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%I #22 Dec 24 2021 13:10:22
%S 1,-1,-1,1,1,-1,-2,-1,0,1,2,1,-1,-3,-4,-4,-2,2,4,4,3,1,-1,-4,-8,-12,
%T -13,-8,0,8,13,12,8,4,1,-1,-5,-13,-25,-37,-41,-33,-13,13,33,41,37,25,
%U 13,5,1,-1,-6,-19,-44,-80,-116,-136,-124,-74,0,74,124,136,116,80,44,19,6,1,-1,-7,-26,-70,-149,-259,-376,-456,-450,-334,-124,124,334,450,456,376,259,149,70,26,7,1
%N Triangle read by rows: row 1 is [1]; for n >= 1, row n gives coefficients of expansion of (-1 - x + x^2 + x^3)*(1 + x + x^2 + x^3)^(n-1) in order of increasing powers of x.
%C The row polynomials can be further factorized, since -3 - x + x^2 + 3*x^3 = -(1-x)*(1+x)^2 and 1 + x + x^2 + x^3 = (1+x)*(1+x^2).
%C The rule for constructing this triangle (ignoring row 0) is the same as that for A008287: each number is the sum of the four numbers immediately above it in the previous row. Here row 1 is [-1, -1, 1, 3] instead of [1, 1, 1, 1].
%C This is a companion to A008287 and A349813.
%H Jack Ramsay, <a href="/A349812/a349812.pdf">On Arithmetical Triangles</a>, The Pulse of Long Island, June 1965 [Mentions application to design of antenna arrays. Annotated scan.]
%e Triangle begins:
%e 1;
%e -1, -1, 1, 1;
%e -1, -2, -1, 0, 1, 2, 1;
%e -1, -3, -4, -4, -2, 2, 4, 4, 3, 1;
%e -1, -4, -8, -12, -13, -8, 0, 8, 13, 12, 8, 4, 1;
%e -1, -5, -13, -25, -37, -41, -33, -13, 13, 33, 41, 37, 25, 13, 5, 1;
%e ...
%p t1:=-1-x+x^2+x^3;
%p m:=1+x+x^2+x^3;
%p lprint([3]);
%p for n from 1 to 12 do
%p w1:=expand(t1*m^(n-1));
%p w4:=series(w1,x,3*n+1);
%p w5:=seriestolist(w4);
%p lprint(w5);
%p od:
%Y Cf. A007318, A008287, A349812, A349813.
%Y The right half of the triangle gives A349816. For the central nonzero entries see A349818.
%K sign,tabf
%O 0,7
%A _N. J. A. Sloane_, Dec 23 2021
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