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Numbers k such that the cube of the sum of digits of k equals the product of digits of k.
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%I #12 Jul 07 2022 11:38:35

%S 0,1,666666,1377789,1377798,1377879,1377897,1377978,1377987,1378779,

%T 1378797,1378977,1379778,1379787,1379877,1387779,1387797,1387977,

%U 1389777,1397778,1397787,1397877,1398777,1555888,1558588,1558858

%N Numbers k such that the cube of the sum of digits of k equals the product of digits of k.

%H Chai Wah Wu, <a href="/A162840/b162840.txt">Table of n, a(n) for n = 0..10000</a>

%F {n: A118880(n)=A007954(n)}. - _R. J. Mathar_, Jul 19 2009

%e 666666 is in the sequence because (1) cubed sum of its digits is (6+6+6+6+6+6)^3 = 46656, (2) the product of its digits is 6*6*6*6*6*6=46656; 46656=46656.

%p A007953 := proc(n) add(d,d=convert(n,base,10)) ; end: A007954 := proc(n) mul(d,d=convert(n,base,10)) ; end: A118880 := proc(n) (A007953(n))^3; end: for n from 1 to 2000000 do if A118880(n) = A007954(n) then printf("%d,\n",n) ; fi; od: # _R. J. Mathar_, Jul 19 2009

%t Select[Range[0,156*10^4],Total[IntegerDigits[#]]^3==Times@@IntegerDigits[#]&] (* _Harvey P. Dale_, Jul 07 2022 *)

%Y Cf. A007954, A118880, A117720, A034710. - _R. J. Mathar_, Jul 19 2009

%K base,nonn

%O 0,3

%A Boris Hostnik (megpplus(AT)siol.net), Jul 14 2009