A128311 Remainder upon division of 2^(n-1)-1 by n.
0, 1, 0, 3, 0, 1, 0, 7, 3, 1, 0, 7, 0, 1, 3, 15, 0, 13, 0, 7, 3, 1, 0, 7, 15, 1, 12, 7, 0, 1, 0, 31, 3, 1, 8, 31, 0, 1, 3, 7, 0, 31, 0, 7, 30, 1, 0, 31, 14, 11, 3, 7, 0, 13, 48, 15, 3, 1, 0, 7, 0, 1, 3, 63, 15, 31, 0, 7, 3, 21, 0, 31, 0, 1, 33, 7, 8, 31, 0, 47, 39, 1, 0, 31, 15, 1, 3, 39, 0, 31, 63
Offset: 1
Keywords
Examples
a(1)=0 since any integer == 0 (mod 1); a(2)=1 since 2^1-1 == 1 (mod 2), a(3)=0 since 3 is a prime > 2, a(4)=3 since 2^3-1 = 7 == 3 (mod 4); a(341)=0 since 341=11*31 is a Sarrus number.
Links
- T. D. Noe, Table of n, a(n) for n = 1..2048
- Wikipedia, Fermat's little theorem
Crossrefs
Programs
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Mathematica
Table[Mod[2^(n-1)-1,n],{n,100}] (* Harvey P. Dale, Dec 22 2012 *)
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PARI
a(n)=(1<<(n-1)-1)%n
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PARI
apply( {A128311(n)=lift(Mod(2,n)^(n-1)-1)}, [1..99]) \\ Much more efficient when n becomes very large. - M. F. Hasler, Mar 13 2020
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Python
def A128311(n): return (pow(2,n-1,n)-1)%n # Chai Wah Wu, Jul 06 2022
Formula
a(n) = M(n-1) - n floor( M(n-1)/n ) = M(n-1) - max{ k in nZ | k <= M(n-1) } where M(k)=2^k-1.
Comments