A334312 Triangle read by rows: T(n,k) = Sum_{i=k..n} A191898(i,k).
1, 2, -1, 3, 0, -2, 4, -1, -1, -1, 5, 0, 0, 0, -4, 6, -1, -2, -1, -3, 2, 7, 0, -1, 0, -2, 3, -6, 8, -1, 0, -1, -1, 2, -5, -1, 9, 0, -2, 0, 0, 0, -4, 0, -2, 10, -1, -1, -1, -4, -1, -3, -1, -1, 4, 11, 0, 0, 0, -3, 0, -2, 0, 0, 5, -10, 12, -1, -2, -1, -2, 2, -1, -1, -2, 4, -9, 2
Offset: 1
Examples
Triangle begins: 1, 2, -1, 3, 0, -2, 4, -1, -1, -1, 5, 0, 0, 0, -4, 6, -1, -2, -1, -3, 2, 7, 0, -1, 0, -2, 3, -6, 8, -1, 0, -1, -1, 2, -5, -1, 9, 0, -2, 0, 0, 0, -4, 0, -2, 10, -1, -1, -1, -4, -1, -3, -1, -1, 4, 11, 0, 0, 0, -3, 0, -2, 0, 0, 5, -10, 12, -1, -2, -1, -2, 2, -1, -1, -2, 4, -9, 2, ...
Links
- Mats Granvik, Mathematica program for recurrence 1
- Mats Granvik, Mathematica program for recurrence 2
Programs
-
Mathematica
nn=14; f[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]]; Flatten[Table[Table[Sum[f[GCD[i, k]], {i, k, n}], {k, 1, n}], {n, 1, nn}]]
Formula
Let: f(n) = Sum_{ d divides n } d*mu(d) = A023900(n), then T(n,k) = Sum_{i=k..n} f(gcd(i,k)).
Recurrence 1:
T(n, 1) = n.
T(n, k) = [n >= k]*[k > 1]*(Sum_{j=0..n-k} Sum_{i=j+1..k-1} (T(k-1,i)-T(k,i)) -Sum_{i=n-k+1..n-1} T(i, k)).
Recurrence 2:
T(n, 1) = n.
T(n, k) = [n >= k]*(Sum_{i=n-k+1..k-1}T(k-1,i)-T(k,i)) + [n >= 2*k]*T(n-k,k).
Comments