OFFSET
0,1
COMMENTS
a(25) = 0, and a(24) cannot exist. The same is true with a(k) and k>25. From A020497, we see that a range of 101 numbers is required to find 24 primes. It is an open question if a(23) exists.
From Robert Israel, Jan 18 2017: (Start)
Dickson's conjecture implies that a(23) does exist.
Let Q = 27926129625869590, and R = 614889782588491410 the product of all primes < 50.
Then for any k, the 23 numbers Q+i+k*R for i = 1, 3, 7, 9, 13, 19, 21, 27, 31, 33, 37, 43, 49, 51, 57, 63, 69, 73, 79, 87, 91, 97, 99 have no prime divisors < 50.
Dickson's conjecture would indicate that there are infinitely many k for which these numbers are all prime, and thus there are 23 primes between Q+k*R and Q+k*R+100. (End)
Heuristics suggest a(23) exists (see above) and has between 20 and 30 digits. There are 192 residue classes mod 23# = 223092870 in which a(23) might fall, all of which are 11 mod 30 and either 3 or 4 mod 7. - Charles R Greathouse IV, Jul 12 2017
EXAMPLE
For n = 1 there is only one prime between 155900 and 156000: 155921.
MAPLE
for n from 1 to 10^5 do
T[n]:= nops(select(isprime, [$10*n+1 ..10*n+9]))
od:
for k from 1 to 10^5-10 do
v:= add(T[k+j], j=0..9):
if not assigned(A[v]) then A[v]:= k fi
od:
seq(A[n], n=0..22); # Robert Israel, Jul 12 2017
MATHEMATICA
Function[s, -1 + Flatten@ Table[FirstPosition[s, n] /. k_ /; MissingQ@ k -> 0, {n, 0, Max@ s}]]@ Table[Count[Range[10 k, 10 k + 100], _?PrimeQ], {k, 0, 10^5}] (* Michael De Vlieger, Jul 12 2017; program writes "-1" for a(23) and a(24). *)
PROG
(PARI) a(n) = my(k=0); while(1, if(primepi(10*k+100)-primepi(10*k)==n, return(k)); k++) \\ Felix Fröhlich, Jul 12 2017
(PARI) a(n)=my(k); while(sum(p=10*k+1, 10*k+99, isprime(p))!=n, k++); k \\ Charles R Greathouse IV, Jul 12 2017
CROSSREFS
KEYWORD
nonn,fini
AUTHOR
Wolfram Hüttermann, Dec 21 2016
EXTENSIONS
Name clarified by FUNG Cheok Yin, Jul 12 2017
STATUS
approved