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%I #34 Sep 02 2022 06:08:38
%S 1,1,-3,-35,-287,-2655,-29315,-381251,-5718975,-97222847,-1847234435,
%T -38791923555,-892214242271,-22305356057375,-602244613549827,
%U -17465093792945795,-541417907581320575,-17866790950183580031,-625337683256425302275,-23137494280487736185507
%N a(n+1) = (a(n) - 2*(n-1)) * (2*n-1), where a(1)=1.
%C The sequence appears when computing constants that encode all odd numbers starting from 3, then from 5, then from 7, etc. The general formula of the constant is a(n) + (2n-1)!!*sqrt(2*Pi*e)*erf(1/sqrt(2)), where n>0. For more information on how to generate the constant please watch the Grime-Haran Numberphile video.
%H James Grime and Brady Haran, <a href="https://www.youtube.com/watch?v=_gCKX6VMvmU">2.920050977316</a>, Numberphile video, Nov 26 2020.
%F Homogeneous recurrence: (-2*n+9)*a(n-4) + (6*n-20)*a(n-3) + (-6*n+12)*a(n-2) + 2*n*a(n-1) - a(n) = 0 with a(1)=a(2)=1, a(3)=-3, a(4)=-35. - _Georg Fischer_, Sep 01 2022
%p A339516 := proc(n)
%p option remember ;
%p if n = 1 then
%p 1;
%p else
%p (2*n-3)*(procname(n-1)-2*(n-2)) ;
%p end if;
%p end proc:
%p seq(A339516(n),n=1..30) ; # _R. J. Mathar_, Aug 24 2022
%t a[1]=1;a[n_]:=(a[n-1]-2(n-2))(2n-3); Array[a,20] (* _Stefano Spezia_, Dec 08 2020 *)
%o (Python)
%o #generate first 50 numbers of the sequence
%o cnt = 50
%o i=0
%o seq = list()
%o seq.append(1)
%o i=1
%o while (i<cnt):
%o seq.append((seq[i-1]-2*(i-1))*(2*i-1))
%o i=i+1
%o print(seq)
%Y Cf. A047907, A093302, A001147.
%K sign
%O 1,3
%A _Kamil Zabkiewicz_, Dec 07 2020
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