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%I #33 Jan 14 2022 07:32:58
%S 1,2,4,7,13,23,40,68,114,189,311,509,830,1350,2192,3555,5761,9331,
%T 15108,24456,39582,64057,103659,167737,271418,439178,710620,1149823,
%U 1860469,3010319,4870816,7881164,12752010,20633205,33385247,54018485,87403766,141422286
%N Let S(1)={1} and, for n>1 let S(n) be the smallest set containing x+1, x+2, and 2*x for each element x in S(n-1). a(n) is the number of elements in S(n).
%F It appears that a(n) = a(n-1) + a(n-2) + n - 3, for n>3.
%F From _R. J. Mathar_, Nov 18 2009: (Start)
%F Apparently: a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>5;
%F and a(n) = A000032(n+1) - n for n>1. (End)
%F From _Ilya Gutkovskiy_, Jul 07 2016: (Start)
%F It appears that the g.f. is x*(1 - x + x^4)/((1 - x)^2*(1 - x - x^2)); and the e.g.f. is phi*exp(phi*x) - exp(-x/phi)/phi - x*(1 + exp(x)) - 1, where phi is the golden ratio. (End)
%F It would be nice to have a proof for any one of these formulas. The others would then presumably follow easily. - _N. J. A. Sloane_, Jul 11 2016
%e Under the indicated set mapping we have {1} -> {2,3} -> {3,4,5,6} -> {4,5,6,7,8,10,12}, ..., so a(2)=2, a(3)=4, a(4)=7, etc.
%o (Python)
%o from itertools import chain, islice
%o def agen(): # generator of terms
%o s = {1}
%o while True:
%o yield len(s)
%o s = set(chain.from_iterable((x+1, x+2, 2*x) for x in s))
%o print(list(islice(agen(), 30))) # _Michael S. Branicky_, Jan 13 2022 after _Chai Wah Wu_ in A123247
%Y Cf. A000032, A122554.
%K nonn
%O 1,2
%A _John W. Layman_, Nov 17 2009
%E a(17)-a(22) from _R. J. Mathar_, Nov 18 2009
%E a(23)-a(35) from _Jinyuan Wang_, Apr 14 2020
%E a(36)-a(38) from _Michael S. Branicky_, Jan 13 2022
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