A331729 Number of Ulam numbers u (A002858) between powers of 2, 2^n < u <= 2^(n+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

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 n-gon.

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, n-ulams], n/2, 1, 1]]!=2]; n], {10^4}]; ulst = ulams; (* Jean-François 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