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

%I A383391 #27 May 06 2025 18:06:59
%S A383391 2924,3794,5564,8294,8414,10064,13454,19304,22154,22814,35684,39974,
%T A383391 40544,40754,41768,46214,49994,52064,56264,60884,63854,65624,68354,
%U A383391 68474,69068,70244,78974,84824,88604,92168,93224,95354,100694,102464,106028,107084,111110,111824
%N A383391 Numbers k such that k^2, (k+1)^2 and (k+2)^2 are all abundant numbers.
%C A383391 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.
%C A383391 Are there odd terms in this sequence? There are none below 2.5*10^10.
%C A383391 From _David A. Corneth_, Apr 26 2025: (Start)
%C A383391 If it exists then it is at least sqrt(A002110(24)/2 * 155925 + 1) - 1 ~= 4.3*10^19.
%C A383391 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.
%C A383391 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)
%C A383391 From _Yifan Xie_, Apr 30 2025: (Start)
%C A383391 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.
%C A383391 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.
%C A383391 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)
%H A383391 Amiram Eldar, <a href="/A383391/b383391.txt">Table of n, a(n) for n = 1..10000</a>
%t A383391 q[n_] := q[n] = DivisorSigma[-1, n^2] > 2; Select[Range[120000], AllTrue[# + {0, 1, 2}, q] &]
%o A383391 (PARI) is1(k) = {my(f = factor(k)); prod(i = 1, #f~, f[i,2] *= 2); sigma(f, -1) > 2;}
%o A383391 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);}
%Y A383391 Subsequence of A381738 and A383390.
%Y A383391 Cf. A002110, A005101, A063734, A096536.
%K A383391 nonn
%O A383391 1,1
%A A383391 _Amiram Eldar_, Apr 25 2025