cp's OEIS Frontend

The primitive terms of A381740. Each term of A381740 is a multiple of a term in this sequence.

The primitive terms of A381738. Each term of A381738 is a multiple of a term in this sequence.

The least odd term is a(105) = 105.

The least term that is coprime to 6 is a(3637276) = 37182145.

If k is a squarefree number that is divisible by 6, 10 or 14, then it is a term. Therefore a lower bound for the asymptotic density of this sequence is 29/(192*zeta(2)) = 0.0918... .

The numbers of terms that do not exceed 10^k, for k = 1, 2, ..., are 2, 8, 99, 972, 9826, 97610, 979190, 9770801, 97650638, 976893969, ... . Apparently, the asymptotic density of this sequence exists and equals 0.0976... .

If k is a term then any multiple of k that is squarefree is a term. The primitive terms are in A381741.

Numbers k such that k^2 is primitive abundant number (A091191).

If p is an odd Mersenne exponent (A174265), then 2^((p-1)/2) * (2^p-1) is a term.

The numbers of terms that do not exceed 10^k, for k = 3, 4, ..., are 5, 47, 459, 4655, 46733, 460693, 4612685, 46177602, ... . Apparently, the asymptotic density of this sequence exists and equals 0.00461... .

Are there numbers k such that k^2, (k+1)^2, (k+2)^2 and (k+3)^2 are all abundant numbers? There are none below 2.5*10^10.

Are there odd terms in this sequence? There are none below 2.5*10^10.

From David A. Corneth, Apr 26 2025: (Start)

If it exists then it is at least sqrt(A002110(24)/2 * 155925 + 1) - 1 ~= 4.3*10^19.

Proof: Exactly two of k, k+1, k+2 and k+3 are odd. Those two are coprime and differ by 2. Let them be m and m+2. Then sigma(m) > 2*m and sigma(m+2) > 2*(m+2). As they are coprime we have sigma(m*(m+2)) > 2*m*2*(m+2) so for a lower bound we look for the smallest odd t that sigma(t) > 4*t. The partial product of p / (p-1) for odd primes p first exceeds 4 when odd primes <= 79 are multiplied so t is divisible by 3 * 5 * 7 * ... * 79. A small search of multiples of this number gives A002110(24)/2 * 155925.

k * (k + 2) >= A002110(24)/2 * 155925 so k * (k + 2) + 1 = (k + 1)^2 >= A002110(24)/2 * 155925 + 1. Taking square roots on both sides and keeping the positive root gives the desired lower bound. (End)

From Yifan Xie, Apr 30 2025: (Start)

Both types of numbers exist, but the constructed ones are too large to be displayed here. For numbers k such that k^2, (k+1)^2, (k+2)^2 and (k+3)^2 are all abundant numbers, choose 4 disjoint subsets of the primes P_1, P_2, P_3 and P_4, and let the product of elements in P_i divide k+i-1. This is achievable because of the Chinese remainder theorem. If P_i contains p_1, ..., p_k, then sigma((k+i-1)^2)/(k+i-1)^2 >= Product_{i=1..k} (p_i+1)/p_i.

We are able to make the right hand side larger than 2 for each i because the infinite product Product_{p is prime} (p+1)/p = Product_{p is prime} (1+1/p) = Sum_{k is squarefree} 1/k diverges, since the squarefree numbers have asymptotic density 6/Pi^2.

For odd terms in this sequence, we can use only the odd primes to construct 3 prime subsets instead, and add a constraint that k == 1 (mod 2) after which the Chinese remainder theorem still applies. (End)

The indices of records of this sequence are the highly composite numbers (A002182) without 2 and 4.