A084768 -id:A084768 - cp's OEIS frontend

A243944 a(n) = A084768(n)^2.

1, 49, 5329, 717409, 106523041, 16735820689, 2727812288881, 456250924320961, 77788137919752001, 13459803510972477169, 2356471368269511061009, 416518496068852312607521, 74207592486779379593752801, 13309569813247406938272432721, 2400816685486139045360488325809

Offset: 0

Paul D. Hanna, Aug 18 2014

nonn

In general, we have the binomial identity:
if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k),
then b(n)^2 = Sum_{k=0..n} (t*(t+1))^k * C(2*k, k)^2 * C(n+k, n-k),
where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2),
and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
Note that the g.f. of A084768 is 1/sqrt(1 - 14*x + x^2).
Limit a(n+1)/a(n) = (7 + 4*sqrt(3))^2 = 97 + 56*sqrt(3).

			G.f.: A(x) = 1 + 49*x + 5329*x^2 + 717409*x^3 + 106523041*x^4 +...

Seiichi Manyama, Table of n, a(n) for n = 0..438

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), A243949 (m=1), A243943 (m=2), this sequence (m=3), A243007 (m=4).

Cf. A084768.

[Evaluate(LegendrePolynomial(n),7)^2 : n in [0..40]]; // G. C. Greubel, May 17 2023

Table[Sum[12^k * Binomial[2*k, k]^2 * Binomial[n+k, n-k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Sep 28 2019 *)
CoefficientList[Series[2*EllipticK[1 - (1-x)^2/(1 - 194*x + x^2)] / (Pi*Sqrt[1 - 194*x + x^2]), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 28 2019 *)
LegendreP[Range[0, 40], 7]^2 (* G. C. Greubel, May 17 2023 *)

{a(n) = sum(k=0, n, 12^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
for(n=0, 20, print1(a(n), ", "))

{a(n) = sum(k=0, n, 3^k * binomial(2*k, k) * binomial(n+k, n-k) )^2}
for(n=0, 20, print1(a(n), ", "))

/* Using AGM: */
{a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 14^2*x +x*O(x^n))), n)}
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 30 2014

[gen_legendre_P(n,0,7)^2 for n in range(41)] # G. C. Greubel, May 17 2023

G.f.: 1 / AGM(1-x, sqrt(1-194*x+x^2)).  - Paul D. Hanna, Aug 30 2014
a(n) = Sum_{k=0..n} 12^k * C(2*k, k)^2 * C(n+k, n-k).
a(n)^(1/2) = Sum_{k=0..n} 3^k * C(2*k, k) * C(n+k, n-k).
a(n) ~ (1+sqrt(3))^(8*n+4) / (sqrt(3) * Pi * n * 2^(4*n+5)). - Vaclav Kotesovec, Sep 28 2019
a(n) = (LegendreP(n, 7))^2. - G. C. Greubel, May 17 2023

A330205 Composite numbers k such that P(k, 7) == 7 (mod k), where P(k, 7) = A084768(k) is the k-th Legendre polynomial evaluated at 7.

Original entry on oeis.org

6, 15, 21, 22, 105, 119, 231, 426, 483, 1290, 1939, 4429, 4450, 4578, 10609, 12999, 14118, 16899, 23262, 26733, 37401, 39858, 82194, 108345, 121335, 127434, 302253, 380757, 724647, 836437, 840147, 1078270, 1522677, 2007411, 15009050, 28913991

Offset: 1

Amiram Eldar, Dec 05 2019

P(p, 7) == 7 (mod p) for all primes p. This is a special case of Schur congruences (see A330203 for references). This sequence consists of the composite numbers for which the congruence holds.

			6 is in the sequence since it is composite and P(6, 7) = 1651609 == 7 (mod 6).

Cf. A008316, A084768, A330203.

