%I #16 Jul 31 2021 22:49:42

%S 221,254,369,411,443,469,495,502,576,595,600,648,658,684,704,711,720,

%T 739,746,753,760,765,767,772,774,779,786,793,811,818,828,835,844,854,

%U 863,866,874,880,884,886,892,893,899,905,910,919,928,929,935,936,937

%N Numbers that are the sum of 6 positive cubes in exactly 3 ways.

%C It appears that this sequence has 1141 terms, the last of which is 26132. - _Donovan Johnson_, Jan 09 2013

%H Donovan Johnson, <a href="/A048931/b048931.txt">Table of n, a(n) for n = 1..1141</a> (terms < 10^7)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CubicNumber.html">Cubic Number.</a>

%e 221 is in the sequence since 221 = 216+1+1+1+1+1 = 125+64+8+8+8+8 = 64+64+64+27+1+1

%t Reap[For[n = 1, n <= 1000, n++, pr = Select[ PowersRepresentations[n, 6, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 3, Print[n, pr]; Sow[n]]]][[2, 1]] (* _Jean-François Alcover_, Jul 31 2013 *)

%o (PARI) mx=10^6; ct=vector(mx); cb=vector(99); for(i=1, 99, cb[i]=i^3); for(i1=1, 99, s1=cb[i1]; for(i2=i1, 99, s2=s1+cb[i2]; if(s2+4*cb[i2]>mx, next(2)); for(i3=i2, 99, s3=s2+cb[i3]; if(s3+3*cb[i3]>mx, next(2)); for(i4=i3, 99, s4=s3+cb[i4]; if(s4+2*cb[i4]>mx, next(2)); for(i5=i4, 99, s5=s4+cb[i5]; if(s5+cb[i5]>mx, next(2)); for(i6=i5, 99, s6=s5+cb[i6]; if(s6>mx, next(2)); ct[s6]++)))))); n=0; for(i=6, mx, if(ct[i]==3, n++; write("b048931.txt", n " " i))) /* Donovan Johnson, Jan 09 2013 */

%Y Cf. A003329, A048929, A048930, A343705, A345512, A345766, A345775, A345815.

%K nonn

%O 1,1

%A _Eric W. Weisstein_

%E Corrected and extended by Larry Reeves (larryr(AT)acm.org), Oct 02 2000