A245095 Triangle read by rows: T(n,k) = A006218(k)*A002865(n-k).

1, 0, 3, 1, 0, 5, 1, 3, 0, 8, 2, 3, 5, 0, 10, 2, 6, 5, 8, 0, 14, 4, 6, 10, 8, 10, 0, 16, 4, 12, 10, 16, 10, 14, 0, 20, 7, 12, 20, 16, 20, 14, 16, 0, 23, 8, 21, 20, 32, 20, 28, 16, 20, 0, 27, 12, 24, 35, 32, 40, 28, 32, 20, 23, 0, 29, 14, 36, 40, 56, 40, 56, 32, 40, 23, 27, 0, 35

Offset

1,3

Comments

Row sums give A006128, n >= 1.

Column 1 is A002865.

Leading diagonal is A006218, n >= 1.

For another version see A221530.

Links

Paolo Xausa, Table of n, a(n) for n = 1..11325 (rows 1..150 of the triangle, flattened)

Example

Triangle begins:
  1;
  0,   3;
  1,   0,  5;
  1,   3,  0,  8;
  2,   3,  5,  0, 10;
  2,   6,  5,  8,  0, 14;
  4,   6, 10,  8, 10,  0, 16;
  4,  12, 10, 16, 10, 14,  0, 20;
  7,  12, 20, 16, 20, 14, 16,  0, 23;
  8,  21, 20, 32, 20, 28, 16, 20,  0, 27;
  12, 24, 35, 32, 40, 28, 32, 20, 23,  0, 29;
  14, 36, 40, 56, 40, 56, 32, 40, 23, 27,  0, 35;
  ...
For n = 6:
  -------------------------
  k   A006218        T(6,k)
  -------------------------
  1      1  *  2   =    2
  2      3  *  2   =    6
  3      5  *  1   =    5
  4      8  *  1   =    8
  5     10  *  0   =    0
  6     14  *  1   =   14
  .         A002865
  -------------------------
So row 6 is [2, 6, 5, 8, 0, 14] and the sum of row 6 is 2+6+5+8+0+14 = 35 equaling A006128(6) = 35.

Mathematica

A245095row[n_]:=Accumulate[DivisorSigma[0, Range[n]]]Reverse[Differences[PartitionsP[Range[-1, n-1]]]]; Array[A245095row, 10] (* Paolo Xausa, Sep 04 2023 *)

Prog

(PARI) a006218(n) = sum(k=1, n, n\k);
a002865(n) = if(n, numbpart(n)-numbpart(n-1), 1);
row(n) = vector(n, i, a006218(i)*a002865(n-i)); \\ Michel Marcus, Jul 18 2014

Crossrefs

Cf. A000005, A000041, A002865, A006128, A006218, A221529, A221530, A245099.

Sequence in context: A119302 A134560 A127626 * A154791 A121440 A348016

Adjacent sequences: A245092 A245093 A245094 * A245096 A245097 A245098

Keyword

nonn,tabl

Author

Omar E. Pol, Jul 14 2014

Status

approved