|
|
|
|
2, 71, 101, 181, 239, 241, 269, 349, 373, 409, 419, 433, 439, 491, 593, 599, 601, 607, 647, 653, 659, 823, 827, 857, 947, 1021, 1031, 1061, 1063, 1091, 1103, 1301, 1427, 1429, 1447, 1451, 1489, 1553, 1559, 1567, 1601, 1607, 1609, 1789, 1867, 1871, 1913, 1999, 2003
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
For every n>=1, A104272(n) >= A080359(n), and the sequence shows where the inequality becomes an equality.
Let prime(m) < a(n)/2 < prime(m+1); then there exist primes p<q such that p is in the interval (2*Prime(m), a(n)) and q is in the interval (a(n), 2*Prime(m+1)).
For example, a(2) = 71, 31 < a(2)/2 < 37 and intervals (62,71), (71,74) contain the primes p = 67 and q = 73 respectively.
Let us call a prime p compatible with another prime q, if the intervals (p/2,q/2) and (p,q], if q>p, (or intervals (q/2,p/2) and (q,p], if q<p) contain the same number of primes. If p is compatible with no other prime, we call it a peculiar prime. The sequence lists the peculiar primes. [Vladimir Shevelev, Apr 25 2012]
|
|
LINKS
|
|
|
FORMULA
|
All solutions of the equation A104272(x)=A080359(x) are x=pi(a(n))-pi(a(n)/2). - Vladimir Shevelev, Apr 25 2012
|
|
EXAMPLE
|
a(2)=71, such that 31<71/2<37, and we see that p=67 is in interval (62, 71) and q=73 is in interval (71, 74).
|
|
MATHEMATICA
|
nn = 200; t = Table[0, {nn+1}]; s = 0;
Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--]; If[s <= nn && t[[s+1]] == 0, t[[s+1]] = k], {k, Prime[3nn]}
];
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[3nn]}
];
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|