A219661 Number of steps to go from (n+1)!-1 to n!-1 with map x -> x - (sum of digits in factorial base representation of x).

1, 2, 5, 19, 83, 428, 2611, 18473, 150726, 1377548, 13851248, 152610108, 1835293041, 23925573979, 335859122743, 5049372125352, 80942722123544, 1378487515335424, 24858383452927384, 473228664468684846

Offset: 1

Author

Antti Karttunen, Dec 03 2012

Keywords

• nonn

Examples

			(1!)-1 (0) is reached from (2!)-1 (1) with one step by subtracting A034968(1) from 1.
(2!)-1 (1) is reached from (3!)-1 (5) with two steps by first subtracting A034968(5) from 5 -> 2, and then subtracting A034968(2) from 2 -> 1.
(3!)-1 (5) is reached from (4!)-1 (23) with five steps by repeatedly subtracting the sum of digits in factorial expansion as: 23 - 6 = 17, 17 - 5 = 12, 12 - 2 = 10, 10 - 3 = 7, 7 - 2 = 5.
Thus a(1)=1, a(2)=2 and a(3)=5.
		

Links

• Hiroaki Yamanouchi, Fast Python-program for computing terms of this sequence

Crossrefs

Row sums of A230420 and A230421.

Cf. A000142, A007623, A219651, A219652, A219662, A219663, A219665, A226061.

Cf. also A213709 (analogous sequence for base-2), A261234 (for base-3).

Programs

Mathematica

Table[Length@ NestWhileList[# - Total@ IntegerDigits[#, MixedRadix[Reverse@ Range[2, 120]]] &, (n + 1)! - 1, # > n! - 1 &] - 1, {n, 0, 8}] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

Scheme

(define (A219661 n) (if (zero? n) n (let loop ((i (-1+ (A000142 (1+ n)))) (steps 1)) (cond ((isA000142? (1+ (A219651 i))) steps) (else (loop (A219651 i) (1+ steps)))))))
(define (isA000142? n) (and (> n 0) (let loop ((n n) (i 2)) (cond ((= 1 n) #t) ((not (zero? (modulo n i))) #f) (else (loop (/ n i) (1+ i)))))))
;; Alternative definition:
(define (A219661 n) (- (A219652 (-1+ (A000142 (1+ n)))) (A219652 (-1+ (A000142 n)))))

Formula

a(n) = A219652((n+1)!-1) - A219652(n!-1).

a(n) = A219662(n) + A219663(n).

Extensions

Terms a(16) - a(20) computed with Hiroaki Yamanouchi's Python-program by Antti Karttunen, Jun 27 2016