranil Ghosh, Jun 21 2017

Scheme

(define (A276008 n) (A059590 (A275727 n)))
;; Standalone program:
(define (A276008 n) (let loop ((n n) (s 0) (f 1) (i 2)) (if (zero? n) s (let ((d (modulo n i))) (if (zero? d) (loop (/ n i) s (* i f) (+ 1 i)) (loop (/ (- n d) i) (+ s f) (* i f) (+ 1 i)))))))

Formula

a(n) = A059590(A275727(n)).