A354531 Numbers k such that 2*(2^k-1) is in A354525.

1, 2, 3, 5, 7, 9, 13, 17, 19, 31, 61, 67, 89, 107, 127, 137, 521, 607, 727

Offset: 1

Author

Jianing Song, Aug 16 2022

Keywords

• nonn
• hard
• more

Comments

Numbers k such that for every prime factor p of 2^k-1 we have gpf(2*(2^k-1)+p) = p.

Numbers k such that for every prime factor p of 2^k-1, 2*(2^k-1)+p is p-smooth.

All terms except 2 are odd: if k is even, then 3 is a factor of 2^k-1, so 3^m = 2*(2^k-1)+3 = 2^(k+1) + 1 => k+1 >= 3^(m-1). The only possible case is (k,m) = (2,2).

Clearly A000043 is a subsequence. The exceptional terms (1, 9, 67, 137, ...) are listed in A354532.

The next term is >= 349. The next composite term, if it exists, is >= 7921 = 89^2.

Examples

			See A354532.
		

Crossrefs

Cf. A000043, A354525, A354532, A354533, A354536.

Programs

PARI

gpf(n) = vecmax(factor(n)[, 1]);
ispsmooth(n,p,{lim=1<<256}) = if(n<=lim, n==1 || gpf(n)<=p, my(N=n/p^valuation(n,p)); forprime(q=2, p, N=N/q^valuation(N,q); if((N<=lim && isprime(N)) || N==1, return(N<=p))); 0); \\ check if n is p-smooth, using brute force if n is too large
isA354531(n,{lim=256},{p_lim=1<<32}) = {
  my(N=2^n-1);
  if(isprime(N), return(1));
  if(n>lim, forprime(p=3, p_lim, if(N%p==0 && !ispsmooth(2*N+p,p), return(0)))); \\ first check if there is a prime factor p <= p_lim of 2^n-1 such that 2*(2^n-1)+p is not p-smooth (for large n)
  my(d=divisors(n));
  for(i=1, #d, my(f=factor(2^d[i]-1)[, 1]); for(j=1, #f, if(!ispsmooth(2*N+f[j],f[j],1<
				

Extensions

a(17)-a(19) from Jinyuan Wang, Jan 21 2025