A002893 -id:A002893 - cp's OEIS frontend

A246065 a(n) = Sum_{k=0..n}C(n,k)^2C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

-1, 1, 9, 39, 177, 927, 5463, 34857, 234657, 1641471, 11820135, 87080265, 653499135, 4979882385, 38441107305, 300027646647, 2364113123073, 18784242756927, 150351698420247, 1211310469545081, 9816017765368671, 79963826730913809, 654504197331971961, 5380270242617370951

Offset: 0

Zhi-Wei Sun, Aug 24 2014

sign

a(n) is always an integer since (2k-1)|C(2k,k) for any nonnegative integer k.
Conjecture: (i) The sequence a(n+1)/a(n) (n = 2,3,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.
(ii) sum_{k=0}^{n-1}a(k) == 0 (mod n^2) for all n > 0. Moreover, for any prime p we have sum_{k=0}^{p-1}a(k) == -p^2*(1+9*(p/3))/2 (mod p^3), where (p/3) is the Legendre symbol.
We are able to prove n | sum_{k=0}^{n-1}a(k). Note also that sum_{k=0}^{n-1}a(k)*9^(n-1-k) = -n^2*A086618(n-1) for all n > 0 since both sides satisfy the same recurrence via the Zeilberger algorithm.
The congruence (0 mod n^2) in (ii) is true, see the formula for A246138 in terms of A005802. - Mark van Hoeij, Nov 07 2023

			a(2) = 9 since Sum_{k=0,1,2}C(2,k)^2*C(2k,k)/(2k-1) = -1 + 8 + 6/3 = 9.

Zhi-Wei Sun, Table of n, a(n) for n = 0..150
Zhi-Wei Sun, A new kind of numbers and their arithmetic properties, preprint, arXiv:1408.5381 [math.NT], 2014-2018.

Cf. A086618, A245769.

a := n -> -hypergeom([-1/2, -n, -n], [1, 1], 4):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Nov 07 2023
ogf := -(1-9*x)^(1/4)*hypergeom([-1/4, 3/4],[1],64*x^3/((1-9*x)*(x-1)^3))/(1-x)^(5/4);
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023

a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k]/(2k-1),{k,0,n}]
Table[a[n],{n,0,20}]

Recurrence (obtained via the Zeilberger algorithm):
9*(n+1)^2*a(n) -(19n^2+58n+63)*a(n+1) + (11n^2+46n+47)*a(n+2)-(n+3)^2*a(n+3) = 0.
a(n) ~ A086618(n)/2 ~ 3^(2*n + 5/2)/(16*Pi*n^2) as n tends to the infinity.
a(n) = (9*(2*n+1)^2*A002893(n) - 4*(n+1)^2*A002893(n+1))/3. - Mark van Hoeij, Nov 07 2023
a(n) = -hypergeom([-1/2, -n, -n], [1, 1], 4). - Peter Luschny, Nov 07 2023

A169714 The function W_5(2n) (see Borwein et al. reference for definition).

Original entry on oeis.org

1, 5, 45, 545, 7885, 127905, 2241225, 41467725, 798562125, 15855173825, 322466645545, 6687295253325, 140927922498025, 3010302779775725, 65046639827565525, 1419565970145097545, 31249959913055650125, 693192670456484513025

Offset: 0

N. J. A. Sloane, Apr 17 2010

nonn

Row sums of the fourth power of A008459. - Peter Bala, Mar 05 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J. M. Borwein, A short walk can be beautiful, 2015.
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, 2010.
Jonathan M. Borwein and Armin Straub, Mahler measures, short walks and log-sine integrals (2012)
Armin Straub, Arithmetic aspects of random walks and methods in definite integration, Ph. D. Dissertation, School Of Science And Engineering, Tulane University, 2012. - From _N. J. A. Sloane_, Dec 16 2012

Cf. A002893, A002895, A169715.

A169714 := proc(n)
    W(5,2*n) ;
end proc: # with W() from A169715, R. J. Mathar, Mar 27 2012

a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^5, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Dec 30 2013, after Peter Bala *)
max = 17; Total /@ MatrixPower[Table[Binomial[n, k]^2, {n, 0, max}, {k, 0, max}], 4] (* Jean-François Alcover, Mar 24 2015, after Peter Bala *)

Sum_{n>=0} a(n)*x^n/n!^2 = (Sum_{n>=0} x^n/n!^2)^5 = BesselI(0, 2*sqrt(x))^5. - Peter Bala, Mar 05 2013
D-finite with recurrence: n^4*a(n) = (35*n^4 - 70*n^3 + 63*n^2 - 28*n + 5)*a(n-1) - (n-1)^2*(259*n^2 - 518*n + 285)*a(n-2) + 225*(n-2)^2*(n-1)^2*a(n-3). - Vaclav Kotesovec, Mar 09 2014
a(n) ~ 5^(2*n+5/2) / (16 * Pi^2 * n^2). - Vaclav Kotesovec, Mar 09 2014

A169715 The function W_6(2n) (see Borwein et al. reference for definition).

