A247349 - OEIS

%I #19 Sep 18 2014 04:52:50

%S 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,

%T 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,

%U 1,4,4,3,2,5,3,3,3,4,4,4,3,2,5,3,3,3,4,3,2,1,4

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

%C "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).

%H Philip Jameson Graber, <a href="http://arxiv.org/abs/1409.4113">Rotation Remainders</a>, arXiv:1409.4113 [math.NT], 2014.

%o (PARI) moveleft(v, m) = {va = v; for (i=1, m, nb = #va; vb = vector(nb, i, if (i<nb, va[i+1], va[1])); va = vb;); va;}

%o newrow(v, m) = {w = moveleft(v, m); concat(w, 1+v[1]);}

%o trg(nn) = {m = 3; v = [1]; for (n=1, nn, for (i=1, #v, print1(v[i], ", ")); print(); v = newrow(v, m););}

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

%K nonn,tabl

%O 1,3

%A _Michel Marcus_, Sep 16 2014