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

Author

Paul Curtz, Jan 18 2013

Keywords

• nonn
• frac
• tabl

Comments

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).

Examples

			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,
  ...
		

Links

• 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.

Crossrefs

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

Programs

Mathematica

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 *)