0,2

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 beforeA069834)/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 isA198631/(A006519?), essentially the fractional Euler numbers 1, -1/2, 0, 1/4, 0,... inA060096.

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. ViaA177427(n) andA191567(n), they yield the Balmer seriesA061037/A061038.

There are three different fractional Euler numbers:

1) The first are 1, -1/2, 0, 1/4, 0, -1/2,... inA060096(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 byWolfdieter LanginA198631(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) ). SeeA097110(n).

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.

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,

...

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 *)

Cf. Second Bernoulli numbersA164555(n)/A027642(n) via Akiyama-Tanigawa algorithm for 1/(n+1),A272263.

Sequence in context:A324251A035364A261734*A143808A294600A247495

Adjacent sequences:A209305A209306A209307*A209309A209310A209311

nonn,frac,tabl

Paul Curtz,Jan 18 2013

approved