A002894 -id:A002894 - cp's OEIS frontend

A337900 The number of walks of length 2n on the square lattice that start from the origin (0,0) and end at the vertex (2,0).

1, 16, 225, 3136, 44100, 627264, 9018009, 130873600, 1914762564, 28210561600, 418151049316, 6230734868736, 93271169290000, 1401915345465600, 21147754404155625, 320042195924198400, 4857445984927644900, 73916947787011560000, 1127482124965160372100

Offset: 1

R. J. Mathar, Sep 29 2020

			a(2) = 16 counts the walks RRRL, RRLR, RLRR, LRRR, RRUD, RRDU, RDRU, RURD, RUDR, RDUR, URRD, DRRU, URDR, DRUR, UDRR, DURR of length 4.

R. J. Mathar, Random Walk on the Square Lattice: Return to (0,0) with or without passing (1,0) (Sep 2020)

Cf. A002894 (at (0,0)), A060150 (at (1,0)), A135389 (at (1,1)), A337901 (at (3,0)), A337902 (at (2,1)).

Cf. A001791.

egf := BesselI(0, 2*x)*BesselI(2, 2*x): ser := series(egf, x, 40):
seq((2*n)!*coeff(ser, x, 2*n), n = 1..19);  # Peter Luschny, Dec 05 2024

a(n) = [A001791(n)]^2.
G.f.: x*4F3(3/2, 3/2, 2, 2; 1, 3, 3; 16*x).
D-finite with recurrence (n-1)^2*(n+1)^2*a(n) - 4*n^2*(2*n-1)^2*a(n-1) = 0.
a(n) = (2n)!*[x^(2n)] BesselI(0, 2x)*BesselI(2, 2x). - Peter Luschny, Dec 05 2024

A352838 Irregular triangle read by rows: T(n, k) is the number of 2n-step closed walks on the square lattice having algebraic area k; n >= 0, 0 <= k <= floor(n^2/4).

Original entry on oeis.org

1, 4, 28, 4, 232, 72, 12, 2156, 1008, 308, 48, 8, 21944, 13160, 5540, 1560, 420, 80, 20, 240280, 168780, 87192, 33628, 11964, 3636, 1200, 264, 72, 12, 2787320, 2168544, 1291220, 610232, 262612, 101976, 40376, 13720, 4900, 1512, 420, 112, 28

Offset: 0

Andrei Zabolotskii, Apr 05 2022

Rows can be extended to negative k with T(n, -k) = T(n, k). Sums of such extended rows give A002894.

T(n, k) is the number of words of length 2n equal to z^k in the Heisenberg group, presented as , where z=[x,y]. In particular, T(n, 0) = A307468(n).

			The table begins:
       1
       4
      28,      4
     232,     72,    12
    2156,   1008,   308,    48,     8
   21944,  13160,  5540,  1560,   420,   80,   20
  240280, 168780, 87192, 33628, 11964, 3636, 1200, 264, 72, 12
     ...
T(2, 0) = 28: the 4-step walks enclosing algebraic area 0 include 16 walks of the form "some two steps, then two steps right back" and 12 walks of the form "some step, step back, a different step, step back".
T(2, 1) = 4: the 4-step walks enclosing algebraic area 1 are the walks around each of the 4 squares touching the origin in the positive direction; cf. A334756(2, 1) = 8, which also counts walks around these squares in the negative direction.

Andrei Zabolotskii, Table of n, a(n) for n = 0..1403 (rows 0..25)
Cédric Béguin, Alain Valette and Andrzej Zuk, On the spectrum of a random walk on the discrete Heisenberg group and the norm of Harper's operator, Journal of Geometry and Physics, 21 (1997), 337-356.
Li Gan, Algebraic Area of Lattice Random Walks and Exclusion Statistics, PhD thesis, Université Paris-Saclay, 2023. See Section 2.1.2, in particular Table 2.1 (divide terms in rows with nonzero A by 2 to get this table).
Stefan Mashkevich and Stéphane Ouvry, Area Distribution of Two-Dimensional Random Walks on a Square Lattice, J. Stat. Phys., 137 (2009), 71-78.
Morteza Mohammad-Noori, Enumeration of closed random walks in the square lattice according to their areas, arXiv:1012.3720 [math.CO], 2010. Published as: Morteza Mohammad-Noori, Enumeration of walks in the square lattice according to their areas, Journal of Combinatorial Mathematics and Combinatorial Computing, 91 (2014), 257-274.

