OFFSET
1,2
COMMENTS
Original name: s(1)*s(2)*...*s(n)*(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 3k-1 for k = 1,2,3,...
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..381
L. Euler, De fractionibus continuis observationes, The Euler Archive, Index Number 123, Section 5
FORMULA
From Peter Bala, Feb 20 2015: (Start)
Recurrence: a(n+1) = 3*a(n) + (3*n - 1)^2*a(n-1) with a(1) = 1 and a(2) = 3.
The triple factorial numbers A008544 satisfy the same second-order recurrence equation. This leads to the continued fraction representation a(n)/A008544(n) = 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + ... + (3*n - 1)^2/(3 )))).
Taking the limit as n -> infinity gives the generalized continued fraction: Sum {k >= 1} (-1)^(k+1)/(3*k - 1) = 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + 11^2/(3 + ... ))))) due to Euler. The alternating sum has the value 1/3*( Pi/sqrt(3) - log(2) ) = A193534. Cf. A024217. (End)
a(n) ~ GAMMA(1/3) * 3^(n-1) * n^(n+1/6) * (Pi - sqrt(3)*log(2)) / (sqrt(2*Pi) * exp(n)). - Vaclav Kotesovec, Feb 21 2015
MAPLE
a[1] := 1: a[2] := 3: for n from 3 to 20 do a[n] := 3*a[n-1]+(3*n-4)^2*a[n-2] end do: seq(a[n], n = 1 .. 20); # Peter Bala, Feb 20 2015
MATHEMATICA
Table[Product[3*k-1, {k, 1, n}] * Sum[(-1)^(k+1)/(3*k-1), {k, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Feb 21 2015 *)
nxt[{n_, a_, b_}]:={n+1, b, 3b+a*(3n-1)^2}; NestList[nxt, {2, 1, 3}, 20][[;; , 2]] (* Harvey P. Dale, Jun 07 2023 *)
PROG
(Magma) I:=[1, 3]; [n le 2 select I[n] else 3*Self(n-1)+(3*n-4)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
New name from Peter Bala, Feb 20 2015
a(18)-a(19) from Vincenzo Librandi, Feb 21 2015
STATUS
approved