A247349 Regular triangle obtained by procedure described in comment in the case of m=3.

1, 1, 2, 2, 1, 2, 2, 1, 2, 3, 3, 2, 1, 2, 3, 2, 3, 3, 2, 1, 4, 2, 1, 4, 2, 3, 3, 3, 2, 3, 3, 3, 2, 1, 4, 3, 3, 2, 1, 4, 3, 2, 3, 3, 3, 4, 3, 2, 3, 3, 3, 3, 2, 1, 4, 3, 3, 3, 3, 2, 1, 4, 4, 3, 2, 5, 3, 2, 1, 4, 4, 3, 2, 5, 3, 3, 3, 4, 4, 4, 3, 2, 5, 3, 3, 3, 4, 3, 2, 1, 4

Offset: 1

Author

Michel Marcus, Sep 16 2014

Keywords

• nonn
• tabl

Comments

"Consider an array of numbers formed by a rotating queue: starting with just the number 1, to obtain the next row we move everything in the last row m steps to the left, with numbers at the front of the row cycling around and appearing at the back. We then append 1 plus the head of the last row to the new row." (from the Introduction of article by P. J. Graber).

Links

• Philip Jameson Graber, Rotation Remainders, arXiv:1409.4113 [math.NT], 2014.

Crossrefs

Cf. A238303 (triangle obtained when m=0).

Programs

PARI

moveleft(v, m) = {va = v; for (i=1, m, nb = #va; vb = vector(nb, i, if (i