A131048 - OEIS

%I #12 Dec 29 2023 10:54:44

%S 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 105,21,7,85,344,588,616,350,168,28,8,171,765,1548,1764,1386,630,252,

%U 36,9

%N (1/3) * (A007318^2 - A007318^(-1)).

%C Left border = A001045: (1, 1, 3, 5, 11, 21, 43, 85, ...).

%C Row sums = (1, 3, 9, 27, ...).

%C Analogous triangles for other powers of P are: A131047, A131049, A131050 and A131051.

%F Let A007318 (Pascal's triangle) = P. then A131048 = (1/3) * (P^2 - 1/P). Delete right border of zeros.

%F From _Peter Bala_, Oct 24 2007: (Start)

%F 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 T(n,n-k) = (1/3)*C(n,k)*(2^k - (-1)^k) = C(n,k)*A001045(k).

%F 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 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 First few rows of the triangle:

%e    1;

%e    1,   2;

%e    3,   3,   3;

%e    5,  12,   6,   4;

%e   11,  25,  30,  10,   5;

%e   21,  66,  75,  60,  15,  6;

%e   43, 147, 231, 175, 105, 21, 7;

%e   ...

%Y Cf. A131047, A131049, A131050, A131051, A001045, A007318.

%Y Cf. A001045, A094440, A132148.

%K nonn,tabl

%O 1,3

%A _Gary W. Adamson_, Jun 12 2007