A236846 Least inverse of A234742: a(n) = minimal k such that when it is remultiplied "upwards", from GF(2)[X] to N, the result is n, and 0 if no such k exists.

0, 1, 2, 3, 4, 0, 6, 7, 8, 5, 0, 11, 12, 13, 14, 0, 16, 0, 10, 19, 0, 9, 22, 0, 24, 25, 26, 15, 28, 0, 0, 31, 32, 29, 0, 0, 20, 37, 38, 23, 0, 41, 18, 0, 44, 0, 0, 47, 48, 21, 50, 0, 52, 0, 30, 55, 56, 53, 0, 59, 0, 61, 62, 27, 64, 0, 58, 67, 0, 0, 0, 0, 40, 73, 74, 43, 76, 49, 46, 0, 0, 17, 82, 0, 36, 0, 0, 87, 88, 0, 0, 35

Offset: 0

Author

Antti Karttunen, Jan 31 2014

Keywords

• nonn

Comments

Apart from zero, each term occurs at most once. 91 is the smallest positive integer not present in this sequence.

Note that in contrast to the reciprocal case, where A234742(n) >= A236837(n) for all n [the former sequence gives the absolute upper bound for the latter], here it is not guaranteed that A234741(n) <= a(n) whenever a(n) > 0. For example, a(25)=25 and A234741(25)=17, and 25-17 = 8. On the other hand, a(75)=43, but A234741(75)=51, and 43-51 = -8.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8192

Crossrefs

Differs from A236847 for the first time at n=91, where a(91)=35, while A236847(91)=91.

A236844 gives the positions of zeros.

Cf. A234742.

Cf. also A236836, A236837.

Programs

Scheme

(define (A236846 n) (let loop ((k n) (minv 0)) (cond ((zero? k) minv) ((= (A234742 k) n) (loop (- k 1) k)) (else (loop (- k 1) minv)))))

Formula

a(n) = minimal k such that A234742(k) = n, and 0 if no such k exists.

For all n, a(n) <= n.