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A246791
Smallest number m such that for 0 <= k < n, np(m+k) = np(m)+k, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).
2
1, 4, 15, 136, 2128, 15453, 479403, 1184231, 10975072, 27112368, 175600366, 2304656281, 14896902677, 59331462112
OFFSET
1,2
COMMENTS
np(m) = A182134(m).
According to the definition, numbers np(a(n)), np(a(n)+1), ..., np(a(n)+n-2), np(a(n)+n-1) are n consecutive numbers in ascending order.
a(15) > 10^12. - Robert Price, Nov 19 2014
See A247116 for the corresponding values of np.
EXAMPLE
a(8) = 1184231 since np(1184231) = 17, np(1184231+1) = 18, ..., np(1184231+6) = 23, np(1184231+7) = 24 are 8 consecutive numbers and 1184231 is the smallest number with this property.
MATHEMATICA
np[n_] := np[n] = Length[Select[Range[Prime[n]+1, Prime[n]^(1 + 1/n)], PrimeQ]]; a[1]=1; a[n_] := a[n] = (For[m = a[n-1], c = Table[np[m+k], {k, 0, n-1}]; c != Range[Min[c], Max[c]], m++]; m); Do[Print[a[n]], {n, 8}]
PROG
(PARI) np(t) = primepi(prime(t)^(1 + 1/t)) - t;
ok(m, n) = {for (k=0, n-1, if (np(m+k) != np(m)+k, return(0)); ); return (1); }
a(n) = {m = 1; while (! ok(m, n), m++); m; } \\ Michel Marcus, Nov 25 2014
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Farideh Firoozbakht, Oct 16 2014
EXTENSIONS
a(9)-a(14) from Robert Price, Nov 19 2014
STATUS
approved