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

%I A209308 #32 Jan 05 2025 19:51:39
%S A209308 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 A209308 16,64,64,128,128,1,8,16,8,32,64,128,32,256,2,2,8,16,64,64,128,64,512,
%U A209308 512,1,2,4,8,32,64,128,16,128,512,1024
%N A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.
%C A209308 1/2^n and successive rows are
%C A209308 1,       1/2,   1/4,   1/8,  1/16,  1/32,   1/64, 1/128, 1/256,...
%C A209308 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 A209308 0,       1/4,   3/8,   3/8,  5/16, 15/64, 21/128,...              = (0 before A069834)/new,
%C A209308 -1/4,   -1/4,     0,   1/4, 25/64, 27/64,...
%C A209308 0,      -1/2,  -3/4, -9/16, -5/32,...
%C A209308 1/2,     1/2, -9/16, -13/8,...
%C A209308 0,      17/8, 51/16,...
%C A209308 -17/8, -17/8,...
%C A209308 0
%C A209308 The first column is A198631/(A006519?), essentially the fractional Euler numbers 1, -1/2, 0, 1/4, 0,...  in A060096.
%C A209308 Numerators b(n): 1, 1, 1, 0, 1, 1, -1, 1, 3, 1, ... .
%C A209308 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 A209308 There are three different Bernoulli numbers:
%C A209308 The first Bernoulli numbers are  1, -1/2, 1/6, 0,... = A027641(n)/A027642(n).
%C A209308 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 A209308 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 A209308 There are three different fractional Euler numbers:
%C A209308 1) The first are  1, -1/2, 0, 1/4, 0, -1/2,... in A060096(n).
%C A209308 Also Akiyama-Tanigawa algorithm for ( 1, 3/2, 7/4, 15/8, 31/16, 63/32,... = A000225(n+1)/A000079(n) ).
%C A209308 2) The second are 1, 1/2, 0, -1/4, 0,  1/2,... , mentioned by _Wolfdieter Lang_ in A198631(n).
%C A209308 3) The third are  0, 1/2, 0, -1/4, 0,  1/2,... , half difference of 2) and 1).
%C A209308 Also Akiyama-Tanigawa algorithm for ( 0, -1/2, -3/4, -7/8, -15/16, -31/32,... =  A000225(n)/A000079(n) ). See A097110(n).
%H A209308 G. C. Greubel, <a href="/A209308/b209308.txt">Table of n, a(n) for n = 0..5049</a>
%H A209308 A. F. Horadam, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/12-3/horadam.pdf">Oresme Numbers</a>, Fibonacci Quarterly, 12, #3, 1974, pp. 267-271.
%e A209308 Triangle begins:
%e A209308   1,
%e A209308   2, 2,
%e A209308   1, 2,  4,
%e A209308   4, 4,  8,  8,
%e A209308   1, 4,  8,  4, 16,
%e A209308   2, 2,  1,  8, 32, 32,
%e A209308   1, 2,  4,  4, 16, 32,  64,
%e A209308   8, 8, 16, 16, 64, 64, 128, 128,
%e A209308   ...
%t A209308 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 A209308 Cf. Second Bernoulli numbers A164555(n)/A027642(n) via Akiyama-Tanigawa algorithm for 1/(n+1), A272263.
%K A209308 nonn,frac,tabl
%O A209308 0,2
%A A209308 _Paul Curtz_, Jan 18 2013