Haskell, 48 bytes
f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]
Uses xnor's "black magic" formula:
$$f(n)=1+6\sum_{a=0}^\infty \left\lfloor \frac{n^2}{3a+1}\right\rfloor - \left\lfloor \frac{n^2}{3a+2}\right\rfloor$$
A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.
Long story short: we count Eisenstein integers of norm \1ドル \le N \le n^2\$, by factoring \$N = (x+y\omega)(x+y\omega^*)\$ into Eisenstein primes and counting how many solutions for \$(x,y)\$ come out of the factorization. We recognize the number of solutions as being equal to
$6ドル \times ((\text{# of divisors of }N \equiv 1\space(\text{mod }3)) - (\text{# of divisors of }N \equiv 2\space(\text{mod }3)))$$
and apply a clever trick to make that really easy to compute for all integers between \1ドル\$ and \$n^2\$ at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.
Haskell, 48 bytes
f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]
Uses xnor's "black magic" formula:
$$f(n)=1+6\sum_{a=0}^\infty \left\lfloor \frac{n^2}{3a+1}\right\rfloor - \left\lfloor \frac{n^2}{3a+2}\right\rfloor$$
A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.
Long story short: we count Eisenstein integers of norm \1ドル \le N \le n^2\$, by factoring \$N = (x+y\omega)(x+y\omega^*)\$ into Eisenstein primes and counting how many solutions for \$(x,y)\$ come out of the factorization. We recognize the number of solutions as being equal to
$6ドル \times ((\text{# of divisors of }N \equiv 1\space(\text{mod }3)) - (\text{# of divisors of }N \equiv 2\space(\text{mod }3)))$$
and apply a clever trick to make that really easy to compute for all integers between \1ドル\$ and \$n^2\$ at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.
Haskell, 48 bytes
f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]
Uses xnor's "black magic" formula:
$$f(n)=1+6\sum_{a=0}^\infty \left\lfloor \frac{n^2}{3a+1}\right\rfloor - \left\lfloor \frac{n^2}{3a+2}\right\rfloor$$
A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.
Long story short: we count Eisenstein integers of norm \1ドル \le N \le n^2\$, by factoring \$N = (x+y\omega)(x+y\omega^*)\$ into Eisenstein primes and counting how many solutions for \$(x,y)\$ come out of the factorization. We recognize the number of solutions as being equal to
$6ドル \times ((\text{# of divisors of }N \equiv 1\space(\text{mod }3)) - (\text{# of divisors of }N \equiv 2\space(\text{mod }3)))$$
and apply a clever trick to make that really easy to compute for all integers between \1ドル\$ and \$n^2\$ at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.
Haskell, 48 bytes
f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]
Uses xnor's "black magic" formula:
xnor's formula $$f(n)=1+6\sum_{a=0}^\infty \left\lfloor \frac{n^2}{3a+1}\right\rfloor - \left\lfloor \frac{n^2}{3a+2}\right\rfloor$$
A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.
Long story short: we count Eisenstein integers of norm 1 ≤ N ≤ n2\1ドル \le N \le n^2\$, by factoring N = (x+yω)(x+yω*)\$N = (x+y\omega)(x+y\omega^*)\$ into Eisenstein primes and counting how many solutions for (x,y)\$(x,y)\$ come out of the factorization. We recognize the number of solutions as being equal to
6 ×ばつ ((# of divisors of N ≡ 1 mod 3) − (# of divisors of N ≡ 2 mod 3)),$6ドル \times ((\text{# of divisors of }N \equiv 1\space(\text{mod }3)) - (\text{# of divisors of }N \equiv 2\space(\text{mod }3)))$$
and apply a clever trick to make that really easy to compute for all integers between 1\1ドル\$ and n2\$n^2\$ at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.
Haskell, 48 bytes
f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]
Uses xnor's "black magic" formula:
A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.
Long story short: we count Eisenstein integers of norm 1 ≤ N ≤ n2, by factoring N = (x+yω)(x+yω*) into Eisenstein primes and counting how many solutions for (x,y) come out of the factorization. We recognize the number of solutions as being equal to
6 ×ばつ ((# of divisors of N ≡ 1 mod 3) − (# of divisors of N ≡ 2 mod 3)),
and apply a clever trick to make that really easy to compute for all integers between 1 and n2 at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.
Haskell, 48 bytes
f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]
Uses xnor's "black magic" formula:
$$f(n)=1+6\sum_{a=0}^\infty \left\lfloor \frac{n^2}{3a+1}\right\rfloor - \left\lfloor \frac{n^2}{3a+2}\right\rfloor$$
A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.
Long story short: we count Eisenstein integers of norm \1ドル \le N \le n^2\$, by factoring \$N = (x+y\omega)(x+y\omega^*)\$ into Eisenstein primes and counting how many solutions for \$(x,y)\$ come out of the factorization. We recognize the number of solutions as being equal to
$6ドル \times ((\text{# of divisors of }N \equiv 1\space(\text{mod }3)) - (\text{# of divisors of }N \equiv 2\space(\text{mod }3)))$$
and apply a clever trick to make that really easy to compute for all integers between \1ドル\$ and \$n^2\$ at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.
Haskell, 48 bytes
f n=1+6*sum[(mod(i+1)3-1)*div(n^2)i|i<-[1..n^2]]
Uses xnor's "black magic" formula:
A proof of its correctness, and an explanation of how xnor managed to express it in 43 bytes of Python, can be found here.
Long story short: we count Eisenstein integers of norm 1 ≤ N ≤ n2, by factoring N = (x+yω)(x+yω*) into Eisenstein primes and counting how many solutions for (x,y) come out of the factorization. We recognize the number of solutions as being equal to
6 ×ばつ ((# of divisors of N ≡ 1 mod 3) − (# of divisors of N ≡ 2 mod 3)),
and apply a clever trick to make that really easy to compute for all integers between 1 and n2 at once. This yields the formula above. Finally, we apply some Python golf magic to end up with the really tiny solution xnor found.