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

Author

Mats Granvik, Apr 22 2020

Keywords

• sign
• tabl

Comments

A334314(n)/A334313(n) = Sum_{k=1..n} T(n,k)/k.

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

Crossrefs

Row sums give A000012.

Cf. A191898, A309229, A023900.

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).