|
Search: id:A142973
|
|
|
| A142973 |
|
a(n) = the smallest positive integer that does not divide any integer k, where the n-th prime <= k <= the (n+1)th prime. |
|
+0 2
|
|
| 4, 6, 4, 6, 5, 6, 4, 6, 10, 4, 10, 6, 4, 6, 9, 10, 7, 10, 6, 5, 8, 6, 9, 11, 6, 4, 6, 5, 6, 16, 6, 10, 4, 15, 4, 10, 11, 6, 11, 9, 7, 12, 5, 6, 4, 18, 14, 6, 5, 6, 8, 7, 12, 10, 8, 9, 4, 9, 6, 4, 14, 22, 6, 5, 6, 21, 10, 12, 5, 6, 8, 16, 11, 10, 6, 10, 16, 6, 10, 12, 8, 12, 5, 8, 6, 9, 14, 6, 4, 6, 15
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
a(n) = A142972(n) + 1.
|
|
LINKS
|
Leroy Quet, Home Page (listed in lieu of email address)
|
|
EXAMPLE
|
The 15th prime is 47 and the 16th prime is 53. So we will consider the integers 47,48,49,50,51,52,53. Now, 1 divides each of these 6 integers. 2 divides 48, 50 and 52. 3 divides 48 and 51. 4 divides 48 and 52. 5 divides 50. 6 divides 48. 7 divides 49. 8 divides 48. But 9 does not divide any integer that is between 47 and 53. So a(15)=9, since 9 is the smallest positive integer that does not divide any integer between 47 and 53.
|
|
PROGRAM
|
Contribution from Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Apr 09 2009: (Start)
(PARI) dividesany(n, m, d)=for(k=n, m, if(k%d==0, return(1))); 0
firstnondiv(n, m)=for(d=2, m+1, if(!dividesany(n, m, d), return(d)))
vector(100, k, firstnondiv(prime(k), prime(k+1))) (End)
|
|
CROSSREFS
|
Cf. A142972.
Sequence in context: A010711 A127018 A083396 this_sequence A084892 A018835 A055166
Adjacent sequences: A142970 A142971 A142972 this_sequence A142974 A142975 A142976
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Leroy Quet Jul 14 2008
|
|
EXTENSIONS
|
More terms from Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Apr 09 2009
|
|
|
Search completed in 0.002 seconds
|