OFFSET
1,5
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
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Farideh Firoozbakht, Oct 13 2014
EXTENSIONS
a(16)-a(22) from Robert Price, Nov 12 2014
STATUS
approved