A055894 Inverse Moebius transform of Pascal's triangle A007318.

1, 1, 1, 2, 2, 2, 2, 3, 3, 2, 3, 4, 8, 4, 3, 2, 5, 10, 10, 5, 2, 4, 6, 18, 22, 18, 6, 4, 2, 7, 21, 35, 35, 21, 7, 2, 4, 8, 32, 56, 78, 56, 32, 8, 4, 3, 9, 36, 87, 126, 126, 87, 36, 9, 3, 4, 10, 50, 120, 220, 254, 220, 120, 50, 10, 4, 2, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11

Offset

0,4

Links

G. C. Greubel, Table of n, a(n) for the first 101 rows, flattened

N. J. A. Sloane, Transforms

Index entries for triangles and arrays related to Pascal's triangle

Example

Triangle starts:
  [0]  1;
  [1]  1,   1;
  [2]  2,   2,   2;
  [3]  2,   3,   3,   2;
  [4]  3,   4,   8,   4,   3;
  [5]  2,   5,  10,  10,   5,   2;
  [6]  4,   6,  18,  22,  18,   6,   4;
  [7]  2,   7,  21,  35,  35,  21,   7,   2;
  [8]  4,   8,  32,  56,  78,  56,  32,   8,   4;
  [9]  3,   9,  36,  87, 126, 126,  87,  36,   9,   3;
  ...

Mathematica

T[n_, k_] := DivisorSum[GCD[k, n], Binomial[n/#, k/#] &]; T[0, 0] = 1; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 02 2015 *)

Prog

(PARI)
T(n, k) = if(n<=0, n==0, sumdiv(gcd(n, k), d, binomial(n/d, k/d) ) );
/* print triangle: */
{ for (n=0, 17, for (k=0, n, print1(T(n, k), ", "); ); print(); ); }
/* Joerg Arndt, Oct 21 2012 */

Crossrefs

Cf. A007318.

Row sums give A055895.

Sequence in context: A335898 A143901 A115263 * A350226 A362136 A224713

Adjacent sequences: A055891 A055892 A055893 * A055895 A055896 A055897

Keyword

nonn,tabl

Author

Christian G. Bower, Jun 09 2000

Status

approved