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A338751
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Primes p such that (p*s) mod q and (p*s) mod r are a pair of twin primes, where q,r,s are the next primes after p.
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2
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3, 37, 53, 563, 877, 1297, 3571, 4327, 6679, 10487, 11047, 19463, 19531, 23557, 26723, 32363, 32957, 34253, 34457, 35527, 84793, 92779, 98317, 113167, 120937, 131933, 133967, 148193, 148457, 160073, 174917, 182627, 205417, 206237, 219787, 220897, 245513, 247759, 257707, 276637, 289129, 290663
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OFFSET
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1,1
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COMMENTS
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Let g1 = q-p, g2 = r-q, g3 = s-r be the gaps between the consecutive primes p,q,r,s. Then p*s = (q-g1)*(q+g2+g3) == -g1*(g2+g3) (mod q) and similarly p*s == -(g1+g2)*g3 (mod r). According to Cramer's conjecture, g1*(g2+g3) < q and (g1+g2)*g3 < r for all but finitely many p: the only exception in this sequence appears to be p=3. When g1*(g2+g3) < q and (g1+g2)*g3 < r we have ((p*s) mod r) - ((p*s) mod q) = (r - (g1+g2)*g3) - (q - g1*(g2+g3)) = g2*(g1-g3+1). This is 2 (allowing the possibility of twin primes) if g2 = 2 and g3 = g1.
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LINKS
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EXAMPLE
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a(3) = 53 is a member because with p=53, q=59, r=61 and s=67 we have (53*67) mod 59 = 11 and (53*67) mod 61 = 13, and 11 and 13 are a pair of twin primes.
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MAPLE
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p2:= 2: p3:= 3: p4:= 5: R:= NULL: count:= 0:
while count < 100 do
p1:= p2; p2:= p3; p3:= p4;
p4:= nextprime(p4);
r2:= (p1*p4) mod p2;
r3:= (p1*p4) mod p3;
if r3 = r2 + 2 and isprime(r2) and isprime(r3) then
R:= R, p1; count:= count+1;
fi
od:
R;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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