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A048943
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Product of divisors of n is a square.
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0
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1, 6, 8, 10, 14, 15, 16, 21, 22, 24, 26, 27, 30, 33, 34, 35, 38, 39, 40, 42, 46, 51, 54, 55, 56, 57, 58, 60, 62, 65, 66, 69, 70, 72, 74, 77, 78, 81, 82, 84, 85, 86, 87, 88, 90, 91, 93, 94, 95, 96, 102, 104, 105, 106, 108, 110, 111, 114, 115, 118, 119, 120, 122, 123, 125
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Contribution from Gerard P. Michon (g.michon(AT)att.net), Oct 10 2010: (Start)
If d is the number of divisors of N, a prime factor of N with multiplicity k in N has a multiplicity kd/2 in the product of all divisors of N (including N itself). Therefore the latter is a square if and only if kd/2 is always even (which is to say that kd is a multiple of 4 for any multiplicity k of a prime factor of N). This happens when one of the following three conditions hold:
- N is a fourth power (all the multiplicities are then multiples of 4 and d is odd).
- N has at least two prime factors with odd multiplicities.
- N has (at least) one prime factor with a multiplicity congruent to 3 modulo 4.
(End)
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LINKS
| Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
G. P. Michon, Divisor Product, Numericana. [From Gerard P. Michon (g.michon(AT)att.net), Oct 10 2010]
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EXAMPLE
| Contribution from Gerard P. Michon (g.michon(AT)att.net), Oct 10 2010: (Start)
a(1) = 1 because it's a fourth power. The product of all divisors of 1 is 1, which is a square.
a(2) = 6 because 2^1.3^1 is the product of two primes with odd multiplicities (1 in both cases). Indeed, the divisor product 1.2.3.6 = 36 is a square.
a(3) = 8 because 2 is a prime factor of 8 with multiplicity 3. Indeed, 1.2.4.8 = 64 is a square.
a(7) = 16 because it's a fourth power; 1.2.4.8.16 = 1024 is the square of 32. (End)
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CROSSREFS
| Sequence in context: A181764 A153032 A086822 * A130763 A120497 A036436
Adjacent sequences: A048940 A048941 A048942 * A048944 A048945 A048946
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KEYWORD
| nonn
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AUTHOR
| Eric Weisstein (eric(AT)weisstein.com)
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