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A208728
Composite numbers n such that b^(n+1) == 1 (mod n) for every b coprime to n.
46
15, 35, 255, 455, 1295, 2703, 4355, 6479, 9215, 10439, 11951, 16211, 23435, 27839, 44099, 47519, 47879, 62567, 63167, 65535, 93023, 94535, 104195, 120959, 131327, 133055, 141155, 142883, 157079, 170819, 196811, 207935, 260831, 283679, 430199, 560735, 576719
OFFSET
1,1
COMMENTS
GCD(b,n)=1 and b^(n+1) == 1 (mod n).
The sequence lists the squarefree composite numbers n such that every prime divisor p of n satisfies (p-1)|(n+1) (similar to Korselt's criterion).
The sequence can be considered as an extension of k-Knödel numbers to k negative, in this case equal to -1.
Numbers n > 3 such that b^(n+2) == b (mod n) for every integer b. Also, numbers n > 3 such that A002322(n) divides n+1. Are there infinitely many such numbers? It seems that such numbers n > 35 have at least three prime factors. - Thomas Ordowski, Jun 25 2017
Proof that 15 and 35 are the only numbers with only two prime factors (and so all others have >= 3): Since n is squarefree and composite, it has at least two prime factors, p and q. If these are the only two, n = p*q. Then the criterion is that (p-1)|(n+1) -> (p-1)|pq+1, and q-1|pq+1. Write pq+1 = j*(p-1) = k*(q-1). Rearranging, p*(j-q)=j+1 and q*(k-p)=k+1. Since j = (pq+1)/(p-1), for large n, j~=q, and k~=p. But we see that p divides j+1~=q, and q divides k+1~=p. For large n this is only possible if p and q are roughly equal, so j-q=k-p=1. Otherwise, we have j+1 >= 2*p and k+1 >= 2*q, and which puts upper bounds on p and q. Enumerating these gives (p,q)=(3,5) and (p,q)=(5,7) as the only solutions. - Alex Meiburg, Oct 03 2024
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Carmichael Number
Eric Weisstein's World of Mathematics, Korselt's Criterion
Eric Weisstein's World of Mathematics, Knödel Numbers
EXAMPLE
6479 is part of the sequence because its prime factors are 11, 19 and 31: (6479+1)/(11-1)=648, (6479+1)/(19-1)=360 and (6479+1)/(31-1)=216.
MAPLE
with(numtheory); P:=proc(n) local d, ok, p;
if issqrfree(n) then p:=factorset(n); ok:=1;
for d from 1 to nops(p) do if frac((n+1)/(p[d]-1))>0 then ok:=0;
break; fi; od; if ok=1 then n; fi; fi; end: seq(P(i), i=5..576719);
MATHEMATICA
Select[Range[2, 576719], SquareFreeQ[#] && ! PrimeQ[#] && Union[Mod[# + 1, Transpose[FactorInteger[#]][[1]] - 1]] == {0} &] (* T. D. Noe, Mar 05 2012 *)
PROG
(PARI) is(n)=if(isprime(n)||!issquarefree(n)||n<3, return(0)); my(f=factor(n)[, 1]); for(i=1, #f, if((n+1)%(f[i]-1), return(0))); 1 \\ Charles R Greathouse IV, Mar 05 2012
KEYWORD
nonn,changed
AUTHOR
Paolo P. Lava, Mar 01 2012
EXTENSIONS
Definition corrected by Thomas Ordowski, Jun 25 2017
STATUS
approved