A011245 a(n) = (n+1)*(2*n+1)*(3*n+1)*(4*n+1).

1, 120, 945, 3640, 9945, 22176, 43225, 76560, 126225, 196840, 293601, 422280, 589225, 801360, 1066185, 1391776, 1786785, 2260440, 2822545, 3483480, 4254201, 5146240, 6171705, 7343280, 8674225, 10178376, 11870145, 13764520, 15877065, 18223920, 20821801, 23688000

Offset

0,2

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = A033593(-n).

G.f.: (1 + 115*x + 355*x^2 + 105*x^3)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009

a(0)=1, a(1)=120, a(2)=945, a(3)=3640, a(4)=9945, a(n) = 5*a(n-1)- 10*a(n-2)+ 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Oct 05 2012

E.g.f.: (1 + 119*x + 353*x^2 + 194*x^3 + 24*x^4)*exp(x). - G. C. Greubel, Mar 04 2020

From Amiram Eldar, Mar 11 2022: (Start)

Sum_{n>=0} 1/a(n) = (4/3 - 3*sqrt(3)/4)*Pi + 12*log(2) - 27*log(3)/4.

Sum_{n>=0} (-1)^n/a(n) = (1 + 4*sqrt(2)/3 - 3*sqrt(3)/2)*Pi - 14*log(2)/3 - 8*sqrt(2)*log(sqrt(2)-1)/3. (End)

Maple

seq( mul(j*n+1, j=1..4), n=0..30); # G. C. Greubel, Mar 04 2020

Mathematica

Table[Times@@(Range[4]n+1), {n, 0, 30}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 120, 945, 3640, 9945}, 30] (* Harvey P. Dale, Oct 05 2012 *)

Prog

(Magma) [&*[s*n+1: s in [1..4]]: n in [0..25]]; // Bruno Berselli, May 23 2011
(PARI) a(n)=24*n^4+50*n^3+35*n^2+10*n+1 \\ Charles R Greathouse IV, May 23 2011
(Sage) [product(j*n+1 for j in (1..4)) for n in (0..30)] # G. C. Greubel, Mar 04 2020
(GAP) List([0..30], n-> (n+1)*(2*n+1)*(3*n+1)*(4*n+1) ); # G. C. Greubel, Mar 04 2020

Crossrefs

Cf. A033593.

Sequence in context: A211471 A250519 A250512 * A213875 A092182 A262387

Adjacent sequences: A011242 A011243 A011244 * A011246 A011247 A011248

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved