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A352480
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First of four consecutive primes p,q,r,s such that p+q+r+s is divisible by A001414(q+r).
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3
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11, 19, 23, 37, 97, 109, 137, 263, 277, 307, 401, 409, 617, 757, 877, 1039, 1187, 1321, 1481, 1567, 1871, 2179, 2333, 2543, 2861, 3371, 3617, 3697, 4649, 4783, 5639, 5651, 6547, 6689, 6779, 6883, 7687, 8807, 9371, 9437, 9767, 10331, 11317, 11777, 11927, 13523, 14503, 15683, 16921, 17291, 19073, 19553
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OFFSET
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1,1
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COMMENTS
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If x is a prime and 2*x-y,2*x-z,2*x+z,2*x+y+32 are consecutive primes with 0 < z < y, then 2*x-y is a term. Thus Dickson's conjecture implies there are infinitely many terms.
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LINKS
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EXAMPLE
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a(3) = 23 is a term because 23, 29, 31, 37 are consecutive primes, and A001414(29+31) = A001414(2^2*3*5) = 12 divides 23+29+31+37 = 120.
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MAPLE
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q:= 2: r:= 3: s:= 5:
R:= NULL: count:= 0:
while count < 50 do
p:= q; q:= r; r:= s; s:= nextprime(s);
a:= add(t[1]*t[2], t = ifactors(q+r)[2]);
if (p+q+r+s) mod a = 0 then count:= count+1; R:= R, p; 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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