A361610 -id:A361610 - cp's OEIS frontend

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

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

Offset: 0

R. J. Mathar, Mar 17 2023

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).

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).

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

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

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

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.

A361609 a(n) = 4^n(1 + (23/8)n + (9/8)*n^2).

Original entry on oeis.org

1, 20, 180, 1264, 7808, 44544, 240640, 1249280, 6291456, 30932992, 149159936, 707788800, 3313500160, 15334375424, 70262980608, 319169757184, 1438814044160, 6442450944000, 28673201668096, 126924873531392, 559101662724096, 2451910929940480, 10709243254538240, 46601700831657984

Offset: 0

R. J. Mathar, Mar 17 2023

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=2) = a(n).

Winston de Greef, Table of n, a(n) for n = 0..1640
Project Euler, Problem 831. Triple Product
Index entries for linear recurrences with constant coefficients, signature (12,-48,64).

Cf. A027471 (k=1), A361610 (k=3), A361608 (k=5).

LinearRecurrence[{12, -48, 64}, {1, 20, 180}, 25] (* or *)
A361609[n_] := 4^n (1 + 23/8 n + 9/8 n^2);
Array[A361609, 25, 0] (* Paolo Xausa, Jan 18 2024 *)

def A361609(n): return (n*(9*n + 23) + 8)<<((n<<1)-3) if n > 1 else 19*n+1 # Chai Wah Wu, Mar 17 2023

G.f.: ( -1-8*x+12*x^2 ) / (4*x-1)^3.
a(n) = 12*a(n-1) -48*a(n-2) +64*a(n-3).
D-finite with recurrence (-9*n^2-5*n+6)*a(n) +4*(9*n^2+23*n+8)*a(n-1)=0.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

A361609 a(n) = 4^n(1 + (23/8)n + (9/8)*n^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

A361609 a(n) = 4^n*(1 + (23/8)*n + (9/8)*n^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

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

A361609 a(n) = 4^n(1 + (23/8)n + (9/8)*n^2).