A264116 Irregular triangle read by rows: T(n,k), n>=1, k>=1, of the alternating sum of entries 1 through k in the n-th row of A235791; the first element of column k is in row k(k+1)/2.
1, 2, 3, 2, 4, 3, 5, 3, 6, 4, 5, 7, 4, 5, 8, 5, 6, 9, 5, 7, 10, 6, 8, 7, 11, 6, 8, 7, 12, 7, 10, 9, 13, 7, 10, 9, 14, 8, 11, 9, 15, 8, 12, 10, 11, 16, 9, 13, 11, 12, 17, 9, 13, 11, 12, 18, 10, 15, 12, 13, 19, 10, 15, 12, 13, 20, 11, 16, 13, 15, 21, 11, 17, 14, 16, 15, 22, 12, 18, 14, 16, 15
Offset: 1
Examples
The data in form of the irregular triangle T(n,k): 1; 2; 3, 2; 4, 3; 5, 3; 6, 4, 5; 7, 4, 5; 8, 5, 6; 9, 5, 7; 10, 6, 8, 7; 11, 6, 8, 7; 12, 7, 10, 9; 13, 7, 10, 9; 14, 8, 11, 9; 15, 8, 12, 10, 11; 16, 9, 13, 11, 12; 17, 9, 13, 11, 12; 18, 10, 15, 12, 13; 19, 10, 15, 12, 13; 20, 11, 16, 13, 15; 21, 11, 17, 14, 16, 15; 22, 12, 18, 14, 16, 15;
Links
- Hartmut F. W. Hoft, Construction process for sigma(15)
- Hartmut F. W. Hoft, Sigma(n) equals area of its symmetric representation
Crossrefs
Programs
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Mathematica
a264116[n_, k_] := Sum[(-1)^(i+1)*Ceiling[(n+1)/i - (i+1)/2], {i,k}] a264116[n_] := Map[a264116[n, #]&, Range[Floor[(Sqrt[8*n+1] - 1)/2]]] Flatten[Map[a264116,Range[22]]] (* data *)
Formula
T(n, k) = Sum_{i=1..k} (-1)^(i+1) A235791(n,i), for n>=1 and 1<=k<=floor((sqrt(8n+1) - 1)/2).
Comments