A090392 Seventh diagonal (m=6) of triangle A084938; a(n) = A084938(n+6,n) = (n^6 + 45*n^5 + 925*n^4 + 11475*n^3 + 92314*n^2 + 413640*n)/720.

0, 720, 1812, 3428, 5768, 9090, 13721, 20069, 28636, 40032, 54990, 74382, 99236, 130754, 170331, 219575, 280328, 354688, 445032, 554040, 684720, 840434, 1024925, 1242345, 1497284, 1794800, 2140450, 2540322, 3001068, 3529938, 4134815

Offset: 0

Author

Philippe Deléham, Jan 31 2004

Keywords

• easy
• nonn

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (7, -21, 35, -35, 21, -7, 1).

Crossrefs

Cf. A084938 A090238.

Programs

Mathematica

LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,720,1812,3428,5768,9090,13721},40] (* Harvey P. Dale, Jul 20 2016 *)

Python

A090392_list, m = [], [1, 5, 18, 58, 177, 461, 0]
for _ in range(1001):
    A090392_list.append(m[-1])
    print(m[-1])
    for i in range(6):
        m[i+1] += m[i] # Chai Wah Wu, Jun 04 2016

Formula

a(n) = A084938(n+6, n) = Sum_{k=0..6} A090238(6, k)*binomial(n, k).

From Chai Wah Wu, Jun 04 2016: (Start)

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n > 6.

G.f.: x*(461*x^5 - 2482*x^4 + 5376*x^3 - 5864*x^2 + 3228*x - 720)/(x - 1)^7. (End)