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A055467
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Nonprime numbers k for which phi(k) + sigma(k) is an integer multiple of the cube of the number of divisors of k.
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4
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1, 95, 99, 121, 125, 159, 287, 319, 415, 447, 511, 543, 654, 671, 703, 767, 799, 831, 895, 959, 1055, 1119, 1247, 1343, 1390, 1495, 1535, 1631, 1727, 1849, 1919, 1983, 2043, 2047, 2060, 2261, 2271, 2335, 2463, 2495, 2559, 2623, 2815, 2828, 2883, 2911
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OFFSET
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1,2
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COMMENTS
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Makowski proved that phi(k) + sigma(k) = k*d(k) if and only if k is a prime (see in Sivaramakrishnan, Chapter I, page 8, Theorem 3). Generally, when phi(k) + sigma(k) = m*d(k) there are special cases in which phi(k) + sigma(k) is divisible by higher powers of the number of divisors d(k).
This sequence is infinite: it includes all the semiprimes p*q such that p == 1 (mod 32), and q == 31 (mod 32). - Amiram Eldar, Mar 25 2024
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REFERENCES
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R. Sivaramakrishnan, Classical Theory of Arithmetic Functions, Marcel Dekker, Inc., New York and Basel, 1989.
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LINKS
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C. A. Nicol, Problem E 1674, The American Mathematical Monthly, Vol. 71, No. 3 (1964), p. 317; Another characterization of prime number, Solutions to Problem E 1674 by Martin J. Cohen and J. A. Fridy, ibid., Vol. 72, No. 2 (1965), pp. 186-187.
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FORMULA
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Integer solutions of phi(x) + sigma(x) = m * d(x)^3 or A000010(x) + A000203(x) = m * A000005(x)^3, where m is an integer.
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EXAMPLE
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95 is a term since it has 4 divisors, phi(95) = 72, sigma(95) = 120, and 72 + 120 = 192 = 3 * 4^3.
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MATHEMATICA
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Select[Range[10000], ! PrimeQ[#] && Mod[EulerPhi[#] + DivisorSigma[1, #], DivisorSigma[0, #]^3] == 0 &] (* Matthew House, Dec 28 2016 *)
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PROG
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(PARI) is(n) = {my(f = factor(n)); f[, 2] != [1]~ && (eulerphi(f) + sigma(f)) % numdiv(f)^3 == 0; } \\ Amiram Eldar, Mar 25 2024
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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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