A015886 a(n) = smallest number k such that sigma(k + n) = sigma(k) + n, or -1 if no such number exists.

1, 2, 3, 2, 3, 2, 5, 74, 3, 2, 3, 2, 5, 4418, 3, 2, 3, 2, 5, 6, 3, 2, 7

Offset: 0

Author

Robert G. Wilson v

Keywords

• nonn
• hard
• nice
• more

Comments

There are solutions to sigma(k)+n=sigma(k+n) whenever n is the difference between two primes (A030173), e.g. k and k+n are primes. There are other values of n that have solutions (see example).

a(23) > 4292000000, if it exists. - Jud McCranie, Jan 05 2000

The sequence begins: 1, 2, 3, 2, 3, 2, 5, 74, 3, 2, 3, 2, 5, 4418, 3, 2, 3, 2, 5, 6, 3, 2, 7, ?, 5, ?, 3, 2, 3, 2, 7, ?, 5, 18, 3, 2, 5, 44, 3, 2, 3, 2, 5, ?, 3, 2, 7, ?, 5, 3315, 3, 2, 7, ?, 5, ?, 3, 2, 3, 2, 7, ?, 5, ?, 3, 2, 5, ?, 3, 2, 3, 2, 7, 18, 5, ?, 3, 2, 5, ?, 3, 2, 7, ?, 5, ?, 3, 2, 13, ?, 7, ?, 5, 32, 3, 2, 5 where the other missing terms (designated by "?") are > 10^9, if they exist. - Jud McCranie, Jan 08 2000

The "other" values of n are the odd n such that n+2 is not prime. For these n, in order for sigma(k) or sigma(n+k) to be odd, either k or n+k must be a square or twice a square. Examples: for n=7, n+k=9^2; for n=13, k=2*47^2 and for n=19, n+k=5^2. Using this idea, it is easy to show that if a(23) exists it is greater than 10^12. - T. D. Noe, Sep 24 2007

Examples

			sigma(74+7) = 121 = sigma(74)+7, so a(7) = 74.
		

Crossrefs

Cf. A000203, A020483, A030173, A054906.

Programs

Mathematica

Table[k=1; While[DivisorSigma[1,k+n] != DivisorSigma[1,k]+n, k++ ]; k, {n,0,22}] (* T. D. Noe, Sep 24 2007 *)

PARI

a(n) = {my(k=1); while(sigma(k+n) != sigma(k) + n, k++); k;} \\ Michel Marcus, May 23 2018

Formula

a(2n) = A020483(n) = A054906(n) - T. D. Noe, Sep 24 2007