OFFSET
1,3
COMMENTS
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)
FORMULA
From Peter Bala, Oct 24 2007: (Start)
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 + ....
T(n,n-k) = (1/3)*C(n,k)*(2^k - (-1)^k) = C(n,k)*A001045(k).
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.
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)
EXAMPLE
First few rows of the triangle:
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, 105, 21, 7;
...
PROG
(PARI) M(n)={my(t=matrix(n+1, n+1, i, j, binomial(i-1, j-1))); (t^2 - t^(-1))[2..n+1, 1..n]/3}
{ my(A=M(10)); for(i=1, #A, print(A[i, 1..i])) } \\ Andrew Howroyd, Sep 23 2025
(PARI) T(n, k)=binomial(n, k)*(2^(n-k) - (-1)^(n-k)) / 3
{ for(n=1, 10, print(vector(n, k, T(n, k-1)))) } \\ Andrew Howroyd, Sep 23 2025
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Gary W. Adamson, Jun 12 2007
STATUS
approved
