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A344020
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A sequence of prime numbers: a(1)=2, a(n+1) is the least prime dividing Product_{i in S} a(i)^2 + Product_{i not in S} a(i)^2, minimized over all subsets S of {1..n}.
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2
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2, 5, 29, 17, 41, 13, 37, 53, 61, 97, 101, 73, 89, 109, 149, 137, 113, 173, 181, 157, 229, 197, 241, 257, 233, 193, 277, 269, 349, 317, 337, 293, 281, 313, 353, 373, 389, 409, 421, 397, 457, 461, 401, 433, 521, 509, 449, 541, 569, 557, 701, 593, 613, 653, 641, 617, 577, 661, 673, 709, 677, 601, 761, 733, 757, 769, 773, 797
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OFFSET
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1,1
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COMMENTS
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All a(i) must be unique and, apart from 2, must be congruent to 1 (mod 4) as p only divides Product_{i in S} a(i)^2 + Product_{i not in S} a(i)^2 if -1 is a quadratic residue modulo p.
Whether all primes congruent to 1 (mod 4) occur in this sequence is unknown.
For n > 1, a(n) >= p, where p is the smallest prime p such that p == 1 (mod 4) and a(2)*a(3)*...*a(n-1) is a nonzero square modulo p. Conjecture: a(n) = p. - Jinyuan Wang and Max Alekseyev, Jul 04 2022
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LINKS
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EXAMPLE
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For n=4 we obtain the 4 partitions with their products: 1 + 2^2 * 5^2 * 29^2 = 84101 = 37 * 2273, 2^2 + 5^2 * 29^2 = 21029 = 17*1237, 5^2 + 2^2 * 29^2 = 3389 and 2^2 * 5^2 + 29^2 = 941. The minimum of the primes dividing these is 17, thus a(4)=17.
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PROG
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(PARI) { A344020_list() = my(a, A, m, p, b, q, z); print1(2, ", "); a = [2]; A=1; while(1, p=5; while( kronecker(A, p)!=1 || p%4!=1, p=nextprime(p+1) ); b=lift(sqrt(A+O(p))*(1+sqrt(-1+O(p)))); z=znprimroot(p); m = nextprime(random(10^6)); q=lift(prod(i=1, #a, Mod(1+x^znlog(Mod(a[i], p), z, p-1), (1-x^(p-1))*Mod(1, m)) )); if( polcoeff(q, znlog(Mod(b, p), z, p-1), x)==0 && polcoeff(q, znlog(Mod(-b, p), z, p-1), x)==0, error("conjecture failed mod", m) ); a=concat(a, [p]); A*=p; print1(p, ", ") ); } \\ Max Alekseyev, Jul 04 2022
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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