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a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)
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%I #46 Dec 15 2023 19:18:29

%S 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,

%T 65,70,75,81,87,93,99,105,112,119,126,133,140,148,156,164,172,180,189,

%U 198,207,216,225,235,245,255,265,275,286,297,308,319,330,342,354,366

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

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

%C 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

%H Vincenzo Librandi, <a href="/A130520/b130520.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,1,-2,1).

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

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

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

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

%F a(n) = floor((n-1)*(n-2)/10). - _Mitch Harris_, Sep 08 2008

%F 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

%F a(n) = A008732(n-5), n > 4. - _R. J. Mathar_, Nov 22 2008

%F a(n) = a(n-5) + n - 4, n > 4. - _Mircea Merca_, Nov 28 2010

%F 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

%F From _Amiram Eldar_, Sep 17 2022: (Start)

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

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

%p seq(floor((n-1)*(n-2)/10), n=0..70); # _G. C. Greubel_, Aug 31 2019

%t Accumulate[Floor[Range[0,70]/5]] (* _Harvey P. Dale_, May 25 2016 *)

%o (Magma) [Round(n*(n-3)/10): n in [0..70]]; // _Vincenzo Librandi_, Jun 25 2011

%o (PARI) a(n) = sum(k=0, n, k\5); \\ _Michel Marcus_, May 13 2016

%o (Sage) [floor((n-1)*(n-2)/10) for n in (0..70)] # _G. C. Greubel_, Aug 31 2019

%o (GAP) List([0..70], n-> Int((n-1)*(n-2)/10)); # _G. C. Greubel_, Aug 31 2019

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

%K nonn,easy

%O 0,7

%A _Hieronymus Fischer_, Jun 01 2007