cp's OEIS Frontend

A238800 Unreduced numerators in triangle that leads to the Euler numbers A198631(n)/A006519(n+1).

%I A238800 #9 Mar 14 2014 11:43:10
%S A238800 1,1,1,-2,1,-3,1,-4,2,1,-5,5,1,-6,9,-10,1,-7,14,-35,1,-8,20,-80,26,1,
%T A238800 -9,27,-150,117,1,-10,35,-250,325,-454,1,-11,44,-385,715,-2497,1,-12,
%U A238800 54,-560,1365,-8172,5914,1,-13
%N A238800 Unreduced numerators in triangle that leads to the Euler numbers A198631(n)/A006519(n+1).
%C A238800 We use the array ASPEC mentioned in A191302:
%C A238800 2, 1,  1,  1,  1,  1,   1,   1,...
%C A238800 2, 3,  4,  5,  6,  7,   8,   9,...
%C A238800 2, 5,  9, 14, 20, 27,  35,  44,...
%C A238800 2, 7, 16, 30, 50, 77, 112, 156,...
%C A238800 with the first upper diagonal of the difference table of the autosequence A198631(n)/A006519(n+1), i.e., 1/2, -1/4, 1/4, -5/8, 13/4, -227/8, 2957/8,...
%C A238800 written by columns:
%C A238800 1/2
%C A238800 1/2,
%C A238800 1/2, -1/4,
%C A238800 1/2, -1/4,
%C A238800 1/2, -1/4, 1/4,
%C A238800 1/2, -1/4, 1/4,
%C A238800 1/2, -1/4, 1/4, -5/8,
%C A238800 1/2, -1/4, 1/4, -5/8,
%C A238800 etc.
%C A238800 Hence, by multiplication of this double triangle by ASPEC, the beginning of the double triangle ESPEC is obtained:
%C A238800 E(0) =     1 =   1
%C A238800 E(1) =   1/2 = 1/2
%C A238800 E(2) =     0 = 1/2 -2/4
%C A238800 E(3) =  -1/4 = 1/2 -3/4
%C A238800 E(4) =     0 = 1/2 -4/4  +2/4
%C A238800 E(5) =   1/2 = 1/2 -5/4  +5/4
%C A238800 E(6) =     0 = 1/2 -6/4  +9/4 -10/8
%C A238800 E(7) = -17/8 = 1/2 -7/4 +14/4 -35/8
%C A238800 E(8) =     0 = 1/2 -8/4 +20/4 -80/8 +26/4.
%C A238800 The terms of the sequence are the reduced numerators. Like A192456(n) for Bernoulli numbers A164555(n)/A027642(n).
%e A238800 a(n) by triangle
%e A238800 1,
%e A238800 1,
%e A238800 1, -2,
%e A238800 1, -3,
%e A238800 1, -4,  2,
%e A238800 1, -5,  5,
%e A238800 1, -6,  9, -10,
%e A238800 1, -7, 14, -35,
%e A238800 1, -8, 20, -80, 26,
%e A238800 etc.
%K A238800 sign,tabf
%O A238800 0,4
%A A238800 _Paul Curtz_, Mar 05 2014