A217769 Least number k > n such that sigma(k) = 2*(k-n), or 0 if no such k exists.

6, 3, 5, 7, 22, 11, 13, 27, 17, 19, 46, 23, 124, 58, 29, 31, 250, 57, 37, 55, 41, 43, 94, 47, 1264, 106, 53, 87, 118, 59, 61, 85, 134, 67, 142, 71, 73, 712, 158, 79, 166, 83, 405, 115, 89, 141, 406, 119, 97, 202, 101, 103, 214, 107, 109, 145, 113, 177, 418, 143

Offset: 0

Author

Jayanta Basu, Mar 28 2013

Keywords

• nonn

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

Examples

			a(4)=22, since 22 is the least number such that sigma(22)=36=2*(22-4).
		

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.

Crossrefs

Cf. A000203, A000396.

Cf. A087998 (negative n).

Programs

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 *)