A089165 -id:A089165 - cp's OEIS frontend

A001850 Central Delannoy numbers: a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k).

1, 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823, 252055236609, 1425834724419, 8079317057869, 45849429914943, 260543813797441, 1482376214227923, 8443414161166173, 48141245001931263

Offset: 0

N. J. A. Sloane

Number of paths from (0,0) to (n,n) in an n X n grid using only steps north, northeast and east (i.e., steps (1,0), (1,1), and (0,1)).
Also the number of ways of aligning two sequences (e.g., of nucleotides or amino acids) of length n, with at most 2*n gaps (-) inserted, so that while unnecessary gappings: - -a a- - are forbidden, both b- and -b are allowed. (If only other of the latter is allowed, then the sequence A000984 gives the number of alignments.) There is an easy bijection from grid walks given by Dickau to such set of alignments (e.g., the straight diagonal corresponds to the perfect alignment with no gaps). - Antti Karttunen, Oct 10 2001
Also main diagonal of array A008288 defined by m(i,1) = m(1,j) = 1, m(i,j) = m(i-1,j-1) + m(i-1,j) + m(i,j-1). - Benoit Cloitre, May 03 2002
So, as a special case of Dmitry Zaitsev's Dec 10 2015 comment on A008288, a(n) is the number of points in Z^n that are L1 (Manhattan) distance <= n from any given point. These terms occur in the crystal ball sequences: a(n) here is the n-th term in the sequence for the n-dimensional cubic lattice. See A008288 for a list of crystal ball sequences (rows or columns of A008288). - Shel Kaphan, Dec 26 2022
a(n) is the number of n-matchings of a comb-like graph with 2*n teeth. Example: a(2) = 13 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has 13 2-matchings: any of the six possible pairs of teeth and {Aa, BC}, {Aa, CD}, {Bb, CD}, {Cc, AB}, {Dd, AB}, {Dd, BC}, {AB, CD}. - Emeric Deutsch, Jul 02 2002
Number of ordered trees with 2*n+1 edges, having root of odd degree, nonroot nodes of outdegree at most 2 and branches of odd length. - Emeric Deutsch, Aug 02 2002
The sum of the first n coefficients of ((1 - x) / (1 - 2*x))^n is a(n-1). - Michael Somos, Sep 28 2003
Row sums of A063007 and A105870. - Paul Barry, Apr 23 2005
The Hankel transform (see A001906 for definition) of this sequence is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
Also number of paths from (0,0) to (n,0) using only steps U = (1,1), H = (1,0) and D =(1,-1), U can have 2 colors and H can have 3 colors. - N-E. Fahssi, Jan 27 2008
Equals row sums of triangle A152250 and INVERT transform of A109980: (1, 2, 8, 36, 172, 852, ...). - Gary W. Adamson, Nov 30 2008
Number of overpartitions in the n X n box (treat a walk of the type in the first comment as an overpartition, by interpreting a NE step as N, E with the part thus created being overlined). - William J. Keith, May 19 2017
Diagonal of rational functions 1/(1 - x - y - x*y), 1/(1 - x - y*z - x*y*z). - Gheorghe Coserea, Jul 03 2018
Dimensions of endomorphism algebras End(R^{(n)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023
a(n) is the number of ways to tile a strip of length n with white squares, black squares, and red dominos, where we must have an equal number of white and black squares. - Greg Dresden and Leo Zhang, Jul 11 2025

			G.f. = 1 + 3*x + 13*x^2 + 63*x^3 + 321*x^4 + 1683*x^5 + 8989*x^6 + ...

