A052509 - OEIS

A052509

Knights-move Pascal triangle: T(n,k), n >= 0, 0 <= k <= n; T(n,0) = T(n,n) = 1, T(n,k) = T(n-1,k) + T(n-2,k-1) for k = 1,2,...,n-1, n >= 2.

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 7, 4, 2, 1, 1, 6, 11, 8, 4, 2, 1, 1, 7, 16, 15, 8, 4, 2, 1, 1, 8, 22, 26, 16, 8, 4, 2, 1, 1, 9, 29, 42, 31, 16, 8, 4, 2, 1, 1, 10, 37, 64, 57, 32, 16, 8, 4, 2, 1, 1, 11, 46, 93, 99, 63, 32, 16, 8, 4, 2, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,5

COMMENTS

Also square array T(n,k) (n >= 0, k >= 0) read by antidiagonals: T(n,k) = Sum_{i=0..k} binomial(n,i).

As a number triangle read by rows, this is T(n,k) = Sum_{i=n-2*k..n-k} binomial(n-k,i), with T(n,k) = T(n-1,k) + T(n-2,k-1). Row sums are A000071(n+2). Diagonal sums are A023435(n+1). It is the reverse of the Whitney triangle A004070. - Paul Barry, Sep 04 2005

Also, twice number of orthants intersected by a generic k-dimensional subspace of R^n [Naiman and Scheinerman, 2017]. - N. J. A. Sloane, Mar 03 2018

LINKS

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened

Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40 (4) (2002) 328-338, Example 1C.

Daniel Q. Naiman and Edward R. Scheinerman, Arbitrage and Geometry, arXiv:1709.07446 [q-fin.MF], 2017 [Contains the square array multiplied by 2].

Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (1998), 98-109. See Tables 5 and 14.

D. J. Price, Some unusual series occurring in n-dimensional geometry, Math. Gaz., 30 (1946), 149-150.

Index entries for triangles and arrays related to Pascal's triangle

FORMULA

T(n, k) = Sum_{m=0..n} binomial(n-k, k-m). - Wouter Meeussen, Oct 03 2002

From Werner Schulte, Feb 15 2018: (Start)

Referring to the square array T(i,j):

G.f. of row n: Sum_{k>=0} T(n,k) * x^k = (1+x)^n / (1-x).

G.f. of T(i,j): Sum_{k>=0, n>=0} T(n,k) * x^k * y^n = 1 / ((1-x)*(1-y-x*y)).

Let a_i(n) be multiplicative with a_i(p^e) = T(i, e), p prime and e >= 0, then Sum_{n>0} a_i(n)/n^s = (zeta(s))^(i+1) / (zeta(2*s))^i for i >= 0.

(End)

T(n, k) = hypergeom([-k, -n + k], [-k], -1). - Peter Luschny, Nov 28 2021

From Jianing Song, May 30 2022: (Start)

Referring to the triangle, G.f.: Sum_{n>=0, 0<=k<=n} T(n,k) * x^n * y^k = 1 / ((1-x*y)*(1-x-x^2*y)).

T(n,k) = 2^(n-k) for ceiling(n/2) <= k <= n. (End)

EXAMPLE

Triangle begins:

[0] 1;

[1] 1, 1;

[2] 1, 2, 1;

[3] 1, 3, 2, 1;

[4] 1, 4, 4, 2, 1;

[5] 1, 5, 7, 4, 2, 1;

[6] 1, 6, 11, 8, 4, 2, 1;

[7] 1, 7, 16, 15, 8, 4, 2, 1;

[8] 1, 8, 22, 26, 16, 8, 4, 2, 1;

[9] 1, 9, 29, 42, 31, 16, 8, 4, 2, 1;

As a square array, this begins:

1 1 1 1 1 1 ...

1 2 2 2 2 2 ...

1 3 4 4 4 4 ...

1 4 7 8 8 8 ...

1 5 11 15 16 ...

1 6 16 26 31 32 ...

MAPLE

a := proc(n::nonnegint, k::nonnegint) option remember: if k=0 then RETURN(1) fi: if k=n then RETURN(1) fi: a(n-1, k)+a(n-2, k-1) end: for n from 0 to 11 do for k from 0 to n do printf(`%d, `, a(n, k)) od: od: # James A. Sellers, Mar 17 2000

with(combinat): for s from 0 to 11 do for n from s to 0 by -1 do if n=0 or s-n=0 then printf(`%d, `, 1) else printf(`%d, `, sum(binomial(n, i), i=0..s-n)) fi; od: od: # James A. Sellers, Mar 17 2000

MATHEMATICA

Table[Sum[Binomial[n-k, k-m], {m, 0, n}], {n, 0, 10}, {k, 0, n}]

T[n_, k_] := Hypergeometric2F1[-k, -n + k, -k, -1];

Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Nov 28 2021 *)

PROG

(PARI) T(n, k)=sum(m=0, n, binomial(n-k, k-m));

for(n=0, 10, for(k=0, n, print1(T(n, k), ", "); ); print(); ); /* show triangle */

(Haskell)

a052509 n k = a052509_tabl !! n !! k

a052509_row n = a052509_tabl !! n

a052509_tabl = [1] : [1, 1] : f [1] [1, 1] where

f row' row = rs : f row rs where