A130520 a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)

0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 81, 87, 93, 99, 105, 112, 119, 126, 133, 140, 148, 156, 164, 172, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265, 275, 286, 297, 308, 319, 330, 342, 354, 366

Offset: 0

Author

Hieronymus Fischer, Jun 01 2007

Keywords

• nonn
• easy

Comments

Complementary with A130483 regarding triangular numbers, in that A130483(n) + 5*a(n) = n*(n+1)/2 = A000217(n).

Given a sequence b(n) defined by variables b(0) to b(5) and recursion b(n) = -(b(n-6) * a(n-2) * (b(n-4) * b(n-2)^3 - b(n-3)^3 * b(n-1)) - b(n-5) * b(n-3) * b(n-1) * (b(n-5) * b(n-2)^2 - b(n-4)^2 * b(n-1)))/(b(n-4) * (b(n-5) * b(n-3)^3 - b(n-4)^3 * b(n-2))). The denominator of b(n+1) has a factor of (b(1) * b(3)^3 - b(2)^3 * b(4))^a(n+1). For example, if b(0) = 2, b(1) = b(2) = b(3) = 1, b(4) = 1+x, b(5) = 4, then the denominator of b(n+1) is x^a(n+1). - Michael Somos, Nov 15 2023

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Crossrefs

Cf. A002264, A002265, A004526, A010872, A010873, A010874, A130481, A130482.

Programs

GAP

List([0..70], n-> Int((n-1)*(n-2)/10)); # G. C. Greubel, Aug 31 2019

Magma

[Round(n*(n-3)/10): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

Maple

seq(floor((n-1)*(n-2)/10), n=0..70); # G. C. Greubel, Aug 31 2019

Mathematica

Accumulate[Floor[Range[0,70]/5]] (* Harvey P. Dale, May 25 2016 *)

PARI

a(n) = sum(k=0, n, k\5); \\ Michel Marcus, May 13 2016

Sage

[floor((n-1)*(n-2)/10) for n in (0..70)] # G. C. Greubel, Aug 31 2019

Formula

a(n) = floor(n/5)*(2*n - 3 - 5*floor(n/5))/2.

a(n) = A002266(n)*(2*n - 3 - 5*A002266(n))/2.

a(n) = A002266(n)*(n -3 +A010874(n))/2.

G.f.: x^5/((1-x^5)*(1-x)^2) = x^5/( (1+x+x^2+x^3+x^4)*(1-x)^3 ).

a(n) = floor((n-1)*(n-2)/10). - Mitch Harris, Sep 08 2008

a(n) = round(n*(n-3)/10) = ceiling((n+1)*(n-4)/10) = round((n^2 - 3*n - 1)/10). - Mircea Merca, Nov 28 2010

a(n) = A008732(n-5), n > 4. - R. J. Mathar, Nov 22 2008

a(n) = a(n-5) + n - 4, n > 4. - Mircea Merca, Nov 28 2010

a(5n) = A000566(n), a(5n+1) = A005476(n), a(5n+2) = A005475(n), a(5n+3) = A147875(n), a(5n+4) = A028895(n). - Philippe Deléham, Mar 26 2013

From Amiram Eldar, Sep 17 2022: (Start)

Sum_{n>=5} 1/a(n) = 518/45 - 2*sqrt(2*(sqrt(5)+5))*Pi/3.

Sum_{n>=5} (-1)^(n+1)/a(n) = 8*sqrt(5)*arccoth(3/sqrt(5))/3 + 92*log(2)/15 - 418/45. (End)