A110076 a(n) is the largest number m such that sigma(m)=10^n, or if there is no such m a(n)=0.

1, 0, 0, 0, 9481, 99301, 997501, 9993001, 99948001, 999795001, 9999750001, 99998670001, 999997950001, 9999986700001, 99999975000001, 999999198750001, 9999999187500001, 99999995096707501, 999999919987500001, 9999999986700000001, 99999499999999800001, 999999999907500000001, 9999999999796009687501

Offset: 0

Author

Farideh Firoozbakht, Jul 31 2005

Keywords

• nonn

Comments

Conjecture: For n>3 a(n) is positive.

For 4 <= n <= 102, a(n) is the product of two distinct primes, but a(103) = a(49)*a(54) and is the product of four distinct primes: 1862645149230957031249999 * 5368709119999999999999999 * 79999999999999999999999999 * 12499999999999999999999999999. - David Wasserman, Nov 18 2008

Examples

			a(12)=999997950001 because sigma(999997950001)=sigma(799999*1249999) =800000*1250000=10^12 and 999997950001 is the largest number with this property(sigma(m)=10^12).
		

Links

• Max Alekseyev, Table of n, a(n) for n = 0..1000
• Max A. Alekseyev, Computing the Inverses, their Power Sums, and Extrema for Euler's Totient and Other Multiplicative Functions. Journal of Integer Sequences, Vol. 19 (2016), Article 16.5.2

Crossrefs

Cf. A110077, A110078.

Programs

Mathematica

a[0] = 1; a[1] = a[2] = a[3] = 0; a[n_] := (For[m = 1, DivisorSigma[ 1, 10^n - m] != 10^n, m++ ];10^n - m); Do[Print[a[n]], {n, 0, 12}]

Extensions

More terms from David Wasserman, Nov 18 2008

Terms a(19) onward from Max Alekseyev, Mar 06 2014