A213553 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = (n-1+h)^3, n>=1, h>=1, and ** = convolution.

1, 10, 8, 46, 43, 27, 146, 142, 118, 64, 371, 366, 334, 253, 125, 812, 806, 766, 658, 466, 216, 1596, 1589, 1541, 1406, 1150, 775, 343, 2892, 2884, 2828, 2666, 2346, 1846, 1198, 512, 4917, 4908, 4844, 4655, 4271, 3646, 2782, 1753, 729, 7942

Offset: 1

Author

Clark Kimberling, Jun 17 2012

Keywords

• nonn
• tabl
• easy

Comments

Principal diagonal: A213554

Antidiagonal sums: A101089

row 1, (1,2,3,...)**(1,8,27,...): A024166

row 2, (1,2,3,...)**(8,27,64,...): (3*k^5 + 30*k^4 + 115*k^3 + 210*k^2 + 122*k)/60

row 3, (1,2,3,...)**(27,64,125,...): (3*k^5 + 45*k^4 + 265*k^3 + 765*k^2 + 542*k)/120

For a guide to related arrays, see A213500.

Examples

			Northwest corner (the array is read by falling antidiagonals):
1.....10....46.....146....371
8.....43....142....366....806
27....118...334....766....1541
64....253...658....1406...2666
125...466...1150...2346...4271
		

Links

• G. C. Greubel, Antidiagonals n = 1..100, flattened

Crossrefs

Cf. A213500.

Programs

GAP

Flat(List([1..12], n-> List([1..n], k-> Binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30 ))); # G. C. Greubel, Jul 31 2019

Magma

[Binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k + n*(3*n^2 +6*n +1))/30: k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 31 2019

Mathematica

(* First program *)
b[n_]:= n; c[n_]:= n^3;
T[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
TableForm[Table[T[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[T[n-k+1, k], {n, 12}, {k, n, 1, -1}]]
r[n_]:= Table[T[n, k], {k, 1, 60}]  (* A213553 *)
d = Table[T[n, n], {n, 1, 40}] (* A213554 *)
s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A101089 *)
(* Second program *)
Table[Binomial[n-k+2, 2]*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jul 31 2019 *)

PARI

t(n,k) = binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30;
for(n=1,12, for(k=1,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Jul 31 2019

Sage

[[binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k + n*(3*n^2 +6*n +1))/30 for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 31 2019

Formula

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) -T(n,k-6).

G.f. for row n: f(x)/g(x), where f(x) = n^3 + (-3*n^3 + 3*n^2 + 3*n + 1)*x + (3*n^3 - 6*n^2 + 4)*x^2 - ((n-1)^3)*x^3 and g(x) = (1 - x)^6.

T(n,k) = k*((3*k^4 - 5*k^2 + 2) + 15*k*(k^2 - 1)*n + 30*(k^2 - 1)*n^2 + 30*(k + 1)*n^3)/60. - G. C. Greubel, Jul 31 2019