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A339056
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Smallest integer k such that d(k^2)/d(k) = 2n-1, where d(k) is the number of divisors of k.
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2
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1, 144, 3600, 1511654400000000, 1587600, 13168189440000, 177844628505600000000, 192099600, 76839840000, 4757720360193884160000, 439167347942400000000, 5037669383908052497858560000000000, 32464832400, 811620810000, 831099709440000
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OFFSET
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1,2
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COMMENTS
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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)).
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.
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REFERENCES
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Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 134-135.
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LINKS
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The IMO Compendium, Problem 3, 39th IMO 1998.
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EXAMPLE
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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].
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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