A373043 a(n) is the number of products P = j*k*n, 1 < j < k < n, such that P is a square.
0, 0, 1, 0, 2, 1, 2, 0, 4, 0, 2, 3, 3, 0, 10, 0, 8, 3, 3, 0, 11, 9, 3, 17, 10, 0, 10, 0, 25, 5, 4, 6, 18, 0, 4, 5, 19, 0, 12, 0, 12, 25, 4, 0, 34, 29, 48, 6, 15, 0, 43, 9, 25, 7, 5, 0, 31, 0, 5, 32, 45, 10, 16, 0, 16, 7, 20, 0, 74, 0, 6, 68, 18, 11, 18, 0, 55, 65, 6, 0
Offset: 4
Keywords
Examples
a(6) = 1: 2*3*6 = 6^2; a(8) = 2: 2*4*8 = 8^2, 3*6*8 = 12^2; a(9) = 1: 2*8*9 = 12^2; a(10) = 2: 2*5*10 = 10^2, 5*8*10 = 20^2.
Links
- David A. Corneth, Table of n, a(n) for n = 4..10000 (terms to n = 2000 from Hugo Pfoertner)
Programs
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PARI
a(n) = my(s=0); for(j=2, n-2, for(k=j+1, n-1, if(issquare(j*k*n), s++))); s
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PARI
a(n) = { my(f = factor(n), c = core(f), res = (issquare(n) && n > 1), u, s); u = sqrtint((n-1)\c); for(i = 1, u, res+=(numdiv(c*i^2)\2); ); res-=u; for(i = u+1, sqrtint((n^2 - 1)\c), d = divisors(c*i^2); s = select(x->x>=n, d); res+=((#d - 2*#s)>>1) ); res } \\ David A. Corneth, May 27 2024 -
Python
from sympy.ntheory.primetest import is_square def A373043(n): return sum(1 for k in range(3,n) for j in range(2,k) if is_square(j*k*n)) # Chai Wah Wu, May 31 2024
Formula
a(n) = 0 for prime n. a(n) > 0 for composite n > 4. Proof: If n is square, then m = i*j*n = 2*8*n = 16*n is square and a(n) > 0. If n is a nonsquare composite, then we can write it as n = j*k where 1 < j < k < n and so j*k*n = j*k*j*k = (j*k)^2 is a square and a(n) > 0 as well. - David A. Corneth, May 27 2024