OFFSET
0,2
COMMENTS
For a right triangle with sides of lengths 8*n^3 + 12*n^2 + 8*n + 2, 4*n^4 + 8*n^3 + 4*n^2, and 4*n^4 + 8*n^3 + 12*n^2 + 8*n + 2, dividing the area by the perimeter gives a(n). - J. M. Bergot, Jul 30 2013
This sequence is the difference between the centered icosahedral (or cuboctahedral) numbers (A005902(n)) and the centered octagonal pyramidal numbers (A000447(n+1)). - Peter M. Chema, Jan 09 2016
a(n) is the sum of the integers in the closed interval (n-1)*n to n*(n+1). - J. M. Bergot, Apr 19 2017
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
G.f.: x*(3 + 8*x + x^2)/(x-1)^4.
a(n) = ceiling(Sum_{i=n^2-(n-1)..n^2+(n-1)} s(i)), for n > 0 and integer i, where s(i) are the real solutions to x = i + sqrt(x), and the summation range excludes the integer solutions which occur where i is an oblong number (A002378). The fractional portion of the summation converges to 2/3 for large n. If s(i) is replaced with i, then the summation equals n^2*(2*n-1) = A015237. - Richard R. Forberg, Oct 15 2014
From Amiram Eldar, May 17 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/6 + 4*log(2) - 4.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/12 - Pi - 2*log(2) + 4. (End)
MATHEMATICA
a[n_]:=2*n^3+n^2; (* Vladimir Joseph Stephan Orlovsky, Dec 21 2008 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 3, 20, 63}, 40] (* Harvey P. Dale, Aug 19 2022 *)
PROG
(Magma) [n^2*(2*n+1): n in [0..50]]; // Vincenzo Librandi, May 01 2011
(PARI) a(n) = ceil(sum(i=n^2-(n-1), n^2+(n-1), if(!issquare(4*i+1), (2*i+1+sqrt(4*i+1))/2, 0))); \\ Michel Marcus, Nov 14 2014, after Richard R. Forberg
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Douglas Winston (douglas.winston(AT)srupc.com), Nov 07 2004
STATUS
approved