

A054800


First term of balanced prime quartets: p(m+1)p(m) = p(m+2)p(m+1) = p(m+3)p(m+2).


63



251, 1741, 3301, 5101, 5381, 6311, 6361, 12641, 13451, 14741, 15791, 15901, 17471, 18211, 19471, 23321, 26171, 30091, 30631, 53611, 56081, 62201, 63691, 71341, 74453, 75521, 76543, 77551, 78791, 80911, 82781, 83431, 84431, 89101, 89381
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OFFSET

1,1


COMMENTS

This sequence is infinite if Dickson's conjecture holds.  Charles R Greathouse IV, Apr 23, 2011
This is actually the complete list of primes starting a CPAP4 (set of 4 consecutive primes in arithmetic progression). It equals A033451 for a(1..24), but it contains a(25) = 74453 which starts a CPAP4 with common difference 18 (the first one with a difference > 6) and therefore is not in A033451.  M. F. Hasler, Oct 26 2018


LINKS

Zak Seidov and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 4000 terms from Seidov)


EXAMPLE

a(1) = 251 = prime(54) = A000040(54) and prime(55)  prime(54) = prime(56)prime(55) = 6.  Zak Seidov, Apr 23 2011


MATHEMATICA

Select[Partition[Prime[Range[9000]], 4, 1], Length[Union[Differences[#]]] == 1&][[All, 1]] (* Harvey P. Dale, Aug 08 2017 *)


PROG

(PARI) p=2; q=3; r=5; forprime(s=7, 1e4, t=sr; if(t==rq&&t==qp, print1(p", ")); p=q; q=r; r=s) \\ Charles R Greathouse IV, Feb 14 2013


CROSSREFS

Cf. A006562, A054801 .. A054840.
Cf. A006560 (first prime to start a CPAPn).
Start of CPAP4 with given common difference (in square brackets): A033451 [6], A033447 [12], A033448 [18], A052242 [24], A052243 [30], A058252 [36], A058323 [42], A067388 [48], A259224 [54], A210683 [60].
Sequence in context: A185941 A215607 A099734 * A033451 A201793 A183840
Adjacent sequences: A054797 A054798 A054799 * A054801 A054802 A054803


KEYWORD

nonn


AUTHOR

Henry Bottomley, Apr 10 2000


STATUS

approved