Select[Range[2000], CompositeQ[#] && Divisible[LegendreP[#, 7] - 7, #] &]

isok(k) = Mod(subst(pollegendre(k), x, 7), k) == 7;
forcomposite (k=1, 10000, if (isok(k), print1(k, ", "))); \\ Michel Marcus, Dec 06 2019

a, b = 1, 7
for n in range(2, 10000):
    a, b = b, ((14*n-7)*b - (n-1)*a)//n
    if (b%n == 7%n) and (not Integer(n).is_prime()): print(n)  # Robin Visser, Aug 18 2023

a(35)-a(36) from Robin Visser, Aug 18 2023

A006442 Expansion of 1/sqrt(1 - 10*x + x^2).

Original entry on oeis.org

1, 5, 37, 305, 2641, 23525, 213445, 1961825, 18205345, 170195525, 1600472677, 15122515985, 143457011569, 1365435096485, 13033485491077, 124715953657025, 1195966908404545, 11490534389896325, 110584004488276645, 1065853221648055025

Offset: 0

N. J. A. Sloane

Number of Delannoy paths from (0,0) to (n,n) with steps U(0,1), H(1,0) and D(1,1) where H can choose from two colors. - Paul Barry, May 25 2005
Number of lattice paths from (0,0) to (n,n) using steps (0,1), (1,1), and two kinds of steps (1,0). - Joerg Arndt, Jul 01 2011
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 09 2022

Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
T. R. S. Walsh, Number of sensed planar maps with n edges and m vertices

Column k=2 of A335333.
Sequences of the form LegendreP(n, 2*m+1): A000012 (m=0), A001850 (m=1), this sequence (m=2), A084768 (m=3), A084769 (m=4).
Cf. A098270, A243943 (a(n)^2).

[Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // G. C. Greubel, May 21 2023

seq(orthopoly[P](n,5), n = 0 .. 20); # Robert Israel, Aug 18 2014

Table[LegendreP[n, 5], {n, 0, 19}] (* Arkadiusz Wesolowski, Aug 13 2012 *)
CoefficientList[Series[1 / Sqrt[1 - 10 x + x^2], {x, 0, 20}], x] (* Vincenzo Librandi, Nov 23 2014 *)

PARI
```
a(n)=subst(pollegendre(n),x,5)
```

/* as lattice paths: same as in A092566 but use */
steps=[[1,0], [1,0], [0,1], [1,1]]; /* note the double [1,0] */
/* Joerg Arndt, Jul 01 2011 */

{a(n)=sum(k=0,n,binomial(n,k)^2*2^k*3^(n-k))} /* Paul D. Hanna */

{a(n) = sum(k=0, n, 2^k * binomial(2*k, k) * binomial(n+k, n-k) )}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 17 2014

[gen_legendre_P(n,0,5) for n in range(41)] # G. C. Greubel, May 21 2023

Legendre polynomial evaluated at 5. - Michael Somos, Dec 04 2001
G.f.: 1/sqrt(1 - 10*x + x^2).
a(n) equals the central coefficient of (1 + 5*x + 6*x^2)^n. - Paul D. Hanna, Jun 03 2003
a(n) equals the (n+1)-th term of the binomial transform of 1/(1-2x)^(n+1). - Paul D. Hanna, Sep 29 2003
a(n) = Sum_{k=0..n} 2^k*binomial(n, k)*binomial(n+k, k). - Benoit Cloitre, Apr 13 2004
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 2^k * 3^(n-k). - Paul D. Hanna, Feb 04 2012
E.g.f.: exp(5*x) * Bessel_I(0, 2*sqrt(6)*x). - Paul Barry, May 25 2005
D-finite with recurrence: n*a(n) - 5*(2n-1)*a(n-1) + (n-1)*a(n-2) = 0 [Eq (4) in the T. D. Noe article]. R. J. Mathar, Jun 26 2012
a(n) ~ (5 + 2*sqrt(6))^n/(2*sqrt(Pi*n)*sqrt(5*sqrt(6) - 12)). - Vaclav Kotesovec, Oct 05 2012
a(n) = hypergeom([-n, n+1], [1], -2). - Peter Luschny, May 23 2014
a(n) = Sum_{k=0..n} 2^k * C(2*k, k) * C(n+k, n-k). - Paul D. Hanna, Aug 17 2014
a(n) = Sum_{k=0..n} (k+1) * 3^k * (-1)^(n-k) * binomial(n,k) * binomial(n+k+1,n) / (n+k+1). - Vladimir Kruchinin, Nov 23 2014
From Peter Bala, Nov 28 2021: (Start)
a(n) = (1/3)*(1/2)^n*Sum_{k >= n} binomial(k,n)^2*(2/3)^k.
a(n) = (1/3)^(n+1)*hypergeom([n+1, n+1], [1], 2/3).
a(n) = (2^n)*hypergeom([-n, -n], [1], 3/2).
a(n) = [x^n] ((x - 1)*(3 - 2*x))^n
a(n) = (1/2)^n*A098270(n). (End)
a(n) = (-1)^n * Sum_{k=0..n} (1/10)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 28 2025
a(n) = Sum_{k=0..floor(n/2)} 6^k * 5^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, Aug 30 2025

