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Numbers n such that A277118(n) = 15.
1

%I #13 Nov 14 2016 13:11:36

%S 13,88,5708,15095,18300,20718,23424,45625,59638,63958,66438,71747,

%T 78239,154992,225235,411285,418434,550190,571439,743873,883958,938420,

%U 989479,1169298,1198004,1256049

%N Numbers n such that A277118(n) = 15.

%C Let p=A001359(n-1). Then n is in the sequence if and only if we have one of constellations of eight consecutive primes: {p=30t+29 (t>=0), p+2,p+8, p+12,p+18, p+24,p+30,p+32}, {p,p+2,p+8,p+14,p+20,p+24,p+30,p+32}, {p,p+2, p+8,p+14, p+18,p+24,p+30,p+32}.

%H Vladimir Shevelev, Peter J. C. Moses, <a href="https://arxiv.org/abs/1610.03385">Constellations of primes generated by twin primes</a>, arXiv:1610.03385 [math.NT], 2016.

%e a(1)=13, then we have one of three mentioned in comment constellations of 8 consecutive primes: p=A001359(12)=149, 151, 157, 163, 167, 173, 179 and 181.

%Y Cf. A001359, A277118, A277119.

%K nonn,more

%O 1,1

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Oct 18 2016