OFFSET
0,7
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).
FORMULA
a(n) = floor(n/5)*(2*n - 3 - 5*floor(n/5))/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)
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 *)
PROG
(Magma) [Round(n*(n-3)/10): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011
(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
(GAP) List([0..70], n-> Int((n-1)*(n-2)/10)); # G. C. Greubel, Aug 31 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Hieronymus Fischer, Jun 01 2007
STATUS
approved