A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9x + 20x^2)^n.

Original entry on oeis.org

1, 9, 121, 1809, 28401, 458649, 7544041, 125700129, 2114588641, 35836273449, 610897146201, 10463745263409, 179939616743121, 3104680678772409, 53721299280288201, 931852905510160449, 16198821321758152641

Offset: 0

Paul D. Hanna, Jun 03 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Vaclav Kotesovec, Asymptotic of a sums of powers of binomial coefficients * x^k, 2012.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Column k=4 of A335333.
Sequences of the form LegendreP(n, 2*m+1): A000012 (m=0), A001850 (m=1), A006442 (m=2), A084768 (m=3), this sequence (m=4).
Cf. A243007 (a(n)^2), A269732.

[Evaluate(LegendrePolynomial(n), 9) : n in [0..40]]; // G. C. Greubel, May 17 2023

Table[SeriesCoefficient[1/Sqrt[1-18*x+x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
LegendreP[Range[0, 40], 9] (* G. C. Greubel, May 17 2023 *)
a[n_] := 4^n*Sum[(5/4)^k*Binomial[n, k]^2, {k, 0, n}];
Table[a[n], {n, 0, 16}]  (* Detlef Meya, May 22 2024 *)

for(n=0,30,print1(subst(pollegendre(n),x,9)","))

{a(n)=sum(k=0, n, binomial(n, k)^2*4^k*5^(n-k))} \\ Paul D. Hanna, Sep 29 2012

[gen_legendre_P(n,0,9) for n in range(41)] # G. C. Greubel, May 17 2023

G.f.: 1/sqrt(1-18*x+x^2).
Also a(n) = (n+1)-th term of the binomial transform of 1/(1-4x)^(n+1).
From Paul Barry, May 25 2005: (Start)
E.g.f.: exp(9*x) * Bessel_I(0, 2*sqrt(20)*x).
a(n) = Sum_{k=0..n} C(n, k)*C(n+k, k)4^k. (End)
D-finite with recurrence: n*a(n) + 9*(1-2*n)*a(n-1) + (n-1)*a(n-2) = 0. - R. J. Mathar, Sep 27 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 4^k * 5^(n-k). - Paul D. Hanna, Sep 29 2012
a(n) ~ sqrt(200 + 90*sqrt(5))*(9 + 4*sqrt(5))^n/(20*sqrt(Pi*n)) = (2 + sqrt(5))^(2*n+1)/(5^(1/4)*2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = hypergeom([-n, n+1], [1], -4). - Peter Luschny, May 23 2014
x*exp(Sum_{n >= 1} a(n)*x^n/n) = x + 9*x^2 + 101*x^3 + 1269*x^4 + ... is an integral power series, the o.g.f. for A269732. - Peter Bala, Jan 25 2018
a(n) = (-1)^n * Sum_{k=0..n} (1/18)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 28 2025
a(n) = Sum_{k=0..floor(n/2)} 20^k * 9^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, Aug 30 2025

A355585 T(j,k) are the numerators s in the representation R = s/t + (2sqrt(3)/Pi)u/v of the resistance between two nodes separated by the distance (j,k) in an infinite triangular lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= floor(j/2) is an irregular triangle read by rows.

