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A143929 Eigentriangle by rows, termwise products of the natural numbers decrescendo and the bisected Fibonacci series.

%I A143929 #22 Feb 01 2025 08:41:37
%S A143929 1,2,1,3,2,3,4,3,6,8,5,4,9,16,21,6,5,12,24,42,55,7,6,15,32,63,110,144,
%T A143929 8,7,18,40,84,165,288,377,9,8,21,48,105,220,432,754,987,10,9,24,56,
%U A143929 126,275,576,1131,1974,2584
%N A143929 Eigentriangle by rows, termwise products of the natural numbers decrescendo and the bisected Fibonacci series.
%C A143929 Row sums = even-indexed Fibonacci terms A001906.
%C A143929 Sum of n-th row terms = rightmost term of next row.
%F A143929 Given A004736: (1; 2,1; 3,2,1; 4,3,2,1; ...), we apply the termwise products of the sequence {A088305(n-1)}_{n>=1} starting (1, 1, 3, 8, 21, ...).
%F A143929 From _Wolfdieter Lang_, Jan 07 2021: (Start)
%F A143929 T(n, m) = 0 if n < m; T(n, 1) = n, for n >= 1, and T(n, m) = F(2*(m-1))*(n-m+1) for n >= m >= 2, with F = A000045 (Fibonacci).
%F A143929 G.f. column m: G(1, x) = x/(1-x)^2, G(m, x) = F(2*(m-1))*x^m/(1-x)^2, for m >= 2. (End)
%F A143929 With offset 0: g.f. of row polynomials R(n, x) := Sum_{m=0..n} t(n, m)*x^m, that is, g.f. of triangle t(n,m) = T(n+1, m+1):
%F A143929 G(z, x) = (1 - x*z)^2 / ((1 - z)^2*(1 - 3*x*z + (x*z)^2)). - _Wolfdieter Lang_, Apr 09 2021
%e A143929 First rows of the triangle T(n, m), n >= 1, m = 1..n:
%e A143929   1;
%e A143929   2, 1;
%e A143929   3, 2,  3;
%e A143929   4, 3,  6,  8;
%e A143929   5, 4,  9, 16,  21;
%e A143929   6, 5, 12, 24,  42,  55;
%e A143929   7, 6, 15, 32,  63, 110, 144;
%e A143929   8, 7, 18, 40,  84, 165, 288, 377;
%e A143929   9, 8, 21, 48, 105, 220, 432, 754, 987;
%e A143929   ...
%e A143929 Example: row 4 = (4, 3, 6, 8) = termwise product of (4, 3, 2, 1) and (1, 1, 3, 8).
%Y A143929 Cf. A000045, A001906, A004736, A088305.
%K A143929 nonn,easy,tabl
%O A143929 1,2
%A A143929 _Gary W. Adamson_, Sep 05 2008