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Cf. A008288, bisection of A026003, A027618, A047665, A052141, A084773, A152250, A109980, A000129, A078057, A241023, A243949.
Main diagonal of A064861.
Column k=2 of A262809 and A263159.

seq(add(multinomial(n+k,n-k,k,k),k=0..n),n=0..20); # Zerinvary Lajos, Oct 18 2006
seq(orthopoly[P](n,3), n=0..100); # Robert Israel, Nov 03 2015

f[n_] := Sum[ Binomial[n, k] Binomial[n + k, k], {k, 0, n}]; Array[f, 21, 0] (* Or *)
a[0] = 1; a[1] = 3; a[n_] := a[n] = (3(2 n - 1)a[n - 1] - (n - 1)a[n - 2])/n; Array[a, 21, 0] (* Or *)
CoefficientList[ Series[1/Sqrt[1 - 6x + x^2], {x, 0, 20}], x] (* Robert G. Wilson v *)
Table[LegendreP[n, 3], {n, 0, 22}] (* Jean-François Alcover, Jul 16 2012, from first formula *)
a[n_] := Hypergeometric2F1[-n, n+1, 1, -1]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Feb 26 2013 *)
a[ n_] := With[ {m = If[n < 0, -1 - n, n]}, SeriesCoefficient[ (1 - 6 x + x^2)^(-1/2), {x, 0, m}]]; (* Michael Somos, Jun 10 2015 *)

a(n):=coeff(expand((1+3*x+2*x^2)^n),x,n);
makelist(a(n),n,0,12); /* Emanuele Munarini, Mar 02 2011 */

{a(n) = if( n<0, n = -1 - n); polcoeff( 1 / sqrt(1 - 6*x + x^2 + x * O(x^n)), n)}; /* Michael Somos, Sep 23 2006 */

{a(n) = if( n<0, n = -1 - n); subst( pollegendre(n), x, 3)}; /* Michael Somos, Sep 23 2006 */

{a(n) = if( n<0, n = -1 - n); n++; subst( Pol(((1 - x) / (1 - 2*x) + O(x^n))^n), x, 1);} /* Michael Somos, Sep 23 2006 */

a(n)=if(n<0, 0, polcoeff((1+3*x+2*x^2)^n, n)) \\ Paul Barry, Aug 22 2007

/* same as in A092566 but use */
steps=[[1,0], [0,1], [1,1]]; /* Joerg Arndt, Jun 30 2011 */

a(n)=sum(k=0,n,binomial(n,k)*binomial(n+k,k)); \\ Joerg Arndt, May 11 2013

my(x='x+O('x^30)); Vec(1/sqrt(1 - 6*x + x^2)) \\ Altug Alkan, Oct 17 2015

# from Nick Hobson.
def f(a, b):
    if a == 0 or b == 0:
        return 1
    return f(a, b - 1) + f(a - 1, b) + f(a - 1, b - 1)
[f(n, n) for n in range(7)]

from gmpy2 import divexact
A001850 = [1, 3]
for n in range(2,10**3):
    A001850.append(divexact(A001850[-1]*(6*n-3)-(n-1)*A001850[-2],n))
# Chai Wah Wu, Sep 01 2014

a = lambda n: hypergeometric([-n, -n], [1], 2)
[simplify(a(n)) for n in range(23)] # Peter Luschny, Nov 19 2014

