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A210196 Triangle of coefficients of polynomials v(n,x) jointly generated with A210195; see the Formula section. 3
1, 1, 4, 1, 8, 8, 1, 12, 24, 16, 1, 16, 48, 64, 32, 1, 20, 80, 160, 160, 64, 1, 24, 120, 320, 480, 384, 128, 1, 28, 168, 560, 1120, 1344, 896, 256, 1, 32, 224, 896, 2240, 3584, 3584, 2048, 512, 1, 36, 288, 1344, 4032, 8064, 10752, 9216, 4608, 1024, 1, 40 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Periodic alternating row sums: 1, -3, 1, -3, 1, -3, ...
For a discussion and guide to related arrays, see A208510.
LINKS
FORMULA
u(n,x) = u(n-1,x) + v(n-1,x) + 1, v(n,x) = 2*x*u(n-1,x) + 2*x*v(n-1,x) + 1, where u(1,x)=1, v(1,x)=1.
Conjecture: T(n,0) = 1 and T(n,k) = 2^(k+1)*binomial(n-1,k) if k>0. - Knud Werner, Jan 10 2022
EXAMPLE
First five rows:
1;
1, 4;
1, 8, 8;
1, 12, 24, 16;
1, 16, 48, 64, 32;
First three polynomials v(n,x): 1, 1 + 4x, 1 + 8x + 8x^2.
MATHEMATICA
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := 2 x*u[n - 1, x] + 2 x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%] (* A210195 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A210196 *)
CROSSREFS
Sequence in context: A327957 A335707 A232816 * A249252 A128414 A192014
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Mar 18 2012
STATUS
approved

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Last modified March 28 09:04 EDT 2024. Contains 371240 sequences. (Running on oeis4.)