OFFSET
1,2
COMMENTS
Conjecture: For any integer n > 1, the term a(n) is the least prime p > 2*n with p dividing a^2 + b^2 - 1 for no 1 <= a < b <= n.
This has been verified for all n = 2..15000.
Note that a^2*(a^2-1)-b^2*(b^2-1) = (a-b)*(a+b)*(a^2+b^2-1). For any positive integers m and n > 1, if k^2*(k^2-1) (k=1..n) are pairwise distinct modulo m, then it is easy to see that m > 2*n.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.
EXAMPLE
a(2) = 5 since the two numbers 1^2*(1^2-1)=0 and 2^2*(2^2-1) = 12 are distinct modulo 5, but they are congruent modulo each of 1,2,3,4.
MATHEMATICA
f[k_]:=f[k]=k^2*(k^2-1);
U[m_, n_]:=U[m, n]=Length[Union[Table[Mod[f[k], m], {k, 1, n}]]]
tab={}; s=1; Do[m=s; Label[bb]; If[U[m, n]==n, s=m; tab=Append[tab, s]; Goto[aa]]; m=m+1; Goto[bb]; Label[aa], {n, 1, 70}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 18 2021
STATUS
approved