

A292270


Sum of all partial fractions in the algorithm used for calculation of A002326(n).


10



1, 1, 4, 1, 13, 25, 36, 1, 38, 81, 12, 26, 124, 121, 196, 1, 103, 73, 324, 42, 224, 175, 91, 147, 232, 14, 676, 170, 303, 841, 900, 1, 264, 1089, 385, 364, 93, 301, 585, 563, 1093, 1681, 44, 355, 152, 118, 83, 484, 1254, 763, 2500, 1043, 156, 2809, 996, 564, 952, 931, 71, 387, 3325, 176, 3124, 1, 649, 4225, 554, 1081
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OFFSET

0,3


COMMENTS

This sequence gives important additional insight into the algorithm for the calculation of A002326 (see A179680 for its description). Let us estimate how many steps are required before (the first) 1 will appear. Note that all partial fractions (which are indeed, integers) are odd residues modulo 2*n+1 from the interval [1,2*n1]. So, if there is no a repetition, then the number of steps does not exceed n. Suppose then that there is a repetition before the appearance of 1. Then for an odd residue k from [1, 2*n1], 2^m_1 == 2^m_2 == k (mod 2*n+1) such that m_2 > m_1. But then 2^(m_2m_1) == 1 (mod 2*n+1). So, since m_2  m_1 < m_2, it means that 1 should appear earlier than the repetition of k, which is a contradiction. So the number of steps <= n. For example, for n=9, 2*n+1 = 19, we have exactly 9 steps with all other odd residues <= 17 modulo 19 appearing before the final 1: 5, 3, 11, 15, 17, 9, 7, 13, 1.
A001122 gives the odd numbers k such that a((k1)/2) = A000290((k1)/2).


LINKS

Antti Karttunen, Table of n, a(n) for n = 0..10001


FORMULA

For all n >= 1, A000196(a((A001122(1+n)1)/2)) = (A001122(1+n)1)/2, in other words, a(A163782(n)) = A000290(A163782(n)).


EXAMPLE

Let n = 9. According to the comment, a(9) = 5 + 3 + 11 + 15 + 17 + 9 + 7 + 13 + 1 = 81.


PROG

(PARI)
A000265(n) = (n >> valuation(n, 2));
A006519(n) = 2^valuation(n, 2);
A292270(n) = { my(x = n+n+1, z = ((1+x)/A006519(1+x)), m = A000265(1+x)); while(m!=1, z += ((x+m)/A006519(x+m)); m = A000265(x+m)); z; };
(Scheme) (define (A292270 n) (let ((x (+ n n 1))) (let loop ((z (/ (+ 1 x) (A006519 (+ 1 x)))) (k 1)) (let ((m (A000265 (+ x k)))) (if (= 1 m) z (loop (+ z (/ (+ x m) (A006519 (+ x m)))) m))))))


CROSSREFS

Cf. A000265, A000290, A001122, A001917, A002326, A006519, A163782, A179382, A179680, A292239, A292265, A292947, A293218, A293219.
Cf. A000225 (gives the positions of ones), A292938 (of squares), A292939 (and the corresponding odd numbers), A292940 (odd numbers corresponding to squares larger than one), A292379 (odd numbers corresponding to squares less than n^2).
Sequence in context: A215502 A144698 A115154 * A051928 A335337 A226906
Adjacent sequences: A292267 A292268 A292269 * A292271 A292272 A292273


KEYWORD

nonn


AUTHOR

Vladimir Shevelev and Antti Karttunen, Oct 05 2017


STATUS

approved