Row lengths are A033638 = A002620 + 1.
Row n ends with 4 * A026741(n) for n > 0.
Row 16 is A178106.
Cf. A307468, A002894, A353091, A385672.
A334756 counts self-avoiding walks only.

z[0, 0, 0, 0] = 1;
z[-1, ] = z[, -1, _] = z[, , -1, ] = z[, , , -1] = 0;
z[m1_, m2_, l1_, l2_] := z[m1, m2, l1, l2] = Expand[z[m1, m2, l1 - 1, l2] + z[m1, m2, l1, l2 - 1] + q^(l2 - l1) z[m1 - 1, m2, l1, l2] + q^(l1 - l2) z[m1, m2 - 1, l1, l2]];
zN[n_] := Sum[z[m, m, n/2 - m, n/2 - m], {m, 0, n/2}];
walks[n_] := Module[{gf = zN[2 n], e}, e = Exponent[gf, q, Max]; CoefficientList[gf q^e, q][[e + 1 ;;]]];
Table[walks[n], {n, 0, 8}]

A119245 Triangle, read by rows, defined by: T(n,k) = (4k+1)binomial(2n+1, n-2k)/(2n+1) for n >= 2k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 14, 20, 1, 42, 75, 9, 132, 275, 54, 1, 429, 1001, 273, 13, 1430, 3640, 1260, 104, 1, 4862, 13260, 5508, 663, 17, 16796, 48450, 23256, 3705, 170, 1, 58786, 177650, 95931, 19019, 1309, 21, 208012, 653752, 389367, 92092, 8602, 252, 1

Offset: 0

Paul D. Hanna, May 10 2006

Closely related to triangle A118919.
Row n contains 1+floor(n/2) terms.
From Peter Bala, Mar 20 2009: (Start)
Combinatorial interpretations of T(n,k):
1) The number of standard tableaux of shape (n-2*k,n+2*k).
2) The entries in column k are (with an offset of 2*k) the number of n-th generation vertices in the tree of sequences with unit increase labeled by 4*k. See [Sunik, Theorem 4]. (End)

			Triangle begins:
     1;
     1;
     2,     1;
     5,     5;
    14,    20,    1;
    42,    75,    9;
   132,   275,   54,   1;
   429,  1001,  273,  13;
  1430,  3640, 1260, 104,  1;
  4862, 13260, 5508, 663, 17; ...

Zoran Sunic, Self describing sequences and the Catalan family tree, Elect. J. Combin., 10 (No. 1, 2003). - _Peter Bala_, Mar 20 2009

Cf. A119244 (eigenvector), A088218, A000108, A000344, A001392; A118919 (variant), A158483; A002057, A002894.

f1 = (1-Sqrt[1-4*x])/(2*x);
DeleteCases[CoefficientList[Normal@Series[f1/(1 - x^2*y*f1^4),{x,0,10},{y,0,5}],{x,y}],0,Infinity]//TableForm  (* Bradley Klee, Feb 26 2018 *)
Table[(1+4*k)/(n+1+2*k)*Binomial[2*n,n+2*k],{n,0,10},{k,0,Floor[n/2]}]//TableForm (* Bradley Klee, Feb 26 2018 *)

T(n,k)=(4*k+1)*binomial(2*n+1,n-2*k)/(2*n+1)

G.f.: A(x,y) = f/(1-x^2*y*f^4), where f=(1-sqrt(1-4*x))/(2*x) is the Catalan g.f. (A000108).
Row sums equal A088218(n) = C(2*n-1,n).
T(n,0) = A000108(n) (the Catalan numbers).
T(n,1) = A000344(n).
T(n,2) = A001392(n).
Sum_{k=0..floor(n/2)} k*T(n,k) = A000346(n-2).
Eigenvector is defined by: A119244(n) = Sum_{k=0..[n\2]} T(n,k)*A119244(k).
...
T(n,k) = (4*k+1)/(n+2*k+1)*binomial(2*n,n+2*k). Compare with A158483. - Peter Bala, Mar 20 2009
T(n,k) = A039599(n, 2*k). - Johannes W. Meijer, Sep 04 2013
A002894(n) = Sum_{k=0..floor(n/2)} (binomial(2k,k)^2)*(4^(n-2*k))*T(n,k). - Bradley Klee, Feb 26 2018

A186416 a(n) = binomial(2n,n)^4/(n+1)^3.

