A234604 Floor of the solutions to c = exp(1 + n/c) for n >= 0, using recursion.

2, 3, 4, 4, 5, 6, 6, 7, 17, 35, 62, 103, 164, 256, 391, 589, 880, 1303, 1919, 2814, 4112, 5993, 8716, 12655, 18353, 26591, 38499, 55710, 80583, 116523, 168453, 243485, 351889, 508506, 734776, 1061672, 1533938, 2216216

Offset: 0

Author

Richard R. Forberg, Dec 28 2013

Keywords

• nonn

Comments

For n = 1 to 7 recursion produces convergence to single valued solutions.

For n >= 8 a dual-valued oscillating recursion persists between two stable values. The floor of the upper value for each n is included here. (The lower values of c are under 6 and approach exp(1) = 2.71828 for large n.)

At large n, the ratio of a(n)/a(n-1) approaches exp(1/exp(1)) = 1.444667861009 with more digits given by A073229.

At n = 0, c = exp(1).

At n = 1, c = 3.5911214766686 = A141251.

At n = 2, c = 4.3191365662914

At n = 3, c = 4.9706257595442

At n = 4, c = 5.5723925978776

At n = 5, c = 6.1383336446072

At n = 6, c = 6.6767832796664

At n = 7, c = 7.1932188286406

The convergence becomes "dual-valued" at n > exp(2) = 7.3890560989 = A072334.

At values of n = 7 and 8 the convergence is noticeably slower than at either larger or smaller values of n.

The recursion at n = exp(2) is only "quasi-stable" where c reluctantly approaches exp(2) = exp(1 + exp(2)/exp(2)) from any starting value, but never reaches it, and is not quite able to hold it if given the solution, due to machine rounding errors.

Crossrefs

Cf. A073229, A141251.

Formula

a(n) = floor(c) for the solutions to c = exp(1 + n/c) at n = 0 to 7, and the floor of the stable upper values of c for n >= 8.

Conjecture: a(n) = floor(e^(-e^(t^2/e^t - t)*t^2 + t + 1)) for all n > 13. - Jon E. Schoenfield, Jan 11 2014

Extensions

Corrected and edited by Jon E. Schoenfield, Jan 11 2014