

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.


8



1, 1, 2, 10, 49, 268, 1505, 9667, 81891, 779193, 7726623, 80770479, 921442854, 11621384700, 159894957124
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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 * A268108 A143147 A317111
Adjacent sequences: A219659 A219660 A219661 * A219663 A219664 A219665


KEYWORD

nonn,base


AUTHOR

Antti Karttunen, Dec 03 2012


STATUS

approved



