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A247349
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Regular triangle obtained by procedure described in comment in the case of m=3.
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0
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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
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OFFSET
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1,3
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COMMENTS
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"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).
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LINKS
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PROG
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(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; }
newrow(v, m) = {w = moveleft(v, m); concat(w, 1+v[1]); }
trg(nn) = {m = 3; v = [1]; for (n=1, nn, for (i=1, #v, print1(v[i], ", ")); print(); v = newrow(v, m); ); }
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CROSSREFS
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Cf. A238303 (triangle obtained when m=0).
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KEYWORD
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AUTHOR
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STATUS
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approved
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