cp's OEIS Frontend

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

%I A084888 #22 Feb 16 2025 08:32:49
%S A084888 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,
%T A084888 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,
%U A084888 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
%N A084888 Number of partitions of n^3 into two squares>0.
%C A084888 a(A050804(n)) = 1.
%H A084888 Reinhard Zumkeller, <a href="/A084888/b084888.txt">Table of n, a(n) for n = 0..1000</a>
%H A084888 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DiophantineEquation2ndPowers.html">Diophantine Equation: 2nd Powers</a>.
%H A084888 Reinhard Zumkeller, <a href="/A084888/a084888.txt">Illustration for A084888 and A000404</a>
%F A084888 a(n) = A025426(A000578(n)).
%e A084888 n=100: 100^3 = 1000000 = 960^2 + 280^2 = 936^2 + 352^2 = 800^2 + 600^2, therefore a(100)=3.
%o A084888 (Haskell)
%o A084888 a084888 = a025426 . a000578  -- _Reinhard Zumkeller_, Jul 18 2012
%o A084888 (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
%o A084888 (Python)
%o A084888 from math import prod
%o A084888 from sympy import factorint
%o A084888 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
%Y A084888 Cf. A000404, A063725, A000578, A000290.
%K A084888 nonn
%O A084888 0,6
%A A084888 _Reinhard Zumkeller_, Jun 10 2003