A219662 Number of times an even number is encountered, when going from (n+1)!-1 to n!-1 using the iterative process described in A219652.

1, 1, 2, 10, 49, 268, 1505, 9667, 81891, 779193, 7726623, 80770479, 921442854, 11621384700, 159894957124

Offset

1,3

Comments

At least for n=7, 8, 9 and 10, a(n) is equal to a(n+1) when taken modulo n.

Links

Antti Karttunen, Table of n, a(n) for n = 1..15

Formula

a(n) = A219661(n) - A219663(n).

Example

(1!)-1 (0) is reached from (2!)-1 (1) with one step by subtracting A034968(1) from 1. Zero is an even number, so a(1)=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. Two is an even number, but one is not, so a(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. Of these only 12 and 10 are even numbers, so a(3)=2.

Prog

(Scheme):
(definec (A219662 n) (if (< n 2) n (let loop ((i (- (A000142 (1+ n)) (A000217 n) 1)) (s 0)) (cond ((isA000142? (1+ i)) (+ s (- 1 (modulo i 2)))) (else (loop (A219651 i) (+ s (- 1 (modulo i 2)))))))))
(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)))))))

Crossrefs

Cf. A219666, A218542, A218543.

Sequence in context: A330543 A052576 A279852 * A355153 A268108 A143147

Adjacent sequences: A219659 A219660 A219661 * A219663 A219664 A219665

Keyword

nonn,base

Author

Antti Karttunen, Dec 03 2012

Status

approved