

A331729


Number of Ulam numbers u (A002858) between powers of 2, 2^n < u <= 2^(n+1).


1



1, 2, 2, 3, 3, 7, 11, 20, 31, 47, 92, 162, 312, 632, 1235, 2460, 4844, 9665, 19335, 38727, 77569, 155729, 310405, 620596, 1240580, 2481645, 4966229, 9926596, 19855760, 39717367, 79428417
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OFFSET

0,2


COMMENTS

Conjecture 1: For all m > 1 there is always at least one Ulam number u(j) such that m < u(j) < 2m.
Conjecture 2: For all m > 4 there is always at least two Ulam numbers u(j), u(j+1) such that m < u(j) < u(j+1) < 2m.
This sequence illustrates how far these conjectures are oversatisfied.
Conjecture 1 implies that Ulam numbers form a complete sequence because u(1) = 1 and 2u(j) >= u(j+1).
Conjecture 2 implies that three consecutive Ulam numbers satisfies the triangle inequality because 2u(j) > u(j+2) > u(j+1) > u(j) and u(j) + u(j+1) > 2u(j) > u(j+2). It further implies that n consecutive Ulam numbers can always form an ngon.


LINKS

Table of n, a(n) for n=0..30.


EXAMPLE

a(6) = 11 because the Ulam numbers between 64 and 128 are (69, 72, 77, 82, 87, 97, 99, 102, 106, 114, 126).


MATHEMATICA

ulams = {1, 2}; Do[AppendTo[ulams, n = Last[ulams]; While[n++; Length[DeleteCases[Intersection[ulams, nulams], n/2, 1, 1]]!=2]; n], {10^4}]; ulst = ulams; (* JeanFrançois Alcover, Sep 08 2011 *)
upi[n_] := Module[{p = 1}, While[ulst[[p]] <= n, p++]; p  1]; Table[upi[2^(n + 1)]  upi[2^n], {n, 0, 16}]


CROSSREFS

Cf. A002858, A199017, A307331.
Sequence in context: A307736 A309713 A153903 * A095741 A252582 A181577
Adjacent sequences: A331726 A331727 A331728 * A331730 A331731 A331732


KEYWORD

nonn,more


AUTHOR

Frank M Jackson, Jan 25 2020


EXTENSIONS

a(20)a(21) from Sean A. Irvine, Feb 29 2020
a(22)a(30) from Amiram Eldar, Aug 22 2020


STATUS

approved



