A339056 -id:A339056 - cp's OEIS frontend

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

1, 144, 324, 400, 784, 1936, 2025, 2500, 2704, 3600, 3969, 4624, 5625, 5776, 7056, 8100, 8464, 9604, 9801, 13456, 13689, 15376, 15876, 17424, 19600, 21609, 21904, 22500, 23409, 24336, 26896, 29241, 29584, 30625, 35344, 39204, 41616, 42849, 44944, 48400, 51984

Offset: 1

Michel Marcus, Oct 07 2012

nonn

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).
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
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
From Bernard Schott, May 29 2020 and Nov 22 2020: (Start)
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].
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:
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).
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).
Some examples:
For numbers with prime signature = (4, 2) (A189988), the ratio is 3 and the smallest such integer is 144 = 2^4 * 3^2.
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.
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.
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(1^2)/d(1) = d(1)/d(1) = 1 an integer, so 1 belongs to the sequence.
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.

Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 134-135.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..100 from Paolo P. Lava)
The IMO Compendium, Problem 3, 39th IMO 1998.
Kin Y. Li, Problem 3, Mathematical Excalibur, Vol. 4, No. 3, Jan.-Mar. 1999.
Index to sequences related to Olympiads.

Cf. A000005, A003593, A025487, A048691.
Subsequences: A189988 (d(k^2)/d(k) = 3), A179746 (d(k^2)/d(k) = 5).
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).

Select[Range[1000], IntegerQ[DivisorSigma[0, #^2]/DivisorSigma[0, #]] &] (* Alonso del Arte, Oct 07 2012 *)
Select[Range[228]^2, Divisible[DivisorSigma[0, #^2], DivisorSigma[0, #]] &] (* Amiram Eldar, May 23 2020 *)

dn2dn(n)= {for (i=1, n, if (denominator(numdiv(i^2)/numdiv(i))==1, print1(i,", ");););}

A339055 Values taken by d(k^2)/d(k) where d(k) is the number of divisors of k and when this ratio is an integer.

Original entry on oeis.org

1, 3, 3, 3, 3, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 5, 3, 3, 3, 3, 3, 3, 5, 5, 5, 3, 3, 5, 3, 5, 3, 3, 3, 3, 3, 5, 5, 3, 3, 5, 5, 5, 3, 3, 3, 3, 5, 3, 3, 3, 5, 3, 3, 3, 5, 5, 5, 5, 3, 3, 3, 3, 3, 5, 5, 5, 5, 3, 3, 5, 3, 5, 5, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 5, 3, 5, 3, 5

Offset: 1

Bernard Schott, Nov 22 2020

nonn

This sequence was the subject of the 3rd problem, proposed by Belarus during the 39th International Mathematical Olympiad in 1998 at Taipei (Taiwan) [see the link IMO].
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); now, by a fine induction, we prove that every positive odd integer is a product of fractions of type (2u+1)/(u+1). Hence, the set of possible integer values of the data coincides with the set of all positive odd integers [see Marcin E. Kuczma reference]. The smallest integers k such that d(k^2)/d(k) = n-th odd integer are in A339056.
a(1) = 1 then from a(2) to a(234) the ratio takes only the values 3 and 5.
a(n) = 3 for numbers k whose prime signature is (4, 2) and the smallest such integer is 144 = 2^4 * 3^2 corresponding to a(2) = 3.
a(n) = 5 for numbers k whose prime signature is (4, 2, 2) and the smallest such integer is 3600 = 2^4 * 3^2 * 5^2 corresponding to a(10) = 5.
a(n) = 9 for numbers k whose prime signature is (4, 4, 2, 2) and the smallest such integer is 1587600 = 2^4 * 3^4 * 5^2 * 7^2 corresponding to a(235) = 9.

			The 4th number k such that d(k^2)/d(k) is an integer is A217584(4) = 400, 400 has 15 divisors and 400^2 = 160000 has 45 divisors, so, a(4) = 45/15 = 3.

Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 134-135.

The IMO Compendium, Problem 3, 39th IMO 1998.
Index to sequences related to Olympiads.

Cf. A000005, A005408, A025487, A048691, A217584, A339056.

for n from 1 to 600 do
q:= tau(n^4)/tau(n^2);
if q = floor(q) then print(q); else fi; od:

Select[DivisorSigma[0, #^2]/DivisorSigma[0, #] & /@ Range[10^5], IntegerQ] (* Amiram Eldar, Nov 23 2020 *)

lista(nn) = {my(q); for (n=1, nn, if (denominator(q=numdiv(n^2)/numdiv(n)) == 1, print1(q, ", ")););}
lista(100000) \\ Michel Marcus, Nov 23 2020

a(n) = A000005(A217584(n)^2)/A000005(A217584(n)).

cp's OEIS Frontend

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

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