A127771 Inverse of number triangle A(n,k) = 1/Euler_phi(n+1) if k <= n <= 2k, 0 otherwise.

1, 0, 1, 0, -1, 2, 0, 1, -2, 2, 0, 0, 0, -2, 4, 0, -1, 2, 0, -4, 2, 0, 0, 0, 0, 0, -2, 6, 0, 1, -2, 2, 0, 0, -6, 4, 0, 0, 0, 0, 0, 0, 0, -4, 6, 0, 0, 0, -2, 4, 0, 0, 0, -6, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, -4, 10, 0, -1, 2, 0, -4, 2, 0, 0, 0, 0, -10, 4

Offset

0,6

Comments

It is conjectured that all elements of the triangle are integers.

Links

Table of n, a(n) for n=0..77.

Example

Triangle begins
  1;
  0,  1;
  0, -1,  2;
  0,  1, -2,  2;
  0,  0,  0, -2,  4;
  0, -1,  2,  0, -4,  2;
  0,  0,  0,  0,  0, -2,  6;
  0,  1, -2,  2,  0,  0, -6,  4;
  0,  0,  0,  0,  0,  0,  0, -4,  6;
  0,  0,  0, -2,  4,  0,  0,  0, -6,  4;
  0,  0,  0,  0,  0,  0,  0,  0,  0, -4,  10;
  0, -1,  2,  0, -4,  2,  0,  0,  0,  0, -10,  4;
  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,   0, -4, 12;
Inverse of the triangle
  1;
  0,  1;
  0, 1/2, 1/2;
  0,  0,  1/2, 1/2;
  0,  0,  1/4, 1/4, 1/4;
  0,  0,   0,  1/2, 1/2, 1/2;
  0,  0,   0,  1/6, 1/6, 1/6,  1/6;
  0,  0,   0,   0,  1/4, 1/4,  1/4,  1/4;
  0,  0,   0,   0,  1/6, 1/6,  1/6,  1/6,  1/6;
  0,  0,   0,   0,   0,  1/4,  1/4,  1/4,  1/4,  1/4;
  0,  0,   0,   0,   0,  1/10, 1/10, 1/10, 1/10, 1/10, 1/10;

Mathematica

rows = 11;
A[n_, k_] := If[k <= n, If[n <= 2 k, 1/EulerPhi[n+1] , 0], 0];
T = Table[A[n, k], {n, 0, rows-1}, {k, 0, rows-1}] // Inverse;
Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Stefano Spezia, Sep 30 2018 *)

Crossrefs

Cf. A000010.

Row sums are A127772.

Sequence in context: A285124 A094238 A127793 * A248806 A118407 A101663

Adjacent sequences: A127768 A127769 A127770 * A127772 A127773 A127774

Keyword

sign,tabl

Author

Paul Barry, Jan 28 2007

Status

approved