A098270 -id:A098270 - cp's OEIS frontend

A001079 a(n) = 10*a(n-1) - a(n-2); a(0) = 1, a(1) = 5.

1, 5, 49, 485, 4801, 47525, 470449, 4656965, 46099201, 456335045, 4517251249, 44716177445, 442644523201, 4381729054565, 43374646022449, 429364731169925, 4250272665676801, 42073361925598085, 416483346590304049

Offset: 0

N. J. A. Sloane

Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(6). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions >1 to the equation x^2=ceiling(x*r*floor(x/r)) where r=sqrt(6). - Benoit Cloitre, Feb 24 2004
a(n) and b(n) (A004189) are the nonnegative proper solutions to the Pell equation a(n)^2 - 6*(2*b(n))^2 = +1, n >= 0. The formula given below by Gregory V. Richardson follows. - Wolfdieter Lang, Jun 26 2013
a(n) are the integer square roots of (A032528 + 1). They are also the values of m where (A032528(m) - 1) has integer square roots. See A122653 for the integer square roots of (A032528 - 1), and see A122652 for the values of m where (A032528(m) + 1) has integer square roots. - Richard R. Forberg, Aug 05 2013
a(n) are also the values of m where floor(2m^2/3) has integer square roots, excluding m = 0. The corresponding integer square roots are given by A122652(n). - Richard R. Forberg, Nov 21 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 24 = 0. - Colin Barker, Feb 09 2014
Dickson on page 384 gives the Diophantine equation "24x^2 + 1 = y^2" and later states "y_{n+1} = 10y_n - y_{n-1}" where y_n is this sequence. - Michael Somos, Jun 19 2023

			Pell equation: n = 0: 1^2 - 24*0^2 = +1, n = 1: 5^2 - 6*(1*2)^2 = 1, n = 2: 49^2 - 6*(2*10)^2 = +1. - _Wolfdieter Lang_, Jun 26 2013
G.f. = 1 + 5*x + 49*x^2 + 485*x^3 + 4801*x^4 + 47525*x^5 + 470449*x^6 + ...

Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163-166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.
L. Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 374.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 281.

T. D. Noe, Table of n, a(n) for n=0..200
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Leonhard Euler, Vollstaendige Anleitung zur Algebra, Zweiter Teil.
Leonhard Euler, De solutione problematum diophanteorum per numeros integros, par. 18.
Tanya Khovanova, Recursive Sequences
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A004189, A001078, A046173, A046172, A036353, A138281, A004189.

I:=[1,5]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 10 2016

A001079 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,5]) ;
    else
        10*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A001079(n),n=0..20) ; # R. J. Mathar, Apr 30 2017

Table[(-1)^n Round[N[Cos[2 n ArcSin[Sqrt[3]]], 50]], {n, 0, 20}] (* Artur Jasinski, Oct 29 2008 *)
a[ n_] := ChebyshevT[n, 5]; (* Michael Somos, Aug 24 2014 *)
CoefficientList[Series[(1-5*x)/(1-10*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
a[n_] := 3^n*Sum[(2/3)^k*Binomial[2*n, 2*k], {k,0,n}]; Flatten[Table[a[n], {n,0,18}]] (* Detlef Meya, May 21 2024 *)

{a(n) = subst(poltchebi(n), 'x, 5)}; /* Michael Somos, Sep 05 2006 */

{a(n) = real((5 + 2*quadgen(24))^n)}; /* Michael Somos, Sep 05 2006 */

{a(n) = n = abs(n); polsym(1 - 10*x + x^2, n)[n+1] / 2}; /* Michael Somos, Sep 05 2006 */

