A323231 - OEIS

A323231

A(n, k) = [x^k] (1/(1-x) + x/(1-x)^n), square array read by descending antidiagonals for n, k >= 0.

%I #26 Dec 15 2024 07:59:03

%S 1,2,1,1,2,1,1,2,2,1,1,2,3,2,1,1,2,4,4,2,1,1,2,5,7,5,2,1,1,2,6,11,11,

%T 6,2,1,1,2,7,16,21,16,7,2,1,1,2,8,22,36,36,22,8,2,1,1,2,9,29,57,71,57,

%U 29,9,2,1,1,2,10,37,85,127,127,85,37,10,2,1

%N A(n, k) = [x^k] (1/(1-x) + x/(1-x)^n), square array read by descending antidiagonals for n, k >= 0.

%D R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 154.

%H G. C. Greubel, <a href="/A323231/b323231.txt">Antidiagonals n = 0..50, flattened</a>

%F A(n, k) = binomial(n + k - 2, k - 1) + 1. Note that binomial(n, n) = 0 if n < 0.

%F A(n, k) = A(k, n) with the exception A(1,0) != A(0,1).

%F A(n, n) = binomial(2*n-2, n-1) + 1 = A323230(n).

%F From _G. C. Greubel_, Dec 27 2021: (Start)

%F T(n, k) = binomial(n-2, k-1) + 1 with T(n, 0) = 1 + [n=1], T(n, n) = 1.

%F T(2*n, n) = A323230(n).

%F Sum_{k=0..n} T(n,k) = n + 1 + 2^(n-2) - [n=0]/4 + [n=1]/2. (End)

%e Array starts:

%e [0] 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...

%e [1] 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ... A040000

%e [2] 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... A000027

%e [3] 1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, ... A000124

%e [4] 1, 2, 5, 11, 21, 36, 57, 85, 121, 166, 221, ... A050407

%e [5] 1, 2, 6, 16, 36, 71, 127, 211, 331, 496, 716, ... A145126

%e [6] 1, 2, 7, 22, 57, 127, 253, 463, 793, 1288, 2003, ... A323228

%e [7] 1, 2, 8, 29, 85, 211, 463, 925, 1717, 3004, 5006, ...

%e [8] 1, 2, 9, 37, 121, 331, 793, 1717, 3433, 6436, 11441, ...

%e [9] 1, 2, 10, 46, 166, 496, 1288, 3004, 6436, 12871, 24311, ...

%e .

%e Read as a triangle (by descending antidiagonals):

%e 1

%e 2, 1

%e 1, 2, 1

%e 1, 2, 2, 1

%e 1, 2, 3, 2, 1

%e 1, 2, 4, 4, 2, 1

%e 1, 2, 5, 7, 5, 2, 1

%e 1, 2, 6, 11, 11, 6, 2, 1

%e 1, 2, 7, 16, 21, 16, 7, 2, 1

%e 1, 2, 8, 22, 36, 36, 22, 8, 2, 1

%e 1, 2, 9, 29, 57, 71, 57, 29, 9, 2, 1

%e .

%e A(0, 1) = C(-1, 0) + 1 = 2 because C(-1, 0) = 1. A(1, 0) = C(-1, -1) + 1 = 1 because C(-1, -1) = 0. Warning: Some computer algebra programs (for example Maple and Mathematica) return C(n, n) = 1 for n < 0. This contradicts the definition given by Graham et al. (see reference). On the other hand this definition preserves symmetry.

%p Binomial := (n, k) -> `if`(n < 0 and n = k, 0, binomial(n,k)):

%p A := (n, k) -> Binomial(n + k - 2, k - 1) + 1:

%p seq(lprint(seq(A(n, k), k=0..10)), n=0..10);

%t T[n_, k_]:= If[k==0, 1 + Boole[n==1], If[k==n, 1, Binomial[n-2, k-1] + 1]];

%t Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* _G. C. Greubel_, Dec 27 2021 *)

%o (Sage)

%o def Arow(n):

%o R.<x> = PowerSeriesRing(ZZ, 20)

%o gf = 1/(1-x) + x/(1-x)^n

%o return gf.padded_list(10)

%o for n in (0..9): print(Arow(n))

%o (Julia)

%o using AbstractAlgebra

%o function Arow(n, len)

%o R, x = PowerSeriesRing(ZZ, len+2, "x")

%o gf = inv(1-x) + divexact(x, (1-x)^n)

%o [coeff(gf, k) for k in 0:len-1] end

%o for n in 0:9 println(Arow(n, 11)) end

%Y Differs from A323211 only in the second term.

%Y Rows include: A040000, A000027, A000124, A050407, A145126, A323228.

%Y Diagonals A(n, n+d): A323230 (d=0), A260878 (d=1), A323229 (d=2).

%Y Antidiagonal sums are A323227(n) if n!=1.

%Y Cf. A007318 (Pascal's triangle).

%K nonn,tabl

%O 0,2

%A _Peter Luschny_, Feb 10 2019