A383391 Numbers k such that k^2, (k+1)^2 and (k+2)^2 are all abundant numbers.

2924, 3794, 5564, 8294, 8414, 10064, 13454, 19304, 22154, 22814, 35684, 39974, 40544, 40754, 41768, 46214, 49994, 52064, 56264, 60884, 63854, 65624, 68354, 68474, 69068, 70244, 78974, 84824, 88604, 92168, 93224, 95354, 100694, 102464, 106028, 107084, 111110, 111824

Offset: 1

Author

Amiram Eldar, Apr 25 2025

Keywords

• nonn

Comments

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)

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Subsequence of A381738 and A383390.

Cf. A002110, A005101, A063734, A096536.

Programs

Mathematica

q[n_] := q[n] = DivisorSigma[-1, n^2] > 2; Select[Range[120000], AllTrue[# + {0, 1, 2}, q] &]

PARI

is1(k) = {my(f = factor(k)); prod(i = 1, #f~, f[i,2] *= 2); sigma(f, -1) > 2;}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, lim, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}