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A186158 Array associated with "the Mysterious B Sequence", by antidiagonals.

%I A186158 #34 Feb 11 2025 08:15:28
%S A186158 18,5,165,3,18,1333,2,8,56,10353,2,5,18,165,78958,1,3,9,38,472,596438,
%T A186158 1,3,6,18,80,1333,4479398,1,2,5,11,32,165,3727,33514643,1,2,4,8,18,56,
%U A186158 333,10353,250104748,1,2,3,6,12,28,96,668,28635,1862945616,1,2,3,5,9,18,45,165,1333,78958
%N A186158 Array associated with "the Mysterious B Sequence", by antidiagonals.
%C A186158 This array represents the solution of Problem 7 in "Unsolved Problems and Rewards" in Links (below). Problem 7 is restated here:
%C A186158 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(2*k-1))^2, V=a(2*k), W=4*b(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+2*d,...) yields b(k)>0 for every k.
%C A186158 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 A186158 Clark Kimberling, <a href="http://faculty.evansville.edu/ck6/integer/unsolved.html">Unsolved Problems and Rewards</a>
%H A186158 Clark Kimberling, <a href="https://web.archive.org/web/2024*/https://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 A186158 Peter Kosinar, <a href="http://web.archive.org/web/20150705033419/http://the-goober.net/misc/MysteriousBSequence.pdf">On The Mysterious B Sequence</a>
%F A186158 Starting with A=(c,c+d,c+2*d,...), put b(0)=a(0) and for k>0, put U=(a(2*k-1))^2, V=a(2*k), W=4*b(k-1), b(k)=V-U/W.
%F A186158 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 A186158 Northwest corner:
%e A186158   18       5     3    2   2   1   1   1   1
%e A186158   165      18    9    6   5   4   3   3   2
%e A186158   1333     56    18   9   6   5   4   3   3
%e A186158   10353    165   38   18  11  8   6   5   4
%e A186158   78958    472   80   32  18  12  9   7   6
%e A186158   596438   1333  165  56  28  18  12  9   8
%e A186158   4479388  3727  333  96  45  26  18  13  10
%e A186158 Column 1 continues with 33514643,250104748,1862945616.
%e A186158 T(1,1)=18 because when (c,d)=(1,2), the only negative number in the sequence B is b(18).
%t A186158 B[0,c_,d_]:=c;
%t A186158 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 A186158 Table[Table[NestWhile[#1+1&,1,B[#1,c,d]>0&],{d,c+1,c+10}],{c,1,5}]//TableForm
%t A186158 (* _Peter J. C. Moses_, Feb 08 2011 *)
%K A186158 nonn,tabl
%O A186158 1,1
%A A186158 _Clark Kimberling_, Feb 15 2011