OFFSET
1,1
COMMENTS
All terms are primes==1 (mod 3).
A modular generalization of Ramanujan numbers, see Section 6 of the Shevelev-Greathouse-Moses paper.
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 3*k+1.
The function pi_(3,1)(n) starts 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3,... with records occurring as specified in A123365/A002476. - R. J. Mathar, Jan 10 2013
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
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.
MAPLE
pimod := proc(m, n, x)
option remember;
a := 0 ;
for k from n to x by m do
if isprime(k) then
a := a+1 ;
end if;
end do:
a ;
end proc:
a := [seq(0, n=1..100)] ;
for x from 1 do
pdiff := pimod(3, 1, x)-pimod(3, 1, x/2) ;
if pdiff+1 <= nops(a) then
v := x+1 ;
n := pdiff+1 ;
if n<v then
a := subsop(n=v, a) ;
print(a) ;
end if;
end if;
end do: # R. J. Mathar, Jan 10 2013
MATHEMATICA
max = 100; pimod[m_, n_, x_] := pimod[m, n, x] = Module[{a = 0}, For[k = n, k <= x, k = k + m, If[PrimeQ[k], a = a + 1]]; a]; a[_] = 0; For[x = 1, x <= max^2, x++, pdiff = pimod[3, 1, x] - pimod[3, 1, x/2]; If[ pdiff + 1 <= max, v = x + 1; n = pdiff + 1; If[ n < v , a[n] = v ] ] ]; Table[a[n], {n, 1, max}] (* Jean-François Alcover, Jan 28 2013, translated and adapted from R. J. Mathar's Maple program *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev and Peter J. C. Moses, Dec 18 2012
STATUS
approved