cp's OEIS Frontend

A159907 Numbers m with half-integral abundancy index, sigma(m)/m = k+1/2 with integer k.

%I A159907 #49 Jun 06 2025 03:30:05
%S A159907 2,24,4320,4680,26208,8910720,17428320,20427264,91963648,197064960,
%T A159907 8583644160,10200236032,21857648640,57575890944,57629644800,
%U A159907 206166804480,17116004505600,1416963251404800,15338300494970880,75462255348480000,88898072401645056,301183421949935616,6219051710415667200
%N A159907 Numbers m with half-integral abundancy index, sigma(m)/m = k+1/2 with integer k.
%C A159907 Obviously, all terms must be even (cf. formula), but e.g. a(9) and a(12) are not divisible by 3. See A007691 for numbers with integral abundancy.
%C A159907 Odd numbers and higher powers of 2 cannot be in the sequence; 6 is in A000396 and thus in A007691, and n=10,12,14,18,20,22 don't have integral 2*sigma(n)/n.
%C A159907 Conjecture: with number 1, multiply-anti-perfect numbers m: m divides antisigma(m) = A024816(m). Sequence of fractions antisigma(m) / m: {0, 0, 10, 2157, 2337, 13101, 4455356, ...}. - _Jaroslav Krizek_, Jul 21 2011
%C A159907 The above conjecture is equivalent to the conjecture that there are no odd multiply perfect numbers (A007691) greater than 1. Proof: (sigma(n)+antisigma(n))/n = (n+1)/2 for all n. If n is even then sigma(n)/n is a half-integer if and only if antisigma(n)/n is an integer. Since all members of this sequence are known to be even, the only way the conjecture can fail is if antisigma(n)/n is an integer, in which case sigma(n)/n is an integer as well. - _Nathaniel Johnston_, Jul 23 2011
%C A159907 These numbers are called hemiperfect numbers. See Numericana & Wikipedia links. - _Michel Marcus_, Nov 19 2017
%H A159907 Max Alekseyev, <a href="/A159907/b159907.txt">Table of n, a(n) for n = 1..130</a>
%H A159907 G. P. Michon, <a href="http://www.numericana.com/answer/numbers.htm#multiperfect">Multiperfect Numbers & Hemiperfect Numbers</a>, Numericana.
%H A159907 Walter Nissen, <a href="http://upforthecount.com/math/abundance.html">Abundancy : Some Resources </a>.
%H A159907 Project Euler, <a href="https://projecteuler.net/problem=241">Problem 241: Perfection Quotients</a>.
%H A159907 Wikipedia, <a href="https://en.wikipedia.org/wiki/Hemiperfect_number">Hemiperfect number</a>
%F A159907 A159907 = { n | 2*A000203(n) is in n*A005408 } = { n | A054024(n) = n/2 }
%e A159907 a(1) = 2 since sigma(2)/2 = (1+2)/2 = 3/2 is of the form k+1/2 with integer k=1.
%e A159907 a(2) = 24 is in the sequence since sigma(24)/24 = (1+2+3+4+6+8+12+24)/24 = (24+12+24)/24 = k+1/2 with integer k=2.
%o A159907 (PARI) isok(n) = denominator(sigma(n,-1)) == 2; \\ _Michel Marcus_, Sep 19 2015
%o A159907 (PARI) forfactored(n=1,10^7, if(denominator(sigma(n,-1))==2, print1(n[1]", "))) \\ _Charles R Greathouse IV_, May 09 2017
%o A159907 (Python)
%o A159907 from fractions import Fraction
%o A159907 from sympy import divisor_sigma as sigma
%o A159907 def aupto(limit):
%o A159907   for k in range(1, limit):
%o A159907     if Fraction(int(sigma(k, 1)), k).denominator == 2:
%o A159907       print(k, end=", ")
%o A159907 aupto(3*10**4) # _Michael S. Branicky_, Feb 24 2021
%Y A159907 Cf. A000203, A088912, A141643 (k=2), A055153 (k=3), A141645 (k=4), A159271 (k=5).
%K A159907 nonn
%O A159907 1,1
%A A159907 _M. F. Hasler_, Apr 25 2009
%E A159907 Terms a(20) onward from _Max Alekseyev_, Jun 05 2025