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A255935 Triangle read by rows: a(n) = Pascal's triangle A007318(n) + A197870(n+1).

%I A255935 #41 Aug 12 2015 14:45:13
%S A255935 0,1,2,1,2,0,1,3,3,2,1,4,6,4,0,1,5,10,10,5,2,1,6,15,20,15,6,0,1,7,21,
%T A255935 35,35,21,7,2,1,8,28,56,70,56,28,8,0,1,9,36,84,126,126,84,36,9,2,1,10,
%U A255935 45,120,210,252,210,120,45,10,0
%N A255935 Triangle read by rows: a(n) = Pascal's triangle A007318(n) + A197870(n+1).
%C A255935 Consider the difference table of a sequence with A000004(n)=0's as main diagonal. (Example: A000045(n).) We call this sequence an autosequence of the first kind.
%C A255935 Based on Pascal's triangle, a(n) =
%C A255935 0,                   T1
%C A255935 1, 2,
%C A255935 1, 2, 0,
%C A255935 1, 3, 3, 2,
%C A255935 etc.
%C A255935 transforms every sequence s(n) in an autosequence of the first kind via the multiplication by the triangle
%C A255935 s0,                  T2
%C A255935 s0, s1,
%C A255935 s0, s1, s2,
%C A255935 s0, s1, s2, s3,
%C A255935 etc.
%C A255935 Examples.
%C A255935 1) s(n) = A198631(n)/A006519(n+1), the second fractional Euler numbers (see A209308). This yields 0*1, 1*1+2*1/2=2, 1*1+2*1/2+0*0=2, 1*1+3*1/2++3*0+2*(-1/4)=2, ... .
%C A255935 The autosequence is 0 followed by 2's or 2*(0,1,1,1,1,1,1,1,... = b(n)).
%C A255935 b(n), the basic autosequence of the first kind, is not in the OEIS (see A140575 and A054977).
%C A255935 2) s(n) = A164555(n)/A027642(n), the second Bernoulli numbers, yields 0,2,2,3,4,5,6,7,... = A254667(n).
%C A255935 Row sums of T1: A062510(n) = 3*A001045(n).
%C A255935 Antidiagonal sums of T1: A111573(n).
%C A255935 With 0's instead of the spaces, every column, i.e.,
%C A255935 0, 1, 1, 1, 1,  1,  1,  1,  1,  1,   1, ...
%C A255935 0, 2, 2, 3, 4,  5,  6,  7,  8,  9,  10, ... = A001477(n) with 0 instead of 1 = A254667(n)
%C A255935 0, 0, 0, 3, 6, 10, 15, 21, 28, 36,  45, ... = A161680(n) with 0 instead of 1
%C A255935 0, 0, 0, 2, 4, 10, 20, 35, 56, 84, 120, ...
%C A255935 etc., is an autosequence of the first kind.
%C A255935 With T(0,0) = 1, it is (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, May 24 2015
%F A255935 a(n) = Pascal's triangle A007318(n) with main diagonal A010673(n) (= period 2: repeat 0, 2) instead of 1's=A000012(n).
%F A255935 a(n) = reversal abs(A140575(n)).
%F A255935 a(n) = A007318(n) + A197870(n+1).
%F A255935 T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 0, T(1,0) = 1, T(1,1) = 2, T(n,k) = 0 if k>n or if k<0 . - _Philippe Deléham_, May 24 2015
%F A255935 G.f.: (-1-2*x*y+x^2*y+x^2*y^2)/((x*y+1)*(x*y+x-1)) - 1. - _R. J. Mathar_, Aug 12 2015
%e A255935 Triangle starts:
%e A255935 0;
%e A255935 1, 2;
%e A255935 1, 2, 0;
%e A255935 1, 3, 3, 2;
%e A255935 1, 4, 6, 4, 0;
%e A255935 1, 5, 10, 10, 5, 2;
%e A255935 1, 6, 15, 20, 15, 6, 0;
%e A255935 ...
%t A255935 a[n_, k_] := If[k == n, 2*Mod[n, 2], Binomial[n, k]]; Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Mar 23 2015 *)
%Y A255935 Cf. A001045, A001477, A010673, A011973, A054977, A062510, A007318, A197870, A006519, A198631, A140575, A209308, A164555, A027642, A111573, A161680, A254667.
%K A255935 nonn,tabl
%O A255935 0,3
%A A255935 _Paul Curtz_, Mar 11 2015