A276090 Left inverse of A276089: For n = sum_{i=1..} d(i)*i! (with each d(i) <= i), a(n) = sum_{j=1..} d(2j-1)*j!.

0, 1, 0, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 4, 5, 6, 7, 6, 7, 6, 7, 0, 1, 0, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 4, 5, 6, 7, 6, 7, 6, 7, 0, 1, 0, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 4, 5, 6, 7, 6, 7, 6, 7, 0, 1, 0, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 4, 5, 6, 7, 6, 7, 6, 7, 0, 1, 0, 1, 0, 1, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 4, 5, 6, 7, 6, 7, 6, 7, 6

Offset

0,7

Comments

This "deaerates" A276089(n) by picking only the digits from the odd positions of its factorial base representation. Of course, when computed for an arbitrary n, those digits, when "compressed" into a(n) are not necessarily valid digits in standard factorial base representation (A007623).

Links

Antti Karttunen, Table of n, a(n) for n = 0..20533

Index entries for sequences related to factorial base representation

Formula

Other identities. For all n >= 0:

a(A276089(n)) = n.

Example

For n = 311 ("22321" in factorial base representation) we pick the digits at odd positions 1, 3 and 5, thus we get a(311) = 2*3! + 3*2! + 1*1! = 19.
For n=373 ("30201"), we pick the digits from those same positions and construct a(373) = 3*3! + 2*2! + 1*1! = 23.

Prog

(MIT/GNU Scheme)
;; Standalone program:
(define (A276090 n) (let loop ((n n) (s 0) (f 1) (i 2) (j 2)) (if (zero? n) s (let ((d (modulo n j))) (loop (floor->exact (/ (/ (- n d) j) (+ 1 j))) (+ s (* f d)) (* i f) (+ 1 i) (+ 2 j))))))

Crossrefs

Left inverse of A276089.

For no apparent reason, the terms a(0)..a(21) are equal to the terms a(3)..a(24) of A118777.

Cf. A007623.

Sequence in context: A071995 A114108 A366912 * A073820 A103509 A361929

Adjacent sequences: A276087 A276088 A276089 * A276091 A276092 A276093

Keyword

nonn,base

Author

Antti Karttunen, Aug 19 2016

Status

approved