A090394 Ninth diagonal (m=8) of triangle A084938; a(n) = A084938(n+8,n) = (n^8 + 84*n^7 + 3346*n^6 + 84840*n^5 + 1550689*n^4 + 21632436*n^3 + 224782284*n^2 + 1377648720*n)/40320.

0, 40320, 95616, 170856, 272584, 409360, 592296, 835702, 1157857, 1581921, 2137005, 2859417, 3794103, 4996303, 6533443, 8487285, 10956358, 14058694, 17934894, 22751550, 28705050, 36025794, 44982850, 55889080, 69106767

Offset

0,2

Links

Chai Wah Wu, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (9, -36, 84, -126, 126, -84, 36, -9, 1).

Formula

a(n) = A084938(n+8, n) = Sum_{k=0..8} A090238(8, k)*binomial(n, k).

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

a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n > 8.

G.f.: x*(29093*x^7 - 212062*x^6 + 663528*x^5 - 1155496*x^4 + 1209824*x^3 - 761832*x^2 + 267264*x - 40320)/(x - 1)^9. (End)

Mathematica

LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {0, 40320, 95616, 170856, 272584, 409360, 592296, 835702, 1157857}, 30] (* Harvey P. Dale, Apr 04 2017 *)

Prog

(Python)
from itertools import islice
def A090394_generator():
    m = [1, 7, 33, 135, 531, 2109, 8411, 29093, 0]
    yield m[-1]
    while True:
        for i in range(8):
            m[i+1]+= m[i]
        yield m[-1]
list(islice(A090394_generator(), 0, 50, 1)) # Chai Wah Wu, Nov 14 2014

Crossrefs

Cf. A084938 A090238.

Sequence in context: A127228 A269126 A168359 * A250055 A321844 A226887

Adjacent sequences: A090391 A090392 A090393 * A090395 A090396 A090397

Keyword

easy,nonn

Author

Philippe Deléham, Jan 31 2004

Status

approved