Original entry on oeis.org

1, 6, 66, 996, 18306, 384156, 8848236, 218040696, 5651108226, 152254667436, 4229523740916, 120430899525096, 3499628148747756, 103446306284890536, 3102500089343886696, 94219208840385966096, 2892652835496484004226, 89662253086458906345036

Offset: 0

N. J. A. Sloane, Apr 17 2010

nonn

Row sums of the fifth power of A008459. - Peter Bala, Mar 05 2013

a(n)/6^(2n) is the probability that two throws of n 6-sided dice will give the same result - Henry Bottomley, Aug 30 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. Bernstein and T. Lange, Two grumpy giants and a baby, in ANTS X, Proc. Tenth Algorithmic Number Theory Symposium, 2013.
J. M. Borwein, A short walk can be beautiful, preprint, Journal of Humanistic Mathematics, Volume 6 Issue 1 (January 2016), pages 86-109.
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Some Arithmetic Properties of Short Random Walk Integrals, FPSAC 2010, San Francisco, USA.
Jonathan M. Borwein and Armin Straub, Mahler measures, short walks and log-sine integrals, preprint, Theoretical Computer Science, Volume 479, 1 April 2013, Pages 4-21.
Armin Straub, Arithmetic aspects of random walks and methods in definite integration, Ph. D. Dissertation, School Of Science And Engineering, Tulane University, 2012. - From _N. J. A. Sloane_, Dec 16 2012

Cf. A002893, A002895, A008459, A169714.

W := proc(n,s)
    local a,ai ;
    if s = 0 then
        return 1;
    end if;
    a := 0 ;
    for ai in combinat[partition](s/2) do
        if nops(ai) <= n then
            af := [op(ai),seq(0,i=1+nops(ai)..n)] ;
            a := a+combinat[numbperm](af)*(combinat[multinomial](s/2,op(ai)))^2 ;
        end if ;
    end do;
    a ;
end proc:
A169715 := proc(n)
    W(6,2*n) ;
end proc: # R. J. Mathar, Mar 27 2012

a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^6, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Dec 30 2013, after Peter Bala *)
max = 17; Total /@ MatrixPower[Table[Binomial[n, k]^2, {n, 0, max}, {k, 0, max}], 5] (* Jean-François Alcover, Mar 24 2015, after Peter Bala *)

Sum_{n>=0} a(n)*x^n/n!^2 = (Sum_{n>=0} x^n/n!^2)^6 = BesselI(0, 2*sqrt(x))^6. - Peter Bala, Mar 05 2013
Recurrence: n^5*a(n) = 2*(2*n-1)*(14*n^4 - 28*n^3 + 28*n^2 - 14*n + 3)*a(n-1) - 4*(n-1)^3*(196*n^2 - 392*n + 255)*a(n-2) + 1152*(n-2)^2*(n-1)^2*(2*n-3)*a(n-3). - Vaclav Kotesovec, Mar 09 2014
a(n) ~ 3^(2*n+3) * 4^(n-1) / (Pi*n)^(5/2). - Vaclav Kotesovec, Mar 09 2014

A244973 a(n) = Sum_{k=0..n} (-1)^kC(n, k)^2C(2k, k), where C(n, k) denotes the binomial coefficient n!/(k!(n-k)!).

