A246783 Smallest number m such that all the n numbers np(m+k-1), 0 < k < n+1 are equal, where np(t) is number of primes p with prime(t) < p < prime(t)^(1+1/t).

1, 1, 1, 1, 67, 67, 67, 67, 67, 54412, 161342, 161342, 1214143, 9915018, 9915018, 68964006, 68964006, 810832784, 19867608968, 52415066804, 119937255921, 272007811177

Offset: 1

Author

Farideh Firoozbakht, Oct 13 2014

Keywords

• nonn
• more

Comments

np(n) = A182134(n).

Conjecture: For every n, a(n) exists.

The np values for distinct terms of the sequence are: 1, 5, 11, 11, 16, 14, ... . It is interesting that for n = 1, 5, 11 & 14, np(a(n)) = n. What is the next term of this sequence (1, 5, 11, 14, ?, ... .)?

The sequence giving the np values of this sequence is A245101.

The sequence {1, 5, 11, 11, 16, 14} commented on above is expanded in A245098.

a(23) > 10^12. - Robert Price, Nov 12 2014

The next term of the sequence (1, 5, 11, 14, ?, ... .) commented on above is >22 and corresponds to t>10^12. - Robert Price, Nov 12 2014

Examples

			a(9) = 67, since all the nine numbers np(67), np(67+1), np(67+2), ..., np(67+8) are equal and 67 is the smallest such number. Note that np(67) = 5.
		

Crossrefs

Cf. A000040, A182134.

Programs

Mathematica

np[n_]:=Length[Select[Range[Prime[n]+1,Prime[n]^(1+1/n)], PrimeQ]];a[n_]:=(For[m=1,Length[Union[Table[np[m+k-1],{k,n}]]]!=1,m++];m);Do[Print[a[n]],{n,15}]

Extensions

a(16)-a(22) from Robert Price, Nov 12 2014