|
| |
|
|
A075365
|
|
Smallest k such that (n+1)(n+2)...(n+k) is divisible by the product of all the primes up to n.
|
|
4
|
|
|
|
0, 2, 3, 2, 5, 4, 7, 6, 5, 5, 11, 10, 13, 12, 11, 10, 17, 16, 19, 18, 17, 16, 23, 22, 21, 20, 19, 18, 29, 28, 31, 30, 29, 28, 27, 26, 37, 36, 35, 34, 41, 40, 43, 42, 41, 40, 47, 46, 45, 44, 43, 42, 53, 52, 51, 50, 49, 48, 59, 58, 61, 60, 59, 58, 57, 56, 67, 66, 65, 64, 71, 70
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
LINKS
|
T. D. Noe, Table of n, a(n) for n=1..1000
|
|
|
FORMULA
|
If p <= n < q, where p and q are consecutive primes, then a(n) = 2p-n, unless n=10. Sketch of proof: a(n) >= 2p-n, to make (n+1)...(n+a(n)) divisible by p. If r is a prime less than p and r does not divide (n+1)...(2p), then r > 2p-n and 2r <= n, so 4p < 3n < 3q. But q/p is known to be < 4/3 for all primes p >= 11.
|
|
|
EXAMPLE
|
a(6) = 4 as (6+1)*(6+2)*(6+3)*(6+4) is divisible by 2*3*5 but (6+1)*(6+2)*(6+3) is not.
|
|
|
MATHEMATICA
|
a[n_] := Module[{div, k, pr}, div=Times@@Prime/@Range[PrimePi[n]]; For[k=0; pr=1, True, k++; pr*=n+k, If[Mod[pr, div]==0, Return[k]]]]
|
|
|
CROSSREFS
|
Cf. A075366, A075367, A075368.
Sequence in context: A072969 A139712 A175856 * A075274 A178144 A135737
Adjacent sequences: A075362 A075363 A075364 * A075366 A075367 A075368
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Sep 20 2002
|
|
|
EXTENSIONS
|
Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), Oct 28 2002
|
|
|
STATUS
|
approved
|
| |
|
|
