How to count lines of cout?

[jonnin]

snipping tool? I use that one mostly.

[Alex009988]

Thanks everyone who shed light on the problem. Especially to jonnin. I'm sorry for being mix up with 115 and 155 and having noticed it. One number confused all system

[lastchance]

Hmm. There's a closed-form expression, in fact. Done by tracing back the dependencies in my iterative formula. However, there must be a quicker argument for choosing n from 2n ...

v_n = ²ⁿC_n

e.g.
v₁ = ²C₁ = 2
v₂ = ⁴C₂ = 6
v₃ = ⁶C₃ = 20
etc.

The ratio of successive terms is v_n/v_n-1 = 2(2n-1)/n

#include <iostream>
using namespace std;

int main()
{
   unsigned long long v = 1;
   for ( int n = 1; n <= 31; n++ )      // overflow danger for higher
   {
      v = v * 2 * ( 2 * n - 1 ) / n;
      cout << "v" << n << ": " << v << '\n';
   }
}

v1: 2
v2: 6
v3: 20
v4: 70
v5: 252
v6: 924
v7: 3432
v8: 12870
v9: 48620
v10: 184756
v11: 705432
v12: 2704156
v13: 10400600
v14: 40116600
v15: 155117520
v16: 601080390
v17: 2333606220
v18: 9075135300
v19: 35345263800
v20: 137846528820
v21: 538257874440
v22: 2104098963720
v23: 8233430727600
v24: 32247603683100
v25: 126410606437752
v26: 495918532948104
v27: 1946939425648112
v28: 7648690600760440
v29: 30067266499541040
v30: 118264581564861424
v31: 465428353255261088

[jonnin]

very nice work. A much better form than mine :(