A059397 Triangle formed by right-bounded rhombus rule, read by rows.
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
Offset: 0
Examples
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).
Links
- W. Klostermeyer et al., A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
Programs
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Maple
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
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 *)
Formula
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
Extensions
More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001
Comments