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%I #26 Mar 01 2016 13:03:41
%S 18,5,165,3,18,1333,2,8,56,10353,2,5,18,165,78958,1,3,9,38,472,596438,
%T 1,3,6,18,80,1333,4479398,1,2,5,11,32,165,3727,33514643,1,2,4,8,18,56,
%U 333,10353,250104748,1,2,3,6,12,28,96,668,28635,1862945616,1,2,3,5,9,18,45,165,1333,78958
%N Array associated with "the Mysterious B Sequence", by antidiagonals.
%C This array represents the solution of Problem 7 in "Unsolved Problems and Rewards" in Links (below). Problem 7 is restated here:
%C For any sequence A=(a(0),a(1),...) of positive real numbers, create a sequence B as follows: let b(0)=a(0) and for k>0, let U=[a(2k-1)]^2, V=a(2k), W=4b(k-1), b(k)=V-U/W, and assume for each k that W is not zero. Determine conditions on c and d for which the arithmetic sequence A=(c,c+d,c+2d,...) yields b(k)>0 for every k.
%C Peter Kosinar found a necessary and sufficient condition to be 0<d<=c. He also proved that if d>c, then the sequence B contains one and only one negative number. The number in row i, column j, is the unique k for which b(k)<0 when c=i and d=i+j.
%H C. Kimberling, <a href="http://faculty.evansville.edu/ck6/integer/unsolved.html">Unsolved Problems and Rewards</a>
%H C. Kimberling, <a href="http://www.fq.math.ca/Papers1/48-4/Kimberling.pdf">Partial sums of generating functions as polynomial sequences</a>, The Fibonacci Quarterly 48 (2010) 327-334. (See Theorem 1.)
%H Peter Kosinar, <a href="http://web.archive.org/web/20150705033419/http://the-goober.net/misc/MysteriousBSequence.pdf">On The Mysterious B Sequence</a>
%F Starting with A=(c,c+d,c+2d,...), put b(0)=a(0) and for k>0, put U=[a(2k-1]^2, V=a(2k), W=4b(k-1), b(k)=V-U/W.
%F For i>=1 and j>=1, put f(i,i+j)=(the index k for which b(k)<0). Then the array, T, is given by T(i,j)=f(i,i+j).
%e Northwest corner:
%e 18.......5.....3....2...2...1...1...1...1
%e 165......18....9....6...5...4...3...3...2
%e 1333.....56....18...9...6...5...4...3...3
%e 10353....165...38...18..11..8...6...5...4
%e 78958....472...80...32..18..12..9...7...6
%e 596438...1333..165..56..28..18..12..9...8
%e 4479388..3727..333..96..45..26..18..13..10
%e Column 1 continues with 33514643,250104748,1862945616.
%e T(1,1)=18 because when (c,d)=(1,2), the only negative number in the sequence B is b(18).
%t B[0,c_,d_]:=c;
%t B[k_,c_,d_]:=B[Mod[k,2],c,d]=c+2d*k-((c+d(-1+2k))^2)/(4B[Mod[k-1,2],c,d]);
%t Table[Table[NestWhile[#1+1&,1,B[#1,c,d]>0&],{d,c+1,c+10}],{c,1,5}]//TableForm
%t (* _Peter J. C. Moses_, Feb 08 2011 *)
%K nonn,tabl
%O 1,1
%A _Clark Kimberling_, Feb 15 2011
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