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A317681 a(n) = smallest m such that sigma(m) = n*m/2.

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%I A317681 #47 Jun 06 2025 03:31:19
%S A317681 1,2,6,24,120,4320,30240,8910720,14182439040,17116004505600,
%T A317681 154345556085770649600,170974031122008628879954060917200710847692800,
%U A317681 141310897947438348259849402738485523264343544818565120000,12749472205565550032020636281352368036406720997031277595140988449695952806020854579200000
%N A317681 a(n) = smallest m such that sigma(m) = n*m/2.
%C A317681 Interleaving of A007539 and A088912.
%C A317681 For even n, a(n) is a multiply perfect number; for odd n it is a hemiperfect number.
%C A317681 Note that 1 is the only number with abundancy 1, and 2 is the only number with abundancy 3/2 (in other words, 1 and 2 are solitary numbers; see A014567). For k >= 4 it is not known whether there are finitely many or infinitely many numbers with abundancy k/2. Also it is not known whether a(n) < a(n+1) always holds.
%C A317681 On the Riemann Hypothesis (RH), a(n) > exp(exp(n/(2*exp(gamma)))), where gamma = 0.5772156649... is the Euler-Mascheroni constant (A001620).
%H A317681 Max Alekseyev, <a href="/A317681/b317681.txt">Table of n, a(n) for n = 2..16</a>
%H A317681 Achim Flammenkamp, <a href="http://wwwhomes.uni-bielefeld.de/achim/mpn.html">The Multiply Perfect Numbers Page</a>
%H A317681 Fred Helenius, <a href="https://web.archive.org/web/20151210102548/http://pw1.netcom.com/~fredh/mpfn/glossary.html">A glossary of MPFN-related terms</a>.
%H A317681 G. P. Michon, <a href="http://www.numericana.com/answer/numbers.htm#multiplicative">Multiplicative functions</a>: Abundancy = sigma(n)/n
%H A317681 G. P. Michon, <a href="http://www.numericana.com/answer/numbers.htm#multiperfect">Multiperfect and hemiperfect numbers</a>
%H A317681 Walter Nissen, <a href="http://upforthecount.com/math/abundance.html">Abundancy: Some Resources </a>
%F A317681 a(2n) = A007539(n), a(2n+1) = A088912(n), n > 0.
%e A317681 a(7) = 4320 since sigma(4320) = 15120 = 7/2*4320 and 4320 is the smallest m such that sigma(m)/m = 7/2.
%t A317681 Nest[Append[#, Block[{m = #1[[-1]] + 1}, While[DivisorSigma[1, m] != #2 m/2, m++]; m]] & @@ {#, Length@ # + 2} &, {1}, 6] (* _Michael De Vlieger_, Aug 05 2018 *)
%o A317681 (PARI) for(n=2, 10, for(m=1, 10^12, if(sigma(m)/m==n/2, print1(m, ", "); break())))
%o A317681 (PARI) a(n) = my(k=1); while (sigma(k) != k*n/2, k++); k; \\ _Michel Marcus_, May 15 2025
%Y A317681 Cf. A007539, A088912, A246454.
%Y A317681 Numbers with abundancy k/2: A000396 (k=4), A141643 (k=5), A005820 (k=6), A055153 (k=7), A027687 (k=8), A141645 (k=9), A046060 (k=10), A159271 (k=11), A046061 (k=12), A160678 (k=13).
%K A317681 nonn,hard
%O A317681 2,2
%A A317681 _Jianing Song_, Aug 04 2018
%E A317681 a(15) = A088912(7) added by _Max Alekseyev_, Jun 05 2025