|
|
A130520
|
|
a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)
|
|
20
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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) = round(n*(n-3)/10) = ceiling((n+1)*(n-4)/10) = round((n^2 - 3*n - 1)/10). - Mircea Merca, Nov 28 2010
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
|
|
|
PROG
|
(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
|
|
|
STATUS
|
approved
|
|
|
|