A361608 a(n) = 7^n*(n+1)*(81*n^4+684*n^3+1401*n^2+434*n+40)/40.

1, 924, 48804, 1337014, 26622288, 437049228, 6295986235, 82489361052, 1005444707211, 11576481361732, 127278262644918, 1346951022678114, 13803666582387682, 137633164619393268, 1340161331495822661, 12782144706910135480, 119711031072135899781, 1103157160378734314700, 10019811250265958667288

Offset

0,2

Comments

The sequences A(n,k) = sum_{j=0..n} sum_{i=0..j} (-1)^(j-i) *binomial(n,j) *binomial(j,i) *binomial(j+k+(k+1)*i,j+k) are C-sequences for fixed integer k, here A(n,k=5)= a(n).

Links

Table of n, a(n) for n=0..18.

Project Euler, Problem 831. Triple Product

R. J. Mathar, On an alternating double sum of a triple product of aerated binomial coefficients, arXiv:2306.08022 (2023)

Index entries for linear recurrences with constant coefficients, signature (42,-735,6860,-36015,100842,-117649)

Formula

G.f.: ( 1+882*x+10731*x^2-40474*x^3+36015*x^4 ) / (7*x-1)^6 .

a(n) = +42*a(n-1) -735*a(n-2) +6860*a(n-3) -36015*a(n-4) +100842*a(n-5) -117649*a(n-6) .

D-finite with recurrence n*(81*n^4+360*n^3-165*n^2-640*n+404)*a(n) -7*(n+1)*(81*n^4+684*n^3+1401*n^2+434*n+40)*a(n-1)=0.

Mathematica

LinearRecurrence[{42, -735, 6860, -36015, 100842, -117649}, {1, 924, 48804, 1337014, 26622288, 437049228}, 20] (* Harvey P. Dale, May 29 2023 *)

Prog

(Python)
def A361608(n): return 7**n*(n*(n*(n*(n*(81*n + 765) + 2085) + 1835) + 474) + 40)//40 # Chai Wah Wu, Mar 17 2023

Crossrefs

Cf. A027471 (k=1), A361609 (k=2), A361610 (k=3).

Sequence in context: A245859 A172224 A283577 * A177304 A022053 A107517

Adjacent sequences: A361605 A361606 A361607 * A361609 A361610 A361611

Keyword

nonn,easy

Author

R. J. Mathar, Mar 17 2023

Status

approved