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

%I A292270 #42 Apr 29 2025 08:54:19
%S A292270 1,1,4,1,13,25,36,1,38,81,12,26,124,121,196,1,103,73,324,42,224,175,
%T A292270 91,147,232,14,676,170,303,841,900,1,264,1089,385,364,93,301,585,563,
%U A292270 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
%N A292270 Sum of all partial fractions in the algorithm used for calculation of A002326(n).
%C A292270 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.
%C A292270 A001122 gives the odd numbers k such that a((k-1)/2) = A000290((k-1)/2).
%H A292270 Antti Karttunen, <a href="/A292270/b292270.txt">Table of n, a(n) for n = 0..10001</a>
%F A292270 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)).
%e A292270 Let n = 9. According to the comment, a(9) = 5 + 3 + 11 + 15 + 17 + 9 + 7 + 13 + 1 = 81.
%o A292270 (PARI)
%o A292270 A000265(n) = (n >> valuation(n, 2));
%o A292270 A006519(n) = 2^valuation(n, 2);
%o A292270 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; };
%o A292270 (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))))))
%Y A292270 Cf. A000265, A000290, A001122, A001917, A002326, A006519, A163782, A179382, A179680, A292239, A292265, A292947, A293218, A293219.
%Y A292270 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).
%K A292270 nonn
%O A292270 0,3
%A A292270 _Vladimir Shevelev_ and _Antti Karttunen_, Oct 05 2017