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A006324
a(n) = n*(n + 1)*(2*n^2 + 2*n - 1)/6.
12
1, 11, 46, 130, 295, 581, 1036, 1716, 2685, 4015, 5786, 8086, 11011, 14665, 19160, 24616, 31161, 38931, 48070, 58730, 71071, 85261, 101476, 119900, 140725, 164151, 190386, 219646, 252155, 288145, 327856, 371536, 419441, 471835, 528990, 591186
OFFSET
1,2
COMMENTS
4-dimensional analog of centered polygonal numbers.
Partial sums ofA000447.-Zak Seidov,May 19 2006
FromJohannes W. Meijer,Jun 27 2009: (Start)
Equals the absolute values of the coefficients that precede the a(n-1) factors of the recurrence relations RR(n) ofA162011.
This sequence enabled the analysis ofA162012andA162013.(End)
Equals the number of integer quadruples (x,y,z,w) such that min(x,y) < min(z,w), max(x,y) < max(z,w), and 0 <= x,y,z,w <= n. -Andrew Woods,Apr 21 2014
For n>3 a(n)=twice the area of an irregular quadrilateral with vertices at the points (C(n,4),C(n+1,4)), (C(n+1,4),C(n+2,4)), (C(n+2,4),C(n+3,4)), and (C(n+3,4),C(n+4,4)). -J. M. Bergot,Jun 14 2014
FORMULA
a(n) = 8*C(n + 2, 4) + C(n + 1, 2).
a(n) = (Sum_{k=1..n} k^5) / (Sum_{k=1..n} k) =A000539(n) /A000217(n). -Alexander Adamchuk,Apr 12 2006
FromJohannes W. Meijer,Jun 27 2009: (Start)
Recurrence relation 0 = Sum_{k=0..5} (-1)^k*binomial(5,k)*a(n-k).
G.f.: (1+6*z+z^2)/(1-z)^5. (End)
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). -Wesley Ivan Hurt,May 02 2021
Sum_{n>=1} 1/a(n) = 6 + 2*sqrt(3)*Pi*tan(sqrt(3)*Pi/2). -Amiram Eldar,Aug 23 2022
a(n) =A053134(n-1) - 4*A002415(n). -Yasser Arath Chavez Reyes,Feb 12 2024
MAPLE
A006324:=n->n*(n + 1)*(2*n^2 + 2*n - 1)/6; seq(A006324(n), n=1..30); #Wesley Ivan Hurt,Jun 14 2014
MATHEMATICA
Table[Sum[k^5, {k, n}]/Sum[k, {k, n}], {n, 40}] (*Alexander Adamchuk,Apr 12 2006 *)
PROG
(Magma) [ n*(n + 1)*(2*n^2 + 2*n - 1)/6: n in [1..30] ]; //Wesley Ivan Hurt,Jun 14 2014
KEYWORD
nonn,easy
AUTHOR
Albert Rich (Albert_Rich(AT)msn.com), Jun 14 1998
EXTENSIONS
Simpler definition fromAlexander Adamchuk,Apr 12 2006
More terms fromZak Seidov
STATUS
approved