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%I #44 Jul 15 2016 21:38:29
%S 1,7,57,182,2057,1068,32318,110443,280182,3626068,23157318,120813568,
%T 123327057,1097376068,11109655182,49925501068,407838170807,
%U 355101282318,3459595983307,15613890344818,365855836217682,110981321985443,2273204469030182,9647724486047943
%N a(n) is the least k such that A051903(k^2+1) = n.
%C Least k such that the largest exponent of a prime in the factorization of k^2+1 is n.
%C Conjecture: for each n > 1, a(n) = A034939(n) or 5^n - A034939(n).
%C 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.
%H Robert Israel, <a href="/A274721/b274721.txt">Table of n, a(n) for n = 1..109</a>
%e 1^2 + 1 = 2.
%e 7^2 + 1 = 2*5^2.
%e 57^2 + 1 = 2*5^3*13.
%e 182^2 + 1 = 5^4 * 53.
%p F:= proc(n) local v, p,w;
%p v:= numtheory:-msqrt(-1,5^n);
%p v:= min(v, 5^n-v);
%p if max(seq(t[2],t=ifactors(v^2+1)[2])) > n then
%p v:= 5^n - v;
%p if max(seq(t[2],t=ifactors(v^2+1)[2])) > n then
%p error "neither %d nor %d works",5^n-v,v fi
%p fi;
%p for p from 13 by 4 while p^n <= v^2+1 do
%p if isprime(p) then
%p w:= numtheory:-msqrt(-1,p^n);
%p w:= min(w, p^n-w);
%p if w < v then
%p if max(seq(t[2],t=ifactors(w^2+1)[2])) = n then
%p v:= w;
%p fi
%p fi
%p fi
%p od;
%p v
%p end proc:
%p F(1):= 1:
%p map(F, [$1..100]);
%Y Cf. A034939, A051903.
%K nonn
%O 1,2
%A _Robert Israel_, Jul 14 2016
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