x='x+O('x^30); Vec((1-5*x)/(1-10*x+x^2)) \\ G. C. Greubel, Dec 20 2017

For all members x of the sequence, 6*x^2 -6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 5) = (S(n, 10)-S(n-2, 10))/2 with S(n, x) := U(n, x/2) and T(n), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(n, 10) = A004189(n+1).
a(n) = sqrt(1+24*A004189(n)^2) (cf. Richardson comment).
a(n)*a(n+3) - a(n+1)*a(n+2) = 240. - Ralf Stephan, Jun 06 2005
Chebyshev's polynomials T(n,x) evaluated at x=5.
G.f.: (1-5*x)/(1-10*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n)= ((5+2*sqrt(6))^n + (5-2*sqrt(6))^n)/2.
a(-n) = a(n).
a(n+1) = 5*a(n) + 2*(6*a(n)^2-6)^(1/2) - Richard Choulet, Sep 19 2007
(sqrt(2)+sqrt(3))^(2*n)=a(n)+A001078(n)*sqrt(6). - Reinhard Zumkeller, Mar 12 2008
a(n+1) = 2*A054320(n) + 3*A138288(n). - Reinhard Zumkeller, Mar 12 2008
a(n) = cosh(2*n* arcsinh(sqrt(2))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^n * cos(2*n* arcsin(sqrt(3))). - Artur Jasinski, Oct 29 2008
a(n) = cos(2*n* arccos(sqrt(3))). - Artur Jasinski, Sep 10 2016
a(n) = A142238(2n-1) = A041006(2n-1) = A041038(2n-1), for all n > 0. - M. F. Hasler, Feb 14 2009
2*a(n)^2 = 3*A122652(n)^2 + 2. - Charlie Marion, Feb 01 2013
E.g.f.: cosh(2*sqrt(6)*x)*exp(5*x). - Ilya Gutkovskiy, Sep 10 2016
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 10*x + sqrt(1 + 96*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 20*x + 4*x^2) is the g.f. of A098270.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - 3/a(n)) = 1/4.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 2/a(n)) = 1/6.
Sum_{n >= 1} 1/(a(n)^2 - 3) = 1/4 - 1/sqrt(24). (End)
a(n) = 3^n*Sum_{k=0..n} (2/3)^k*binomial(2*n, 2*k). - Detlef Meya, May 21 2024

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

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

A084773 Coefficients of 1/sqrt(1-12x+4x^2); also, a(n) is the central coefficient of (1+6x+8x^2)^n.

Original entry on oeis.org

1, 6, 52, 504, 5136, 53856, 575296, 6225792, 68026624, 748832256, 8291791872, 92255680512, 1030537089024, 11550176206848, 129824329777152, 1462841567576064, 16518691986407424, 186887008999047168, 2117944490818011136, 24038305911245635584, 273199990096465494016

Offset: 0

Paul D. Hanna, Jun 10 2003

nonn

Number of Delannoy paths from (0,0) to (n,n) with steps U(0,1), H(1,0) and D(1,1) where H and D can choose from two colors each (or where one step is monochrome and the other two are bicolored). - Paul Barry, May 30 2005
2^n*P_n(3), where P_n is the n-th Legendre polynomial. 2^n*LegendreP(n,k) yields the central coefficients of (1 + 2*k*x + (k^2-1)*x^2)^n, with g.f. 1/sqrt(1 -4*k*x +4*x^2) and e.g.f. exp(2*k*x)*BesselI(0, 2*sqrt(k^2-1)*x). - Paul Barry, May 30 2005
Diagonal of rational functions 1/(1 - x - 2*y - 2*x*y), 1/(1 - x - 2*y*z - 2*x*y*z), 1/(1 - 2*x - y*z - 2*x*y*z). - Gheorghe Coserea, Jul 07 2018

			G.f.: 1/sqrt(1-2*b*x+(b^2-4*c)*x^2) yields central coefficients of (1+b*x+c*x^2)^n.

Harvey P. Dale, Table of n, a(n) for n = 0..939
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.

Sequences of the form 2^n*LegendreP(n, 2*m+1): A000079 (m=0), this sequence (m=1), A098270 (m=2).
Cf. A001850, A052141, A059474.
See A152254 for another interpretation.

[2^n*Evaluate(LegendrePolynomial(n), 3): n in [0..40]]; // G. C. Greubel, May 21 2023

a := n -> 2^n*hypergeom([-n, -n], [1], 2):
seq(simplify(a(n)), n=0..20); # Peter Luschny, May 20 2015

CoefficientList[Series[1/Sqrt[(1-12x+4x^2)],{x,0,20}],x] (* Harvey P. Dale, Dec 13 2017 *)
Table[2^n*LegendreP[n,3], {n,0,40}] (* G. C. Greubel, May 21 2023 *)

for(n=0,30,t=polcoeff((1+6*x+8*x^2)^n,n,x); print1(t","))

[2^n*gen_legendre_P(n,0,3) for n in range(41)] # G. C. Greubel, May 21 2023

a(n) = 2*A052141(n) = 2^n * A001850(n), n>0.
From Paul Barry, May 30 2005: (Start)
E.g.f.: exp(6*x)*Bessel_I(0, 2*sqrt(8)*x).
a(n) = Sum_{k=0..floor(n/2)} C(n, k)*C(2(n-k), n)*(-1)^k*3^(n-2*k). (End)
D-finite with recurrence: n*a(n) + 6*(-2*n+1)*a(n-1) + 4*(n-1)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012
G.f.: G(0)/2, where G(k) = 1 + 1/( 1 - x*(6-2*x)*(2*k+1)/(x*(6-2*x)*(2*k+1) + (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
a(n) = 2^n*hyper2F1(-n, -n, 1, 2). - Peter Luschny, May 20 2015
a(n) = A059474(n,n). - Alois P. Heinz, Oct 05 2017
a(n) ~ 2^(n - 5/4) * (1 + sqrt(2))^(2*n + 1) / sqrt(Pi*n). - Vaclav Kotesovec, Jul 11 2018

