A191567 -id:A191567 - cp's OEIS frontend

A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

1, 2, 2, 1, 2, 4, 4, 4, 8, 8, 1, 4, 8, 4, 16, 2, 2, 1, 8, 32, 32, 1, 2, 4, 4, 16, 32, 64, 8, 8, 16, 16, 64, 64, 128, 128, 1, 8, 16, 8, 32, 64, 128, 32, 256, 2, 2, 8, 16, 64, 64, 128, 64, 512, 512, 1, 2, 4, 8, 32, 64, 128, 16, 128, 512, 1024

Offset: 0

Paul Curtz, Jan 18 2013

1/2^n and successive rows are
1,       1/2,   1/4,   1/8,  1/16,  1/32,   1/64, 1/128, 1/256,...
1/2,     1/2,   3/8,   1/4,  5/32,  3/32,  7/128,  1/32,...       = A000265/A075101, the Oresme numbers n/2^n. Paul Curtz, Jan 18 2013 and May 11 2016
0,       1/4,   3/8,   3/8,  5/16, 15/64, 21/128,...              = (0 before A069834)/new,
-1/4,   -1/4,     0,   1/4, 25/64, 27/64,...
0,      -1/2,  -3/4, -9/16, -5/32,...
1/2,     1/2, -9/16, -13/8,...
0,      17/8, 51/16,...
-17/8, -17/8,...
0
The first column is A198631/(A006519?), essentially the fractional Euler numbers 1, -1/2, 0, 1/4, 0,...  in A060096.
Numerators b(n): 1, 1, 1, 0, 1, 1, -1, 1, 3, 1, ... .
Coll(n+1) - 2*Coll(n) = -1/2, -5/8, -1/2, -11/32, -7/32, -17/128, -5/64, -23/512, ... = -A075677/new, from Collatz problem.
There are three different Bernoulli numbers:
The first Bernoulli numbers are  1, -1/2, 1/6, 0,... = A027641(n)/A027642(n).
The second Bernoulli numbers are 1,  1/2, 1/6, 0,... = A164555(n)/A027642(n). These are the binomial transform of the first one.
The third Bernoulli numbers are  1,   0,  1/6, 0,... = A176327(n)/A027642(n), the half sum. Via A177427(n) and A191567(n), they yield the Balmer series A061037/A061038.
There are three different fractional Euler numbers:
1) The first are  1, -1/2, 0, 1/4, 0, -1/2,... in A060096(n).
Also Akiyama-Tanigawa algorithm for ( 1, 3/2, 7/4, 15/8, 31/16, 63/32,... = A000225(n+1)/A000079(n) ).
2) The second are 1, 1/2, 0, -1/4, 0,  1/2,... , mentioned by Wolfdieter Lang in A198631(n).
3) The third are  0, 1/2, 0, -1/4, 0,  1/2,... , half difference of 2) and 1).
Also Akiyama-Tanigawa algorithm for ( 0, -1/2, -3/4, -7/8, -15/16, -31/32,... =  A000225(n)/A000079(n) ). See A097110(n).

			Triangle begins:
  1,
  2, 2,
  1, 2,  4,
  4, 4,  8,  8,
  1, 4,  8,  4, 16,
  2, 2,  1,  8, 32, 32,
  1, 2,  4,  4, 16, 32,  64,
  8, 8, 16, 16, 64, 64, 128, 128,
  ...

G. C. Greubel, Table of n, a(n) for n = 0..5049
A. F. Horadam, Oresme Numbers, Fibonacci Quarterly, 12, #3, 1974, pp. 267-271.

Cf. Second Bernoulli numbers A164555(n)/A027642(n) via Akiyama-Tanigawa algorithm for 1/(n+1), A272263.

max = 10; t[0, k_] := 1/2^k; t[n_, k_] := t[n, k] = (k + 1)*(t[n - 1, k] - t[n - 1, k + 1]); denoms = Table[t[n, k] // Denominator, {n, 0, max}, {k, 0, max - n}]; Table[denoms[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 05 2013 *)

A177427 Numerators of the Inverse Akiyama-Tanigawa transform of the aerated even-indexed Bernoulli numbers 1, 0, 1/6, 0, -1/30, 0, 1/42, ...

Original entry on oeis.org

1, 1, 13, 7, 149, 157, 383, 199, 7409, 7633, 86231, 88331, 1173713, 1197473, 1219781, 620401, 42862943, 43503583, 279379879, 283055551, 57313183, 19328341, 449489867, 1362695813, 34409471059, 34738962067, 315510823603, 45467560829, 9307359944587, 9382319148907, 293103346860157, 147643434162641, 594812856101039, 54448301591149

Offset: 0

Paul Curtz, May 07 2010

These are the numerators of the first row of a Table T(n,k) which contains the even-indexed Bernoulli numbers in the first column: T(2n,0) = A000367(n)/A002445(n), T(2n+1,0)=0, and which generates rows with the Akiyama-Tanigawa transform. (Because the first column is given, the algorithm is an inverse Akiyama-Tanigawa transform.)

These are the absolute values of the numerators of the Taylor expansion of sinh(log(x+1))*log(x+1)at x=0. - Gary Detlefs, Aug 31 2011

			The table T(n,k) of fractions generated by the Akiyama-Tanigawa transform, with the column T(n,0) equal to Bernoulli(n) for even n and equal to 0 for odd n, starts in row n=0 as:
1, 1, 13/12, 7/6, 149/120, 157/120, 383/280, 199/140, ...
0, -1/6, -1/4, -3/10, -1/3, -5/14, -3/8, -7/18, -2/5, -9/22, ...
1/6, 1/6, 3/20, 2/15, 5/42, 3/28, 7/72, 4/45, 9/110, 5/66, ...
0, 1/30, 1/20, 2/35, 5/84, 5/84, 7/120, 28/495, 3/55, 15/286, ...
-1/30, -1/30, -3/140, -1/105, 0, 1/140, 49/3960, 8/495, ...
0, -1/42, -1/28, -4/105, -1/28, -29/924, -7/264, -28/1287, -87/5005, ...
1/42, 1/42, 1/140, -1/105, -5/231, -9/308, -343/10296, -1576/45045, ...

L. A. Medina, V. H. Moll, E. S. Rowland, Iterated primitives of logarithmic powers, arXiv:0911.1325, arXiv:0911.1325 [math.NT], 2009-2010.
D. Merlini, R. Sprugnoli, M. C. Verri, The Akiyama-Tanigawa Transformation, Integers, 5 (1) (2005) #A05.

Cf. A177690 (denominators).

t[n_, 0] := BernoulliB[n]; t[1, 0]=0; t[n_, k_] := t[n, k] = (t[n, k-1] + (k-1)*t[n, k-1] - t[n+1, k-1])/k; Table[t[0, k], {k, 0, 33}] // Numerator (* Jean-François Alcover, Aug 09 2012 *)

From Groux Roland, Jan 07 2011: (Start)
T(0,k) = H(k)/2 + 1/(k+1) with H(k) harmonic number of order k.
T(0,k)= -(1/2)*(k+1)*Integral_{x=0..1} x^n*log(x*(1-x)) dx.
G.f.: Sum_{k>=0} T(0,k) x^k = (x-2)*(log(1-x))/(2*x*(1-x)). (End)
T(1,n) = -A191567(n)/A061038(n+2) = -A060819(n)/A145979(n). - Paul Curtz, Jul 19 2011
(T(1,n))^2 = A181318(n)/A061038(n+2). - Paul Curtz, Jul 19 2011, index corrected by R. J. Mathar, Sep 09 2011

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A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

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