A182456 a(0)=1; for n>0, a(n) = ( a(n-1) mod (n+3) )*(n+3).
1, 4, 20, 12, 35, 24, 54, 40, 77, 60, 104, 84, 135, 112, 170, 144, 209, 180, 252, 220, 299, 264, 350, 312, 405, 364, 464, 420, 527, 480, 594, 544, 665, 612, 740, 684, 819, 760, 902, 840, 989, 924, 1080, 1012, 1175, 1104, 1274, 1200, 1377, 1300, 1484
Offset: 0
Examples
a(6) = (a(5) mod 9) * 9 = (24 mod 9) * 9 = 6*9 = 54.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
Crossrefs
Cf. A182455.
Programs
-
Mathematica
CoefficientList[Series[(-1 - 3*x - 14*x^2 + 14*x^3 + 8*x^4 - 8*x^5)/((x - 1)^3*(1 + x)^2), {x,0,50}], x] (* G. C. Greubel, Feb 25 2017 *) RecurrenceTable[{a[0]==1,a[n]==Mod[a[n-1],n+3](n+3)},a,{n,50}] (* Harvey P. Dale, Oct 21 2018 *) -
PARI
my(x='x+O('x^50)); Vec((-1-3*x-14*x^2+14*x^3+8*x^4-8*x^5)/((x-1)^3*(1+x)^2)) \\ G. C. Greubel, Feb 25 2017 -
Python
a=1 for n in range(1, 55): print(a, end=",") a = (a%(n+3)) * (n+3)
Formula
From Alexander R. Povolotsky, May 01 2012: (Start)
for the same sequence with index starting from 1 instead of 0, i.e. k=1,2,...
a(k+1) = (k+3)^2 - (k+3)*a(k)/(k+2).
G.f.: (-1-3*x-14*x^2+14*x^3+8*x^4-8*x^5)/((x-1)^3*(1+x)^2). (End)