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A064158 Integers m such that (x1*x2*..xk)^(x1+x2+..xk) = (x1+x2+..xk)^(x1*x2*..xk) where x1x2..xk are the digits of m in base 10.

%I A064158 #22 Aug 27 2025 09:48:42
%S A064158 1,2,3,4,5,6,7,8,9,22,112,121,123,132,211,213,231,312,321,1124,1142,
%T A064158 1214,1241,1412,1421,2114,2141,2411,4112,4121,4211,11125,11133,11152,
%U A064158 11215,11222,11251,11313,11331,11512,11521,12115,12122,12151,12212
%N A064158 Integers m such that (x1*x2*..xk)^(x1+x2+..xk) = (x1+x2+..xk)^(x1*x2*..xk) where x1x2..xk are the digits of m in base 10.
%C A064158 With the exception of only 112, 121, and 211, each term of this sequence satisfies (sum of digits) equals (product of digits). For 112, 121, and 211, the sum of the digits is 4, the product of the digits is 2, and the terms qualify because 2^4 equals 4^2. - _Harvey P. Dale_, Sep 30 2011
%H A064158 Harvey P. Dale, <a href="/A064158/b064158.txt">Table of n, a(n) for n = 1..179</a>
%F A064158 Union of {112, 121, 211} and A034710. - _Jason Yuen_, Aug 27 2025
%e A064158 22 belongs to the sequence because (2*2)^(2+2) = (2+2)^(2*2).
%t A064158 okQ[n_]:=Module[{idn=IntegerDigits[n],t,p}, t= Times@@idn; p=Total[idn];t^p==p^t]; Select[Range[12500],okQ]
%Y A064158 Cf. A007953, A007954, A034710.
%K A064158 easy,nonn,base,changed
%O A064158 1,2
%A A064158 _Felice Russo_, Sep 14 2001
%E A064158 More terms from _Jason Earls_, Dec 04 2001