A260310 Pairs with balanced sums of prime divisors (A008472) and inverse prime divisors (A069359), ordered by larger members.

3, 8, 7, 16, 11, 18, 7, 27, 17, 45, 29, 50, 41, 54, 53, 60, 31, 64, 71, 84, 29, 99, 107, 132, 61, 147, 41, 153, 131, 162, 53, 207, 157, 220, 113, 225, 179, 228, 239, 240, 131, 242, 79, 243, 73, 245, 127, 255, 127, 256, 229, 264, 223, 280, 113, 297, 199, 315, 73, 325, 317, 336, 181, 338, 43, 343, 269, 348

Offset: 1

Author

Juri-Stepan Gerasimov, Jul 22 2015

Keywords

• nonn

Comments

Consider pairs (x,y) of numbers where sum(p|x) p + sum(q|y) q = x*sum(p|x) 1/p + y*sum(q|y) 1/q where p, q are primes and sum(p|x) p > sum(q|y) q.

Or, pairs of numbers x and y where A008472(x) + A008472(y) = A069359(x) + A069359(y) where A008472(x) > A008472(y).

A001222(a(2n -1)) = 1 and A001222(a(2n)) >= 3.

For the vast majority of the time, a(2n-1) is prime. There seems to be about 1 pair per decade.

Conjecture: a(2n) < a(2n+2) for all n>0, but there are many times (1/10.84) that a(2n) + 1 = a(2n+2).

Conjecture: if a(2n-1) is prime then a(2n) is composite and vice versa. And when a(2n-1) is composite, it is congruent to 0 (mod 6). - Robert G. Wilson v, Jul 22 2015

The first conjecture appears to be satisfied because if both x and y were prime then the sum of the A008472 were the sum of the two primes and the sum of the A069359 were two. - R. J. Mathar, Aug 03 2015

Examples

			3 and 8 is first pair of this sequence because A008472(3) + A008472(8) = 3 + 2 = 5 is equal to A069359(3) + A069359(8) = 1 + 4 = 5.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..19812

Crossrefs

Cf. A001222, A008472, A069359.

Programs

Mathematica

f[n_] := f[n] = Block[{fi = FactorInteger[n][[All, 1]]}, {Plus @@ fi, n*Plus @@ (1/fi)}] /; n > 0; k =3; lst = {}; While[ k < 400, j = 2; While[ j < k, If[ f[k][[1]] + f[j][[1]] == f[k][[2]] + f[j][[2]] && f[k][[1]] != f[k][[2]], AppendTo[lst, {j,k}]]; j++]; k++]; lst // Flatten (* Robert G. Wilson v, Jul 22 2015 *)

Extensions

Corrected and edited by Robert G. Wilson v, Jul 22 2015