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A055468
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Composite numbers k for which phi(k) + sigma(k) is an integer multiple of the 4th power of the number of divisors of k.
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4
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121, 125, 511, 767, 895, 1535, 1919, 2047, 2559, 2815, 3071, 3199, 3327, 3455, 3711, 3839, 4223, 4351, 4479, 4607, 4735, 4863, 5262, 5631, 5726, 5759, 5902, 5966, 6014, 6527, 7167, 7295, 7423, 7679, 7807, 8063, 9599, 9727, 9819, 9983, 10239
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OFFSET
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1,1
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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 128), and q == 127 (mod 128). - 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)^4 or A000010(x) + A000203(x) = m * A000005(k)^4, where m is an integer.
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EXAMPLE
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511 is a term since it has 4 divisors, phi(511) = 432, sigma(511) = 592, and 432 + 592 = 1024 = 4 * 4^4.
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MATHEMATICA
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Select[Range[10000], CompositeQ[#] && Divisible[EulerPhi[#] + DivisorSigma[1, #], DivisorSigma[0, #]^4] &] (* Amiram Eldar, Mar 25 2024 *)
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PROG
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(PARI) is(n)=my(f=factor(n)); (eulerphi(f)+sigma(f))%numdiv(f)^4==0 && !isprime(n) \\ Charles R Greathouse IV, Mar 01 2017
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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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