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Eigentriangle by rows, termwise products of the natural numbers decrescendo and the bisected Fibonacci series.
1

%I #18 Mar 30 2024 23:08:49

%S 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 8,7,18,40,84,165,288,377,9,8,21,48,105,220,432,754,987,10,9,24,56,

%U 126,275,576,1131,1974,2584

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

%C Row sums = even-indexed Fibonacci terms A001906.

%C Sum of n-th row terms = rightmost term of next row.

%F 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 From _Wolfdieter Lang_, Jan 07 2021: (Start)

%F 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 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 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 G(z, x) = (1 - x*z)^2 / ((1 - z)^2*(1 - 3*x*z + (x*z)^2)). - _Wolfdieter Lang_, Apr 09 2021

%e First rows of the triangle T(n, m), n >= 1, m = 1..n:

%e 1;

%e 2, 1;

%e 3, 2, 3;

%e 4, 3, 6, 8;

%e 5, 4, 9, 16, 21;

%e 6, 5, 12, 24, 42, 55;

%e 7, 6, 15, 32, 63, 110, 144;

%e 8, 7, 18, 40, 84, 165, 288, 377;

%e 9, 8, 21, 48, 105, 220, 432, 754, 987;

%e ...

%e Example: row 4 = (4, 3, 6, 8) = termwise product of (4, 3, 2, 1) and (1, 1, 3, 8).

%Y Cf. A000045, A001906, A004736, A088305.

%K nonn,easy,tabl

%O 1,2

%A _Gary W. Adamson_, Sep 05 2008