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A234604
Floor of the solutions to c = exp(1 + n/c) for n >= 0, using recursion.
1
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
0,1
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.
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
CROSSREFS
Sequence in context: A356989 A356994 A274687 * A236346 A306890 A116549
KEYWORD
nonn
AUTHOR
Richard R. Forberg, Dec 28 2013
EXTENSIONS
Corrected and edited by Jon E. Schoenfield, Jan 11 2014
STATUS
approved