A346943 a(n) = a(n-1) + n*(n+1)*a(n-2) with a(0)=1, a(1)=1.

1, 1, 7, 19, 159, 729, 7407, 48231, 581535, 4922325, 68891175, 718638075, 11465661375, 142257791025, 2550046679775, 36691916525775, 730304613424575, 11958031070311725, 261722208861516375, 4805774015579971875, 114729101737416849375, 2334996696935363855625

Offset: 0

Author

Vaclav Kotesovec, Aug 08 2021, following a suggestion from John M. Campbell

Keywords

• nonn
• easy

Comments

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)

Links

• Leonhard Euler, E123: De fractionibus continuis observationes, originally published in Commentarii academiae scientiarum Petropolitanae 11, 1750, pp. 32-81, reprinted in Opera Omnia: Series 1, Volume 14, 291-349.

Crossrefs

Cf. A000246, A000932, A024167.

Programs

Mathematica

RecurrenceTable[{a[n] == a[n-1] + n*(n+1)*a[n-2], a[0]==1, a[1]==1}, a, {n,0,20}]
nmax = 20; CoefficientList[Series[(-2 + Pi + 2*Pi*x + 4*Sqrt[1 - x^2] + 2*x*(-2 + Sqrt[1 - x^2]) - 4*(1 + 2*x) * ArcSin[Sqrt[1 - x]/Sqrt[2]]) / (2*(1 - x)^(5/2) * (1 + x)^(3/2)), {x, 0, nmax}], x] * Range[0, nmax]!

Formula

a(n) ~ n! * (Pi - 2) * n^(3/2) / sqrt(2*Pi).

a(n) ~ (Pi - 2) * n^(n+2) / exp(n).

E.g.f. A(x) satisfies the differential equation -6*A(x) - (6*x + 1)*A'(x) + (1 - x^2)*A''(x) = 0, A(0)=1, A'(0)=1.

E.g.f.: (-2 + Pi + 2*Pi*x + 4*sqrt(1-x^2) + 2*x*(-2+sqrt(1-x^2)) - 4*(1+2*x) * arcsin(sqrt(1-x)/sqrt(2))) / (2*(1-x)^(5/2) * (1+x)^(3/2)).