

A186158


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


1



18, 5, 165, 3, 18, 1333, 2, 8, 56, 10353, 2, 5, 18, 165, 78958, 1, 3, 9, 38, 472, 596438, 1, 3, 6, 18, 80, 1333, 4479398, 1, 2, 5, 11, 32, 165, 3727, 33514643, 1, 2, 4, 8, 18, 56, 333, 10353, 250104748, 1, 2, 3, 6, 12, 28, 96, 668, 28635, 1862945616, 1, 2, 3, 5, 9, 18, 45, 165, 1333, 78958
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OFFSET

1,1


COMMENTS

This array represents the solution of Problem 7 in "Unsolved Problems and Rewards" in Links (below). Problem 7 is restated here:
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(2k1)]^2, V=a(2k), W=4b(k1), b(k)=VU/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.
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.


REFERENCES

C. Kimberling, Partial sums of generating functions as polynomial sequences, The Fibonacci Quarterly 48 (2010) 327334. (See Theorem 1.)


LINKS

Table of n, a(n) for n=1..65.
C. Kimberling, Unsolved Problems and Rewards
C. Kimberling, Partial sums of generating functions as polynomial sequences (abstract), The Fibonacci Quarterly 48 (2010).
Peter Kosinar, On The Mysterious B Sequence


FORMULA

Starting with A=(c,c+d,c+2d,...), put b(0)=a(0) and for k>0, put U=[a(2k1]^2, V=a(2k), W=4b(k1), b(k)=VU/W.
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).


EXAMPLE

Northwest corner:
18.......5.....3....2...2...1...1...1...1
165......18....9....6...5...4...3...3...2
1333.....56....18...9...6...5...4...3...3
10353....165...38...18..11..8...6...5...4
78958....472...80...32..18..12..9...7...6
596438...1333..165..56..28..18..12..9...8
4479388..3727..333..96..45..26..18..13..10
Column 1 continues with 33514643,250104748,1862945616.
T(1,1)=18 because when (c,d)=(1,2), the only negative number in the sequence B is b(18).


MATHEMATICA

B[0, c_, d_]:=c;
B[k_, c_, d_]:=B[Mod[k, 2], c, d]=c+2d*k((c+d(1+2k))^2)/(4B[Mod[k1, 2], c, d]);
Table[Table[NestWhile[#1+1&, 1, B[#1, c, d]>0&], {d, c+1, c+10}], {c, 1, 5}]//TableForm
(* by Peter J. C. Moses, Feb 08 2011 *)


CROSSREFS

Sequence in context: A040312 A214893 A065909 * A038642 A040311 A221351
Adjacent sequences: A186155 A186156 A186157 * A186159 A186160 A186161


KEYWORD

nonn,tabl


AUTHOR

Clark Kimberling, Feb 15 2011


STATUS

approved