Original entry on oeis.org

1, -1, -1, 17, -65, 49, 881, -5489, 12223, 42785, -479951, 1746271, 440881, -39651457, 212039855, -326783183, -2817155137, 23175692033, -68726927071, -128775914225, 2285692892785, -10156877725985, 6169206210815, 196882990135745, -1274770281690575

Offset: 0

Zhi-Wei Sun, Jul 08 2014

Zhi-Wei Sun introduced this sequence in arXiv:1407.0967. For any prime p > 5, he proved that Sum_{k=1..p-1} a(k)/k^2 == 0 (mod p) and Sum_{k=1..p-1} a(k)/k == 0 (mod p^2). This is quite similar to Wolstenholme's congruences Sum_{k=1..p-1} 1/k^2 == 0 (mod p) and Sum_{k=1..p-1} 1/k == 0 (mod p^2) for any prime p > 3.
Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n) - a(n))/(p*n)^3 is always a p-adic integer.
The author proved a weaker version of this in arXiv:1610.03384. - Zhi-Wei Sun, Nov 12 2016

			a(3) = 17 since C(3,0)^2*C(2*0,0) - C(3,1)^2*C(2,1) + C(3,2)^2*C(4,2) - C(3,3)^2*C(6,3) = 1 - 18 + 54 - 20 = 17.

Zhi-Wei Sun, Table of n, a(n) for n = 0..200
V. J. Guo, G.-S. Mao and H. Pan, Proof of a conjecture involving Sun polynomials, J. Difference Equ. Appl., 22(2016), no. 8, 1184-1197; also arXiv:1511.04005 [math.NT], 2015.
Zhi-Wei Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, arXiv:1407.0967 [math.NT], 2014-2016.
Zhi-Wei Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, Ramanujan J. 40 (2016), no. 3, 511-533.
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1016.03384 [math.NT], 2016.

Cf. A000172, A002893, A275027, A277640.

a := n -> hypergeom([1/2, -n, -n], [1, 1], -4):
seq(simplify(a(n)), n = 0..24);  # Peter Luschny, Mar 16 2025

s[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](-1)^k,{k,0,n}]
Table[s[n],{n,0,20}]

a(n) = sum(k=0, n, (-1)^k*binomial(n,k)^2*binomial(2*k,k)); \\ Michel Marcus, Nov 13 2016

Recurrence (obtained via the Zeilberger algorithm):
(n+3)^2*(4n+5)*a(n+3) + (20n^3+125n^2+254n+165)*a(n+2) + (76n^3+399n^2+678n+375)*a(n+1) - 25*(n+1)^2*(4n+9)*a(n) = 0.
Lim_sup_{n->oo} |a(n)|^(1/n) = 5. - Vaclav Kotesovec, Jul 13 2014
a(n) = Sum_{k = 0..n} (-1)^(n-k)*C(n,2*k)^2*C(2*k,k) = Sum_{k = 0..n} (-1)^(n-k)*C(n,k)*C(n,2*k)*C(n-k,k). - Zhi-Wei Sun, Nov 12 2016
Conjecture: a(n) = Sum_{k = 0..n} binomial(n, k)*b(k), where b(n) = Sum_{k = 0..n} (-1)^k*binomial(n, k)^2*binomial(2*k, n). [Added Mar 16 2025: this conjecture can be verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package for Maple]. - Peter Bala, Jul 19 2024

A246459 a(n) = Sum_{k=0..n} C(n,k)^2C(2k,k)(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 7, 55, 465, 4047, 35673, 316521, 2819295, 25173855, 225157881, 2016242265, 18070920255, 162071863425, 1454320387575, 13055422263255, 117237213829953, 1053070838993151, 9461217421304505, 85019389336077225, 764113545253570191, 6868417199986308129

Offset: 0

Zhi-Wei Sun, Aug 26 2014

nonn

Zhi-Wei Sun proved that for any n > 0 we have Sum_{k=0..n-1} a(k) = n^2*A086618(n-1), and (Sum_{k=0..n-1}a(k,x))/n is a polynomial with integer coefficients, where a(k,x) = sum_{j=0..k}C(k,j)^2*C(2j,j)*(2j+1)*x^j.

			a(2) = 55 since Sum_{k=0,1,2} C(2,k)^2*C(2k,k)(2k+1) = 1 + 8*3 + 6*5 = 55.

Zhi-Wei Sun, Table of n, a(n) for n = 0..160
Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

Cf. A086618, A245769, A246065, A246138.

A246459:=n->add(binomial(n,k)^2*binomial(2*k,k)*(2*k+1), k=0..n): seq(A246459(n), n=0..20); # Wesley Ivan Hurt, Aug 26 2014

a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](2k+1),{k,0,n}]
Table[a[n],{n,0,20}]

Recurrence (obtained via the Zeilberger algorithm): 9*(n+1)^2*a(n) - (19*n^2+74*n+87)*a(n+1) + (n+3)*(11*n+29)*a(n+2) - (n+3)^2*a(n+3) = 0.
a(n) ~ 3^(2*n+1/2) / Pi. - Vaclav Kotesovec, Aug 27 2014
a(n) = (4*n+3)*A002893(n)/3. - Mark van Hoeij, Nov 12 2023

A288456 Chebyshev coefficients of density of states of honeycomb lattice.

