A339056 Smallest integer k such that d(k^2)/d(k) = 2n-1, where d(k) is the number of divisors of k.

1, 144, 3600, 1511654400000000, 1587600, 13168189440000, 177844628505600000000, 192099600, 76839840000, 4757720360193884160000, 439167347942400000000, 5037669383908052497858560000000000, 32464832400, 811620810000, 831099709440000

Offset: 1

Author

Bernard Schott, Nov 25 2020

Keywords

• nonn

Comments

This sequence is related to the 3rd problem of the 39th International Mathematical Olympiad in 1998 at Taipei (Taiwan) (see link IMO).

As the set of integer values of the ratio d(k^2)/d(k) is exactly the set of all positive odd integers (see Marcin E. Kuczma reference), there exists, for each odd number, a smallest number k for which d(k^2)/d(k) = 2n-1.

All terms are perfect squares and if a number k is such that d(k^2)/d(k) = m, then all numbers that have the same prime signature as k give the same ratio m (see examples below); nevertheless, numbers with other prime signatures can also give this same ratio m (see example a(4)).

More results found by Amiram Eldar:

a(16) > 3*10^46,

a(17) = 13194538987069440000,

a(18) = 74219281802265600000000,

a(19) = 31164973305898534502400000000000000,

a(20) = 440046121805632742400000000,

a(21) = 439167347942400000000,

a(22) > 3*10^46.

Examples

			All numbers k with prime signature = [4, 2] give a ratio d(k^2)/d(k) = (9*5)/(5*3) = 3, and the smallest one is a(2) = 2^4*3^2 = 144.
All numbers k with prime signature = [4, 2, 2] give a ratio d(k^2)/d(k) = (9*5*5)/(5*3*3) = 5, and the smallest one is a(3) = 2^4*3^2*5^2 = 3600.
All numbers k with prime signature = [16, 10, 8] or [24, 12, 6] or [38, 10, 6] give the same ratio d(k^2)/d(k) = (33*21*17)/(17*11*9) = (49*25*13)/(25*13*7) = (77*21*13)/(39*11*7) = 7, but the smallest one is a(4) = 1511654400000000 = 2^16*3^10*5^8 < 2^24*3^12*5^6 < 2^38*3^10*5^6.
The successive prime signatures of the first ten terms are [], [4, 2], [4, 2, 2], [16, 10, 8], [4, 4, 2, 2], [16, 8, 4, 2], [16, 10, 8, 6], [4, 4, 2, 2, 2], [8, 4, 4, 2, 2], [28, 14, 4, 2, 2].
		

References

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

Links

• The IMO Compendium, Problem 3, 39th IMO 1998.
• 39th International Mathematical Olympiad, Problems and solutions, Taiwan, July 1998.
• Index to sequences related to Olympiads.

Crossrefs

Cf. A000005, A025487, A048691, A339055.

Subsequence of A217584.

Programs

PARI

isok(k, n) = numdiv(k^2)/numdiv(k) == n;
a(n) = my(k=1, m=2*n-1); while (!isok(k^2, m), k++); k^2; \\ Michel Marcus, Nov 26 2020

Extensions

a(12) corrected by Amiram Eldar, Nov 26 2020