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A182456 a(0)=1; for n>0, a(n) = ( a(n-1) mod (n+3) )*(n+3). 1

%I

%S 1,4,20,12,35,24,54,40,77,60,104,84,135,112,170,144,209,180,252,220,

%T 299,264,350,312,405,364,464,420,527,480,594,544,665,612,740,684,819,

%U 760,902,840,989,924,1080,1012,1175,1104,1274,1200,1377,1300,1484

%N a(0)=1; for n>0, a(n) = ( a(n-1) mod (n+3) )*(n+3).

%H G. C. Greubel, <a href="/A182456/b182456.txt">Table of n, a(n) for n = 0..1000</a>

%F For k>0, a(2*k) = A182455(2*k+3)-1, a(2k+1) = A182455(2k).

%F From Alexander R. Povolotsky, May 01 2012: (Start)

%F for the same sequence with index starting from 1 instead of 0, i.e. k=1,2,...

%F a(k+1) = (k+3)^2 -((k+3)*a(k))/(k+2).

%F G.f.: (-1-3*x-14*x^2+14*x^3+8*x^4-8*x^5)/((x-1)^3*(1+x)^2). (End)

%e a(6) = (a(5) mod 9) * 9 = (24 mod 9) * 9 = 6*9 = 54.

%t 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 *)

%t RecurrenceTable[{a[0]==1,a[n]==Mod[a[n-1],n+3](n+3)},a,{n,50}] (* _Harvey P. Dale_, Oct 21 2018 *)

%o (Python)

%o a=1

%o for n in range(1, 55):

%o print(a, end=",")

%o a = (a%(n+3)) * (n+3)

%o (PARI) 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

%Y Cf. A182455.

%K nonn,easy

%O 0,2

%A _Alex Ratushnyak_, Apr 30 2012

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Last modified January 20 13:06 EST 2022. Contains 350472 sequences. (Running on oeis4.)