A384362 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = Sum_{i=0..kn} 2^i Sum_{j=0..i} (-1)^j * binomial(i,j) * binomial(i-j,n)^k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 10, 4, 1, 1, 74, 148, 8, 1, 1, 730, 13540, 2440, 16, 1, 1, 9002, 2308756, 3087368, 42256, 32, 1, 1, 133210, 632363044, 10208479240, 778026256, 752800, 64, 1, 1, 2299754, 253970683348, 69754997963528, 52520969994256, 207633589664, 13660480, 128, 1

Offset: 0

Seiichi Manyama, May 27 2025

			Square array begins:
  1,  1,     1,         1,              1, ...
  1,  2,    10,        74,            730, ...
  1,  4,   148,     13540,        2308756, ...
  1,  8,  2440,   3087368,    10208479240, ...
  1, 16, 42256, 778026256, 52520969994256, ...

Columns k=0..2 give A000012, A000079, A098270.
Rows n=0..1 give A000012, A004123(k+1).
Cf. A262809, A384364.

a(n, k) = sum(i=0, k*n, 2^i*sum(j=0, i, (-1)^j*binomial(i, j)*binomial(i-j, n)^k));

A(n,k) = (1/3) * Sum_{j>=0} (2/3)^j * binomial(j,n)^k.

A059473 Triangle T(n, k) is coefficient of z^nw^k in 1/(1 - 2z - 2w - 2z*w) read by rows in order 00, 10, 01, 20, 11, 02, ...

Original entry on oeis.org

1, 2, 2, 4, 10, 4, 8, 32, 32, 8, 16, 88, 148, 88, 16, 32, 224, 536, 536, 224, 32, 64, 544, 1696, 2440, 1696, 544, 64, 128, 1280, 4928, 9344, 9344, 4928, 1280, 128, 256, 2944, 13504, 31936, 42256, 31936, 13504, 2944, 256, 512, 6656, 35456, 100736, 167072, 167072, 100736, 35456, 6656, 512

Offset: 0

N. J. A. Sloane, Feb 03 2001

			[0]  1;
[1]  2,   2;
[2]  4,  10,    4;
[3]  8,  32,   32,    8;
[4] 16,  88,  148,   88,   16;
[5] 32, 224,  536,  536,  224,  32;
[6] 64, 544, 1696, 2440, 1696, 544, 64;
...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Column k = 0 gives A000079.

T(n, n) gives A098270.

read transforms; SERIES2(1/(1-2*z-2*w-2*z*w),x,y,12): SERIES2TOLIST(%,x,y,12);
# Alternative:
T := (n, k) -> 2^n*binomial(n, k)*hypergeom([-k, -n + k], [-n], -1/2):
for n from 0 to 10 do seq(simplify(T(n, k)), k = 0 .. n) end do; # Peter Luschny, Nov 26 2021

T[n_, k_] := Sum[2^(n + k - j)*Binomial[n, j]*Binomial[n + k - j, n], {j, 0, n}]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 04 2017 *)

G.f.: 1/(1 - 2*z - 2*w - 2*z*w).
T(n, k) = Sum_{j=0..n} 2^(n + k - j)*binomial(n, j)*binomial(n + k - j, n). - G. C. Greubel, Oct 04 2017
T(n, k) = 2^n*binomial(n, k)*hypergeom([-k, k - n], [-n], -1/2). - Peter Luschny, Nov 26 2021

A157077 Triangle read by rows, coefficients of the Legendre polynomials P(n, x) times 2^n: T(n, k) = 2^n * [x^k] P(n, x), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 2, -2, 0, 6, 0, -12, 0, 20, 6, 0, -60, 0, 70, 0, 60, 0, -280, 0, 252, -20, 0, 420, 0, -1260, 0, 924, 0, -280, 0, 2520, 0, -5544, 0, 3432, 70, 0, -2520, 0, 13860, 0, -24024, 0, 12870, 0, 1260, 0, -18480, 0, 72072, 0, -102960, 0, 48620, -252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756

