OFFSET
1,1
COMMENTS
As noted in A007530, p is of the form p = 6*k-1, so p+2 = 6*k+1, p+6 = 6*k+5, p+8 = 6*k+7. The sum is therefore s = p+(p+2)+(p+6)+(p+8) = 12*(2*k+1). From the same mod 30 formula in A007530, p is also of the form p = 10*m+1, so the sum of the primes in the quadruplet is also of the form s = 5*(8*m+4). Watching the factors of 12 and 5 in these two factorizations of the sum, the factor 60 (least common multiple) is in some sense the optimum, and this is the rationale to look for primes that are a 60th of the sum.
The primes that are generated by a 60th of the sum are: 7, 13, 139, 1049, 4481, 8147, 11047, 11411, 13049, 17191,...
The indices of the p in the sequence of quadruplets (A007530) are 3, 4, 8, 15, 29, 43, 47, 49, 52, 61, 64, 82, 85, 92, 96, 104, 112,...
REFERENCES
A. Bartholome, J. Rung, H. Kern: Zahlentheorie für Einsteiger, vieweg Verlag, 5. Auflage, Wiesbaden, 2006
F. Ischebeck: Einladung zur Zahlentheorie, B.I. Wissenschaftsverlag, Mannheim-Leipzig-Wien-Zürich, 1992
H. Tietze: Gelöste und ungelöste mathematische Probleme aus alter und neuer Zeit, Band 1, dtv Wissenschaft, Muenchen, 1984
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
EXAMPLE
p=11, 11+13+17+19 = 1*60, 1 is not prime, 11 is not a term of the sequence.
p=101: 101+103+107+109 = 7*60, 7 = prime(4), 101 = prime(26) is a term.
p=191: 191+193+197+199 = 13*60, 13 = prime(6), 191 = prime(43) is a term.
MATHEMATICA
Reap[For[n = 1, n < 150000, n++, p = Prime[n]; If[PrimeQ[p + 2] && PrimeQ[p + 6] && PrimeQ[p + 8] && PrimeQ[(p + 4)/15], Sow[p]]]][[2, 1]](* Jean-François Alcover, May 07 2012 *)
Select[Partition[Prime[Range[150000]], 4, 1], Differences[#]=={2, 4, 2}&&PrimeQ[ Total[#]/60]&][[All, 1]] (* Harvey P. Dale, Apr 09 2018 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 01 2010
STATUS
approved