A144679 a(n) = [n/5 + 1]*[n/5 + 2]*(3*n - 10*[n/5] + 3)/6, where [.] = floor.
1, 2, 3, 4, 5, 8, 11, 14, 17, 20, 26, 32, 38, 44, 50, 60, 70, 80, 90, 100, 115, 130, 145, 160, 175, 196, 217, 238, 259, 280, 308, 336, 364, 392, 420, 456, 492, 528, 564, 600, 645, 690, 735, 780, 825, 880, 935, 990, 1045, 1100, 1166, 1232, 1298, 1364, 1430, 1508, 1586, 1664
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- James Cummings, Daniel Král', Florian Pfender, Konrad Sperfeld, Andrew Treglown, and Michael Young, Monochromatic triangles in three-coloured graphs, Journal of Combinatorial Theory B 103 no. 4 (2013) 489-503 (also: arXiv:1206.1987).
- Brian O'Sullivan and Thomas Busch, Spontaneous emission in ultra-cold spin-polarised anisotropic Fermi seas, arXiv 0810.0231v1 [quant-ph], 2008. [Eq (10b), lambda=5]
- Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,2,-4,2,0,0,-1,2,-1).
Programs
-
Magma
R
:=PowerSeriesRing(Integers(), 60); Coefficients(R!( 1/((1-x)*(1-x^5))^2 )); // G. C. Greubel, Oct 18 2021 -
Maple
n:=80; lambda:=5; S10b:=[]; for ii from 0 to n do x:=floor(ii/lambda); snc:=1/6*(x+1)*(x+2)*(3*ii-2*x*lambda+3); S10b:=[op(S10b),snc]; od: S10b; A144679 := proc(n) option remember; local k; sum(THN5(n-k),k=0..4) end: THN5:= proc(n) option remember; THN5(n):= binomial(floor(n/5)+3,3) end: seq(A144679(n), n=0..57); # Johannes W. Meijer, May 20 2011
-
Mathematica
LinearRecurrence[{2,-1,0,0,2,-4,2,0,0,-1,2,-1}, {1,2,3,4,5,8,11,14,17,20,26,32}, 60] (* Jean-François Alcover, Nov 22 2017 *) CoefficientList[Series[1/((x-1)^4(x^4+x^3+x^2+x+1)^2),{x,0,100}],x] (* Harvey P. Dale, Aug 29 2021 *)
-
PARI
apply( {A144679(n)=(3*n+3-10*n\=5)*(n+1)*(n+2)\6}, [0..55]) \\ M. F. Hasler, Jun 25 2021
-
Sage
def A144679_list(prec): P.
= PowerSeriesRing(ZZ, prec) return P( 1/((1-x)*(1-x^5))^2 ).list() A144679_list(60) # G. C. Greubel, Oct 18 2021
Formula
From Johannes W. Meijer, May 20 2011: (Start)
a(n-4) + a(n-3) + a(n-2) + a(n-1) + a(n) = A122047(n+2).
G.f.: 1/((1-x)^4*(1 + x + x^2 + x^3 + x^4)^2). (End)
a(n) = r*A000292(q+1) + (5-r)*A000292(q) = (n + 2r + 1)*(q + 2)*(q + 1)/6, where A000292(q) = binomial(q+2,3), r = (n+1) mod 5, q = (n+1-r)/5. - M. F. Hasler, Jun 25 2021
Comments