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Numbers of the form 2^a + 3^b = 2^c + 3^d for nonnegative a, b, c, d and a <> c.
3

%I #24 Oct 28 2022 07:15:44

%S 5,11,17,35,259

%N Numbers of the form 2^a + 3^b = 2^c + 3^d for nonnegative a, b, c, d and a <> c.

%C No more terms < 10^4000. - _David Wasserman_, Feb 07 2005

%H Douglas Edward Iannucci, <a href="https://arxiv.org/abs/1907.03347">On duplicate representations as 2^x+3^y for nonnegative integers x and y</a>, arXiv:1907.03347 [math.NT], 2019.

%e a(5) = 256 + 3 = 2^8 + 3^1 = 259 = 16 + 243 = 2^4 + 3^5.

%t (* Max exponent 8 is sufficient to get all known solutions *)

%t f[_] = 0; Reap[Do[n = 2^a + 3^b; f[n] = f[n] + 1; If[f[n] == 1, ab[n] = {a, b}, If[f[n] >= 2, Print[HoldForm[2^x + 3^y] /. Thread[{x, y} -> ab[n]], " = ", HoldForm[2^x + 3^y] /. Thread[{x, y} -> {a, b}], " = ", n]; Sow[n]]], {a, 0, 8}, {b, 0, 8}]][[2, 1]] // Union (* _Jean-François Alcover_, Apr 29 2014 *)

%K nonn,fini,full

%O 1,1

%A _Frank Ellermann_, Jul 11 2003