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A272200
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Bisection of primes congruent to 1 modulo 3 (A002476), depending on the corresponding A001479 entry being congruent to 1 modulo 3 or not. Here the first case.
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4
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13, 19, 43, 61, 97, 103, 109, 127, 157, 163, 181, 193, 241, 277, 283, 331, 349, 373, 379, 409, 433, 463, 487, 499, 523, 601, 607, 619, 631, 661, 673, 691, 727, 733, 757, 769, 787, 811, 859, 883, 937, 967, 991
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OFFSET
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1,1
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COMMENTS
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The other primes congruent to 1 modulo 3 are given in A272201.
Each prime == 1 (mod 3) has a unique representation A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) = A001479(m+1) and B(m) = A001480(m+1), m >= 1 (see also A001479). The present sequence gives such primes corresponding to A(m+1) == 1 (mod 3). The ones corresponding to A(m+1) not == 1 (mod 3) (the complement) are given in A272201.
This bisection of the primes from A002476 is needed in the formula for the coefficients of the q-expansion (q = exp(2*Pi*i*z), Im(z) > 0) of the modular cusp form (eta(6*z))^4|_{z=z(q)} = Eta64(q) with Dedekind's eta function. See A000727 which gives the coefficients of the q-expansion of F(q) = Eta64(q^(1/6))/q^(1/6) = (Product_{m>=0} (1 - q^m))^4. The coefficients F(q) = Sum_{n>=0} f(6*n+1)*q^n are given in the Finch link on p. 5, using multiplicativity. For primes congruent to 1 modulo 6 the formula involves x_p and y_p which are the present A and B for prime p == 1 (mod 3).
See also the p-defects of the elliptic curve y^2 = x^3 + 1, related to (eta(6*z))^4, given in A272198 with another (equivalent) way to find the coefficients of the Eta64(q) expansion, hence those of F(q).
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LINKS
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FORMULA
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This sequence collects the 1 (mod 3) primes p(m) = A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) == 1 (mod 3), for m >= 1. A(m) = A001479(m+1).
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MAPLE
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filter:= proc(n) local S, x, y;
if not isprime(n) then return false fi;
S:= remove(hastype, [isolve(x^2+3*y^2=n)], negative);
subs(S[1], x) mod 3 = 1
end proc:
select(filter, [seq(i, i=7..1000, 6)]); # Robert Israel, Apr 29 2019
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MATHEMATICA
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filterQ[n_] := Module[{S, x, y}, If[!PrimeQ[n], Return[False]]; S = Solve[x > 0 && y > 0 && x^2 + 3 y^2 == n, Integers]; Mod[x /. S[[1]], 3] == 1];
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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