Original entry on oeis.org

1, -3, -15, 141, -1503, 9117, -46959, 2349, 1947969, -26479299, 166125105, -476958771, -7411008159, 122517898461, -1220344831791, 7016585864301, -14334148360575, -334610402172291, 4919241139007601, -42532841711020275, 172482611175249057, 717799148664446493, -24646866746992333551

Offset: 0

Yen-Lee Loh, Jun 16 2017

sign

This is the sequence of integers z^n g_n for n=0,2,4,6,... where g_n are the coefficients in the Chebyshev polynomial expansion of the density of states of the honeycomb lattice (z=3), g(w) = 1 / (Pi*sqrt(1-w^2)) * Sum_{n>=0} (2-delta_n) g_n T_n(w). Here |w| <= 1 and delta is the Kronecker delta.
The Chebyshev coefficients, g_n, are related to the number of walks on the lattice that return to the origin, W_n, as g_n = Sum_{k=0..n} a_{nk} z^{-k} W_k, where z is the coordination number of the lattice and a_{nk} are the coefficients of Chebyshev polynomials such that T_n(x) = Sum_{k=0..n} a_{nk} x^k.
The author was unable to obtain a closed form for z^n g_n.

Yen Lee Loh, A general method for calculating lattice Green functions on the branch cut, arXiv:1706.03083 [math-ph], 2017.

Related to numbers of walks returning to origin, W_n, on honeycomb lattice (A002893).

Whon[n_] := If[OddQ[n], 0,
   Sum[Binomial[n/2, j]^2 Binomial[2 j, j], {j, 0, n/2}]];
ank[n_, k_] := SeriesCoefficient[ChebyshevT[n, x], {x, 0, k}];
zng[n_] := Sum[ank[n, k]*3^(n - k)*Whon[k], {k, 0, n}];
Table[zng[n], {n,0,50}]

A328807 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) is Sum_{i=0..n} binomial(n,i)*Sum_{j=0..i} binomial(i,j)^k.

Original entry on oeis.org

1, 1, 3, 1, 3, 8, 1, 3, 9, 20, 1, 3, 11, 27, 48, 1, 3, 15, 45, 81, 112, 1, 3, 23, 93, 195, 243, 256, 1, 3, 39, 225, 639, 873, 729, 576, 1, 3, 71, 597, 2583, 4653, 3989, 2187, 1280, 1, 3, 135, 1665, 11991, 32133, 35169, 18483, 6561, 2816

Offset: 0

Seiichi Manyama, Oct 28 2019

T(n,k) is the constant term in the expansion of (1 + Product_{j=1..k-1} (1 + x_j) + Product_{j=1..k-1} (1 + 1/x_j))^n for k > 0.

For fixed k > 0 is T(n,k) ~ (2^k + 1)^(n + (k-1)/2) / (2^((k-1)^2/2) * sqrt(k) * (Pi*n)^((k-1)/2)). - Vaclav Kotesovec, Oct 28 2019

			Square array begins:
     1,   1,   1,    1,     1,      1, ...
     3,   3,   3,    3,     3,      3, ...
     8,   9,  11,   15,    23,     39, ...
    20,  27,  45,   93,   225,    597, ...
    48,  81, 195,  639,  2583,  11991, ...
   112, 243, 873, 4653, 32133, 260613, ...

Seiichi Manyama, Antidiagonals n = 0..100, flattened

Columns k=0..5 give A001792, A000244, A026375, A002893, A328808, A328809.
Main diagonal gives A328810.
Cf. A309010, A328747, A328748.

T[n_, k_] := Sum[Binomial[n, i] * Sum[Binomial[i, j]^k, {j, 0, i}], {i, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 06 2021 *)

A109673 Hexagonal pyramid related to Prouhet-Tarry problem.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 2, 8, 8, 2, 1, 8, 15, 8, 1, 2, 8, 8, 2, 1, 2, 1, 1, 3, 3, 1, 3, 15, 24, 15, 3, 3, 24, 60, 60, 24, 3, 1, 15, 60, 93, 60, 15, 1, 3, 24, 60, 60, 24, 3, 3, 15, 24, 15, 3, 1, 3, 3, 1, 1, 4, 6, 4, 1, 4, 24, 52, 52, 24, 4, 6, 52, 160, 228, 160, 52, 6, 4, 52

Offset: 0

Philippe Deléham, Aug 07 2005

Entries of slices [n,n] in A109672, read by rows.

