

A237043


Numbers n such that 2^n  1 is not squarefree, but 2^d  1 is squarefree for every proper divisor d of n.


12



6, 20, 21, 110, 136, 155, 253, 364, 602, 657, 812, 889, 979, 1081
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OFFSET

1,1


COMMENTS

Primitive elements of A049094: the elements of A049094 are precisely the positive multiples of members of this sequence.
If p^2 divides 2^n  1 for some odd prime p, then by definition the multiplicative order of 2 mod p^2 divides n. The multiplicative order of 2 mod p^2 is p times the multiplicative order of 2 mod p unless p is a Wieferich prime, in which case the two orders are identical. Hence either p is a Wieferich prime or p*log_2(p+1) <= n. This should allow finding larger members of this sequence.  Charles R Greathouse IV, Feb 04 2014
If n is in the sequence and m>1 then m*n is not in the sequence. Because n is a proper divisor of m*n and 2^n1 is not squarefree.  Farideh Firoozbakht, Feb 11 2014
a(15)>=1207.  Max Alekseyev, Sep 28 2015


LINKS

Table of n, a(n) for n=1..14.


PROG

(PARI) default(factor_add_primes, 1);
isA049094(n)=my(f=factor(n>>valuation(n, 2))[, 1], N, o); for(i=1, #f, if(n%(f[i]1) == 0, return(1))); N=2^n1; fordiv(n, d, f=factor(2^d1)[, 1]; for(i=1, #f, if(d==n, return(!issquarefree(N))); o=valuation(N, f[i]); if(o>1, return(1)); N/=f[i]^o))
is(n)=fordiv(n, d, if(isA049094(d), return(d==n))); 0
(PARI) \\ Simpler but slow
is(n)=fordiv(n, d, if(!issquarefree(2^d1), return(d==n))); 0


CROSSREFS

Cf. A049094, A005420, A065069, A282631, A282632.
Sequence in context: A222604 A112809 A289659 * A243905 A062017 A103678
Adjacent sequences: A237040 A237041 A237042 * A237044 A237045 A237046


KEYWORD

nonn,hard,more


AUTHOR

Charles R Greathouse IV, Feb 02 2014


EXTENSIONS

a(14) from Charles R Greathouse IV, Sep 21 2015, following Womack's factorization of 2^9911.


STATUS

approved



