%I #12 Nov 16 2024 17:32:51

%S 0,0,0,1,0,2,0,2,4,0,5,2,0,3,3,2,0,1,0,0,4,2,4,4,2,0,2,0,2,3,1,7,0,3,

%T 0,6,2,1,6,3,0,7,0,3,0,6,5,4,0,2,3,0,8,6,6,2,0,4,4,0,6,5,2,0,2,3,3,8,

%U 0,7,2,6,5,5,1,2,6,3,9,8,0,3,0,0,3,4,3,3,0,0,10,5,2,8,1,6,4,0,12,5

%N Number of ordered pairs (k, m) with 0 < m < n and |k| < prime(m)^2 such that (k+prime(m+1)^2)/(k+prime(m)^2) = (k+prime(n+1)^2)/(k+prime(n)^2).

%C Conjecture: Let k, m and n be integers with 0 < m < n and -2*prime(m) < k <= 2*prime(m). If (k+prime(m+1)^2)/(k+prime(m)^2) = (k+prime(n+1)^2)/(k+prime(n)^2), then we must have k = -9, m = 5 and n = 11.

%C We have verified this for n up to 2000.

%C The conjecture essentially implies that for each k = -3..10 all the ratios (prime(n+1)^2+k)/(prime(n)^2+k) (n = 1,2,...) are pairwise distinct. We have verified that for any k = -1, 1 the ratios (prime(n+1)^2+k)/(prime(n)^2+k) (n = 1..110000) are indeed pairwise distinct.

%H Zhi-Wei Sun, <a href="/A263399/b263399.txt">Table of n, a(n) for n = 1..150</a>

%e a(4) = 1 since (prime(5)^2-13)/(prime(4)^2-13) = (11^2-13)/(7^2-13) = 3 = (7^2-13)/(5^2-13) = (prime(4)^2-13)/(prime(3)^2-13).

%e a(12) = 2 since (prime(13)^2+35)/(prime(12)^2+35) = (41^2+35)/(37^2+35) = 11/9 = (19^2+35)/(17^2+35) = (prime(8)^2+35)/(prime(7)^2+35), and (prime(13)^2-511)/(prime(12)^2-511) = (41^2-511)/(37^2-511) = 15/11 = (31^2-511)/(29^2-511) = (prime(11)^2-511)/(prime(10)^2-511). Note that 35 = 2*prime(7)+1.

%e a(22) = 2 since (prime(23)^2-85)/(prime(22)^2-85) = (83^2-85)/(79^2-85) = 21/19 = (43^2-85)/(41^2-85) = (prime(14)^2-85)/(prime(13)^2-85), and (prime(23)^2-4081)/(prime(22)^2-4081) = (83^2-4081)/(79^2-4081) = 13/10 = (73^2-4081)/(71^2-4081) = (prime(21)^2-4081)/(prime(20)^2-4081). Note that -85 = -2*prime(13)-3.

%t p[n_]:=p[n]=Prime[n]

%t f[k_,n_]:=f[k,n]=(k+p[n+1]^2)/(k+p[n]^2)

%t Do[r=0;Do[If[f[k,m]==f[k,n],r=r+1],{m,1,n-1},{k,1-p[m]^2,p[m]^2-1}];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000040, A000290, A001248.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Oct 16 2015