Original entry on oeis.org

0, 1, 8, -2, 27, -5, 928, -70, 16, 11249, -2671, 123, 46872, -34354, 5992, -438, 1792225, -445535, 28075, -10303, 23152256, -5824226, 1168304, -178754, 38336, 100685835, -25547957, 5343755, -885717, 101355, 3970817992, -338056246, 72962904, -12914726, 1825464, -386166

Offset: 0

Hugo Pfoertner, Jul 09 2022

The distance vector (j,k) is defined in an oblique coordinate system with an angle of 120 degrees between the axes, see e.g. A307012.
Atkinson and Steenwijk (1999) (see links in A211074) provided a generalization of the method used to calculate the resistance between two arbitrary nodes in an infinite square lattice of one-ohm resistors to infinite triangular lattices. Similar to the square lattice, the integral describing the resistance distance between nodes can exactly be represented by an expression of the form given in the name of this sequence with integer coefficients. Atkinson and Steenwijk, page 489, provided results for j <= 3 found by evaluation of the integral (17) (given below) and application of Mathematica's "Simplify" function.
R(j,k) = (1/Pi) * Integral_{y=0..Pi/2} (1 - exp(-|j-k|*x)*cos((j+k)*y)) / (sinh(x)*cos(y)) dy, with x = arccosh(2/cos(y)-cos(y)).
It would be useful to know whether, since the publication cited, a recurrence analogous to that known for the square lattice (used in A355565) for determining the coefficients has also been found for the triangular lattice.
The results in this sequence were found by systematic parameter variation of u and v and continued fraction expansion of the difference from the exact value of the integral for the resistance distance to determine s/t.

			The triangle begins:
          0;
          1;
          8,        -2;
         27,        -5;
        928,       -70,      16;
      11249,     -2671,     123;
      46872,    -34354,    5992,    -438;
    1792225,   -445535,   28075,  -10303;
   23152256,  -5824226, 1168304, -178754,  38336;
  100685835, -25547957, 5343755, -885717, 101355;
. The combined triangles used to calculate the resistances are:
   \ j                0              |                 1               |
   k\---------- s/t ----------- u/v -|----------- s/t ----------- u/v -|
   0|           0/1             0/ 1 |             .               .   |
   1|           1/3             0/ 1 |             .               .   |
   2|           8/3            -2/ 1 |           -2/3             1/ 1 |
   3|          27/1           -24/ 1 |           -5/1             5/ 1 |
   4|         928/3          -280/ 1 |          -70/1            64/ 1 |
   5|       11249/3         -3400/ 1 |        -2671/3           808/ 1 |
   6|       46872/1       -212538/ 5 |       -34354/3         51929/ 5 |
   7|     1792225/3      -2708944/ 5 |      -445535/3        673429/ 5 |
   8|    23152256/3    -244962336/35 |     -5824226/3      61623224/35 |
   9|   100685835/1   -3195918288/35 |    -25547957/1     810930216/35 |
  10|  3970817992/3  -42013225014/35 |   -338056246/1    2146081719/ 7 |
  11| 52514317745/3 -111125508824/ 7 | -13481564911/3  142641647567/35 |
.
continued
   \ j             2              |               3            |
   k\-------- s/t ---------- u/v -|--------- s/t -------- u/v -|
   4|        16/1          -14/ 1 |           .            .   |
   5|       123/1         -111/ 1 |           .            .   |
   6|      5992/3        -9054/ 5 |       -438/1       1989/5  |
   7|     28075/1      -127303/ 5 |     -10303/3      15576/5  |
   8|   1168304/3    -12361214/35 |    -178754/3    1891328/35 |
   9|   5343755/1   -169618717/35 |    -885717/1   28113999/35 |
  10|  72962904/1  -2315951182/35 |  -12914726/1   81986531/ 7 |
  11| 993810715/1 -31545031729/35 | -184858117/1 5867671888/35 |
.
continued
   \ j           4             |             5           |
   k\------- s/t -------- u/v -|------- s/t ------- u/v -|
   8|    38336/3    -405592/35 |         .           .   |
   9|   101355/1   -3217136/35 |         .           .   |
  10|  1825464/1  -57942922/35 |  -386166/1  12257507/35 |
  11| 28123355/1 -892677136/35 | -3085317/1  97932579/35 |
.
Using the terms for (j,k) = (10,5) with {s, t, u, v} = {-386166, 1, 12257507, 35} the resistance is R = T(10,5)/A355586(10,5) + (2*sqrt(3)/Pi) * A355587(10,5)/A355588(10,5) = -386166/1 + (2*sqrt(3)/Pi)*12257507/35 = 0.731139136228538824636... . This equals the integral for the resistance distance R(j,k) after substitution of j=10 and k=5.

