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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(n-1). 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
OFFSET
1,2
FORMULA
It appears that a(n) = a(n-1) + a(n-2) + n - 3, for n>3.
From R. J. Mathar, Nov 18 2009: (Start)
Apparently: a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>5;
and a(n) = A000032(n+1) - n for n>1. (End)
From Ilya Gutkovskiy, Jul 07 2016: (Start)
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))
print(list(islice(agen(), 30))) # Michael S. Branicky, Jan 13 2022 after Chai Wah Wu in A123247
CROSSREFS
Sequence in context: A051013 A128609 A335471 * A374764 A374632 A114832
KEYWORD
nonn
AUTHOR
John W. Layman, Nov 17 2009
EXTENSIONS
a(17)-a(22) from R. J. Mathar, Nov 18 2009
a(23)-a(35) from Jinyuan Wang, Apr 14 2020
a(36)-a(38) from Michael S. Branicky, Jan 13 2022
STATUS
approved