OFFSET
0,1
COMMENTS
a(0) = 6 corresponds to the smallest perfect number.
Is n = 144 the first number for which a(n) = 0? - T. D. Noe, Mar 28 2013
No, a(144) = 95501968. - Giovanni Resta, Mar 28 2013
We can instead compute k - sigma(k)/2 for increasing k, which is computationally much faster. In this case, we stop computing when all n have been found for a range of numbers. - T. D. Noe, Mar 28 2013
Also, the first number whose deficiency is 2n. This is the even bisection of A082730. Hence, the first number in the following sequences: A000396, A191363, A125246, A141548, A125247, A101223, A141549, A141550, A125248, A223608, A223607, A223606. - T. D. Noe, Mar 29 2013
10^12 < a(654) <= 618970019665683124609613824. - Donovan Johnson, Jan 04 2014
LINKS
Donovan Johnson, Table of n, a(n) for n = 0..653 (first 241 terms from T. D. Noe)
Nichole Davis, Dominic Klyve and Nicole Kraght, On the difference between an integer and the sum of its proper divisors, Involve, Vol. 6 (2013), No. 4, 493-504; DOI: 10.2140/involve.2013.6.493.
EXAMPLE
a(4)=22, since 22 is the least number such that sigma(22)=36=2*(22-4).
MATHEMATICA
Table[Min[Select[Range[2000], DivisorSigma[1, #] == 2*(# - i) &]], {i, 0, 60}]
nn = 144; t = Table[0, {nn}]; k = 0; While[k++; Times @@ t == 0, s = (2*k - DivisorSigma[1, k])/2; If[s >= 0 && s < nn && IntegerQ[s] && t[[s + 1]] == 0, t[[s + 1]] = k]]; t (* T. D. Noe, Mar 28 2013 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jayanta Basu, Mar 28 2013
STATUS
approved