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A260310
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Pairs with balanced sums of prime divisors (A008472) and inverse prime divisors (A069359), ordered by larger members.
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4
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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
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1,1
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COMMENTS
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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.
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
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LINKS
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EXAMPLE
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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