OFFSET
0,3
COMMENTS
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.
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:
0, -1/6, -1/4, -3/10, -1/3, -5/14, -3/8, -7/18, ...
1/6, 1/6, 3/20, 2/15, 5/42, 3/28, 7/72, 4/45, 9/110, ...
0, 1/30, 1/20, 2/35, 5/84, 5/84, 7/120, 28/495, ...
-1/30, -1/30, -3/140, -1/105, 0, 1/140, 49/3960, ...
0, -1/42, -1/28, -4/105, -1/28, -29/924, ...
1/42, 1/42, 1/140, -1/105, -5/231, -9/308, -343/10296, ...
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.
Also, numerators of (3*n + 1)/12. - Bruno Berselli, Apr 13 2018
Also, numerators of (3*n + 1)/4. - Altug Alkan, Apr 17 2018
LINKS
M. Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), #00.2.9.
D. Merlini, R. Sprugnoli, M. C. Verri, The Akiyama-Tanigawa Transformation, Integers, 5 (1) (2005) #A05
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).
FORMULA
From R. J. Mathar, Jan 06 2011: (Start)
a(n) = 2*a(n-4) - a(n-8).
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)
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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Curtz, Apr 23 2010
STATUS
approved