Greatest numbers in each slice (central numbers) form A002893 : 1, 3, 15, 93, 639, ...

			Slice [0, 0]:
... 1 ...
Slice [1,1]:
... 1 1 ...
.. 1 3 1 ...
... 1 1 ...
Slice [2,2]:
.... 1 2 1 ...
... 2 8 8 2 ...
.. 1 8 15 8 1 ...
... 2 8 8 2 ...
.... 1 2 1 ....
Slice [3,3]:
...... 1 3 3 1 .....
.... 3 15 24 15 3 ...
... 3 24 60 60 24 3 ...
.. 1 15 60 93 60 15 1 ...
... 3 24 60 60 24 3 ...
.... 3 15 24 15 3 ....
...... 1 3 3 1 ....

Sum of terms in slice [n, n] = 3^(2n); example : 1+2+1+2+8+15+8+1+2+8+8+2+1+2+1 = 3^4 = 81 for the slice [2, 2].

A141057 Number of Abelian cubes of length 3n over an alphabet of size 3. An Abelian cube is a string of the form x x' x'' with |x| = |x'| = |x''| and x is a permutation of x' and x''.

Original entry on oeis.org

1, 3, 27, 381, 6219, 111753, 2151549, 43497891, 912018123, 19671397617, 434005899777, 9754118112951, 222621127928109, 5147503311510927, 120355825553777043, 2841378806367492381, 67648182142185172683, 1622612550613755130497, 39178199253650491044441

Offset: 0

Jeffrey Shallit, Aug 01 2008

nonn

Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for primes p >= 5 and positive integers n and k. Extending the sequence to negative n via a(-n) = Sum_{k = 0..n} C(-n,k)^3 * Sum_{j = 0..k} C(k,j)^3 produces the sequence [-1, 255, -53893, 14396623, -4388536251, 1461954981315, -518606406878589, ...] that appears to satisfy the same supercongruences. - Peter Bala, Apr 27 2022

			a(1) = 3 as the Abelian cubes are aaa, bbb, ccc.
G.f.: A(x) = 1 + 3*x + 27*x^2/2!^3 + 381*x^3/3!^3 + 6219*x^4/4!^3 +...
A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 +...]^3. - _Paul D. Hanna_

Alois P. Heinz, Table of n, a(n) for n = 0..250

Cf. A000172 (Franel numbers), A002893.

a:= proc(n) option remember; `if`(n<3, [1, 3, 27][n+1],
     ((567*n^6-3213*n^5+7083*n^4-7920*n^3+4968*n^2-1680*n+240)*a(n-1)
      -3*(3*n-4)*(63*n^5-399*n^4+1039*n^3-1380*n^2+920*n-240)*a(n-2)
      +729*(21*n^2-35*n+15)*(n-2)^4*a(n-3))/(n^4*(21*n^2-77*n+71)))
    end:
seq(a(n), n=0..20); # Alois P. Heinz, May 25 2013
A141057_list := proc(len) series(hypergeom([], [1, 1], x)^3, x, len);
seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
A141057_list(19); # Peter Luschny, May 31 2017

a[n_] := Sum[Binomial[n, k]^3 HypergeometricPFQ[{-k, -k, -k}, {1, 1}, -1], {k, 0, n}]; Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Jun 27 2019 *)

{a(n)=if(n<0,0,n!^3*polcoeff(sum(m=0,n,x^m/m!^3+x*O(x^n))^3,n))}

{a(n)=sum(k=0,n,binomial(n,k)^3*sum(j=0,k,binomial(k,j)^3))}

N=33; x='x+O('x^N)
Vec(serlaplace(serlaplace(serlaplace(sum(n=0,N,x^n/(n!^3)))^3))) /* show terms */

a(n) = sum of (n!/(n1)! (n2)! (n3!))^3 over all nonnegative n1, n2, n3 such that n1+n2+n3 = n.
G.f.: Sum_{n>=0} a(n)*x^n/n!^3 = [ Sum_{n>=0} x^n/n!^3 ]^3. - Paul D. Hanna, Jan 19 2011
a(n) = Sum_{k=0..n} C(n,k)^3 * Sum_{j=0..k} C(k,j)^3 = Sum_{k=0..n} C(n,k)^3*A000172(k). - Paul D. Hanna, Jan 20 2011
a(n) ~ 3^(3*n+2) / (4 * Pi^2 * n^2). - Vaclav Kotesovec, Sep 04 2014
a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^3. - Peter Luschny, May 31 2017

Extended by Paul D. Hanna, Jan 19 2011

Offset corrected by Alois P. Heinz, May 25 2013

A087457 Number of odd length roads between any adjacent nodes in virtual optimal chordal ring of degree 3 (the length of chord < number of nodes/2).

