A176672 a(2*n) = 1 + 6*n, a(2*n+1) = A165367(n).

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, 19, 79, 41, 85, 22, 91, 47, 97, 25, 103, 53, 109, 28, 115, 59, 121, 31, 127, 65, 133, 34, 139, 71, 145, 37, 151, 77, 157, 40, 163, 83, 169, 43, 175, 89, 181, 46, 187, 95, 193

Offset: 0

Author

Paul Curtz, Apr 23 2010

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A061038, A145979.

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