|
%I #30 Aug 09 2022 14:16:05
%S 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,
%T 16,64,64,128,128,1,8,16,8,32,64,128,32,256,2,2,8,16,64,64,128,64,512,
%U 512,1,2,4,8,32,64,128,16,128,512,1024
%N Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.
%C 1/2^n and successive rows are
%C 1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128, 1/256,...
%C 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
%C 0, 1/4, 3/8, 3/8, 5/16, 15/64, 21/128,... = (0 before A069834)/new,
%C -1/4, -1/4, 0, 1/4, 25/64, 27/64,...
%C 0, -1/2, -3/4, -9/16, -5/32,...
%C 1/2, 1/2, -9/16, -13/8,...
%C 0, 17/8, 51/16,...
%C -17/8, -17/8,...
%C 0
%C The first column is A198631/(A006519?), essentially the fractional Euler numbers 1, -1/2, 0, 1/4, 0,... in A060096.
%C Numerators b(n): 1, 1, 1, 0, 1, 1, -1, 1, 3, 1, ... .
%C 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.
%C There are three different Bernoulli numbers:
%C The first Bernoulli numbers are 1, -1/2, 1/6, 0,... = A027641(n)/A027642(n).
%C The second Bernoulli numbers are 1, 1/2, 1/6, 0,... = A164555(n)/A027642(n). These are the binomial transform of the first one.
%C 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.
%C There are three different fractional Euler numbers:
%C 1) The first are 1, -1/2, 0, 1/4, 0, -1/2,... in A060096(n).
%C Also Akiyama-Tanigawa algorithm for ( 1, 3/2, 7/4, 15/8, 31/16, 63/32,... = A000225(n+1)/A000079(n) ).
%C 2) The second are 1, 1/2, 0, -1/4, 0, 1/2,... , mentioned by _Wolfdieter Lang_ in A198631(n).
%C 3) The third are 0, 1/2, 0, -1/4, 0, 1/2,... , half difference of 2) and 1).
%C Also Akiyama-Tanigawa algorithm for ( 0, -1/2, -3/4, -7/8, -15/16, -31/32,... = A000225(n)/A000079(n) ). See A097110(n).
%H G. C. Greubel, <a href="/A209308/b209308.txt">Table of n, a(n) for n = 0..5049</a>
%H A. F. Horadam, <a href="http://www.fq.math.ca/Scanned/12-3/horadam.pdf">Oresme Numbers</a>, Fibonacci Quarterly, 12, #3, 1974, pp. 267-271.
%e Triangle begins:
%e 1,
%e 2, 2,
%e 1, 2, 4,
%e 4, 4, 8, 8,
%e 1, 4, 8, 4, 16,
%e 2, 2, 1, 8, 32, 32,
%e 1, 2, 4, 4, 16, 32, 64,
%e 8, 8, 16, 16, 64, 64, 128, 128,
%e ...
%t 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 *)
%Y Cf. Second Bernoulli numbers A164555(n)/A027642(n) via Akiyama-Tanigawa algorithm for 1/(n+1), A272263.
%K nonn,frac,tabl
%O 0,2
%A _Paul Curtz_, Jan 18 2013
|
