|
| |
|
|
A137678
|
|
Consider the consecutive composite numbers between prime(n) and prime(n+1). Letting k=prime(n+1)-prime(n)-1, a(n) is the number of these numbers that have all primes factors less than k.
|
|
1
|
|
|
|
0, 0, 0, 1, 0, 1, 0, 0, 2, 0, 2, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 4, 1, 0, 0, 3, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 1, 3, 0, 0, 0, 3, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 4, 0, 0, 0, 0, 0, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,9
|
|
|
COMMENTS
|
Grimm's conjecture states that for k consecutive composite numbers there are k distinct prime numbers such that each prime divides one of the composite numbers. As pointed out by Grimm, for a composite number c having a prime factor p>=k, we associate p with c. Hence there are only a(n) numbers remaining between prime(n) and prime(n+1) for which we need to associate a distinct prime. Puzzle 430 gives a heuristic algorithm for finding those a(n) primes. According to Puzzle 430, the largest known value of a(n) is 10, which occurs between the primes 31397 and 31469.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(4)=1 because of the 3 numbers between 7 and 11 (8=2^3, 9=3^2, 10=2*5), only 8 has all of its prime factors less than 3. Similarly, a(9)=2 because of the 5 numbers between 23 and 29, only 24 and 27 have all prime factors less than 5.
|
|
|
MATHEMATICA
|
Table[p1=Prime[i]; p2=Prime[i+1]; cnt=0; k=p2-p1-1; Do[If[FactorInteger[n][[ -1, 1]]<k, cnt++ ], {n, p1+1, p2-1}]; cnt, {i, 150}]
|
|
|
PROG
|
(PARI) a(n) = my(p = prime(n), q = nextprime(p+1)); sum(c=p+1, q-1, vecmax(factor(c)[, 1]) < q-p-1); \\ Michel Marcus, Sep 30 2017
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
