A024217 a(n) = ( Product {k = 1..n} 3*k - 2 ) * ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).
1, 3, 25, 222, 3166, 47016, 951544, 19827408, 520029520, 13952218560, 449559799360, 14756761434240, 563961412362880, 21893890640563200, 968019931702297600, 43385863589508249600, 2178487766250586470400, 110704921777161066700800, 6222745685273069016064000
Offset: 1
Links
- L. Euler, De fractionibus continuis observationes, The Euler Archive, Index Number 123, Section 5
Programs
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Magma
I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1)+(3*n-5)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015
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Maple
a[1] := 1: a[2] := 3: for n from 3 to 20 do a[n] := 3*a[n-1]+(3*n-5)^2*a[n-2] end do: seq(a[n], n = 1 .. 20); # Peter Bala, Feb 20 2015
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Mathematica
Table[Product[3*k-2,{k,1,n}] * Sum[(-1)^(k+1)/(3*k-2),{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Feb 21 2015 *)
Formula
From Peter Bala, Feb 20 2015: (Start)
a(n) = A007559(n) * Sum {k = 1..n} (-1)^(k+1)/(3*k - 2).
Recurrence: a(n+1) = 3*a(n) + (3*n - 2)^2*a(n-1) with a(1) = 1 and a(2) = 3.
The triple factorial numbers A007559 also satisfy this second-order recurrence equation. This leads to the continued fraction representation a(n)/A007559(n) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + ... + (3*n - 2)^2/(3 ))))).
Taking the limit as n -> infinity gives the generalized continued fraction: Sum {k >= 1} (-1)^(k+1)/(3*k - 2) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + 10^2/(3 + ... ))))) due to Euler.
a(n) ~ sqrt(2*Pi) * (sqrt(3)*Pi + 3*log(2)) * 3^(n-2) * n^(n-1/6) / (GAMMA(1/3) * exp(n)). - Vaclav Kotesovec, Feb 21 2015
Extensions
New name from Peter Bala, Feb 20 2015
a(18)-a(19) from Vincenzo Librandi, Feb 21 2015
Comments