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A263399
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).
2
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, 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, 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
OFFSET
1,6
COMMENTS
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.
We have verified this for n up to 2000.
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.
LINKS
EXAMPLE
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).
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.
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.
MATHEMATICA
p[n_]:=p[n]=Prime[n]
f[k_, n_]:=f[k, n]=(k+p[n+1]^2)/(k+p[n]^2)
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 16 2015
STATUS
approved