int main(){
long long int d_min = 100000000;
long long int l_min = 100000000;
long long int r_min = 100000000;
long long int sa;
long long int wa;
for (long long int i = 1; i < 10000000000; i++) {
sa = 3*i + 1;
wa = (sa + 1) * i + 1;
if (is_penta(sa) && is_penta(wa)) {
d_min = min(d_min, sa);
l_min = penta(i);
r_min = penta(i+1);
}
}
printf("d_min: %lld\n", d_min);
printf("l_min: %lld\n", l_min);
printf("r_min: %lld\n", r_min);
return 0;
}
/*
* T(n) = n*(n+1)/2
* P(m) = m*(3m-1)/2
* H(k) = k*(2k-1)
*
* If T(n) == P(m) == H(k), it is easy to see that k=(n+1)/2
* So, for all odd n, we have a corresponding H(k) where k = (n+1)/2
*
* We now have to find m where T(n) == P(m) and n is odd.
* One way of doing this is to verify for all T(n), where n is odd, if T(n) is
* also pentagonal.
* Given,
* T(n) = m(3*m - 1)/2
* => 2*T(n) = 3m^2 - m
* => 3m^2 - m - 2T(n) = 0
*
* Solving the above quadratic equation for integral m using:
* m = (-b +/- sqrt(b^2 - 4ac))/2a, where
* a = 3, b = -1, c = -2T(n), we get the required value of m
*
* m = [ -(-1) +/- sqrt((-1)^2 - 4*3*(-2T(n)) ] / 2(3)
* m = [ 1 + sqrt(1 + 24T(n)) ] / 6
* We can ignore the second root since it turns out to be a negative number
*/
int
main()
{
int n;
uint64_t tn;
for (n = 3; ;n += 2) {
tn = T(n);
if (is_penta(tn)) {
if (tn == 40755) {
continue;
}
printf("n-%d T(n)-%lu\n", n, tn);
break;
}
}
return (0);
}
int compare_pentas(int j, int k)
{
return is_penta(k-j) &&
is_penta(k+j) ;
}
