A114550 Decimal expansion of the constant Sum_{n>=0} 1/A112373(n), where the partial quotients of the continued fraction A114551 satisfy A114551(2n) = A112373(n) and A114551(2n+1) = A112373(n+1)/A112373(n).

2, 5, 8, 4, 4, 0, 1, 7, 2, 4, 0, 1, 9, 7, 7, 6, 7, 2, 4, 8, 1, 2, 0, 7, 6, 1, 4, 7, 1, 5, 3, 3, 3, 1, 3, 4, 2, 1, 1, 2, 3, 8, 2, 0, 9, 0, 4, 6, 7, 9, 6, 9, 0, 0, 0, 3, 1, 3, 4, 3, 8, 5, 8, 3, 9, 6, 7, 5, 4, 4, 8, 2, 9, 8, 9, 1, 8, 6, 7, 9, 6, 3, 6, 1, 4, 0, 8, 8, 7, 4, 6, 9, 7, 7, 8, 0, 1, 8, 6, 9, 6, 4, 2, 7, 2

Offset: 1

Author

Paul D. Hanna, Dec 08 2005

Keywords

• cons
• nonn

Comments

A112373 is defined by the recurrence: let b(n) = A112373(n), then

b(n) =(b(n-1)^3 + b(n-1)^2)/b(n-2) for n>=2 with b(0)=b(1)=1.

Thus the sum of unit fractions 1/A112373(n) converges rapidly.

Examples

			2.584401724019776724812076147153331342112382090467969...
= Sum_{n>=0} 1/A112373(n) = 1/1 +1/1 +1/2 +1/12 +1/936 +1/68408496 +...
= [2;1,1,2,2,6,12,78,936,73086,68408496,...] (continued fraction).
		

Links

• Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, arXiv:1507.00063 [math.NT], 2015 and J. Int. Seq. 18 (2015) # 15.8.4.

Crossrefs

Cf. A112373, A114551 (continued fraction), A114552.

Programs

Mathematica

dm = 5; digits = 105;
b[n_] := b[n] = If[n < 2, 1, (b[n - 1]^3 + b[n - 1]^2)/b[n - 2]];
s[m_] := s[m] = N[Sum[1/b[n], {n, 0, m}], digits + 5];
s[m = dm];
s[m += dm];
While[RealDigits[s[m]] != RealDigits[s[m - dm]], m += dm];
RealDigits[s[m], 10, digits][[1]] (* Jean-François Alcover, Sep 30 2019 *)
c[0]=2; c[1] = c[2] = 1; c[n_] := c[n] = c[n-1] c[n-2] + Mod[n, 2] c[n-2];
RealDigits[FromContinuedFraction[c /@ Range[0, 14]], 10, 105][[1]] (* Jean-François Alcover, Oct 01 2019 *)