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Cubes for which the product of the digits is a cube > 0.
1

%I #23 Dec 01 2023 05:26:19

%S 1,8,24389,226981,9393931,11239424,17373979,36264691,66923416,

%T 94818816,348913664,435519512,463684824,549353259,555412248,743677416,

%U 3929352552,4982686912,5526456832,11329982936,12374478297,12681938368,15142552424

%N Cubes for which the product of the digits is a cube > 0.

%D Felice Russo, A set of new Smarandache Functions, Sequences and conjectures in number theory, American Research Press, Lupton USA.

%H David A. Corneth, <a href="/A067070/b067070.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Harry J. Smith)

%e 24389 is in the sequence because (1) it is a cube and (2) the product of its digits is 2*4*3*8*9, = 1728 which is a cube > 0.

%t pdcQ[n_]:=Module[{pd=Times@@IntegerDigits[n]},pd>0&&IntegerQ[ Surd[ pd,3]]]; Select[Range[3000]^3,pdcQ] (* _Harvey P. Dale_, Jun 01 2015 *)

%o (PARI) ProdD(x)= { local(p=1); while (x>9 && p>0, p*=x%10; x\=10); return(p*x) }

%o iscube(x)= { if (x==0, return(1)); f=factor(x)~; for(i=1, length(f), if (t=f[2, i]%3, return(0))); return(1); }

%o { n=0; for (m=1, 10^10, p=ProdD(m^3); if (p && iscube(p), n++; print1(m^3, ", "); if (n==100, return)) ) } \\ _Harry J. Smith_, May 04 2010

%o (PARI)

%o first(n) = {

%o my(res = List(), c, f, vp, i);

%o for(i = 1, oo,

%o c = i^3;

%o vp = vecprod(digits(c));

%o if(vp == 0,

%o next

%o );

%o f = factor(vp);

%o if(gcd(f[,2])%3 == 0,

%o listput(res, c);

%o if(#res >= n,

%o return(res)

%o )

%o )

%o )

%o } \\ _David A. Corneth_, Dec 01 2023

%Y Cf. A000578, A007954, A052045, A052382.

%K nonn,base

%O 1,2

%A _Amarnath Murthy_, Jan 05 2002

%E More terms from _Sascha Kurz_, Mar 23 2002

%E One further term from Luc Stevens (lms022(AT)yahoo.com), May 03 2006

%E Edited by _R. J. Mathar_, Aug 08 2008

%E Offset changed from 0 to 1 by _Harry J. Smith_, May 04 2010