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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. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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)

REFERENCES

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

LINKS

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.

EXAMPLE

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.

MATHEMATICA

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 *)

PROG

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

CROSSREFS

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).

Sequence in context: A320457 A154051 A335543 * A030633 A189988 A232892

Adjacent sequences:  A217581 A217582 A217583 * A217585 A217586 A217587

KEYWORD

nonn

AUTHOR

Michel Marcus, Oct 07 2012

STATUS

approved

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Last modified June 25 08:27 EDT 2021. Contains 345453 sequences. (Running on oeis4.)