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 A274721 a(n) is the least k such that A051903(k^2+1) = n. 1
 1, 7, 57, 182, 2057, 1068, 32318, 110443, 280182, 3626068, 23157318, 120813568, 123327057, 1097376068, 11109655182, 49925501068, 407838170807, 355101282318, 3459595983307, 15613890344818, 365855836217682, 110981321985443, 2273204469030182, 9647724486047943 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Least k such that the largest exponent of a prime in the factorization of k^2+1 is n. Conjecture: for each n > 1, a(n) = A034939(n) or 5^n - A034939(n). For any n > 1, -1 has two square roots mod 5^n; at least one of these is not a square root of -1 mod 5^(n+1).  If v is this number, v < 5^n so v^2 < 25^n.  v^2+1 might be divisible by p^(n+1) for p = 13 or 17, or a square root of -1 mod 13^n or 17^n might be smaller than v, but that seems very unlikely.  Thus the conjecture. LINKS Robert Israel, Table of n, a(n) for n = 1..109 EXAMPLE 1^2 + 1 = 2. 7^2 + 1 = 2*5^2. 57^2 + 1 = 2*5^3*13. 182^2 + 1 = 5^4 * 53. MAPLE F:= proc(n) local v, p, w;   v:= numtheory:-msqrt(-1, 5^n); v:= min(v, 5^n-v); if max(seq(t, t=ifactors(v^2+1))) > n then     v:= 5^n - v;     if max(seq(t, t=ifactors(v^2+1))) > n then          error "neither %d nor %d works", 5^n-v, v fi fi; for p from 13 by 4 while p^n <= v^2+1 do     if isprime(p) then      w:= numtheory:-msqrt(-1, p^n);      w:= min(w, p^n-w);      if w < v then         if max(seq(t, t=ifactors(w^2+1))) = n then            v:= w;         fi      fi     fi od; v end proc: F(1):= 1: map(F, [\$1..100]); CROSSREFS Cf. A034939, A051903. Sequence in context: A110830 A304691 A218428 * A180817 A201438 A202510 Adjacent sequences:  A274718 A274719 A274720 * A274722 A274723 A274724 KEYWORD nonn AUTHOR Robert Israel, Jul 14 2016 STATUS approved

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Last modified October 18 14:11 EDT 2021. Contains 348068 sequences. (Running on oeis4.)