Let p(n) be the n-th prime congruent to 1 mod 4. Then a(n) = the least m for which m^2+1=p(n)*k^2 has a solution.
A094048
Let p(n) be the n-th prime congruent to 1 mod 4. Then a(n) = the least m for which m^2+1=p(n)*k^2 has a solution.
Terms
- a(0) =2a(1) =18a(2) =4a(3) =70a(4) =6a(5) =32a(6) =182a(7) =29718a(8) =1068a(9) =500a(10) =5604a(11) =10a(12) =8890182a(13) =776a(14) =1744a(15) =113582a(16) =4832118a(17) =1118a(18) =1111225770a(19) =1764132a(20) =14a(21) =1710a(22) =23156a(23) =71011068a(24) =16a(25) =82a(26) =8920484118a(27) =1063532a(28) =2482a(29) =126862368
External references
- oeis: A094048