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Smallest number m such that for 0 < k < n+1, np(m+k-1) = np(m)-k+1, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).
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%I #26 Dec 22 2014 22:11:33

%S 1,7,25,25,181,208,208,1867,14345,19609,40918,40918,620326,2552265,

%T 2552265,7225612,7225612,16679492,33772734,33772734,33772734,

%U 620326386,1516416904,1516416904,4764006481,5272314878,21423652192

%N Smallest number m such that for 0 < k < n+1, np(m+k-1) = np(m)-k+1, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).

%C np(m) = A182134(m).

%C According to the definition, numbers np(a(n)), np(a(n)+1), ..., np(a(n)+n-2) and np(a(n)+n-1) are n consecutive numbers in descending order.

%C a(34) > 10^12. - _Robert Price_, Dec 07 2014

%C See A251736 for the corresponding values of np.

%H Robert Price, <a href="/A246792/b246792.txt">Table of n, a(n) for n = 1..33</a>

%e a(15) = 2552265, since np(2552265) = 24, np(2552265+1) = 23 , ..., np(2552265+13) = 11, np(2552265+14) = 10 are 15 consecutive numbers in descending order.

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

%o (PARI) np(n) = primepi(prime(n)^(1+1/n))-n;

%o isok(m, n) = {for (k=1, n, if (np(m+k-1) != np(m)-k+1, return (0));); return (1);}

%o a(n) = {m = 1; while (! isok(m, n), m++); m;} \\ _Michel Marcus_, Dec 07 2014

%Y Cf. A000040, A182134, A246791, A251736.

%K nonn

%O 1,2

%A _Farideh Firoozbakht_, Oct 16 2014

%E a(18)-a(33) from _Robert Price_, Dec 07 2014