Offset: 0

Roger L. Bagula, Feb 22 2009

			The term order is Q(x) = a_0 + a_1*x + ... + a_n*x^n. The coefficients of the first few polynomials in this order are:
{1},
{0, 2},
{-2, 0, 6},
{0, -12, 0, 20},
{6, 0, -60, 0, 70},
{0, 60, 0, -280, 0, 252},
{-20, 0, 420, 0, -1260, 0, 924},
{0, -280, 0, 2520, 0, -5544, 0, 3432},
{70, 0, -2520, 0, 13860, 0, -24024, 0, 12870},
{0, 1260, 0, -18480, 0, 72072, 0, -102960, 0, 48620},
{-252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756}.
.
From _Jon E. Schoenfield_, Jul 04 2022: (Start)
As a right-aligned triangle:
                                                             1;
                                                     0,      2;
                                                 -2, 0,      6;
                                         0,     -12, 0,     20;
                                      6, 0,     -60, 0,     70;
                              0,     60, 0,    -280, 0,    252;
                         -20, 0,    420, 0,   -1260, 0,    924;
                  0,    -280, 0,   2520, 0,   -5544, 0,   3432;
              70, 0,   -2520, 0,  13860, 0,  -24024, 0,  12870;
        0,  1260, 0,  -18480, 0,  72072, 0, -102960, 0,  48620;
  -252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756. (End)

Paul W. Haggard, Some applications of Legendre numbers, International Journal of Mathematics and Mathematical Sciences, vol. 11, Article ID 538097, 8 pages, 1988. See Table 3 p. 412.
Eric Weisstein's World of Mathematics, Legendre Polynomial

Cf. A100258, A126869, A000984, A005430, A000079, A069835, A084773, A098269, A098270.

with(orthopoly):with(PolynomialTools): seq(print(CoefficientList (2^n*P(n, x), x,termorder=forward)),n=0..10); # Peter Luschny, Dec 18 2014

Table[CoefficientList[2^n*LegendreP[n, x], x], {n, 0, 10}]; Flatten[%]

tabl(nn) = for (n=0, nn, print(Vecrev(2^n*pollegendre(n)))); \\ Michel Marcus, Dec 18 2014

def A157077_row(n):
    if n==0: return [1]
    T = [c[0] for c in (2^n*gen_legendre_P(n, 0, x)).coefficients()]
    return [0 if is_odd(n+k) else T[k//2] for k in (0..n)]
for n in range(9): print(A157077_row(n)) # Peter Luschny, Dec 19 2014

Row sums are 2^n.
From Peter Luschny, Dec 19 2014: (Start)
T(n,0) = A126869(n).
T(n,n) = A000984(n).
T(n,1) = (-1)^floor(n/2)*A005430(floor(n/2)+1) if n is odd else 0.
Let Q(n, x) = 2^n*P(n, x).
Q(n,0) = (-1)^floor(n/2)*A126869(floor(n/2)) if n is even else 0.
Q(n,1) = A000079(n).
Q(n,2) = A069835(n).
Q(n,3) = A084773(n).
Q(n,4) = A098269(n).
Q(n,5) = A098270(n). (End)
From Fabián Pereyra, Jun 30 2022: (Start)
n*T(n,k) = 2*(2*n-1)*T(n-1,k-1) - 4*(n-1)*T(n-2,k).
T(n,k) = (-1)^floor((n-k)/2)*binomial(n+k,k)*binomial(n,floor((n-k)/2))*(1+(-1)^(n-k))/2.
O.g.f.: A(x,t) = 1/sqrt(1-4*x*t+4*x^2) = 1 + (2*t)*x + (-2+6*t^2)*x^2 + (-12*t+20*t^3)*x^3 + (6-60*t^2+70*t^4)*x^4 + .... (End)

Name clarified and edited by Peter Luschny, Dec 18 2014

cp's OEIS Frontend

A001079 a(n) = 10*a(n-1) - a(n-2); a(0) = 1, a(1) = 5.

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

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A084773 Coefficients of 1/sqrt(1-12*x+4*x^2); also, a(n) is the central coefficient of (1+6*x+8*x^2)^n.

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A384362 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = Sum_{i=0..k*n} 2^i * Sum_{j=0..i} (-1)^j * binomial(i,j) * binomial(i-j,n)^k.

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A059473 Triangle T(n, k) is coefficient of z^n*w^k in 1/(1 - 2*z - 2*w - 2*z*w) read by rows in order 00, 10, 01, 20, 11, 02, ...

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A157077 Triangle read by rows, coefficients of the Legendre polynomials P(n, x) times 2^n: T(n, k) = 2^n * [x^k] P(n, x), n >= 0, 0 <= k <= n.

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A084773 Coefficients of 1/sqrt(1-12x+4x^2); also, a(n) is the central coefficient of (1+6x+8x^2)^n.

A384362 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where A(n,k) = Sum_{i=0..kn} 2^i Sum_{j=0..i} (-1)^j * binomial(i,j) * binomial(i-j,n)^k.

A059473 Triangle T(n, k) is coefficient of z^nw^k in 1/(1 - 2z - 2w - 2z*w) read by rows in order 00, 10, 01, 20, 11, 02, ...