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Least number of positive cubes needed to represent n!.
1

%I #10 Feb 25 2015 23:32:58

%S 1,1,2,6,3,5,5,4,4,3,3,3,3,4,3,3,3,3,3,3,3

%N Least number of positive cubes needed to represent n!.

%F a(n)=A002376(n!).

%e a(4)=3 because 4!=24=2^3+2^3+2^3.

%e a(0)=1 because 0!=1=1^3.

%e a(1)=1 because 1!=1=1^3.

%e a(2)=2 because 2!=2=1^3+1^3.

%e a(3)=6 because 3!=6=1^3+1^3+1^3+1^3+1^3+1^3.

%e a(4)=3 because 4!=24=2^3+2^3+2^3.

%e a(5)=5 because 5!=120=1^3+3^3+3^3+4^3+1^3.

%e a(6)=5 because 6!=720=4^3+6^3+6^3+6^3+2^3.

%e a(7)=4 because 7!=5040=1^3+5^3+17^3+1^3.

%e a(8)=4 because 8!=40320=2^3+10^3+34^3+2^3.

%e a(9)=3 because 9!=362880=52^3+56^3+36^3.

%e a(10)=3 because 10!=3628800=96^3+140^3+4^3.

%e a(11)=3 because 11!=39916800=222^3+303^3+105^3.

%e a(12)=3 because 12!=479001600=214^3+777^3+47^3.

%e a(13)=4 because 13!=6227020800=106^3+255^3+1838^3+33^3.

%e a(14)=3 because 14!=87178291200=1344^3+4392^3+312^3.

%e a(15)=3 because 15!=1307674368000=2040^3+10908^3+1092^3.

%e a(16)=3 because 16!=20922789888000=8400^3+27040^3+8240^3.

%e a(17)=3 because 17!=355687428096000=22848^3+69984^3+9984^3.

%e a(18)=3 because 18!=6402373705728000=54060^3+184080^3+18900^3.

%e From _Donovan Johnson_, May 17 2010: (Start)

%e a(19)=3 because 19!=121645100408832000=131040^3+331200^3+436320^3.

%e a(20)=3 because 20!=2432902008176640000=87490^3+1034430^3+1098440^3.

%e (End)

%o (PARI) a(n,up,dw,k)=local(i,m);if(k==1,if(n==round(sqrtn(n,3))^3,return(1),return(-1)),forstep(i=up,dw,-1,m=n-i^3;if(a(m,min(i,floor(sqrtn(m,3))),ceil(sqrtn(m/(k-1),3)),k-1)==1,return(1)))) for(n=0,18,for(k=1,9,if(a(n!,floor(sqrtn(n!,3)),ceil(sqrtn(n!/k,3)),k)==1,print1(k", ");break))) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 01 2007

%Y Cf. A000142, A002376, A002325, A003072, A003327, A003328, A003329.

%K nonn

%O 0,3

%A _Hugo Pfoertner_, Jun 22 2003

%E More terms from _David W. Wilson_, Jun 23 2003

%E More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 01 2007

%E a(19)-a(20) from _Donovan Johnson_, May 17 2010