

A185006


Ramanujan primes R_(4,1)(n): a(n) is the smallest number such that if x >= a(n), then pi_(4,1)(x)  pi_(4,1)(x/2) >= n, where pi_(4,1)(x) is the number of primes==1 (mod 4) <= x.


3



13, 37, 41, 89, 97, 109, 149, 229, 233, 241, 257, 277, 281, 317, 349, 397, 401, 409, 421, 433, 569, 593, 601, 641, 653, 661, 709, 757, 761, 821, 929, 937, 941, 953, 977, 997, 1009, 1021, 1049, 1061, 1093, 1097, 1117, 1193, 1213, 1237, 1249
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OFFSET

1,1


COMMENTS

All terms are primes==1 (mod 4).
A general conception of generalized Ramanujan numbers, see in Section 6 of the Shevelev, Greathouse IV, & Moses link.
We conjecture that for all n >= 1, a(n) <= A104272(3*n). This conjecture is based on observation that, if interval (x/2, x] contains >= 3*n primes, then at least n of them are of the form 4*k+1.


LINKS

Table of n, a(n) for n=1..47.
Vladimir Shevelev, Charles R. Greathouse IV, Peter J. C. Moses, On intervals (kn, (k+1)n) containing a prime for all n>1, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. arXiv:1212.2785


FORMULA

lim(a(n)/prime(4*n)) = 1 as n tends to infinity.


MATHEMATICA

Table[1 + NestWhile[#1  1 &, A104272[[3 k]], Count[Mod[Select[Range@@{Floor[#1/2 + 1], #1}, PrimeQ], 4], 1] >= k &], {k, 1, 10}] using the code nn = 1000; A104272 = Table[0, {nn}]; s = 0; Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s]; If[s < nn, A104272[[s + 1]] = k], {k, Prime[3*nn]}]; A104272 = A104272 + 1 (* T. D. Noe, Nov 15 2010 *)


CROSSREFS

Cf. A104272, A185005, A185004, A185007.
Sequence in context: A301857 A220462 A280997 * A285887 A063913 A119705
Adjacent sequences: A185003 A185004 A185005 * A185007 A185008 A185009


KEYWORD

nonn


AUTHOR

Vladimir Shevelev and Peter J. C. Moses, Dec 18 2012


STATUS

approved