See A211074 for more references and links (with alternatives).

D. Atkinson and F. J. van Steenwijk, Infinite resistive lattices, Am. J. Phys. 67 (1999), 486-492.
R. J. Mathar, Recurrence for the Atkinson-Steenwijk Integrals for Resistors in the Infinite Triangular Lattice, viXra:2208.0111 (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].

A355586 are the corresponding denominators t.
A355587 and A355588 are u and v.
Cf. A307012 (discussion of oblique coordinate system).
Cf. A084768 (when divided by 3 apparently gives the difference between successive values of s/t in column 0).
Cf. A355565, A355566, A355567 (similar problem for the square lattice).

Rtri(n,p)={my(alphat(beta)=acosh(2/cos(beta)-cos(beta))); intnum (beta=0, Pi/2, (1 - exp (-abs(n-p) * alphat(beta))*cos((n+p)*beta)) / (cos(beta)*sinh(alphat(beta)))) / Pi};
searchr (target, maxn=1000000, maxd=10, maxrat=1000, minn=0, mind=1) = {my (Rcons=2*sqrt(3)/Pi, delta=oo); for (d=mind, maxd, my(PP=Rcons/d); for (nn=minn, maxn, foreach ([-nn,nn], n, my (P=PP*n, T=target-P, Q = bestappr(T,maxrat), D=abs(target-P-Q)); if(D

\\ Alternative method using a recurrence; calculates triangle of s/t
jk(j,k) = {my(jj=j,kk=k); if(k<1,jj=j-k+1;kk=2-k); my(km=(jj+1)/2); if(kk>km, kk=2*km-kk); [jj,kk]};
D(n) = subst(pollegendre(n), 'x, 7);
ST(nend) = {my(nmax=nend+1, N=matrix(nmax,(nmax+1)\2)); for (n=2, nmax, N[n,1]=(1/3) * sum(k=0,n-2,D(k))); for (n=3, nmax, N[n,2] = (1/2)*(6*N[n-1,1] - 2*N[jk(n-1,2)[1],jk(n-1,2)[2]] - N[n-2,1] - N[n,1])); for (n=5, nmax, for (m=3, (n+1)\2, N[n,m] = 6*N[jk(n-1,m-1)[1],jk(n-1,m-1)[2]] - N[jk(n-1,m)[1],jk(n-1,m)[2]] - N[jk(n-2,m-1)[1],jk(n-2,m-1)[2]] - N[jk(n-2,m-2)[1],jk(n-2,m-2)[2]] - N[jk(n-1,m-2)[1],jk(n-1,m-2)[2]] - N[jk(n,m-1)[1],jk(n,m-1)[2]] )); N};
ST(11)

T(n,0)/A355586(n,0) = T(n-1,0)/A355586(n-1,0) + A084768(n-1)/3 for n>=1 (conjectured).

