A103292 Numbers k such that sigma(2^k - 1) >= 2(2^k - 1), i.e., the number 2^k - 1 is perfect or abundant.

12, 24, 36, 40, 48, 60, 72, 80, 84, 90, 96, 108, 120, 132, 140, 144, 156, 160, 168, 180, 192, 200, 204, 210, 216, 220, 228, 240, 252, 264, 270, 276, 280, 288, 300, 312, 320, 324, 330, 336, 348, 360, 372, 384, 396, 400, 408, 420, 432, 440, 444, 450, 456, 468, 480, 492, 504

Offset: 1

Author

Max Alekseyev, Jan 28 2005

Keywords

• nonn

Comments

Numbers k that 2^k - 1 is in A023196.

Are there any odd terms? This is a subsequence of A103291. Is the number 1 the only term where they differ? This is so if there is no least deficient number of the form 2^n-1 besides 1.

For each n in the sequence, 2n is also in the sequence: sigma[2^(2n)-1] = sigma[(2^n+1)(2^n-1)] >= (2^n+1)*sigma(2^n-1) because for each divisor d|2^n-1 there is (at least) the divisor (2^n+1)d |[(2^n+1)(2^n-1)]. Inserting sigma(2^n-1) >=2(2^n-1) yields (2^n+1)*sigma(2^n-1)>=(2^n+1)*2*(2^n-1)=2*[2^(2n)-1] qed. - R. J. Mathar, Aug 07 2007

From David Wasserman, May 16 2008: (Start)

Odd members exist. One such n is the lcm of the first 4416726 members of A139686, which has 6864499 digits. To show that n is a member, it's not necessary to exactly compute sigma(2^n-1).

The function f(x) = sigma(x)/x is multiplicative and has the property that for any a, b > 1, f(ab) > f(a). So it suffices to find some y such that f(y) >= 2 and y divides 2^n-1. In this case, y is the product of the first 4416726 members of A014663 and has 35260810 digits. (A014663(4416726) = 278379727.)

To see that this works, note that if a divides b, then 2^a-1 divides 2^b-1. For 1 <= i <= 4416726, A014663(i) divides 2^A139686(i)-1 by definition and A139686(i) divides n, so 2^A139686(i)-1 divides 2^n-1 and therefore A014663(i) divides 2^n-1. Then we can compute that f(y) = Product_{i = 1..4416726} (1 + 1/A014663(i)) is > 2.

The members of A014663 are the only primes that can divide 2^n-1 with n odd. Any powers of these primes are also possible divisors.

By including powers, we can construct a much smaller y. I found a y with 7057382 digits, omega(y) = 969004 and bigomega(y) = 969440. This y is close to the minimum possible. The least n such that y divides 2^n-1 is an odd number with 1472897 digits.

However, minimizing y is not the way to minimize n. We can get a smaller n by skipping primes p such that the order of 2 mod p is divisible by a large prime. This increases the number and size of the prime factors needed to make f(y) >= 2 and the time needed to find them.

The least odd n that I've found has 28375 digits. The corresponding y has 305621222 digits, omega(y) = 31903142 and bigomega(y) = 32796897. To find these prime factors, I searched up to A014663(96433108) = 7154804519.

I believe that the smallest odd member has between 10000 and 20000 digits, but the largest lower bound I can prove has 8 digits: f(p^i) is bounded above by 1 + 1/(p-1) and Product_{i=1..c} (1 + 1/(A014663(i)-1)) < 2 if c < 968858, so y must be at least Product_{i=1..968858} A014663(i), which has 7054790 digits.

Then n must be large enough that 2^n-1 >= y, yielding a lower bound of 23435503. I don't see any way to increase this significantly. (End)

Links

• Amiram Eldar, Table of n, a(n) for n = 1..135

Crossrefs

Cf. A103288, A103289, A103291, A023196.

Cf. A014663, A139686.

Programs

PARI

for(i=1,1000,n=2^i-1;if(sigma(n)>=2*n,print(i)));

Extensions

Extended to a(32) by R. J. Mathar, Aug 07 2007

Terms from a(33) onwards from David Wasserman, May 16 2008