Let p(n) be the n-th prime congruent to 1 mod 4. Then a(n) = the least k for which m^2+1=p(n)*k^2 has a solution.
A094049
Let p(n) be the n-th prime congruent to 1 mod 4. Then a(n) = the least k for which m^2+1=p(n)*k^2 has a solution.
Terms
- a(0) =1a(1) =5a(2) =1a(3) =13a(4) =1a(5) =5a(6) =25a(7) =3805a(8) =125a(9) =53a(10) =569a(11) =1a(12) =851525a(13) =73a(14) =149a(15) =9305a(16) =385645a(17) =85a(18) =82596761a(19) =126985a(20) =1a(21) =113a(22) =1517a(23) =4574225a(24) =1a(25) =5a(26) =535979945a(27) =63445a(28) =145a(29) =7170685
External references
- oeis: A094049