A168264 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.)
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
Offset: 1
Keywords
Examples
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.
Links
- Anonymous?, Polynomial Calculator
- Eric Weisstein's World of Mathematics, Distinct Prime Factors
- G. Xiao, WIMS server, Factoris (both expands and factors polynomials)
Crossrefs
Formula
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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