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 A159907 Numbers n with half-integral abundancy index, sigma(n)/n = k+1/2 with integer k. 19

%I

%S 2,24,4320,4680,26208,8910720,17428320,20427264,91963648,197064960,

%T 8583644160,10200236032,21857648640,57575890944,57629644800,

%U 206166804480,17116004505600,1416963251404800,15338300494970880

%N Numbers n with half-integral abundancy index, sigma(n)/n = k+1/2 with integer k.

%C Obviously, all a(k) 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 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 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 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 These numbers are called hemiperfect numbers. See Numericana & Wikipedia links. - _Michel Marcus_, Nov 19 2017

%H G. P. Michon, <a href="http://www.numericana.com/answer/numbers.htm#multiperfect">Multiperfect Numbers & Hemiperfect Numbers</a>, Numericana.

%H Walter Nissen, <a href="http://upforthecount.com/math/abundance.html">Abundancy : Some Resources </a>.

%H Project Euler, <a href="https://projecteuler.net/problem=241">Problem 241: Perfection Quotients</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Hemiperfect_number">Hemiperfect number</a>

%F A159907 = { n | 2*A000203(n) is in n*A005408 } = { n | A054024(n) = n/2 }

%e a(1) = 2 since sigma(2)/2 = (1+2)/2 = 3/2 is of the form k+1/2 with integer k=1.

%e 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.

%p with(numtheory);

%p P:=proc(i)

%p local a,n;

%p for n from 2 to i do

%p a:=(n+1)/2-sigma(n)/n; if a=trunc(a) then print(n); fi;

%p od;

%p end:

%p P(10000000000); # _Paolo P. Lava_, Dec 12 2011

%o (PARI) isok(n) = denominator(sigma(n,-1)) == 2; \\ _Michel Marcus_, Sep 19 2015

%o (PARI) forfactored(n=1,10^7, if(denominator(sigma(n,-1))==2, print1(n[1]", "))) \\ _Charles R Greathouse IV_, May 09 2017

%Y Cf. A000203, A088912, A141643 (k=2), A055153 (k=3), A141645 (k=4), A159271 (k=5).

%K nonn

%O 1,1

%A _M. F. Hasler_, Apr 25 2009

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Last modified November 16 07:06 EST 2018. Contains 317258 sequences. (Running on oeis4.)