Original entry on oeis.org

1, 5, 31, 213, 1551, 11723, 90945, 719253, 5773279, 46889355, 384487665, 3177879675, 26442188865, 221278343445, 1860908156031, 15717475208853, 133256583398655, 1133591857814363, 9672323357640129, 82752014457666363, 709719620585186529, 6100394753270329605

Offset: 1

B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Oct 23 2003

nonn

			a(1)=1; a(2)=9*a(1)-2*2=9-4=5; a(3)=9*5-2*7=31; a(4)=9*31-2*33=213; etc

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, see page number?

Seiichi Manyama, Table of n, a(n) for n = 1..1052 (terms 1..100 from T. D. Noe)

Cf. A086617, A086618, A002893.

a := 1; s := 0; for k from 1 to 10 do for i from 0 to k do ss := ((2*(i))!/((i)!*(i+1)!))*((k)!/((i)!*(k-i)!))^2; s := s+ss; od; a := (9*a-2*s); s := 0; od;
# Alternative:
a := n -> hypergeom([1/2, -n, -n], [1, 1], 4)/3;
seq(simplify(a(n)), n = 1..22);  # Peter Luschny, Nov 06 2023

Table[Sum[Binomial[n,k]^2*Binomial[2k,k],{k,0,n}]/3,{n,1,20}] (* Vaclav Kotesovec, Oct 14 2012 *)

a(n) = sum(k=0, n, binomial(n,k)^2*binomial(2*k,k))/3; \\ Michel Marcus, May 10 2020

a(1) = 1; a(n) = 9*a(n-1) - 2*A086618(n), where A086618(n) = Sum_{k=0..n} Catalan(n)*binomial(n, k)^2, and Catalan(n) = (2*n)!/(n!*(n+1)!). - Michael Somos
a(n) = A002893(n)/3 = (1/3)*Sum_{k=0..n}binomial(n,k)^2*binomial(2k,k). - Philippe Deléham, Sep 14 2008
Recurrence: n^2*a(n) = (10*n^2-10*n+3)*a(n-1) - 9*(n-1)^2*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ 3^(2*n+1/2)/(4*Pi*n). - Vaclav Kotesovec, Oct 14 2012
G.f.: (hypergeom([1/3, 1/3],[1],-27*x*(x-1)^2/(9*x-1)^2)/(1-9*x)^(2/3)-1)/3.  - Mark van Hoeij, May 14 2013
G.f.: G(0)/(6*x*(1-9*x)^(2/3) ) -1/(3*x), where G(k)= 1 + 1/(1 - 3*(3*k+1)^2*x*(1-x)^2/(3*(3*k+1)^2*x*(1-x)^2 - (k+1)^2*(1-9*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013
a(n) = hypergeom([1/2, -n, -n], [1, 1], 4) / 3. - Peter Luschny, Nov 06 2023

cp's OEIS Frontend

A246065 a(n) = Sum_{k=0..n}C(n,k)^2*C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

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A169714 The function W_5(2n) (see Borwein et al. reference for definition).

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A169715 The function W_6(2n) (see Borwein et al. reference for definition).

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A244973 a(n) = Sum_{k=0..n} (-1)^k*C(n, k)^2*C(2*k, k), where C(n, k) denotes the binomial coefficient n!/(k!*(n-k)!).

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A246459 a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

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A288456 Chebyshev coefficients of density of states of honeycomb lattice.

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A328807 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) is Sum_{i=0..n} binomial(n,i)*Sum_{j=0..i} binomial(i,j)^k.

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A109673 Hexagonal pyramid related to Prouhet-Tarry problem.

A246065 a(n) = Sum_{k=0..n}C(n,k)^2C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

A244973 a(n) = Sum_{k=0..n} (-1)^kC(n, k)^2C(2k, k), where C(n, k) denotes the binomial coefficient n!/(k!(n-k)!).

A246459 a(n) = Sum_{k=0..n} C(n,k)^2C(2k,k)(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).