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a(n) = n / A088444(n), where A088444(n) is the smallest divisor d of n such that all intervals [(k-1)*d+1:k*d] contain at least one prime, 1<=k<=n/d; a(1)=1.
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%I #9 May 08 2022 10:34:17

%S 1,1,1,2,1,3,1,4,3,2,1,4,1,2,5,4,1,6,1,5,7,2,1,8,5,2,3,4,1,6,1,4,3,2,

%T 7,6,1,2,3,8,1,7,1,4,9,2,1,8,7,10,3,4,1,9,11,8,3,2,1,12,1,2,9,8,13,11,

%U 1,4,3,14,1,12,1,2,15,4,11,13,1,16,9,2,1,14,17,2,3,11,1,18,13,4,3,2,5

%N a(n) = n / A088444(n), where A088444(n) is the smallest divisor d of n such that all intervals [(k-1)*d+1:k*d] contain at least one prime, 1<=k<=n/d; a(1)=1.

%C Greatest divisor d of n such that all d intervals [(k-1)*n/d+1:k*n/d] contain at least one prime, 1<=k<=d; a(1)=1.

%C The scatter plot shows some interesting features. - _Antti Karttunen_, May 08 2022

%H Antti Karttunen, <a href="/A088445/b088445.txt">Table of n, a(n) for n = 1..20000</a>

%H Antti Karttunen, <a href="/A088445/a088445.txt">Data supplement: n, a(n) computed for n = 1..65537</a>

%H <a href="/index/Pri#gaps">Index entries for primes, gaps between</a>

%F a(n) = n / A088444(n).

%o (PARI)

%o aicalop(d,u) = { for(k=1,u,for(i=1+((k-1)*d),k*d,if(isprime(i),break); if(i==(k*d),return(0)))); (1); }; \\ All Intervals Contain At Least One Prime.

%o A088444(n) = if(1==n,n,fordiv(n,d,if(aicalop(d,n/d),return(d))); (0));

%o A088445(n) = (n/A088444(n)); \\ _Antti Karttunen_, May 08 2022

%Y Cf. A088444.

%K nonn,look

%O 1,4

%A _Reinhard Zumkeller_, Sep 30 2003