a(n) = P_n(3), where P_n is n-th Legendre polynomial.
G.f.: 1 / sqrt(1 - 6*x + x^2).
a(n) = a(n-1) + 2*A002002(n) = Sum_{j} A063007(n, j). - Henry Bottomley, Jul 02 2001
Dominant term in asymptotic expansion is binomial(2*n, n)/2^(1/4)*((sqrt(2) + 1)/2)^(2*n + 1)*(1 + c_1/n + c_2/n^2 + ...). - Michael David Hirschhorn
a(n) = Sum_{i=0..n} (A000079(i)*A008459(n, i)) = Sum_{i=0..n} (2^i * C(n, i)^2). - Antti Karttunen, Oct 10 2001
a(n) = Sum_{k=0..n} C(n+k, n-k)*C(2*k, k). - Benoit Cloitre, Feb 13 2003
a(n) = Sum_{k=0..n} C(n, k)^2 * 2^k. - Michael Somos, Oct 08 2003
a(n - 1) = coefficient of x^n in A120588(x)^n if n>=0. - Michael Somos, Apr 11 2012
G.f. of a(n-1) = 1 / (1 - x / (1 - 2*x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / ...)))))). - Michael Somos, May 11 2012
INVERT transform is A109980. BINOMIAL transform is A080609. BINOMIAL transform of A006139. PSUM transform is A089165. PSUMSIGN transform is A026933. First backward difference is A110170. - Michael Somos, May 11 2012
E.g.f.: exp(3*x)*BesselI(0, 2*sqrt(2)*x). - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..n} C(2*n-k, n)*C(n, k). - Paul Barry, Apr 23 2005
a(n) = Sum_{k>=n} binomial(k, n)^2/2^(k+1). - Vladeta Jovovic, Aug 25 2006
a(n) = a(-1 - n) for all n in Z. - Michael Somos, Sep 23 2006
D-finite with recurrence: a(-1) = a(0) = 1; n*a(n) = 3*(2*n-1)*a(n-1) - (n-1)*a(n-2). Eq (4) in T. D. Noe's article in JIS 9 (2006) #06.2.7.
Define general Delannoy numbers by (i,j > 0): d(i,0) = d(0,j) = 1 =: d(0,0) and d(i,j) = d(i-1,j-1) + d(i-2,j-1) + d(i-1,j). Then a(k) = Sum_{j >= 0} d(k,j)^2 + d(k-1,j)^2 = A026933(n)+A026933(n-1). This is a special case of the following formula for general Delannoy numbers: d(k,j) = Sum_{i >= 0, p=0..n} d(p, i) * d(n-p, j-i) + d(p-1, i) * d(n-p-1, j-i-1). - Peter E John, Oct 19 2006
Coefficient of x^n in (1 + 3*x + 2*x^2)^n. - N-E. Fahssi, Jan 11 2008
a(n) = A008288(A046092(n)). - Philippe Deléham, Apr 08 2009
G.f.: 1/(1 - x - 2*x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ... (continued fraction). - Paul Barry, May 28 2009
G.f.: d/dx log(1/(1 - x*A001003(x))). - Vladimir Kruchinin, Apr 19 2011
G.f.: 1/(2*Q(0) + x - 1) where Q(k) = 1 + k*(1-x) - x - x*(k + 1)*(k + 2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n) = Sum_{k=0..n} C(n,k) * C(n+k,k). - Joerg Arndt, May 11 2013
G.f.: G(0), where G(k) = 1 + x*(6 - x)*(4*k + 1)/(4*k + 2 - 2*x*(6-x)*(2*k + 1)*(4*k + 3)/(x*(6 - x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k - 1)/(x*(6 - x)*(2*k - 1) + 2*(k + 1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k + 1)/(x*(6 - x)*(2*k + 1) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
a(n)^2 = Sum_{k=0..n} 2^k * C(2*k, k)^2 * C(n+k, n-k) = A243949(n). - Paul D. Hanna, Aug 17 2014
a(n) = hypergeom([-n, -n], [1], 2). - Peter Luschny, Nov 19 2014
a(n) = Sum_{k=0..n/2} C(n-k,k) * 3^(n-2*k) * 2^k * C(n,k). - Vladimir Kruchinin, Jun 29 2015
a(n) = A049600(n, n-1).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). Cf. A126086 and A274668. - Peter Bala, Jan 15 2020
a(n) ~ c * (3 + 2*sqrt(2))^n / sqrt(n), where c = 1/sqrt(4*Pi*(3*sqrt(2)-4)) = 0.572681... (Banderier and Schwer, 2005). - Amiram Eldar, Jun 07 2020
a(n+1) = 3*a(n) + 2*Sum_{l=1..n} A006318(l)*a(n-l). [Eq. (1.16) in Qi-Shi-Guo (2016)]
a(n) ~ (1 + sqrt(2))^(2*n+1) / (2^(5/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Jan 09 2023
a(n-1) + a(n) = A241023(n) for n >= 1. - Peter Bala, Sep 18 2024
a(n) = Sum_{k=0..n} C(n+k, 2*k) * C(2*k, k). - Greg Dresden and Leo Zhang, Jul 11 2025

New name and reference Sep 15 1995

Formula and more references from Don Knuth, May 15 1996

A355565 T(j,k) are the numerators s in the representation R = s/t + (2/Pi)*u/v of the resistance between two nodes separated by the distance vector (j,k) in an infinite square lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= j, is a triangle read by rows.

