A338939 a(n) is the number of partitions n = a + b such that a*b is a perfect square.
0, 1, 0, 1, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 1, 1, 1, 1, 0, 3, 0, 1, 0, 1, 2, 3, 0, 1, 1, 3, 0, 1, 0, 3, 1, 1, 1, 1, 1, 3, 1, 1, 0, 1, 1, 1, 0, 1, 0, 5, 1, 3, 1, 1, 1, 1, 0, 3, 0, 3, 1, 1, 0, 1, 4, 1, 0, 3, 0, 3, 0, 1, 1, 3, 2, 1, 0, 3, 0, 3, 0, 3, 0, 1, 4, 1, 1, 1, 1, 3, 1, 1, 0, 1, 1, 1, 1, 1, 0, 5
Offset: 1
Keywords
Examples
n = 10 = 1 + 9 = 2 + 8 = 5 + 5 with 1*9 = 3^2 and 2*8 = 4^2 and 5*5 = 5^2. Then a(10) = 3. Also 10 = 2^1 * 5^1. So t=1, a_1=1 and a(n) = A = 2*1+1 = 3.
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000
Programs
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Maple
A338939:=n->map[fold=(`+`,0)](i->if(issqr(i*(n-i)),1,0),[$1..1/2*n]); seq(A338939(n),n=1..100); # Felix Huber, Oct 02 2024
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Mathematica
a[n_] := Count[IntegerPartitions[n, {2}], ?(IntegerQ @ Sqrt[Times @@ #] &)]; Array[a, 100] (* _Amiram Eldar, Nov 22 2020 *) -
PARI
a(n)=my(c=0);for(i=1,n-1,if((2*i<=n)&&issquare(i*(n-i)),c++));c for(n=1,100,print1(a(n),", ")) \\ Derek Orr, Nov 18 2020
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PARI
a(n) = sum(i=1, n-1, (2*i<=n) && issquare(i*(n-i))); \\ Michel Marcus, Dec 21 2020
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PARI
first(n) = {my(res = vector(n)); for(i = 1, n, d = divisors(i^2); for(i = (#d + 1)\2, #d, c = d[i] + d[#d + 1 - i]; if(c <= n, res[c]++ , next(2) ) ) ); res } \\ David A. Corneth, Dec 21 2020 -
Python
from sympy.abc import x,y from sympy.solvers.diophantine.diophantine import diop_quadratic def A338939(n): return len(diop_quadratic(x*(n-x)-y**2))>>2 # Chai Wah Wu, Aug 21 2024
Formula
Let n = 2^t * p_1^a_1 * p_2^a_2 * ... * p_r^a_r * q_1^b_1 * q_2^b_2 *...* q_s^b_s with t >= 0, a_i >= 0 for i = 1..r, where p_i == 1 (mod 4) and q_j == -1 (mod 4) for j = 1..s. Further, let A = (2a_1 + 1)*(2a_2 + 1)*...*(2a_r + 1). Then a(n) = (A-1)/2 for odd n and a(n) = A for even n.
Comments