cp's OEIS Frontend

A249861 a(n) are values of k that yield "record-breaking" integer sequence lengths for the recursion: b(i) = 3*(b(i-1) + b(i-2))/2, with b(0) = 1 and b(1) = k.

%I A249861 #22 Aug 11 2015 01:26:00
%S A249861 1,5,37,101,229,485,2533,6629,23013,88549,219621,481765,1006053,
%T A249861 3103205,7297509,24074725,158292453,1232034277,3379517925,7674485221,
%U A249861 282552392165,1382064019941,5780110531045,14576203553253,84944947730917,647894901152229
%N A249861 a(n) are values of k that yield "record-breaking" integer sequence lengths for the recursion: b(i) = 3*(b(i-1) + b(i-2))/2, with b(0) = 1 and b(1) = k.
%C A249861 For all such b(i) sequences there is a value of i = Nmax, such that for i<=Nmax the sequence produces integers and for i > Nmax the sequence produces rational non-integers. Thus the "length" of the integer portion of b(i) is Nmax + 1 (when including b(0)).
%C A249861 Nmax is a function of the integer k > 0, written as Nmax(k).
%C A249861 Let B(i,k) be the array of all b(i) sequences, where k also serves as column index.
%C A249861 By definition k=1 is record-breaking. Nmax(1) = 3.
%C A249861 B(i,1) has four integer elements: 1,1,3,6.
%C A249861 Nmax(2k) = 1 for all k. Only b(0) and b(1) are integers when k is even.
%C A249861 A record-breaking b(i) has k = r such that Nmax(r) > Nmax(k) for all k < r.
%C A249861 Nmax(k) varies in an irregular pattern, tied to powers of 2 as explained below. This is also true of b(i) sequence families with other nonzero values of b(0).
%C A249861 (This contrasts with the family where b(0) = 0. In that case the record-breaking values k are simply the powers of 2, since each record-breaking Nmax is +1 greater in length than the prior such Nmax.)
%C A249861 Here, the values of Nmax for record-breaking b(i) lengths starting at k = 1 are 3, 6, 7, 8, 9, 12, 13, 15, 17, 18, 19, 20, 22, 23, 25, 28, 31, 32, 33, 39, 41, 43, 44, 47, 50, 54, 56, 59, 60, 62, 63, 64, 65, 69, 70, 71, 72, 73, 74, 76, 80, 83, 87, 89, 90, ...
%C A249861 B(i,k) follows these rules, regardless of irregularity in Nmax:
%C A249861 Rule 1:  If B(i,k) is an integer then B(i, k + 2^(i-1)) is an integer.
%C A249861 Rule 2:  If k = r1 and k = r2 yield two consecutive record-breaking b(i) sequences, then r2 = r1 + 2^(Nmax(r1) - 1).
%C A249861 Thus the index of the next record-breaking sequence can be found without searching.
%C A249861 Rule 3: B(i,k + 2^(i-1)) - B(i,k) = A083858(i) for all k, regardless of whether the B(i,k) sequence elements selected are integers.
%C A249861 Conjecture: Rules 1 & 2 apply for all such b(i) sequences of the form: b(i) = m*(b(i-1) + b(i-2))/2, b(0) = j, b(1) = k, for j >= 0 and odd integers m > 2.
%C A249861 Rule 3, for any odd m, generalizes as: B(i,k + 2^(i-1)) - B(i,k), for all k, equals sequences of the form  f(i) = m*f(i-1) + 2m*f(i-2), f(0) = 0, f(1) = 1.
%C A249861 It appears that a simple rule for finding the next record-breaking sequence may not exist for b(i) sequences of the form: b(i) = 2*(b(i-1) + b(i-2))/3, b(0) = 1, b(1) = 1. A factor 2 must be "occasionally" applied to the predicted power of 3.
%C A249861 Also note, (a(n)-1)/4 is an integer. The first seven nonzero values of this reduced form are the same as A113828, but otherwise no relationship.
%F A249861 a(n) = a(n-1) + 2^(Nmax(a(n-1)) - 1), where Nmax at a(n-1) (i.e., the prior record-breaking k value) must be found by calculation of b(i) and "observation".
%e A249861 Subtracting 1 from Nmax gives the exponents of 2 needed to generate a(n) using the formula above,  as:
%e A249861 a(1) = 1 (by definition)
%e A249861 a(2) = 1 + 2^(3-1) = 5
%e A249861 a(3) = 5 + 2^(6-1) = 37
%e A249861 a(4) = 37 + 2^(7-1) = 101
%e A249861 ... etc.
%Y A249861 Cf. A083858.
%K A249861 nonn,more
%O A249861 1,2
%A A249861 _Richard R. Forberg_, Nov 07 2014