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A089587
a(n) is the smallest integer k, 0<k<n, that most often satisfies the condition: k^m > k^(m+1) (modulo n) as m varies from 1 to n-1, for n>2, with a(1)=0 and a(2)=1.
1
0, 1, 2, 3, 2, 5, 4, 3, 7, 8, 3, 5, 9, 4, 8, 11, 2, 13, 11, 3, 16, 5, 6, 5, 21, 9, 19, 18, 16, 8, 16, 23, 9, 6, 16, 31, 26, 11, 20, 23, 16, 11, 36, 27, 34, 12, 6, 29, 43, 41, 26, 35, 16, 37, 43, 25, 49, 49, 3, 53, 47, 16, 32, 57, 38, 4, 37, 39, 12, 16, 57, 61, 37, 26, 61, 49, 67, 47, 55
OFFSET
1,3
COMMENTS
The values at a(2^n+1), n>=0, are given by A089588: {1,2,2,7,2,9,38,79,2,220,..} and it is conjectured that a(n)=2 only when n is a Fermat number 2^(2^j)+1.
FORMULA
a(2^(2^n)+1)=2; a(2^n-1)=2^(n-1); a(4^n-2)=4^(n-1) (conjectures).
EXAMPLE
a(7)=4 since 4 is the smallest number between 1 and 6 that has the maximum number of decreasing power residues mod 7:
n=7: k=1..6, m=1..7
k^m (mod 7).. #>'s
1=1=1=1=1=1=1:0
2<4>1<2<4>1<2:2
3>2<6>4<5>1<3:3
4>2>1<4>2>1<4:4 <- max of 4 >'s first occurs at k=4
5>4<6>2<3>1<5:3
6>1<6>1<6>1<6:3
PROG
(PARI) a(n) = local(high, lowK, count); high = 0; for (k = 1, n - 1, count = 0; for (m = 1, n - 1, if (k^m%n > k^(m+1)%n, count++)); if (count > high, high = count; lowK = k)); if(n==2, 1, lowK) \\ David Wasserman, Sep 20 2005 [a(2) corrected by Georg Fischer, Jun 20 2022]
CROSSREFS
Sequence in context: A205706 A141658 A350169 * A278327 A067316 A210221
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 09 2003
STATUS
approved