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A217584 Numbers k such that d(k^2)/d(k) is an integer, where d(k) is the number of divisors of k.

%I A217584 #105 Nov 27 2020 11:18:37
%S A217584 1,144,324,400,784,1936,2025,2500,2704,3600,3969,4624,5625,5776,7056,
%T A217584 8100,8464,9604,9801,13456,13689,15376,15876,17424,19600,21609,21904,
%U A217584 22500,23409,24336,26896,29241,29584,30625,35344,39204,41616,42849,44944,48400,51984
%N A217584 Numbers k such that d(k^2)/d(k) is an integer, where d(k) is the number of divisors of k.
%C A217584 The ratio d(k^2)/d(k) is: 1 for the number 1, 3 for numbers of the form p^4*q^2, 5 for numbers of the form p^4*q^2*r^2 (p, q, r being different primes).
%C A217584 Primes can't be in the sequence. A prime p has two divisors, while p^2 has three divisors: 1, p, p^2. - _Alonso del Arte_, Oct 07 2012
%C A217584 All the terms are squares since d(m) is odd if and only if m is a square, so d(k^2) is odd and since d(k)|d(k^2), d(k) is also odd, so k is a square. The ratio d(k^2)/d(k) can take values other than 1, 3, and 5: 1587600 is the least term with a ratio 9, and 192099600 is the least term with a ratio 15. - _Amiram Eldar_, May 23 2020
%C A217584 From _Bernard Schott_, May 29 2020 and Nov 22 2020: (Start)
%C A217584 This sequence comes from the 3rd problem, proposed by Belarus, during the 39th International Mathematical Olympiad in 1998 at Taipei (Taiwan) [see the link IMO].
%C A217584 If the prime signature of k is (u_1, u_2, ... , u_q) then d(k^2)/d(k) = Product_{i=1..q} (2*u_i+1)/(u_i+1). Two results:
%C A217584 1) If k is a term such that d(k^2)/d(k) = m, then all numbers that have the same prime signature of k are also terms and give the same ratio (see examples below).
%C A217584 2) The set of the integer values of the ratio d(k^2)/d(k) is exactly the set of all positive odd integers (see Marcin E. Kuczma reference).
%C A217584 Some examples:
%C A217584 For numbers with prime signature = (4, 2) (A189988), the ratio is 3 and the smallest such integer is 144 = 2^4 * 3^2.
%C A217584 For numbers with prime signature = (4, 2, 2) (A179746), the ratio is 5 and the smallest such integer is 3600 = 2^4 * 3^2 * 5^2.
%C A217584 For numbers with prime signature = (4, 4, 2, 2) the ratio is 9 and the smallest such integer is 1587600 = 2^4 * 3^4 * 5^2 * 7^2.
%C A217584 For numbers with prime signature = (8, 4, 4, 2, 2) the ratio is 17 and the smallest such integer is 76839840000 = 2^8 * 3^4 * 5^4 * 7^2 * 11^2 (found by _David A. Corneth_ with other prime signatures). (End)
%D A217584 Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 134-135.
%H A217584 Amiram Eldar, <a href="/A217584/b217584.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..100 from Paolo P. Lava)
%H A217584 The IMO Compendium, <a href="https://imomath.com/othercomp/I/Imo1998.pdf">Problem 3</a>, 39th IMO 1998.
%H A217584 Kin Y. Li, <a href="http://www.math.ust.hk/excalibur/v4_n3.pdf">Problem 3</a>, Mathematical Excalibur, Vol. 4, No. 3, Jan.-Mar. 1999.
%H A217584 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%e A217584 d(1^2)/d(1) = d(1)/d(1) = 1 an integer, so 1 belongs to the sequence.
%e A217584 144^2 has 45 divisors: 1, 2, 3, 4, 6, 8, 9, 12, ..., 20736, while 144 has 15 divisors: 1, 2, 3, 4, 6, 8, 9, 12, ..., 144; 45/15 = 3 and so 144 is in the sequence.
%t A217584 Select[Range[1000], IntegerQ[DivisorSigma[0, #^2]/DivisorSigma[0, #]] &] (* _Alonso del Arte_, Oct 07 2012 *)
%t A217584 Select[Range[228]^2, Divisible[DivisorSigma[0, #^2], DivisorSigma[0, #]] &] (* _Amiram Eldar_, May 23 2020 *)
%o A217584 (PARI) dn2dn(n)= {for (i=1, n, if (denominator(numdiv(i^2)/numdiv(i))==1, print1(i,", ");););}
%Y A217584 Cf. A000005, A003593, A025487, A048691.
%Y A217584 Subsequences: A189988 (d(k^2)/d(k) = 3), A179746 (d(k^2)/d(k) = 5).
%Y A217584 Cf. A339055 (values taken by d(a(n)^2)/d(a(n))), A339056 (smallest k such that d(k^2)/d(k) = n-th odd).
%K A217584 nonn
%O A217584 1,2
%A A217584 _Michel Marcus_, Oct 07 2012