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A154774
Numbers n such that 9900n^2 is the average of a twin prime pair.
2
10, 14, 15, 25, 60, 74, 76, 87, 127, 129, 130, 140, 171, 196, 207, 224, 259, 263, 302, 314, 315, 319, 333, 337, 377, 389, 451, 454, 470, 491, 508, 518, 549, 568, 574, 589, 592, 624, 629, 636, 690, 696, 729, 748, 753, 769, 770, 771, 781, 791, 802, 833, 844
OFFSET
1,1
COMMENTS
Inspired by Zak Seidov's post to the SeqFan list, cf. link: This yields A154674 as 9900 a(n)^2. Indeed, if N/11 is a square, then N=11 m^2 and this can't be the average of a twin prime pair unless m=30a (considering N+1 mod 2,3,5 and N-1 mod 5).
LINKS
Zak Seidov, "A154676", Jan 15 2009
FORMULA
a(n) = sqrt(A154674(n)/9900).
MATHEMATICA
Select[Range[1000], AllTrue[9900#^2+{1, -1}, PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Jan 19 2019 *)
PROG
(PARI) for(i=1, 999, isprime(9900*i^2+1) & isprime(9900*i^2-1) & print1(i", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Jan 15 2009
STATUS
approved