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A059397
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Triangle formed by right-bounded rhombus rule, read by rows.
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3
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1, 1, 1, 1, 2, 3, 1, 3, 7, 6, 1, 4, 12, 18, 16, 1, 5, 18, 37, 53, 40, 1, 6, 25, 64, 120, 148, 109, 1, 7, 33, 100, 227, 369, 430, 297, 1, 8, 42, 146, 385, 760, 1146, 1244, 836, 1, 9, 52, 203, 606, 1391, 2518, 3519, 3656, 2377, 1, 10, 63, 272, 903, 2346, 4900, 8188
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OFFSET
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0,5
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COMMENTS
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LINKS
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W. Klostermeyer et al., A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
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FORMULA
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Each entry is sum of 3 entries above it in previous row and the entry directly above two rows back (provided the entries are properly aligned).
G.f.=G(t,z)=g(tz)/(1-zg(tz)), where g(z)=(1-z-z^2-sqrt((1+z-z^2)(1-3z-z^2)))/(2z^2). - Emeric Deutsch, Sep 03 2007
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EXAMPLE
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If triangle is reflected in the vertical axis it looks like this:
1
1 1
3 2 1
6 7 3 1
16 18 12 4 1
and now the rhombus rule is clearly visible (e.g. 18 = 6 + 7 + 3 + 2).
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MAPLE
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g:=proc(z) options operator, arrow: (1/2-(1/2)*z-(1/2)*z^2-(1/2)*sqrt((1+z-z^2)*(1-3*z-z^2)))/z^2 end proc: G:=simplify(g(t*z)/(1-z*g(t*z))): Gser:=simplify(series(G, z=0, 13)): for n from 0 to 10 do P[n]:=sort(coeff(Gser, z, n)) end do: for n from 0 to 10 do seq(coeff(P[n], t, j), j=0..n) end do; # yields sequence in triangular form - Emeric Deutsch, Sep 03 2007
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MATHEMATICA
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max = 10; g[z_] := (1 - z - z^2 - Sqrt[(1 + z - z^2)*(1 - 3*z - z^2)])/(2 z^2); s = Series[g[t*z]/(1 - z*g[t*z]), {z, 0, max}, {t, 0, max}] // Normal; t[n_, k_] := SeriesCoefficient[s, {z, 0, n}, {t, 0, k}]; t[0, 0] = 1; Table[t[n, k], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 16 2014, after Emeric Deutsch *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001
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STATUS
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approved
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