A059397 - OEIS

%I #12 Jan 05 2025 19:51:36

%S 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,

%T 109,1,7,33,100,227,369,430,297,1,8,42,146,385,760,1146,1244,836,1,9,

%U 52,203,606,1391,2518,3519,3656,2377,1,10,63,272,903,2346,4900,8188

%N Triangle formed by right-bounded rhombus rule, read by rows.

%C T(n,n)=A128720(n). Mirror image of A132276. - _Emeric Deutsch_, Sep 03 2007

%H W. Klostermeyer et al., <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/35-4/klostermeyer.pdf">A Pascal rhombus</a>, Fibonacci Quarterly, 35 (1976), 318-328.

%F 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).

%F 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

%e If triangle is reflected in the vertical axis it looks like this:

%e 1

%e 1 1

%e 3 2 1

%e 6 7 3 1

%e 16 18 12 4 1

%e and now the rhombus rule is clearly visible (e.g. 18 = 6 + 7 + 3 + 2).

%p 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

%t 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_ *)

%Y A variation on A059317. Row sums give A059398.

%Y Cf. A128720, A132276.

%K nonn,tabl,easy

%O 0,5

%A _N. J. A. Sloane_, Jan 29 2001

%E More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001