Original entry on oeis.org

1, 2, 48, 2500, 192080, 18670176, 2125170432, 270968717448, 37634544090000, 5588044012339360, 875419364366134016, 143310129125665075392, 24338673855047938317568, 4264316875814353400000000, 767401591466550107174400000, 141345980472409642279275210000, 26569505644587874058090478570000

Offset: 0

Emanuele Munarini, Feb 21 2011

Cf. A000108, A002894, A000888, A002897, A186414, A186415.

A186416 := proc(n) binomial(2*n,n)^4/(n+1)^3 ; end proc: # R. J. Mathar, Feb 23 2011

Table[Binomial[2n,n]^4/(n+1)^3,{n,0,40}]

makelist(binomial(2*n,n)^4/(n+1)^3,n,0,40);

G.f.: 4F3(1/2,1/2,1/2,1/2;2,2,2;256*x), where nFm(...;..;.) denotes a generalized hypergeometric series.

a(n) = (A000108(n))^3*A000984(n). - R. J. Mathar, Feb 23 2011

A248586 a(n) = Sum_{i=0..n} C(n,i)*C(2i,i)^2.

Original entry on oeis.org

1, 5, 45, 521, 6733, 92385, 1316865, 19274925, 287694285, 4359037985, 66837293545, 1034774126325, 16149186405025, 253737607849445, 4009771017244485, 63681603585696321, 1015763347140335565, 16264070907887454465

Offset: 0

R. J. Mathar, Oct 09 2014

Seiichi Manyama, Table of n, a(n) for n = 0..815

Cf. A002894 (inverse binomial transform), A002893.

Table[Sum[Binomial[n, k] Binomial[2k, k]^2, {k, 0, n}],{n,0,100}] (* Emanuele Munarini, Oct 28 2016 *)

makelist(sum(binomial(n,k)*binomial(2*k,k)^2,k,0,n),n,0,12); /* Emanuele Munarini, Oct 28 2016 */

a(n) = sum(i=0, n, binomial(n,i)*binomial(2*i,i)^2); \\ Michel Marcus, Oct 09 2014

a(n) = Sum_{i=0..n} A007318(n,i)*A002894(i).
Conjecture: n^2*a(n) +(-19*n^2+19*n-5)*a(n-1) +35*(n-1)^2*a(n-2) -17*(n-1)*(n-2)*a(n-3)=0.
G.f.: LegendreP(-1/2, (1+15x)/(1-17x)) /[sqrt(1-17x)*sqrt(1-x)]. [Corrected by Robert Israel, Oct 28 2016]
From Emanuele Munarini, Oct 28 2016: (Start)
a(n) = hypergeometric(1/2,1/2,-n;1,1;-16).
G.f.: A(t) = (2/Pi)*(ellipticK(16*t/(1-t))/(1-t)).
Diff. eq. satisfied by the g.f.: t*(1-t)*(1-18*t+17*t^2)*A''(t)+(1-t)*(1-37*t+68*t^2)*A'(t)-(34*t^2-35*t+5)*A(t)=0.
Remark: the conjectured recurrence for the coefficients a(n) comes from this diff. eq. for A(t).
(End)
a(n) ~ 17^(n+1)/(16*Pi*n). - Vaclav Kotesovec, Oct 30 2016

A268147 A double binomial sum involving absolute values.

Original entry on oeis.org

0, 16, 512, 12288, 262144, 5242880, 100663296, 1879048192, 34359738368, 618475290624, 10995116277760, 193514046488576, 3377699720527872, 58546795155816448, 1008806316530991104, 17293822569102704640, 295147905179352825856, 5017514388048998039552

Offset: 0

Richard P. Brent, Jan 27 2016

A fast algorithm follows from Theorem 1 of Brent et al. article.

Colin Barker, Table of n, a(n) for n = 0..800
Richard P. Brent, Hideyuki Ohtsuka, Judy-anne H. Osborn, Helmut Prodinger, Some binomial sums involving absolute values, arXiv:1411.1477v2 [math.CO], 2016.
Index entries for linear recurrences with constant coefficients, signature (32,-256).

