|
%I #21 Jul 31 2021 22:49:34
%S 158,165,184,228,235,247,256,261,268,273,275,280,282,284,287,291,294,
%T 306,310,313,317,324,331,332,343,345,347,350,352,362,371,373,376,378,
%U 380,385,387,388,392,395,399,404,406,408,418,425,430,432,436,437,441
%N Numbers that are the sum of 6 positive cubes in exactly 2 ways.
%C It appears that this sequence has 1094 terms, the last of which is 21722. - _Donovan Johnson_, Jan 09 2013
%H Donovan Johnson, <a href="/A048930/b048930.txt">Table of n, a(n) for n = 1..1094</a> (terms < 10^7)
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CubicNumber.html">Cubic Number.</a>
%e 158 is in the sequence since 158 = 64+64+27+1+1+1 = 125+8+8+8+8+1
%t Reap[For[n = 1, n <= 500, n++, pr = Select[ PowersRepresentations[n, 6, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 2, 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]==2, n++; write("b048930.txt", n " " i))) /* Donovan Johnson, Jan 09 2013 */
%Y Cf. A003329, A048927, A048929, A048931, A345511, A345774, A345814.
%K nonn
%O 1,1
%A _Eric W. Weisstein_
%E Terms corrected by _Donovan Johnson_, Jan 09 2013
|
