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A086484
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Let k be the largest number such that n is a k-th power; then a(n) is the least positive number m such that m + n is a (k+1)st power.
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1
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1, 2, 1, 4, 4, 3, 2, 8, 18, 6, 5, 4, 3, 2, 1, 16, 8, 7, 6, 5, 4, 3, 2, 1, 2, 10, 54, 8, 7, 6, 5, 32, 3, 2, 1, 28, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 64, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 162, 18, 17, 16, 15, 14, 13, 12, 11, 10
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OFFSET
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1,2
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LINKS
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EXAMPLE
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For n = 3, k = 1 as 3 = 3^1 is a 1st power (the gcd of the exponents is 1). Hence we look for m such that m + 3 is a (1+1)th power, i.e., a 2nd power. The next 2nd power after 3 is 4 and so m + 3 = 4 so a(n) = m = 4-3 = 1.
For n = 36, k = 2 as 36 = 2^2 * 3^2 is a 2nd power (the gcd of the exponents is 2). Hence we look for m such that m + 36 is a (2+1)th power, i.e., a 3rd power. The next 3rd power after 36 is 64 and so m + 36 = 64 so a(n) = m = 64 - 36 = 28. (End)
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PROG
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(PARI) print1(1", "); for (n = 2, 100, f = factor(n); g = f[1, 2]; for (i = 2, matsize(f)[1], g = gcd(g, f[i, 2])); x = sqrtn(n, g+1); print1(round(ceil(x))^(g + 1) - n, ", ")); \\ David Wasserman, Mar 07 2005
(PARI) a(n) = {my(f = factor(n), k = gcd(f[, 2])); (sqrtnint(n, k+1) + 1)^(k + 1) - n} \\ David A. Corneth, Sep 24 2023
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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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