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%I #16 Sep 24 2023 05:45:57
%S 1,2,9,8,625,18,117649,128,6561,1250,25937424601,72,23298085122481,
%T 235298,5625,32768,48661191875666868481,13122,
%U 104127350297911241532841,5000,1058841,51874849202,907846434775996175406740561329,1152
%N If n = (p_1)^(m_1)...(p_k)^(m_k) then a(n) = (p_1)^((p_1)^(m_1) - 1)...(p_k)((p_k)^(m_k) - 1).
%C These integers are refactorable: the number of divisors divides the number itself.
%H Amiram Eldar, <a href="/A036879/b036879.txt">Table of n, a(n) for n = 1..388</a>
%H Simon Colton, <a href="http://www.cs.uwaterloo.ca/journals/JIS/colton/joisol.html">Refactorable Numbers - A Machine Invention</a>, J. Integer Sequences, Vol. 2 (1999), Article 99.1.2.
%H Simon Colton, <a href="http://web.archive.org/web/20070831060523/http://www.dai.ed.ac.uk/homes/simonco/research/hr/">HR - Automatic Theory Formation in Pure Mathematics</a>. [Wayback Machine copy]
%F (p_1)^(m_1)...(p_k)^(m_k) -> (p_1)^((p_1)^(m_1) - 1)...(p_k)((p_k)^(m_k) - 1).
%e a(6) = 18 because 6 = 2^(1)3^(1) -> 2^(2^(1) - 1)3^(3^(1) - 1) = 18.
%t f[p_, e_] := p^(p^e-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 24] (* _Amiram Eldar_, Sep 24 2023 *)
%o (PARI) a(n) = my(f = factor(n)); for (i=1, #f~, f[i,2] = f[i,1]^f[i,2] - 1); factorback(f); \\ _Michel Marcus_, Dec 08 2014
%Y Cf. A033950.
%K nonn,mult
%O 1,2
%A Simon Colton (simonco(AT)cs.york.ac.uk)
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