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L. Euler, <a href="https://scholarlycommons.pacific.edu/euler-works/index.2.html123/">E123: De fractionibus continuis observationes</a>, originally published in Commentarii academiae scientiarum Petropolitanae 11, 1750, pp. 32-81, reprinted in Opera Omnia: Series 1, Volume 14, 291-349.
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From Peter Bala, Dec 09 2024: (Start)
b(n) := A000246(n+2) = (n+2)!/2^(n+1) * binomial(n+1, floor((n+1)/2)) satisfies the same second-order recurrence as a(n) with the initial conditions b(0) = 1 and b(1) = 3. This leads to the finite continued fraction a(n)/b(n) = 1/(1 + 2/(1 + 6/(1 + ... + n*(n+1)/1). Letting n tend to infinity gives the continued fraction representation 1/(1 + 2/(1 + 6/(1 + ... + n*(n+1)/(1 + ...) = Pi/2 - 1, due to Euler - see paragraph 31, p. 48. (End)
L. Euler, <a href="https://scholarlycommons.pacific.edu/euler-works/index.2.html">E123</a>, : De fractionibus continuis observationes, </a>, originally published in Commentarii academiae scientiarum Petropolitanae 11, 1750, pp. 32-81, reprinted in Opera Omnia: Series 1, Volume 14, 291-349.
From Peter Bala, Dec 09 2024: (Start)
b(n) := A000246(n+2) = (n+2)!/2^(n+1) * binomial(n+1, floor((n+1)/2)) satisfies the same second-order recurrence as a(n) with the initial conditions b(0) = 1 and b(1) = 3. This leads to the finite continued fraction a(n)/b(n) = 1/(1 + 2/(1 + 6/(1 + ... + n*(n+1)/1). Letting n tend to infinty gives the continued fraction representation 1/(1 + 2/(1 + 6/(1 + ... + n*(n+1)/(1 + ...) = Pi/2 - 1, due to Euler - see paragraph 31, p. 48. (End)
From Peter Bala, Dec 10 2024:(Start)
2/Pi = 1 - 1/(2 + 2/(1 + 6/(1 + 12/(1 + 20/(1 + ... + n*(n+1)/(1 + ...), a continued fraction representation due to Euler. See A346943.
Equals 1 - (1/2)*Sum_{n >= 0} A005566(n)*(-1/4)^n. (End)
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The behavior of a(n) for large n is given by lim_limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + n^2/(9 + ...))))) = log(2) - (1 - 1/2 + 1/3 - 1/4); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
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