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 A159907 Numbers n with half-integral abundancy index, sigma(n)/n = k+1/2 with integer k. 24
 2, 24, 4320, 4680, 26208, 8910720, 17428320, 20427264, 91963648, 197064960, 8583644160, 10200236032, 21857648640, 57575890944, 57629644800, 206166804480, 17116004505600, 1416963251404800, 15338300494970880 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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. 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. 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 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 These numbers are called hemiperfect numbers. See Numericana & Wikipedia links. - Michel Marcus, Nov 19 2017 LINKS G. P. Michon, Multiperfect Numbers & Hemiperfect Numbers, Numericana. Walter Nissen, Abundancy : Some Resources . Project Euler, Problem 241: Perfection Quotients. Wikipedia, Hemiperfect number FORMULA A159907 = { n | 2*A000203(n) is in n*A005408 } = { n | A054024(n) = n/2 } EXAMPLE a(1) = 2 since sigma(2)/2 = (1+2)/2 = 3/2 is of the form k+1/2 with integer k=1. 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. MAPLE with(numtheory); P:=proc(i) local a, n; for n from 2 to i do   a:=(n+1)/2-sigma(n)/n; if a=trunc(a) then print(n); fi; od; end: P(10000000000); # Paolo P. Lava, Dec 12 2011 PROG (PARI) isok(n) = denominator(sigma(n, -1)) == 2; \\ Michel Marcus, Sep 19 2015 (PARI) forfactored(n=1, 10^7, if(denominator(sigma(n, -1))==2, print1(n[1]", "))) \\ Charles R Greathouse IV, May 09 2017 (Python) from fractions import Fraction from sympy import divisor_sigma as sigma def aupto(limit):   for k in range(1, limit):     if Fraction(int(sigma(k, 1)), k).denominator == 2:       print(k, end=", ") aupto(3*10**4) # Michael S. Branicky, Feb 24 2021 CROSSREFS Cf. A000203, A088912, A141643 (k=2), A055153 (k=3), A141645 (k=4), A159271 (k=5). Sequence in context: A355561 A059332 A000794 * A242484 A088912 A342573 Adjacent sequences:  A159904 A159905 A159906 * A159908 A159909 A159910 KEYWORD nonn AUTHOR M. F. Hasler, Apr 25 2009 STATUS approved

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Last modified November 26 05:48 EST 2022. Contains 358353 sequences. (Running on oeis4.)