A332596 Number of quadrilateral regions in a "frame" of size n X n (see Comments in A331776 for definition), divided by 8.

0, 1, 10, 26, 63, 107, 189, 294, 455, 627, 891, 1202, 1650, 2121, 2719, 3392, 4292, 5239, 6470, 7832, 9463, 11129, 13205, 15460, 18164, 20919, 24130, 27572, 31679, 35945, 40977, 46340, 52384, 58511, 65421, 72718, 81104, 89589, 98989, 108860, 120062, 131551

Offset: 1

Author

Scott R. Shannon and N. J. A. Sloane, Mar 02 2020

Keywords

• nonn

Comments

See A331776 for many other illustrations.

Theorem. Let z(n) = Sum_{i, j = 1..n, gcd(i,j)=1} (n+1-i)*(n+1-j) (this is A115004) and z_2(n) = Sum_{i, j = 1..n, gcd(i,j)=2} (n+1-i)*(n+1-j) (this is A331761). Then, for n >= 2, 8*a(n) = 4*z(n) - 8*z_2(n) + 8*n^2 - 36*n + 24. - Scott R. Shannon and N. J. A. Sloane, Mar 06 2020

Links

• Jinyuan Wang, Table of n, a(n) for n = 1..1000
• Scott R. Shannon, Colored illustration for a(3) = 10 (there are 8*10 quadrilaterals).

Crossrefs

Cf. A115004, A331761, A331776, A332594, A332595.

Programs

Maple

V := proc(m, n, q) local a, i, j; a:=0;
for i from 1 to m do for j from 1 to n do
if gcd(i, j)=q then a:=a+(m+1-i)*(n+1-j); fi; od: od: a; end;
f := n -> if n=1 then 0 else 8*n^2 - 36*n + 24 + 4*V(n,n,1) 8*V(n, n, 2); fi;
[seq(f(n)/8, n=1..60)]; # N. J. A. Sloane, Mar 10 2020

PARI

a(n) = sum(i=1, n, sum(j=1, n, if(gcd(i, j)==1, (n+1-i)*(n+1-j), 0)))/2 - sum(i=1, n, sum(j=1, n, if(gcd(i, j)==2, (n+1-i)*(n+1-j), 0))) + n^2 - 9*n/2 + 3; \\ Jinyuan Wang, Aug 07 2021

Python

from sympy import totient
def A332596(n): return 0 if n == 1 else ((n-1)*(n-4) - sum(totient(i)*(n+1-i)*(2*n+2-7*i) for i in range(2,n//2+1)) + sum(totient(i)*(n+1-i)*(2*n+2-i) for i in range(n//2+1,n+1)))//2 # Chai Wah Wu, Aug 16 2021

Formula

For n > 1, a(n) = ((n-1)*(n-4) - Sum_{i=2..floor(n/2)} (n+1-i)*(2*n+2-7*i)*phi(i) + Sum_{i=floor(n/2)+1..n} (n+1-i)*(2*n+2-i)*phi(i))/2. - Chai Wah Wu, Aug 16 2021

Extensions

More terms from N. J. A. Sloane, Mar 10 2020