OFFSET
0,6
COMMENTS
There are several sequences of integers of the form floor(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,9,11,12,15,16,24).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,0,0,1,-3,3,-1).
FORMULA
a(n) = round((2*n+1)*(2*n^2 + 2*n - 21)/192).
a(n) = floor((n-1)*(2*n^2 + 5*n - 15)/96).
a(n) = ceiling((n+2)*(2*n^2 - n - 18)/96).
a(n) = round(n*(n+4)*(2*n - 5)/96).
a(n) = a(n-16) + (n+1)*(n-16) + 90, n > 15.
G.f.: x^4*(1 - x + x^2) / ( (1+x)*(x^2+1)*(x^4+1)*(x-1)^4 ). - R. J. Mathar, Dec 06 2010
a(n)= 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-8) - 3*a(n-9) + 3*a(n-10) - a(n-11). - R. J. Mathar, Dec 06 2010
EXAMPLE
a(16) = 0 + 0 + 0 + 0 + 1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 + 10 + 12 + 14 + 16 = 90.
MAPLE
seq(round(n*(n+4)*(2*n-5)/96), n=0..50)
MATHEMATICA
Accumulate[Table[Floor[n^2/16], {n, 0, 60}]] (* Harvey P. Dale, Dec 13 2010 *)
PROG
(Magma) [Round((2*n+1)*(2*n^2+2*n-21)/192): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011
(PARI) concat([0, 0, 0, 0], Vec((1-x+x^2)/((1+x)*(x^2+1)*(x^4+1)*(x-1)^4)+O(x^99))) \\ Charles R Greathouse IV, Oct 18 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mircea Merca, Dec 04 2010
STATUS
approved