

A168043


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(n1). a(n) is the number of elements in S(n).


3



1, 2, 4, 7, 13, 23, 40, 68, 114, 189, 311, 509, 830, 1350, 2192, 3555, 5761, 9331, 15108, 24456, 39582, 64057, 103659, 167737, 271418, 439178, 710620, 1149823, 1860469, 3010319, 4870816, 7881164, 12752010, 20633205, 33385247, 54018485, 87403766, 141422286
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OFFSET

1,2


LINKS



FORMULA

It appears that a(n) = a(n1) + a(n2) + n  3, for n>3.
Apparently: a(n) = 3*a(n1)  2*a(n2)  a(n3) + a(n4) for n>5;
and a(n) = A000032(n+1)  n for n>1. (End)
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)
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


EXAMPLE

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.


PROG

(Python)
from itertools import chain, islice
def agen(): # generator of terms
s = {1}
while True:
yield len(s)
s = set(chain.from_iterable((x+1, x+2, 2*x) for x in s))


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



