A088445 - OEIS

A088445

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.

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, 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, 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

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OFFSET

1,4

COMMENTS

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.

The scatter plot shows some interesting features. - Antti Karttunen, May 08 2022

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..20000

Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Index entries for primes, gaps between

FORMULA

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

PROG

(PARI)

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.

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

A088445(n) = (n/A088444(n)); \\ Antti Karttunen, May 08 2022

CROSSREFS

Cf. A088444.

Sequence in context: A079786 A032451 A214494 * A280696 A293247 A020653

Adjacent sequences: A088442 A088443 A088444 * A088446 A088447 A088448

KEYWORD

nonn,look

AUTHOR

Reinhard Zumkeller, Sep 30 2003

STATUS

approved