A335333 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2(2k+1)*x + x^2).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 13, 1, 1, 7, 37, 63, 1, 1, 9, 73, 305, 321, 1, 1, 11, 121, 847, 2641, 1683, 1, 1, 13, 181, 1809, 10321, 23525, 8989, 1, 1, 15, 253, 3311, 28401, 129367, 213445, 48639, 1, 1, 17, 337, 5473, 63601, 458649, 1651609, 1961825, 265729, 1

Offset: 0

Seiichi Manyama, Jun 02 2020

			Square array begins:
  1,    1,     1,      1,      1,       1, ...
  1,    3,     5,      7,      9,      11, ...
  1,   13,    37,     73,    121,     181, ...
  1,   63,   305,    847,   1809,    3311, ...
  1,  321,  2641,  10321,  28401,   63601, ...
  1, 1683, 23525, 129367, 458649, 1256651, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Eric Weisstein's World of Mathematics, Legendre Polynomial.

Columns k=0..4 give A000012, A001850, A006442, A084768, A084769.
Rows n=0..6 give A000012, A005408, A003154(n+1), A160674, A144124, A335338, A144126.
Main diagonal gives A331656.
T(n,n-1) gives A331657.
Cf. A307883, A307884.

T[n_, k_] := LegendreP[n, 2*k + 1]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 03 2021 *)

PARI
```
T(n, k) = pollegendre(n, 2*k+1);
```

T(n,k) is the coefficient of x^n in the expansion of (1 + (2*k+1)*x + k*(k+1)*x^2)^n.
T(n,k) = Sum_{j=0..n} k^j * (k+1)^(n-j) * binomial(n,j)^2.
T(n,k) = Sum_{j=0..n} k^j * binomial(n,j) * binomial(n+j,j).
n * T(n,k) = (2*k+1) * (2*n-1) * T(n-1,k) - (n-1) * T(n-2,k).
T(n,k) = P_n(2*k+1), where P_n is n-th Legendre polynomial.
From Seiichi Manyama, Aug 30 2025: (Start)
T(n,k) = (-1)^n * Sum_{j=0..n} (1/(2*(2*k+1)))^(n-2*j) * binomial(-1/2,j) * binomial(j,n-j).
T(n,k) = Sum_{j=0..floor(n/2)} (k*(k+1))^j * (2*k+1)^(n-2*j) * binomial(n,2*j) * binomial(2*j,j).
E.g.f. of column k: exp((2*k+1)*x) * BesselI(0, 2*sqrt(k*(k+1))*x). (End)

A300946 Rectangular array A(n, k) = (-1)^k*hypergeom([-k, k + n/2 - 1/2], [1], 4) with row n >= 0 and k >= 0, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 19, 1, 5, 33, 239, 1, 7, 51, 387, 3011, 1, 9, 73, 587, 4737, 38435, 1, 11, 99, 847, 7123, 59523, 496365, 1, 13, 129, 1175, 10321, 89055, 761121, 6470385, 1, 15, 163, 1579, 14499, 129367, 1135005, 9854211, 84975315

Offset: 0

Peter Luschny, Mar 16 2018

			Array starts:
[0] 1,  1,  19,  239,  3011,  38435,  496365,  6470385, ... [A299864]
[1] 1,  3,  33,  387,  4737,  59523,  761121,  9854211, ... [A299507]
[2] 1,  5,  51,  587,  7123,  89055, 1135005, 14660805, ... [A245926]
[3] 1,  7,  73,  847, 10321, 129367, 1651609, 21360031, ... [A084768]
[4] 1,  9,  99, 1175, 14499, 183195, 2351805, 30539241, ... [A245927]
[5] 1, 11, 129, 1579, 19841, 253707, 3284737, 42924203, ...
[6] 1, 13, 163, 2067, 26547, 344535, 4508877, 59402397, ...

Cf. A299864, A299507, A245926, A084768, A245927.

Cf. A300945.

Arow[n_, len_] := Table[(-1)^k Hypergeometric2F1[-k, k + n/2 - 1/2, 1, 4], {k, 0, len}]; Table[Print[Arow[n, 7]], {n, 0, 6}];

A331656 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+k,k) * n^k.

