A130483 a(n) = Sum_{k=0..n} (k mod 5) (Partial sums of A010874).
0, 1, 3, 6, 10, 10, 11, 13, 16, 20, 20, 21, 23, 26, 30, 30, 31, 33, 36, 40, 40, 41, 43, 46, 50, 50, 51, 53, 56, 60, 60, 61, 63, 66, 70, 70, 71, 73, 76, 80, 80, 81, 83, 86, 90, 90, 91, 93, 96, 100, 100, 101, 103, 106, 110, 110, 111, 113, 116, 120, 120, 121, 123, 126, 130, 130
Offset: 0
Links
- Shawn A. Broyles, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).
Programs
-
GAP
a:=[0,1,3,6,10,10];; for n in [7..71] do a[n]:=a[n-1]+a[n-5]-a[n-6]; od; a; # G. C. Greubel, Aug 31 2019
-
Magma
I:=[0,1,3,6,10,10]; [n le 6 select I[n] else Self(n-1) + Self(n-5) - Self(n-6): n in [1..71]]; // G. C. Greubel, Aug 31 2019
-
Maple
seq(coeff(series(x*(1+2*x+3*x^2+4*x^3)/((1-x^5)*(1-x)), x, n+1), x, n), n = 0..70); # G. C. Greubel, Aug 31 2019
-
Mathematica
Accumulate[Mod[Range[0,70],5]] (* or *) Accumulate[PadRight[{},70,{0,1,2,3,4}]] (* Harvey P. Dale, Nov 11 2016 *)
-
PARI
a(n) = sum(k=0, n, k % 5); \\ Michel Marcus, Apr 28 2018
-
Sage
def A130483_list(prec): P.
= PowerSeriesRing(ZZ, prec) return P(x*(1+2*x+3*x^2+4*x^3)/((1-x^5)*(1-x))).list() A130483_list(70) # G. C. Greubel, Aug 31 2019
Formula
G.f.: x*(1 + 2*x + 3*x^2 + 4*x^3)/((1-x^5)*(1-x)).
From Wesley Ivan Hurt, Jul 23 2016: (Start)
a(n) = a(n-5) - a(n-6) for n>5; a(n) = a(n-5) + 10 for n>4.
a(n) = 10 + Sum_{k=1..4} k*floor((n-k)/5). (End)
a(n) = ((n mod 5)^2 - 3*(n mod 5) + 4*n)/2. - Ammar Khatab, Aug 13 2020
Comments