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%I #20 Jun 08 2023 08:49:45
%S 1,2,2,4,3,3,6,6,4,4,9,8,8,5,5,12,12,10,10,6,6,16,15,15,12,12,7,7,20,
%T 20,18,18,14,14,8,8,25,24,24,21,21,16,16,9,9,30,30,28,28,24,24,18,18,
%U 10,10,36,35,35,32,32,27,27,20,20,11,11,42,42,40,40,36,36,30,30,22,22,12,12
%N Triangle read by rows, matrix product A128179 * A000012.
%C Row sums give A000292, starting (1, 4, 10, 20, 35, 56, 84, ...).
%F Regarded as an array by antidiagonals A(i, j) = degree in q of q-Fibonacci number F(i+2, j-1) where F(1, k) = F(2, k) = 1, F(n, k) = F(n-1, k) + q^(n+k-2) * F(n-2, k). - _Michael Somos_, Jun 08 2011
%e First few rows of the triangle are:
%e 1;
%e 2, 2;
%e 4, 3, 3;
%e 6, 6, 4, 4;
%e 9, 8, 8, 5, 5;
%e 12, 12, 10, 10, 6, 6;
%e 16, 15, 15, 12, 12, 7, 7;
%e ...
%e First few rows of the array are:
%e 1, 2, 3, 4, 5, 6, 7, 8, ...
%e 2, 3, 4, 5, 6, 7, 8, 9, ...
%e 4, 6, 8, 10, 12, 14, 16, 18, ...
%e 6, 8, 10, 12, 14, 16, 18, 20, ...
%e 9, 12, 15, 18, 21, 24, 27, 30, ...
%e ...
%e A(3, 4) = 10 because F(5, 3) = 1 + q^4 + q^5 + q^6 + q^10. A(4, 4) = 12 because F(6, 3) = 1 + q^4 + q^5 + q^6 + q^7 + q^10 + q^11 + q^12.
%o (PARI) {T(n, k) = (n - k + 2)\2 * ((n + k + 1)\2)} /* _Michael Somos_, Jun 08 2011 */
%Y Cf. A000292, A128179.
%K nonn,tabl
%O 1,2
%A _Gary W. Adamson_, Mar 11 2007
%E a(19) = 10 inserted and more terms from _Georg Fischer_, Jun 08 2023
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