

A128708


Let j(n) be the Jacobsthal function, A048669. Then a(n) is the number of times that the gap j(n) appears between consecutive numbers <= n+1 and coprime to n.


2



1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 2, 8, 1, 3, 1, 2, 2, 1, 1, 4, 5, 1, 9, 2, 1, 2, 1, 16, 2, 1, 2, 6, 1, 1, 2, 4, 1, 2, 1, 2, 6, 1, 1, 8, 7, 5, 2, 2, 1, 9, 2, 4, 2, 1, 1, 4, 1, 1, 6, 32, 2, 2, 1, 2, 2, 2, 1, 12, 1, 1, 10, 2, 2, 2, 1, 8, 27, 1, 1, 4, 2, 1, 2, 4, 1, 6, 2, 2, 2, 1, 2, 16, 1, 7, 6, 10
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OFFSET

1,4


COMMENTS

Differs from A087653 starting at n=35. For prime n and e>0, a(n^e)=n^(e1). The closelyrelated sequence A128707 satisfies the inequality a(n)*A128707(n) <= n1, with equality for prime n. If m is the squarefree kernel of n (A007947), then a(n)/a(m) = n/m.


LINKS

T. D. Noe, Table of n, a(n) for n=1..10000


EXAMPLE

The numbers coprime to 15 are 1,2,4,7,8,11,13,14,16,17,19,22,... Observe that the differences are periodic: 1,2,3,1,3,2,1,2,1,2,3,... The maximum value is 3, which occurs twice in the first period. Hence a(15)=2.


MATHEMATICA

JacobsthalCount[n_] := Module[{g, d, mx}, g=Select[Range[n+1], GCD[n, # ]==1&]; d=Rest[g]Most[g]; mx=Max@@d; Count[d, mx]]; Table[JacobsthalCount[n], {n, 100}]


CROSSREFS

Sequence in context: A000190 A003557 A073752 * A087653 A295666 A322020
Adjacent sequences: A128705 A128706 A128707 * A128709 A128710 A128711


KEYWORD

nonn


AUTHOR

T. D. Noe, Mar 24 2007


STATUS

approved