Original entry on oeis.org

1, 3, 37, 847, 28401, 1256651, 69125869, 4548342975, 348434664769, 30463322582899, 2993348092318101, 326572612514776079, 39170287549040392369, 5123157953193993402171, 725662909285939100555101, 110662236267661479984580351, 18077209893508013563092846849

Offset: 0

Ilya Gutkovskiy, Jan 23 2020

Seiichi Manyama, Table of n, a(n) for n = 0..321
Eric Weisstein's World of Mathematics, Legendre Polynomial

Main diagonal of A335333.

Cf. A001850, A006442, A084768, A084769, A110129, A331657.

Join[{1}, Table[Sum[Binomial[n, k] Binomial[n + k, k] n^k, {k, 0, n}], {n, 1, 16}]]
Table[SeriesCoefficient[1/Sqrt[1 - 2 (2 n + 1) x + x^2], {x, 0, n}], {n, 0, 16}]
Table[LegendreP[n, 2 n + 1], {n, 0, 16}]
Table[Hypergeometric2F1[-n, n + 1, 1, -n], {n, 0, 16}]

a(n) = {sum(k=0, n, binomial(n,k) * binomial(n+k,k) * n^k)} \\ Andrew Howroyd, Jan 23 2020

a(n) = central coefficient of (1 + (2*n + 1)*x + n*(n + 1)*x^2)^n.
a(n) = [x^n] 1 / sqrt(1 - 2*(2*n + 1)*x + x^2).
a(n) = n! * [x^n] exp((2*n + 1)*x) * BesselI(0,2*sqrt(n*(n + 1))*x).
a(n) = Sum_{k=0..n} binomial(n,k)^2 * n^k * (n + 1)^(n - k).
a(n) = P_n(2*n+1), where P_n is n-th Legendre polynomial.
a(n) ~ exp(1/2) * 4^n * n^(n - 1/2) / sqrt(Pi). - Vaclav Kotesovec, Jan 28 2020
From Seiichi Manyama, Aug 30 2025: (Start)
a(n) = (-1)^n * Sum_{k=0..n} (1/(2*(2*n+1)))^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k).
a(n) = Sum_{k=0..floor(n/2)} (n*(n+1))^k * (2*n+1)^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)

A331657 a(n) = Sum_{k=0..n} (-1)^(n - k) * binomial(n,k) * binomial(n+k,k) * n^k.

Original entry on oeis.org

1, 1, 13, 305, 10321, 458649, 25289461, 1666406209, 127779121345, 11178899075537, 1098961472475901, 119937806278590321, 14389588419704763409, 1882432013890951832425, 266678501426944160023653, 40673387011956179149166849, 6644919093900517186643470081

Offset: 0

Ilya Gutkovskiy, Jan 23 2020

Seiichi Manyama, Table of n, a(n) for n = 0..322
Eric Weisstein's World of Mathematics, Legendre Polynomial

Cf. A001850, A006442, A084768, A084769, A110129, A331656, A335333.

[&+[(-1)^(n-k)*Binomial(n,k)*Binomial(n+k,k)*n^k:k in [0..n]]:n in [0..16]]; // Marius A. Burtea, Jan 23 2020

Join[{1}, Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[n + k, k] n^k, {k, 0, n}], {n, 1, 16}]]
Table[SeriesCoefficient[1/Sqrt[1 - 2 (2 n - 1) x + x^2], {x, 0, n}], {n, 0, 16}]
Table[LegendreP[n, 2 n - 1], {n, 0, 16}]
Table[(-1)^n Hypergeometric2F1[-n, n + 1, 1, n], {n, 0, 16}]

a(n) = {sum(k=0, n, (-1)^(n - k) * binomial(n,k) * binomial(n+k,k) * n^k)} \\ Andrew Howroyd, Jan 23 2020

