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%I #30 Apr 20 2018 01:03:12
%S 1,1,7,5,13,4,19,11,25,7,31,17,37,10,43,23,49,13,55,29,61,16,67,35,73,
%T 19,79,41,85,22,91,47,97,25,103,53,109,28,115,59,121,31,127,65,133,34,
%U 139,71,145,37,151,77,157,40,163,83,169,43,175,89,181,46,187,95,193
%N a(2*n) = 1 + 6*n, a(2*n+1) = A165367(n).
%C Motivation: Start an array from a left column of fractions 0, 1/6, 0, -1/30, 0, ... = A176327(.)/A176592(.), which is zero followed by the Bernoulli numbers from B_2 onwards.
%C Construct more columns of the array by iteration of the Akiyama-Tanigawa algorithm working backwards through the rows of the table. In our case, the array starts with column indices k>=0:
%C 0, -1/6, -1/4, -3/10, -1/3, -5/14, -3/8, -7/18, ...
%C 1/6, 1/6, 3/20, 2/15, 5/42, 3/28, 7/72, 4/45, 9/110, ...
%C 0, 1/30, 1/20, 2/35, 5/84, 5/84, 7/120, 28/495, ...
%C -1/30, -1/30, -3/140, -1/105, 0, 1/140, 49/3960, ...
%C 0, -1/42, -1/28, -4/105, -1/28, -29/924, ...
%C 1/42, 1/42, 1/140, -1/105, -5/231, -9/308, -343/10296, ...
%C The fractions of the top row are -A060819(n)/A145979(n). The current sequence contains essentially the difference between numerator and denominator of each fraction, a(2)=6+1, a(3)=4+1, a(4)=10+3, ... The sum of numerator and denominator is essentially A060819.
%C Also, numerators of (3*n + 1)/12. - _Bruno Berselli_, Apr 13 2018
%C Also, numerators of (3*n + 1)/4. - _Altug Alkan_, Apr 17 2018
%H M. Kaneko, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL3/KANEKO/AT-kaneko.html">The Akiyama-Tanigawa algorithm for Bernoulli numbers</a>, J. Integer Sequences, 3 (2000), #00.2.9.
%H D. Merlini, R. Sprugnoli, M. C. Verri, <a href="http://www.emis.de/journals/INTEGERS/papers/f5/f5.Abstract.html">The Akiyama-Tanigawa Transformation</a>, Integers, 5 (1) (2005) #A05
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,2,0,0,0,-1).
%F From _R. J. Mathar_, Jan 06 2011: (Start)
%F a(n) = 2*a(n-4) - a(n-8).
%F G.f.: (1 + x + 7*x^2 + 5*x^3 + 11*x^4 + 2*x^5 + 5*x^6 + x^7) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2). (End)
%F a(n) = (2*(3*n + 1)*(11 + 5*(-1)^n) + (6*n + 5 + 3*(-1)^n)*(1 - (-1)^n)*(-1)^((2*n + 3 + (-1)^n)/4))/32. - _Luce ETIENNE_, Jan 28 2015
%Y Cf. A061038, A145979.
%K nonn,easy
%O 0,3
%A _Paul Curtz_, Apr 23 2010
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