A236846 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

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

FORMULA

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

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

PROG

(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)))))

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.