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A084355
Least number of positive cubes needed to represent n!.
1
1, 1, 2, 6, 3, 5, 5, 4, 4, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3
OFFSET
0,3
FORMULA
a(n)=A002376(n!).
EXAMPLE
a(4)=3 because 4!=24=2^3+2^3+2^3.
a(0)=1 because 0!=1=1^3.
a(1)=1 because 1!=1=1^3.
a(2)=2 because 2!=2=1^3+1^3.
a(3)=6 because 3!=6=1^3+1^3+1^3+1^3+1^3+1^3.
a(4)=3 because 4!=24=2^3+2^3+2^3.
a(5)=5 because 5!=120=1^3+3^3+3^3+4^3+1^3.
a(6)=5 because 6!=720=4^3+6^3+6^3+6^3+2^3.
a(7)=4 because 7!=5040=1^3+5^3+17^3+1^3.
a(8)=4 because 8!=40320=2^3+10^3+34^3+2^3.
a(9)=3 because 9!=362880=52^3+56^3+36^3.
a(10)=3 because 10!=3628800=96^3+140^3+4^3.
a(11)=3 because 11!=39916800=222^3+303^3+105^3.
a(12)=3 because 12!=479001600=214^3+777^3+47^3.
a(13)=4 because 13!=6227020800=106^3+255^3+1838^3+33^3.
a(14)=3 because 14!=87178291200=1344^3+4392^3+312^3.
a(15)=3 because 15!=1307674368000=2040^3+10908^3+1092^3.
a(16)=3 because 16!=20922789888000=8400^3+27040^3+8240^3.
a(17)=3 because 17!=355687428096000=22848^3+69984^3+9984^3.
a(18)=3 because 18!=6402373705728000=54060^3+184080^3+18900^3.
From Donovan Johnson, May 17 2010: (Start)
a(19)=3 because 19!=121645100408832000=131040^3+331200^3+436320^3.
a(20)=3 because 20!=2432902008176640000=87490^3+1034430^3+1098440^3.
(End)
PROG
(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
KEYWORD
nonn
AUTHOR
Hugo Pfoertner, Jun 22 2003
EXTENSIONS
More terms from David W. Wilson, Jun 23 2003
More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 01 2007
a(19)-a(20) from Donovan Johnson, May 17 2010
STATUS
approved