OFFSET
1,3
COMMENTS
If n is of the form 4^i*(8j+7) (where i>=0, j>=0) then n^3 is not in the sequence because n^3 is of the mentioned form so n^3 is in A004215 hence according to the definition n^3 is not in this sequence (see formula for A004215). Hence 7^3, 15^3, 23^3, 28^3, 31^3, 39^3, ... are not in the sequence. Is there a number n such that n^3 is not in the sequence but n is not of the form 4^i*(8j+7)? - Farideh Firoozbakht, Nov 23 2006
A number n^3 belongs to this sequence if and only if n is sum of three squares. Proof is immediate from Catalan's identity (x^2 + y^2 + z^2)^3 = x^2*(3*z^2 - x^2 - y^2)^2 + y^2*(3*z^2 - x^2 - y^2)^2 + z^2*(z^2 - 3*x^2 - 3*y^2)^2. - Artur Jasinski, Dec 09 2006
If n = a^2 + b^2 + c^2, then n^3 = (n*a)^2 + (n*b)^2 + (n*c)^2. Conversely, suppose there were an n such that n^3 is in A000378 but n is not. Then n must be of form 4^k*(8i+7). But n^3 would also be of the form 4^k*(8i+7) and thus n^3 would not be in A000378, contradicting the original assumption. This argument is easily extended to all odd powers, i.e., n^(2k+1) is in A000378 iff n is in A000378. - Ray Chandler, Feb 03 2009
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..2000
EXAMPLE
125 is in the sequence because
125 = 5^3 = 0^2 + 2^2 + 11^2
= 0^2 + 5^2 + 10^2
= 3^2 + 4^2 + 10^2
= 5^2 + 6^2 + 8^2.
27 = 3^3 = 1^2 + 1^2 + 5^2, so 27 is a term.
125 = 5^3 = 0^2 + 2^2 + 11^2, so 125 is a term.
216 = 6^3 = 2^2 + 4^2 + 14^2, so 216 is a term.
MATHEMATICA
Select[Range[0, 50]^3, SquaresR[3, # ] > 0 &] (* Ray Chandler, Nov 23 2006 *)
PROG
(PARI) isA125084(n)={ local(cnt, a, b) ; cnt=0 ; a=0; while(a^2<=n, b=0 ; while(b<=a && a^2+b^2<=n, if(issquare(n-a^2-b^2), return(1) ) ; b++ ; ) ; a++ ; ) ; return(0) ; } { for(n=1, 300, if(isA125084(n^3), print1(n^3, ", ") ; ) ; ) ; } \\ R. J. Mathar, Nov 23 2006
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Nov 20 2006, Nov 21 2006, Nov 22 2006
EXTENSIONS
STATUS
approved