A173690 Partial sums of round(n^2/5).
0, 0, 1, 3, 6, 11, 18, 28, 41, 57, 77, 101, 130, 164, 203, 248, 299, 357, 422, 494, 574, 662, 759, 865, 980, 1105, 1240, 1386, 1543, 1711, 1891, 2083, 2288, 2506, 2737, 2982, 3241, 3515, 3804, 4108, 4428, 4764, 5117, 5487, 5874, 6279, 6702, 7144, 7605, 8085, 8585
Offset: 0
Examples
a(5) = round(1/5) + round(4/5) + round(9/5) + round(16/5) + round(25/5) = 0 + 1 + 2 + 3 + 5 = 11.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..2000
- 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,-3,1,0,1,-3,3,-1).
Crossrefs
Cf. A008738.
Programs
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Maple
A173690 := proc(n) add( round(i^2/5),i=0..n) ; end proc: # R. J. Mathar, Jan 10 2011
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Mathematica
Accumulate[Round[Range[0,50]^2/5]] (* or *) LinearRecurrence[{3,-3,1,0,1,-3,3,-1},{0,0,1,3,6,11,18,28},60] (* Harvey P. Dale, Mar 16 2022 *) -
PARI
a(n)=(2*n^3+3*n^2+n+6)\30 \\ Charles R Greathouse IV, May 30 2011
Formula
a(n) = Sum_{k=0..n} round(k^2/5);
a(n) = round((2*n^3 + 3*n^2 + n)/30);
a(n) = floor((2*n^3 + 3*n^2 + n + 6)/30);
a(n) = ceiling((2*n^3 + 3*n^2 + n - 6)/30);
a(n) = a(n-5) + (n-2)^2 + 2, n > 4;
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-5) - 3*a(n-6) + 3*a(n-7) - a(n-8), n > 7.
G.f.: x^2*(x+1)*(x^2 - x + 1) / ( (x^4 + x^3 + x^2 + x + 1)*(x-1)^4 ).
Comments