A084888 Number of partitions of n^3 into two squares>0.

0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 2, 0, 0, 2, 0, 0, 0, 2, 1, 0, 2, 0, 0, 0, 0, 3, 2, 0, 0, 2, 0, 0, 1, 0, 2, 0, 0, 2, 0, 0, 2, 2, 0, 0, 0, 2, 0, 0, 0, 0, 4, 0, 2, 2, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 8, 0, 0, 2, 0, 0, 0, 1, 2, 2, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 8, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 1, 0, 3, 2

Offset

0,6

Comments

a(A050804(n)) = 1.

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Eric Weisstein's World of Mathematics, Diophantine Equation: 2nd Powers.

Reinhard Zumkeller, Illustration for A084888 and A000404

Formula

a(n) = A025426(A000578(n)).

Example

n=100: 100^3 = 1000000 = 960^2 + 280^2 = 936^2 + 352^2 = 800^2 + 600^2, therefore a(100)=3.

Prog

(Haskell)
a084888 = a025426 . a000578  -- Reinhard Zumkeller, Jul 18 2012
(PARI) a(n)=my(f=factor(n^3)); (prod(i=1, #f~, if(f[i, 1]%4==1, f[i, 2]+1, f[i, 2]%2==0||f[i, 1]<3))-issquare(n)+1)\2 \\ Charles R Greathouse IV, May 18 2016
(Python)
from math import prod
from sympy import factorint
def A084888(n): return ((m:=prod(1 if p==2 else (3*e+1 if p&3==1 else (3*e+1)&1) for p, e in factorint(n).items()))+((((~n**3 & n**3-1).bit_length()&1)<<1)-1 if m&1 else 0))>>1 # Chai Wah Wu, May 17 2023

Crossrefs

Cf. A000404, A063725, A000578, A000290.

Sequence in context: A092303 A329343 A063725 * A091400 A129448 A239003

Adjacent sequences: A084885 A084886 A084887 * A084889 A084890 A084891

Keyword

nonn

Author

Reinhard Zumkeller, Jun 10 2003

Status

approved