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A175777 Partial sums of floor(n^2/16). 2
0, 0, 0, 0, 1, 2, 4, 7, 11, 16, 22, 29, 38, 48, 60, 74, 90, 108, 128, 150, 175, 202, 232, 265, 301, 340, 382, 427, 476, 528, 584, 644, 708, 776, 848, 924, 1005, 1090, 1180, 1275, 1375, 1480, 1590, 1705, 1826, 1952, 2084, 2222, 2366, 2516, 2672 (list; graph; refs; listen; history; text; internal format)
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

Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,0,0,1,-3,3,-1).

Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.

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

Sequence in context: A217838 A212367 A225088 * A098574 A212366 A309838

Adjacent sequences:  A175774 A175775 A175776 * A175778 A175779 A175780

KEYWORD

nonn,easy

AUTHOR

Mircea Merca, Dec 04 2010

STATUS

approved

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Last modified August 4 19:37 EDT 2020. Contains 336202 sequences. (Running on oeis4.)