Cf. A000984, A002894, A166337.

a:= proc(n) option remember;
      16*`if`(n<2, n, n*a(n-1)/(n-1))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Jan 29 2016

Table[n*16^n, {n, 0, 20}] (* Jean-François Alcover, Oct 24 2016 *)
LinearRecurrence[{32,-256},{0,16},20] (* Harvey P. Dale, Jul 19 2018 *)

a(n) = sum(k=-n,n, sum(l=-n,n,binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k-l)^2));

concat(0, Vec(16*x/(1-16*x)^2 + O(x^20))) \\ Colin Barker, Feb 11 2016

a(n)=n*16^n \\ Charles R Greathouse IV, May 10 2016

a(n) = Sum_{k=-n..n} (Sum_{l=-n..n} binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k-l)^2).
From Colin Barker, Feb 11 2016: (Start)
a(n) = n*16^n.
a(n) = 32*a(n-1)-256*a(n-2) for n>1.
G.f.: 16*x / (1-16*x)^2.
(End)

A275653 a(n) = binomial(4n,2n)binomial(3n,2*n).

Original entry on oeis.org

1, 18, 1050, 77616, 6370650, 554822268, 50199951984, 4664758248000, 442077195513690, 42533571002422500, 4141601026094832300, 407220411993767798400, 40363606408574136870000, 4028061310168832261158176, 404311537318239680601595200, 40785601782042745410592271616

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)* binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
We also have Sum_{k = 0..4*n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
Compare with the identities
Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)* binomial(2*n,n) = A275655(n)
Sum_{k = 0..n} (-1)^(n+k)*binomial(8*n + k,8*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(8*n,4*n)* binomial(5*n,2*n)*binomial(2*n,n)/binomial(6*n,3*n).
See also A275652, A275654 and A275655.

Cf. A001448, A002894, A005809, A275652, A275654, A275655.

seq((4*n)!*(3*n)!/(n!*(2*n)!^3), n = 0..20);

Table[Binomial[4 n, 2 n] Binomial[3 n, 2 n], {n, 0, 15}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = (4*n)!*(3*n)!/(n!*(2*n)!^3).
a(n) = A001448(n) * A005809(n).
Recurrence: a(n) = 3*(3*n - 1)*(3*n - 2)*(4*n - 1)*(4*n - 3)/(n^2*(2*n - 1)^2) * a(n-1).
a(n) = [x^n] ((1 + x)^2/(1 - x))^(2*n) * [x^n] (1 + x)^(3*n) = [x^n] G(x)^(6*n) where G(x) = 1 + 3*x + 38*x^2 + 1150*x^3 + 47099^x^4 + 2264968*x^5 + 120311611*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^6, where F(x) = 1 + 3*x + 92*x^2 + 4579*x^3 + 282605*x^4 + 19698991*x^5 + 1484923315*x^6 + ... appears to have integer coefficients.
a(n) ~ sqrt(3/2)*108^n/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(3*n-k-1,n-k)*binomial(4*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(4*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)

A275655 a(n) = binomial(6n,3n)binomial(2n,n).

Original entry on oeis.org

1, 40, 5544, 972400, 189290920, 39089615040, 8385425017200, 1847301025078080, 415026659401497000, 94660194875011205440, 21850091031597537252544, 5092815839064962373499680, 1196622940864849837505171824, 283073284848591452381449360000

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k) *binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
We also note that Sum_{k = 0..6*n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k, k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n). See also A275652, A275653 and A275654.

Cf. A000984, A002894, A066802, A275652, A275653, A275654.

seq((6*n)!*(2*n)!/((3*n)!*n!)^2, n = 0..20);

Table[Binomial[6 n, 3 n] Binomial[2 n, n], {n, 0, 13}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = (6*n)!*(2*n)!/((3*n)!*n!)^2.
a(n) = A066802(n) * A000984(n).
Recurrence: a(n) = 16*(2*n - 1)^2*(6*n - 1)*(6*n - 5)/(n^2*(3*n - 1)*(3*n - 2)) * a(n-1).
a(n) = [x^(3*n)] (1 + x)^(6*n) * [x^n] (1 + x)^(2*n) = [x^n] G(x)^(8*n) where G(x) = 1 + 5*x + 159*x^2 + 11690*x^3 + 1160817*x^4 + 135123516*x^5 + 17357714116*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^8, where F(x) = 1 + 5*x + 359*x^2 + 42270*x^3 + 6182313*x^4 + 1021669966*x^5 + 182605696304*x^6 + ... appears to have integer coefficients.
a(n) ~ 256^n/(sqrt(3)*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(5*n-k-1,n-k)*binomial(6*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(6*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)

