OFFSET
1,2
COMMENTS
T(n, k) is the k-subdiagonal sum of the matrix M(n) whose permanent is A322277(n).
FORMULA
O.g.f.: x*(- 1 + 2*y + 3*y^2 - 2*y^3 + 2*x*(- 1 + y^2) + x^4*(- 1 + 3*y^2) + x^2*(- 6 + 6*y + 2*y^2 - 6*y^3) + x^3*(- 2 + 4*y + 2*y^2 - 4*y^3))/((- 1 + x)^4*(1 + x)^2*(- 1 + y)^3*(1 + y)).
E.g.f.: (1/4)*exp(- x - y)*(- exp(2*x)*x + exp(2*y)*(x + 2*y) + 2*exp(2*(x + y))*(3*x^2 + x^3 - y - x*(- 2 + y + y^2))).
T(n, k) = (1/2)*n*(n^2 - k^2) if n and k are both even; T(n, k) = (1/2)*n*(n^2 - k^2 + 1) if n is even and k is odd; T(n, k) = (1/2)*(n*(n^2 - k^2 + 1) - 2*k) if n is odd and k is even; T(n, k) = (1/2)*(n*(n^2 - k^2 + 2) - 2*k) if n and k are both odd.
Diagonal: T(n, n-1) = A325657(n).
1st column: T(n, 0) = A317614(n).
EXAMPLE
The triangle T(n, k) begins:
---+-----------------------------
n\k| 0 1 2 3 4
---+-----------------------------
1 | 1
2 | 4 4
3 | 15 14 7
4 | 32 32 24 16
5 | 65 64 53 42 21
...
For n = 3 the matrix M(3) is
1, 2, 3
6, 5, 4
7, 8, 9
and therefore T(3, 0) = 1 + 5 + 9 = 15, T(3, 1) = 6 + 8 = 14, and T(3, 2) = 7.
MAPLE
a:=(n, k)->(1/4)*(2*(- 1 + (- 1)^n)*k - 2*k^2*n + n*(2 + (- 1)^(1+k) + (- 1)^(1 + n) + 2*n^2)): seq(seq(a(n, k), k=0..n-1), n=1..11);
MATHEMATICA
T[n_, k_]:=(1/4)*(2*(- 1 + (- 1)^n)*k - 2*k^2*n + n*(2 + (- 1)^(1+k) + (- 1)^(1 + n) + 2*n^2)); Flatten[Table[T[n, k], {n, 1, 11}, {k, 0, n-1}]]
PROG
(GAP) Flat(List([1..11], n->List([0..n-1], k->(1/4)*(2*(- 1 + (- 1)^n)*k - 2*k^2*n + n*(2 + (- 1)^(1+k) + (- 1)^(1 + n) + 2*n^2)))));
(Magma) [[(1/4)*(2*(- 1 + (- 1)^n)*k - 2*k^2*n + n*(2 + (- 1)^(1+k) + (- 1)^(1 + n) + 2*n^2)): k in [0..n-1]]: n in [1..11]];
(PARI) T(n, k) = (1/4)*(2*(- 1 + (- 1)^n)*k - 2*k^2*n + n*(2 + (- 1)^(1+k) + (- 1)^(1 + n) + 2*n^2));
tabl(nn) = for(n=1, nn, for(k=0, n-1, print1(T(n, k), ", ")); print);
tabl(11) \\ yields sequence in triangular form
CROSSREFS
KEYWORD
AUTHOR
Stefano Spezia, May 13 2019
STATUS
approved