

A260310


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


4



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
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OFFSET

1,1


COMMENTS

Consider pairs (x,y) of numbers where sum(px) p + sum(qy) q = x*sum(px) 1/p + y*sum(qy) 1/q where p, q are primes and sum(px) p > sum(qy) q.
For the vast majority of the time, a(2n1) 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(2n1) is prime then a(2n) is composite and vice versa. And when a(2n1) 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


LINKS



EXAMPLE



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 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



