0,2

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).

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 (noe(AT)sspectra.com), Sep 24 2007

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

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

Table[k=1; While[DivisorSigma[1, k+n] != DivisorSigma[1, k]+n, k++ ]; k, {n, 22}] - T. D. Noe (noe(AT)sspectra.com), Sep 24 2007

Cf. A000203, A030173.

Sequence in context: A065559 A087317 A086489 this_sequence A108656 A164962 A119880

Adjacent sequences: A015883 A015884 A015885 this_sequence A015887 A015888 A015889

nonn,hard,nice

Robert G. Wilson v (rgwv(AT)rgwv.com)

a(23) > 4292000000, if it exists - Jud McCranie (j.mccranie(AT)comcast.net), Jan 05 2000

Jud McCranie (j.mccranie(AT)comcast.net), Jan 08 2000 reports that 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.