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A131048 (1/3) * (A007318^2 - A007318^(-1)).

%I A131048 #12 Dec 29 2023 10:54:44
%S A131048 1,1,2,3,3,3,5,12,6,4,11,25,30,10,5,21,66,75,60,15,6,43,147,231,175,
%T A131048 105,21,7,85,344,588,616,350,168,28,8,171,765,1548,1764,1386,630,252,
%U A131048 36,9
%N A131048 (1/3) * (A007318^2 - A007318^(-1)).
%C A131048 Left border = A001045: (1, 1, 3, 5, 11, 21, 43, 85, ...).
%C A131048 Row sums = (1, 3, 9, 27, ...).
%C A131048 Analogous triangles for other powers of P are: A131047, A131049, A131050 and A131051.
%F A131048 Let A007318 (Pascal's triangle) = P. then A131048 = (1/3) * (P^2 - 1/P). Delete right border of zeros.
%F A131048 From _Peter Bala_, Oct 24 2007: (Start)
%F A131048 O.g.f.: 1/(1 - (2*x + 1)*t + (x^2 + x - 2)*t^2) = 1 + (1 + 2*x)*t + (3 + 3*x + 3*x^2)*t^2 + ....
%F A131048 T(n,n-k) = (1/3)*C(n,k)*(2^k - (-1)^k) = C(n,k)*A001045(k).
%F A131048 The row polynomials R(n,x) := Sum_{k = 0..n} T(n,n-k)*x^(n-k) = (1/3)*((x + 2)^n - (x - 1)^n) and have the divisibility property R(n,x) divides R(m,x) in the polynomial ring Z[x] if n divides m.
%F A131048 The polynomials R(n,-x), n >= 2, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re x = 1/2 in the complex plane. Compare with A094440. (End)
%e A131048 First few rows of the triangle:
%e A131048    1;
%e A131048    1,   2;
%e A131048    3,   3,   3;
%e A131048    5,  12,   6,   4;
%e A131048   11,  25,  30,  10,   5;
%e A131048   21,  66,  75,  60,  15,  6;
%e A131048   43, 147, 231, 175, 105, 21, 7;
%e A131048   ...
%Y A131048 Cf. A131047, A131049, A131050, A131051, A001045, A007318.
%Y A131048 Cf. A001045, A094440, A132148.
%K A131048 nonn,tabl
%O A131048 1,3
%A A131048 _Gary W. Adamson_, Jun 12 2007