Original entry on oeis.org

0, 1, 0, 2, -1, 0, 17, -4, 1, 0, 40, -49, 6, -1, 0, 401, -140, 97, -8, 1, 0, 1042, -1569, 336, -161, 10, -1, 0, 11073, -4376, 4321, -660, 241, -12, 1, 0, 29856, -48833, 13342, -9681, 1144, -337, 14, -1, 0, 325441, -136488, 160929, -33188, 18929, -1820, 449, -16, 1, 0

Offset: 0

Hugo Pfoertner, Jul 07 2022

The recurrence given by Cserti (2000), page 5, (32) is used to calculate the resistance between two arbitrarily spaced nodes in an infinite square lattice whose edges are replaced by one-ohm resistors. The lower triangle, including the diagonal, in Table I of Atkinson and Steenwijk (1999), page 487, is reproduced. The solution to the resistor grid problem shown in the xkcd Web Comic #356 "Nerd Sniping", provided in A211074, is the special case (j,k) = (2,1).

Using the terms of A280079 and A280317 as pairs of grid indices leads to strictly increasing resistances, i.e., R(A280079(m),A280317(m)) > R(A280079(i),A280317(i)) for m > i. This implies that for grid points on the same radius the resistance increases with the circumferential angle between 0 and Pi/4. The further dependence of the resistance along the circumferential angle with a fixed radius results from symmetry. - Hugo Pfoertner, Aug 31 2022

			The triangle begins:
     0;
     1,     0;
     2,    -1,   0;
    17,    -4,   1,    0;
    40,   -49,   6,   -1,  0;
   401,  -140,  97,   -8,  1,  0;
  1042, -1569, 336, -161, 10, -1, 0
.
The combined triangles used to calculate the resistances are:
  \  k      0       |        1        |       2       |      3       |
   \    s/t     u/v |    s/t    u/v   |  s/t      u/v |  s/t    u/v  |
  j \---------------|-----------------|---------------|--------------|
  0 |   0       0   |     .      .    |   .        .  |   .      .   |
  1 |   1/2     0   |    0      1     |   .        .  |   .      .   |
  2 |   2      -2   |   -1/2    2     |  0        4/3 |   .      .   |
  3 |  17/2   -12   |   -4     23/3   |  1/2      2/3 |  0     23/15 |
  4 |  40    -184/3 | - 49/2   40     |  6    -118/15 | -1/2   12/5  |
  5 | 401/2  -940/3 | -140    3323/15 | 97/2 -1118/15 | -8    499/35 |
.
continued:
  \ k     4       |      5       |
   \  s/t   u/v   | s/t    u/v   |
  j \-------------|--------------|
  0 |  .     .    |  .      .    |
  1 |  .     .    |  .      .    |
  2 |  .     .    |  .      .    |
  3 |  .     .    |  .      .    |
  4 | 0   176/105 |  .      .    |
  5 | 1/2  20/21  | 0    563/315 |
.
E.g., the resistance for a node distance vector (4,1) is R = T(4,1)/A131406(5,2) + (2/Pi)*A355566(4,1)/A355567(4,1) = -49/2 + (2/Pi)*40/1 = 80/Pi - 49/2.

See A211074 for more references and links.

