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A232932
The least positive integer k such that Kronecker(D/k) = -1 where D runs through all negative fundamental discriminants (-A003657).
9
2, 3, 3, 5, 2, 7, 2, 11, 5, 13, 3, 2, 7, 3, 2, 5, 2, 3, 3, 11, 2, 2, 5, 7, 3, 2, 13, 5, 3, 2, 7, 3, 11, 2, 11, 2, 7, 11, 7, 2, 3, 2, 5, 3, 2, 5, 3, 3, 5, 2, 11, 2, 13, 5, 5, 2, 5, 3, 2, 7, 2, 3, 2, 2, 5, 13, 2, 3, 2, 5, 17, 3, 2, 7, 3, 3, 5, 2, 13, 2, 7, 5, 19, 2, 3, 11, 3, 2, 5, 2, 3, 3, 7, 2, 5, 2, 5, 11, 5, 3, 2, 5, 3, 2, 11, 2, 3, 7, 2, 2, 11, 7, 3, 2, 5, 3, 2, 5, 3, 3, 2, 11, 2, 19, 5, 5, 2, 3, 2, 17, 3, 2, 7, 2, 3, 3, 13, 2, 5, 2, 5, 11, 7, 3, 2, 7, 3, 13, 2, 3, 5, 2, 2
OFFSET
1,1
COMMENTS
From Jianing Song, Feb 14 2019: (Start)
a(n) is necessarily prime. Otherwise, if a(n) is not prime, we have (D/p) = 0 or 1 for all prime divisors p of a(n), so (D/a(n)) must be 0 or 1 too, a contradiction.
a(n) is the least inert prime in the imaginary quadratic field with discriminant D, D = -A003657(n). (End)
LINKS
S. R. Finch, Average least nonresidues, December 4, 2013. [Cached copy, with permission of the author]
FORMULA
a(n) = A306220(A003657(n)). - Jianing Song, Feb 14 2019
EXAMPLE
A003657(4) = 8, (-8/5) = -1, (-8/3) = 1 and (-8/2) = 0, so a(4) = 5.
MATHEMATICA
nMax = 200; FundamentalDiscriminantQ[n_] := n != 1 && (Mod[n, 4] == 1 || ! Unequal[Mod[n, 16], 8, 12]) && SquareFreeQ[n/2^IntegerExponent[n, 2]]; discrims = Select[-Range[4 nMax], FundamentalDiscriminantQ]; f[d_] := For[k = 1, True, k++, If[FreeQ[{0, 1}, KroneckerSymbol[d, k]], Return[k] ] ]; a[n_] := f[discrims[[n]]]; Table[a[n], {n, 1, nMax}] (* Jean-François Alcover, Nov 05 2016, after Robert G. Wilson v *)
CROSSREFS
Sequence in context: A324799 A069461 A329071 * A252502 A063256 A229703
KEYWORD
nonn
AUTHOR
Steven Finch, Dec 02 2013
EXTENSIONS
Name simplified by Jianing Song, Feb 14 2019
STATUS
approved