

A165959


Size of the range of the Ramanujan Prime Corollary, 2*A168421(n)  A104272(n).


6



2, 3, 5, 5, 5, 11, 3, 7, 3, 9, 5, 11, 7, 9, 7, 11, 15, 13, 27, 25, 21, 15, 13, 11, 5, 17, 7, 3, 11, 9, 15, 9, 21, 13, 3, 15, 13, 7, 5, 15, 11, 11, 17, 15, 27, 21, 15, 13, 7, 21, 19, 15, 9, 3, 17, 15, 7, 7, 7, 9, 9, 17, 15, 11, 9, 5, 5, 21, 17, 11, 7, 15, 9
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OFFSET

1,1


COMMENTS

All but the first term is odd because A104272 has only one even term, 2. Because of all primes > 2 are odd, 1 can be subtracted from each term.
If this sequence has an infinite number of terms in which a(n) = 3, then the twin prime conjecture can be proved.
R_n is the sequence A104272(n) and k = pi(R_n)= A000720(R_n) with i>k.
By comparing the fractions we can see that (p_(i+1)p_i)/(2*sqrt(p_i)) and a(n)/(2*sqrt(p_k)) are < 1 for all n > 0, in fact a(n)/(1.8*sqrt(p_k)) < 1 for all n > 0. When taking into account numbers in A182873(n) and A190874(n) to sqrt(R_n) we see that A182873(n)/(A190874(n)*sqrt(R_n)) < 1 for all n > 1.


LINKS



FORMULA



EXAMPLE

A168421(19) = 127, A104272(19) = 227; so a(19) = 2*A168421(19)  A104272(19) = 254  227 = 27. Note: for n = 20, 21, 22, 23 A168421(n) = 127. Because A168421 remains the same for these n and A104272 increases, the size of the range for a(n) for these n decreases. Note: a(18) = 2*97  181 = 194  181 = 13. This is nearly half a(19). The actual gap betweens A104272(19) and the next prime, 229, is 2.


MATHEMATICA

nn = 100; R = Table[0, {nn}]; s = 0;
Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s]; If[s < nn, R[[s + 1]] = k], {k, Prime[3 nn]}];
A104272 = R + 1; t = Table[0, {nn}];
Do[m = PrimePi[2 n]  PrimePi[n]; If[0 < m <= nn, t[[m]] = n], {n, 15 nn}];
A168421 = NextPrime[Join[{1}, t]] // Most;


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



