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A168264
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For all sufficiently high values of k, d(n^k) > d(m^k) for all m < n. (Let k, m, and n represent positive integers only.)
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8
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1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 420, 840, 1260, 1680, 2310, 4620, 9240, 13860, 18480, 27720, 30030, 60060, 120120, 180180, 240240, 360360, 510510, 1021020, 2042040, 3063060, 4084080, 6126120, 9699690, 19399380, 38798760, 58198140
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OFFSET
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1,2
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COMMENTS
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d(n) is the number of divisors of n (A000005(n)).
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LINKS
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G. Xiao, WIMS server, Factoris (both expands and factors polynomials)
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FORMULA
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If the canonical factorization of n into prime powers is Product p^e(p), then the formula for the number of divisors of the k-th power of n is Product_p (ek + 1). (See also A146289, A146290.)
For two positive integers m and n with different prime signatures, let j be the largest exponent of k for which m and n have different coefficients, after the above formula for each integer is expanded as a polynomial. Let m_j and n_j denote the corresponding coefficients. d(n^k) > d(m^k) for all sufficiently high values of k if and only if n_j > m_j.
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EXAMPLE
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Since the exponents in 1680's prime factorization are (4,1,1,1), the k-th power of 1680 has (4k+1)(k+1)^3 = 4k^4 + 13k^3 + 15k^2 + 7k + 1 divisors. Comparison with the analogous formulas for all smaller members of A025487 shows the following:
a) No number smaller than 1680 has a positive coefficient in its "power formula" for any exponent larger than k^4.
b) The only power formula with a k^4 coefficient as high as 4 is that for 1260 (4k^4 + 12k^3 + 13k^2 + 6k + 1).
c) The k^3 coefficient for 1680 is higher than for 1260.
So for all sufficiently high values of k, d(1680^k) > d(m^k) for all m < 1680.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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