OFFSET
1,2
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Dead link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).
FORMULA
a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(10 + 3*n*(n+5))/20160.
This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=5. - Alexander R. Povolotsky, May 17 2008
G.f.: x*(x^2+4*x+1) / (1-x)^9. - Colin Barker, Apr 23 2015
Sum_{n>=1} 1/a(n) = -162*sqrt(21/5)*Pi*tan(sqrt(35/3)*Pi/2) - 136269/100. - Amiram Eldar, Jan 26 2022
MATHEMATICA
Table[Binomial[n+5, 6]*(3*n^2+15*n+10)/28, {n, 1, 30}] (* G. C. Greubel, Dec 01 2018 *)
Nest[Accumulate, Range[40]^3, 5] (* Harvey P. Dale, Feb 06 2023 *)
PROG
(PARI) a(n)=sum(t=1, n, sum(s=1, t, sum(l=1, s, sum(j=1, l, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1))))))) \\ Alexander R. Povolotsky, May 17 2008
(PARI) Vec(-x*(x^2+4*x+1)/(x-1)^9 + O(x^100)) \\ Colin Barker, Apr 23 2015
(PARI) a(n) = binomial(n+5, 6)*(3*n^2+15*n+10)/28 \\ Charles R Greathouse IV, Apr 23 2015
(Magma) [Binomial(n+5, 6)*(3*n^2+15*n+10)/28: n in [1..30]]; // G. C. Greubel, Dec 01 2018
(Sage) [binomial(n+5, 6)*(3*n^2+15*n+10)/28 for n in (1..30)] # G. C. Greubel, Dec 01 2018
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
EXTENSIONS
Edited by Ralf Stephan, Dec 16 2004
STATUS
approved