|
%I #51 Sep 01 2018 10:17:06
%S 1,-2,4,3,-8,-12,16,36,9,-32,-96,-54,64,240,216,27,-128,-576,-720,
%T -216,256,1344,2160,1080,81,-512,-3072,-6048,-4320,-810,1024,6912,
%U 16128,15120,4860,243,-2048,-15360,-41472,-48384,-22680,-2916,4096,33792,103680,145152,90720,20412,729,-8192,-73728,-253440
%N Triangle read by rows: T(0,0) = 1; T(n,k) = -2*T(n-1,k) + 3*T(n-2,k-1) for 0 <= k <= floor(n/2); T(n,k)=0 for n or k < 0.
%C The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle in A303901 ((3-2x)^n).
%C The coefficients in the expansion of 1/(1-3x+2x^2) are given by the sequence generated by the row sums.
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 394-396.
%H Zagros Lalo, <a href="/A302747/a302747.pdf">Left-justified triangle</a>
%e Triangle begins:
%e .
%e n | k = 0 1 2 3 4 5 6
%e ---+-----------------------------------------------------
%e 0 | 1
%e 1 | -2
%e 2 | 4 3
%e 3 | -8 -12
%e 4 | 16 36 9
%e 5 | -32 -96 -54
%e 6 | 64 240 216 27
%e 7 | -128 -576 -720 -216
%e 8 | 256 1344 2160 1080 81
%e 9 | -512 -3072 -6048 -4320 -810
%e 10 | 1024 6912 16128 15120 4860 243
%e 11 | -2048 -15360 -41472 -48384 -22680 -2916
%e 12 | 4096 33792 103680 145152 90720 20412 729
%e 13 | -8192 -73728 -253440 -414720 -326592 -108864 -10206
%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, -2 t[n - 1, k] + 3 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten
%o (PARI) T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, -2*T(n-1,k) + 3*T(n-2,k-1)));
%o tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ _Michel Marcus_, May 10 2018
%Y Row sums give A014983.
%Y Cf. A303901, A303941.
%K tabf,easy,sign
%O 0,2
%A _Zagros Lalo_, May 04 2018
|
