A060184 Triangle of generalized sum of divisors function, read by rows.
1, 0, 1, 2, 0, -1, 1, 2, 1, 2, 0, 0, 1, 2, 1, 1, -2, 0, 1, 3, 1, 5, 6, 0, 0, -1, -1, 2, 1, 5, 5, -2, 0, -2, -3, 2, 2, 9, 10, 0, 1, 4, 3, 0, 4, 0, 2, 9, 9, -3, 1, 3, -2, -7, 2, 0, 3, 14, 16, 0, 2, 6, -1, -9, 2, 0, 3, 15, 17, -2, 1, 8, 19, 10, -6, 4, 0, -1, 0, 15, 22, 0, 1, 9, 21, 7, -13, 2, 0, -2, -4, 11, 20, -4, 2, 15, 33, 14, -15, 3, 0, -4, -10, 10, 28, 0, 3
Offset: 1
Examples
Triangle turned on its side begins:
1 0 2 -1 2 0 2 -2 3 0 2 ...
1 0 1 2 1 1 1 6 -1 ...
1 0 1 0 5 -1 5 ...
Links
- P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, Proc. London Math. Soc., 19 (1919), 75-113; Coll. Papers II, pp. 303-341.
Programs
-
Mathematica
max = 27(*rows*); t[n_, k_] := Module[{m, mm, q, s}, mm = Array[m, k]; s = Sum[q^Total[mm]/Times @@ (1+q^mm), Evaluate[Sequence @@ Transpose[{mm, Join[{1}, Most[mm]+1], max-Range[k-1, 0, -1]}]]]; SeriesCoefficient[s, {q, 0, n}]]; Table[Print[an = Table[t[n, k], {k, Floor[(Sqrt[8*n+1]-1)/2], 1, -1}]]; an, {n, 1, max}] // Flatten (* Jean-François Alcover, Jan 21 2014 *)
Formula
G.f. for k-th diagonal (the k-th row of the sideways triangle shown in the example): Sum_{ m_1 < m_2 < ... < m_k} q^(m_1+m_2+...+m_k)/((1+q^m_1)*(1+q^m_2)*...*(1+q^m_k)) = Sum_n T(n, k)*q^n.
Extensions
More terms from Vladeta Jovovic, Sep 20 2007
Comments