A231724 a(n) = the difference between the n-th node of the infinite trunk of the factorial beanstalk (A219666(n)) and the greatest integer (A219655(n)) which is as many A219651-iteration steps distanced from the root (zero); a(n) = A219655(n) - A219666(n).

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

Offset: 0

Author

Antti Karttunen, Nov 13 2013

Keywords

• nonn

Comments

For all n, the following holds: A219653(n) <= A219666(n) <= A219655(n). This sequence gives the distance of the node n in the infinite trunk of factorial beanstalk (A219666(n)) from the right (greater) edge of the A219654(n) wide window which it at that point must pass through.

This sequence relates to the factorial base representation (A007623) in the same way as A218604 relates to the binary system and similar remarks apply here.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..3149

Crossrefs

Cf. A231723, A230409, A219662 & A219663.

Programs

Scheme

(define (A231724 n) (- (A219655 n) (A219666 n)))

Formula

a(n) = A219655(n) - A219666(n).

A219654(n) = a(n) + A231723(n) + 1.