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A263724
Least prime p = prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2 + q^2, where q > prime(n+3) is also prime.
1
373, 653, 1997, 1901, 2309, 3389, 4373, 5381, 6701, 8069, 10589, 12269, 18269, 18461, 19541, 22973, 24821, 29021, 32909, 38261, 46589, 45869, 50021, 53549, 56909, 66029, 69389, 77261, 87629, 93581, 102101, 107741, 118901, 128981, 131837, 145517, 152909, 159869, 170021, 188261, 184901, 196661, 214469, 229781, 237821, 252509, 277157, 281429, 291101, 305933, 317693, 333029, 344021, 359981, 370661, 387341, 395069, 418349, 460949
OFFSET
2,1
COMMENTS
The corresponding prime q is in A263725.
The prime p exists for all n > 1 under Schinzel's Hypothesis H; see Sierpinski (1988), p. 221.
If q = prime(n+4), then p is in A133559 (prime sums of squares of 5 consecutive primes). The converse holds if a(n) != a(m) when n != m (which holds if a(n) < a(n+1), as appears to be true).
REFERENCES
W. Sierpinski, Elementary Theory of Numbers, 2nd English edition, revised and enlarged by A. Schinzel, Elsevier, 1988; see p. 221.
EXAMPLE
The primes 373 = 3^2 + 5^2 + 7^2 + 11^2 + 13^2, 653 = 5^2 + 7^2 + 11^2 + 13^2 + 17^2, and 1997 = 7^2 + 11^2 + 13^2 + 17^2 + 37^2 lead to a(1) = 373, a(2) = 653, and a(3) = 1997.
MATHEMATICA
Table[k = 4;
While[p = Sum[Prime[n + j]^2, {j, 0, 3}] + Prime[n + k]^2; ! PrimeQ[p],
k++]; p, {n, 2, 60}]
CROSSREFS
Sequence in context: A142395 A182573 A142921 * A133559 A229499 A023313
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Oct 24 2015
STATUS
approved