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%I #22 Aug 22 2020 09:45:44
%S 1,2,2,3,3,7,11,20,31,47,92,162,312,632,1235,2460,4844,9665,19335,
%T 38727,77569,155729,310405,620596,1240580,2481645,4966229,9926596,
%U 19855760,39717367,79428417
%N Number of Ulam numbers u (A002858) between powers of 2, 2^n < u <= 2^(n+1).
%C Conjecture 1: For all m > 1 there is always at least one Ulam number u(j) such that m < u(j) < 2m.
%C 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.
%C This sequence illustrates how far these conjectures are oversatisfied.
%C Conjecture 1 implies that Ulam numbers form a complete sequence because u(1) = 1 and 2u(j) >= u(j+1).
%C 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.
%e a(6) = 11 because the Ulam numbers between 64 and 128 are (69, 72, 77, 82, 87, 97, 99, 102, 106, 114, 126).
%t 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 *)
%t upi[n_] := Module[{p = 1}, While[ulst[[p]] <= n, p++]; p - 1]; Table[upi[2^(n + 1)] - upi[2^n], {n, 0, 16}]
%Y Cf. A002858, A199017, A307331.
%K nonn,more
%O 0,2
%A _Frank M Jackson_, Jan 25 2020
%E a(20)-a(21) from _Sean A. Irvine_, Feb 29 2020
%E a(22)-a(30) from _Amiram Eldar_, Aug 22 2020
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