A306641 A(n,k) = Sum_{j=0..n} (kn)!/(j! (n-j)!)^k, square array A(n,k) read by antidiagonals, for n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 4, 4, 4, 1, 12, 36, 8, 5, 1, 48, 900, 400, 16, 6, 1, 240, 45360, 94080, 4900, 32, 7, 1, 1440, 3855600, 60614400, 11988900, 63504, 64, 8, 1, 10080, 493970400, 82065984000, 114144030000, 1704214512, 853776, 128, 9

Offset: 0

Seiichi Manyama, Mar 02 2019

Columns are the number of maximal chains to multiples of J in the graded poset of (2 X n) antimagic squares. See Stanley. - Arnav Krishna, Jan 13 2023

			Square array begins:
   1,  1,     1,          1,               1, ...
   2,  2,     4,         12,              48, ...
   3,  4,    36,        900,           45360, ...
   4,  8,   400,      94080,        60614400, ...
   5, 16,  4900,   11988900,    114144030000, ...
   6, 32, 63504, 1704214512, 249344297250048, ...

R. P. Stanley, Enumerative Combinatorics, Vol I, Exercise 53, p. 540.

Seiichi Manyama, Antidiagonals n = 0..49, flattened

Columns 0-3 give A000027(n+1), A000079, A002894, A306642, A345646.
Rows 0-1 give A000012, A208529(n+2).
Main diagonal gives A306644.

A337902 The number of walks of length 2n+1 on the square lattice that start from the origin (0,0) and end at the vertex (2,1).

Original entry on oeis.org

3, 50, 735, 10584, 152460, 2208492, 32207175, 472780880, 6982113996, 103673813880, 1546866469148, 23179817220000, 348690679038000, 5263441096145400, 79698007774092375, 1210159553338375200, 18422202264818467500, 281089726445607849000

Offset: 1

R. J. Mathar, Sep 29 2020

			a(1)=3 represents 3 walks of length 3: RRU, URR and RUR.

Cf. A002894 (at (0,0)), A060150 (at (1,0)), A135389 (at (1,1)), A337900 (at (2,0)), A337901 (at (3,0))

a(n) =  binomial(2*n+1,n-1)*binomial(2*n+1,n) = A002054(n)*A001700(n).
G.f.: 3*x*3F2(2,5/2,5/2; 3,4; 16*x).
D-finite with recurrence  (n-1)*(n+2)*(n+1)*a(n) -4*n*(2*n+1)^2*a(n-1)=0.
A135389(n) = 2*A060150(n+1) +2*a(n).

cp's OEIS Frontend

A337900 The number of walks of length 2n on the square lattice that start from the origin (0,0) and end at the vertex (2,0).

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A352838 Irregular triangle read by rows: T(n, k) is the number of 2n-step closed walks on the square lattice having algebraic area k; n >= 0, 0 <= k <= floor(n^2/4).

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A119245 Triangle, read by rows, defined by: T(n,k) = (4*k+1)*binomial(2*n+1, n-2*k)/(2*n+1) for n >= 2*k >= 0.

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A186416 a(n) = binomial(2n,n)^4/(n+1)^3.

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A248586 a(n) = Sum_{i=0..n} C(n,i)*C(2i,i)^2.

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A268147 A double binomial sum involving absolute values.

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A275653 a(n) = binomial(4*n,2*n)*binomial(3*n,2*n).

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A275655 a(n) = binomial(6*n,3*n)*binomial(2*n,n).

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A119245 Triangle, read by rows, defined by: T(n,k) = (4k+1)binomial(2n+1, n-2k)/(2n+1) for n >= 2k >= 0.

A275653 a(n) = binomial(4n,2n)binomial(3n,2*n).

A275655 a(n) = binomial(6n,3n)binomial(2n,n).

A306641 A(n,k) = Sum_{j=0..n} (kn)!/(j! (n-j)!)^k, square array A(n,k) read by antidiagonals, for n >= 0, k >= 0.