OFFSET
0,3
COMMENTS
Partial sums of A036410.
There are several sequences of integers of the form ceiling(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,8,10,11,12, 14,15,16,19,20,23,24).
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 (2,0,-1,-1,0,2,-1).
FORMULA
a(n) = round((2*n+1)*(2*n^2 + 2*n + 41)/144).
a(n) = floor((n+1)*(2*n^2 + n + 41)/72).
a(n) = ceiling((2*n^3 + 3*n^2 + 42*n)/72).
a(n) = a(n-12) + (n+1)*(n-12) + 61.
G.f.: x*(1-x^2+x^4) / ( (1+x)*(1+x+x^2)*(x-1)^4 ). - R. J. Mathar, Jun 22 2011
EXAMPLE
a(12) = 0 + 1 + 1 + 1 + 2 + 3 + 3 + 5 + 6 + 7 + 9 + 11 + 12 = 61.
MAPLE
seq(floor((n+1)*(2*n^2+n+41)/72), n=0..50)
MATHEMATICA
Accumulate[Ceiling[Range[0, 50]^2/12]] (* or *) LinearRecurrence[{2, 0, -1, -1, 0, 2, -1}, {0, 1, 2, 3, 5, 8, 11}, 60] (* Harvey P. Dale, Apr 16 2023 *)
PROG
(Magma) [Round((2*n+1)*(2*n^2+2*n+41)/144): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011
(PARI) a(n)=(n+1)*(2*n^2+n+41)\72 \\ Charles R Greathouse IV, Jul 06 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mircea Merca, Dec 05 2010
STATUS
approved