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

%I A182456 #27 Aug 17 2025 21:11:06
%S A182456 1,4,20,12,35,24,54,40,77,60,104,84,135,112,170,144,209,180,252,220,
%T A182456 299,264,350,312,405,364,464,420,527,480,594,544,665,612,740,684,819,
%U A182456 760,902,840,989,924,1080,1012,1175,1104,1274,1200,1377,1300,1484
%N A182456 a(0)=1; for n>0, a(n) = ( a(n-1) mod (n+3) )*(n+3).
%H A182456 G. C. Greubel, <a href="/A182456/b182456.txt">Table of n, a(n) for n = 0..1000</a>
%F A182456 For k>0, a(2*k) = A182455(2*k+3)-1, a(2*k+1) = A182455(2*k).
%F A182456 From _Alexander R. Povolotsky_, May 01 2012: (Start)
%F A182456 for the same sequence with index starting from 1 instead of 0, i.e. k=1,2,...
%F A182456 a(k+1) = (k+3)^2 - (k+3)*a(k)/(k+2).
%F A182456 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 A182456 a(6) = (a(5) mod 9) * 9 = (24 mod 9) * 9 = 6*9 = 54.
%t A182456 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 A182456 RecurrenceTable[{a[0]==1,a[n]==Mod[a[n-1],n+3](n+3)},a,{n,50}] (* _Harvey P. Dale_, Oct 21 2018 *)
%o A182456 (Python)
%o A182456 a=1
%o A182456 for n in range(1, 55):
%o A182456     print(a, end=",")
%o A182456     a = (a%(n+3)) * (n+3)
%o A182456 (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
%Y A182456 Cf. A182455.
%K A182456 nonn,easy
%O A182456 0,2
%A A182456 _Alex Ratushnyak_, Apr 30 2012