Rainer Rosenthal, Table of n, a(n) for n = 0..135, rows 0..15 of triangle, flattened.
J. Cserti, Application of the lattice Green's function for calculating the resistance of infinite networks of resistors, arXiv:cond-mat/9909120 [cond-mat.mes-hall], 1999-2000.
Hugo Pfoertner, Grid points sorted by increasing R values, (2022).
Hugo Pfoertner, PARI program for inverse problem, (2022). Finds the grid point [x,y] that leads to the best approximation of a given resistance distance R (ohms) between [0,0] and [x,y].
Physics Stack Exchange, On this infinite grid of resistors, what's the equivalent resistance? Answer by user PBS, Apr 21 2018.
Rainer Rosenthal, Maple program

A131406 are the corresponding denominators t, with indices shifted by 1.
A355566 and A355567 are u and v.
Cf. A025547, A025550, A089165, A211074, A355953, A355955, A356201, A356202.
Cf. A355585, A355586, A355587, A355588 (same problem for the infinite triangular lattice).
Cf. A280079, A280317.

Maple
```
See link.
```

alphas[beta_] :=
Log[2 - Cos[beta] + Sqrt[3 + Cos[beta]*(Cos[beta] - 4)]];
Rsqu[n_, p_] :=
Simplify[(1/Pi)*
   Integrate[(1 - Exp[-Abs[n]*alphas[beta]]*Cos[p*beta])/
     Sinh[alphas[beta]], {beta, 0, Pi}]];
Table[Rsqu[n, k], {n, 0, 4}, {k, 0, n}] // TableForm (* Hugo Pfoertner, Aug 21 2022, calculates R, after Atkinson and Steenwijk *)

R(m,p,x=pi) = {if (m==0 && p==0, return(0)); if (m==1 && p==0, return(1/2)); if (m==1 && p==1, return(2/x)); if(m==p, my(mm=m-1); return(R(mm,mm)*4*mm/(2*mm+1) - R(mm-1,mm-1)*(2*mm-1)/(2*mm+1))); if (p==(m-1), my(mm=m-1); return(2*R(mm,mm) - R(mm,mm-1))); if (p==0, my(mm=m-1); return(4*R(mm,0) - R(mm-1,0) - 2*R(mm,1))); if (p0, my(mm=m-1); return(4*R(mm,p) - R(mm-1,p) - R(mm,p+1) - R(mm,p-1)))};
for(j=0,9,for(k=0,j,my(q=pi*R(j,k,pi));print1(numerator(polcoef(q,1,pi)),", "));print())

The resistance for the distance vector (j,k) is R(j,k) = T(j,k)/(1+mod(j+k,2)) +(2/Pi)*A355566(j,k)/A355567(j,k), avoiding the use of A131406.
From Rainer Rosenthal, Aug 04 2022: (Start)
R(0,0) = 0; R(1,0) = 1/2.
R(n,n) = R(n-1,n-1) + (2/Pi)/(2*n-1) for n >= 1.
R(j,k) = R(k,j) and R(-j,k) = R(j,k).
4*R(j,k) = R(j-1,k) + R(j+1,k) + R(j,k-1) + R(j,k+1) for (j,k) != (0,0).
(End)
T(j+1,0) = A089165(j)/(1 + mod(j,2)) for j >= 0. - Hugo Pfoertner, Aug 21 2022

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).

Row sums are the central Delannoy numbers (A001850). Column 0 yields A110170 (first differences of the central Delannoy numbers). sum(k*T(n,k),k=0..n)=A089165(n-1) (n>=1; partial sums of the central Delannoy numbers).

cp's OEIS Frontend

A001850 Central Delannoy numbers: a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k).

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A355565 T(j,k) are the numerators s in the representation R = s/t + (2/Pi)*u/v of the resistance between two nodes separated by the distance vector (j,k) in an infinite square lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= j, is a triangle read by rows.

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A297193 Irregular triangle read by rows: rows are partial alternating sums of rows of A297191.

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A110169 Triangle read by rows: T(n,k) (0<=k<=n) is the number of Delannoy paths of length n that start with exactly k (1,1) steps.

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