A292270 Sum of all partial fractions in the algorithm used for calculation of A002326(n).

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

Offset: 0

Author

Vladimir Shevelev and Antti Karttunen, Oct 05 2017

Keywords

• nonn

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*n-1]. So, if there is no 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*n-1], 2^m_1 == 2^m_2 == k (mod 2*n+1) such that m_2 > m_1. But then 2^(m_2-m_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((k-1)/2) = A000290((k-1)/2).

Examples

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

Links

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

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).

Programs

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))))))

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)).