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%I #30 Apr 08 2019 03:08:06
%S 3,1,2,2,-1,2,3,-3,1,2,4,-6,4,-1,2,5,-10,10,-5,1,2,6,-15,20,-15,6,-1,
%T 2,7,-21,35,-35,21,-7,1,2,8,-28,56,-70,56,-28,8,-1,2,9,-36,84,-126,
%U 126,-84,36,-9,1,2,10,-45,120,-210,252,-210,120,-45,10,-1,2
%N Triangle read by rows, giving coefficients of extended Jacobsthal recurrence.
%C Sequence identical to half its p-th differences from the second term.
%C This sequence is the second of a family after A135356.
%C This triangle looks like a Pascal's triangle without first column, and with signs and with additional right diagonal consisting of 2's. - _Michel Marcus_, Apr 07 2019
%F Every row sums to 3. - _Jean-François Alcover_, Apr 04 2019 (further to a remark e-mailed by _Paul Curtz_).
%e Triangle begins
%e 3; : A000244 = 1, 3, 9, 27, ... is the main sequence
%e 1, 2; : A001045 = 0, 1, 1, 3, ... is the main sequence
%e 2, -1, 2; : 0, 0, (A007910 = 1, 2, 3, ... ) is the main sequence
%e 3, -3, 1, 2; : 0, 0, 0, 1, 3, 6, 10, 17, ... is the main sequence
%e 4, -6, 4, -1, 2; : A134987 = 0, 0, 0, 0, 1, ... is the main sequence
%e ...
%e See signatures of linear recurrence of corresponding sequences.
%t T[n_, k_] := T[n, k] = Which[n == 0, 3, k == n, 2, k == 0, n, k == n-1, (-1)^k, True, T[n-1, k] - T[n-1, k-1]];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Apr 06 2019 *)
%Y Cf. A000244, A001045, A007910, A134977 (sum of antidiagonals), A134987, A135356.
%K sign,tabl
%O 0,1
%A _Paul Curtz_, Feb 01 2008
%E In agreement with author, T(0, 0) = 3 and offset 0 by _Michel Marcus_, Apr 06 2019
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