a(n) = central coefficient of (1 + (2*n - 1)*x + n*(n - 1)*x^2)^n.
a(n) = [x^n] 1 / sqrt(1 - 2*(2*n - 1)*x + x^2).
a(n) = n! * [x^n] exp((2*n - 1)*x) * BesselI(0,2*sqrt(n*(n - 1))*x).
a(n) = Sum_{k=0..n} binomial(n,k)^2 * n^k * (n - 1)^(n - k).
a(n) = P_n(2*n-1), where P_n is n-th Legendre polynomial.
a(n) = (-1)^n * 2F1(-n, n + 1; 1; n).
a(n) ~ 4^n * n^(n - 1/2) / (exp(1/2) * sqrt(Pi)). - Vaclav Kotesovec, Jan 26 2020
From Seiichi Manyama, Aug 30 2025: (Start)
a(n) = (-1)^n * Sum_{k=0..n} (1/(2*(2*n-1)))^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k).
a(n) = Sum_{k=0..floor(n/2)} ((n-1)*n)^k * (2*n-1)^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)

A299507 a(n) = (-1)^n*hypergeom([-n, n], [1], 4).

Original entry on oeis.org

1, 3, 33, 387, 4737, 59523, 761121, 9854211, 128772609, 1694927619, 22437369633, 298419470979, 3984500221569, 53376363001731, 717044895641121, 9656091923587587, 130310873022310401, 1761872309456567811, 23861153881099854369, 323634591584064809859

Offset: 0

Peter Luschny, Mar 16 2018

nonn

Cf. A299864, A245926, A084768, A245927, A253283.

Cf. A300945, A300946.

seq(simplify( (-1)^n*hypergeom([-n, n], [1], 4)), n = 0..20); # Peter Bala, Apr 18 2024

a[n_] := (-1)^n Hypergeometric2F1[-n, n, 1, 4]; Table[a[n], {n, 0, 19}]

From Vaclav Kotesovec, Jul 05 2018: (Start)
Recurrence: n*(2*n-3)*a(n) = 2*(14*n^2 - 28*n + 11)*a(n-1) - (n-2)*(2*n-1)*a(n-2).
a(n) ~ 2^(-3/2) * 3^(1/4) * (7 + 4*sqrt(3))^n / sqrt(Pi*n). (End)
From Peter Bala, Apr 18 2024: (Start)
a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k-1, k-1)*3^k = R(n, 3) for n >= 1, where R(n, x) denotes the n-th row polynomial of A253283.
a(n) = 3*n* hypergeom([1 - n, n + 1], [2], -3) for n >= 1.
a(n) = (1/2)*(LegendreP(n, 7) - LegendreP(n-1, 7)) for n >= 1.
a(n) = [x^n] ( (1 - x)/(1 - 4*x) )^n.
It follows that the Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
G.f.: (sqrt(x^2 - 14*x + 1) - x + 1)/(2*sqrt(x^2 - 14*x + 1)) = 1 + 3*x + 33*x^2 + 387*x^3 + .... (End)

cp's OEIS Frontend

A243944 a(n) = A084768(n)^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

PARI

SageMath

Formula

A330205 Composite numbers k such that P(k, 7) == 7 (mod k), where P(k, 7) = A084768(k) is the k-th Legendre polynomial evaluated at 7.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Sage

Extensions

A006442 Expansion of 1/sqrt(1 - 10*x + x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

SageMath

Formula

A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9*x + 20*x^2)^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

SageMath

Formula

A355585 T(j,k) are the numerators s in the representation R = s/t + (2*sqrt(3)/Pi)*u/v of the resistance between two nodes separated by the distance (j,k) in an infinite triangular lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= floor(j/2) is an irregular triangle read by rows.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

PARI

PARI

Formula

A335333 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2*(2*k+1)*x + x^2).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9x + 20x^2)^n.

A355585 T(j,k) are the numerators s in the representation R = s/t + (2sqrt(3)/Pi)u/v of the resistance between two nodes separated by the distance (j,k) in an infinite triangular lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= floor(j/2) is an irregular triangle read by rows.

A335333 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2(2k+1)*x + x^2).