

A278515


Number of steps required to iterate map k > A255131(k) when starting from k = (A000196(n)+1)^2 before n is reached, or 0 if n is not reached.


2



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

0,7


LINKS

Antti Karttunen, Table of n, a(n) for n = 0..10200


EXAMPLE

For n=15, we start iterating from the next larger square, (⌊√15⌋+1)^2 = 16, and in just a single step (16  A002828(16) = 15) we land to n, so a(15) = 1.
For n=16, we start iterating from the next larger square, which is 25, and thus we have 25 > A255131(25) = 24, 24 > A255131(24) = 21, 21 > A255131(21) = 18, 18 > A255131(18) = 16, thus four steps were required to reach 16, and a(16) = 4.
For n=17, when we do the same iteration, we see that we will pass by it, thus a(17) = 0.


PROG

(Scheme) (define (A278515 n) (let* ((r (A000196 n)) (end (A000290 r))) (let loop ((k (A000290 (+ 1 r))) (s 0)) (cond ((= k n) s) ((<= k end) 0) (else (loop (A255131 k) (+ 1 s)))))))


CROSSREFS

Cf. A000196, A000290, A002828, A255131.
Cf. A005563 (positions of ones), A276573 (of nonzero terms).
Sequence in context: A217540 A226861 A185643 * A285709 A080101 A025895
Adjacent sequences: A278512 A278513 A278514 * A278516 A278517 A278518


KEYWORD

nonn


AUTHOR

Antti Karttunen, Nov 28 2016


STATUS

approved



