A295113 -id:A295113 - cp's OEIS frontend

A295112 a(n) = Sum_{k=0..n} binomial(n,2k)binomial(2k,k)/(2k-1).

-1, -1, 1, 5, 13, 29, 63, 139, 317, 749, 1827, 4575, 11699, 30419, 80161, 213573, 574253, 1556077, 4244835, 11647151, 32122231, 88995879, 247573565, 691246369, 1936445619, 5441165699, 15331341373, 43308322049, 122624939677, 347957102909, 989335822559, 2818200111867

Offset: 0

Zhi-Wei Sun, Nov 14 2017

As binomial(2*k,k) = 2*(2*k-1)*A000108(k-1) for all k = 1,2,..., we see that a(n) is always an odd integer. Clearly, a(n) > 0 for all n > 1. a(n) can be viewed as an analog of Motzkin numbers, which should have some combinatorial interpretations.
Conjecture: The sequence a(n+1)/a(n) (n = 5,6,...) is strictly increasing with limit 3, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 9,10,...) is strictly decreasing to the limit 1.
See also A295113 for a conjecture involving the current sequence.

			a(3) = 5 since binomial(3,2*0)*binomial(2*0,0)/(2*0-1) + binomial(3,2*1)*binomial(2*1,1)/(2*1-1) = -1 + 3*2 = 5.

Zhi-Wei Sun, Table of n, a(n) for n = 0..400
Min Bian, Olivia X. M. Yao, Yan Zhang, and Alina F. Y. Zhao, Proof of a conjecture of Z. W. Sun, Contributions to Discrete Mathematics (2019) Vol. 14, No. 1, 214-221.
Zhi-Wei Sun, Conjectures involving arithmetical sequences, arXiv:1208.2683 [math.CO], 2012-2013; in: Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu), Proc. 6th China-Japan Seminar (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258.

Cf. A000108, A001006, A167022, A295113.

a := n -> -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4):
seq(simplify(a(n)), n=0..31); # Peter Luschny, Nov 15 2017

W[n_]:=W[n]=Sum[Binomial[n,2k]Binomial[2k,k]/(2k-1),{k,0,n/2}]; Table[W[n],{n,0,35}]
a[n_] := -AppellF1[-n, -1/2, -1/2, 1, 2, -2]; Table[a[n], {n,0,31}] (* Peter Luschny, Nov 15 2017 *)

Via the Zeilberger algorithm we have the recurrence: (n+3)*a(n+3) = (3n+7)*a(n+2) + (n-5)*a(n+1) - 3*(n+1)*a(n) for any nonnegative integer n.
a(n) = -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4). - Peter Luschny, Nov 15 2017
a(n) ~ 3^(n + 3/2) / (4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Nov 15 2017
From Mélika Tebni, Sep 03 2025: (Start)
G.f.: -sqrt((1 + x)*(1 - 3*x)) / (1 - x)^2.
a(n) = -Sum_{k=0..n} Sum_{j=0..k} A167022(j). (End)

A295132 a(n) = (2/n)Sum_{k=1..n} (2k+1)M(k)^2 where M(k) is the Motzkin number A001006(k).

Original entry on oeis.org

6, 23, 90, 432, 2286, 13176, 80418, 513764, 3400518, 23167311, 161640554, 1150633512, 8332048638, 61232315553, 455830692210, 3432015694314, 26101221114582, 200295455169015, 1549473966622602, 12074304397434552, 94713783502786686, 747454269790900728

Offset: 1

Zhi-Wei Sun, Nov 15 2017

nonn

Sun (2014) conjectures that for any prime p > 3 we have Sum_{k = 0..p-1} M(k)^2 == (2 - 6*p)(p/3) (mod p^2) and Sum_{k = 0..p - 1} k*M(k)^2 == (9*p - 1)(p/3) (mod p^2), where (p/3) is the Legendre symbol.

Sun (2018) proves that a(n) is always an integer.

			a(2) = 23 since (2/2)*Sum_{k=1..2} (2k + 1)*M(k)^2 = (2*1 + 1)*M(1)^2 + (2*2 + 1)*M(2)^2 = 3*1^2 + 5*2^2 = 23.

Zhi-Wei Sun, Table of n, a(n) for n = 1..200
Zhi-Wei Sun, Congruences involving generalized central trinomial coefficients, Sci. China Math. 57(2014), no.7, 1375-1400.
Zhi-Wei Sun, Arithmetic properties of Delannoy numbers and Schroder numbers, J. Number Theory 183(2018), 146-171.
Zhi-Wei Sun, On Motzkin numbers and central trinomial coefficients, arXiv:1801.08905 [math.CO], 2018.

Cf. A001006, A005043, A295112, A295113.

h := k -> (4*k+2)*hypergeom([(1-k)/2,-k/2],[2],4)^2:
a := proc(n) add(simplify(h(k)),k=1..n): if % mod n = 0 then %/n else -1 fi end:
seq(a(n), n=1..25); # Peter Luschny, Nov 16 2017

M[n_] := M[n] = Sum[Binomial[n, 2k] Binomial[2k, k]/(k + 1), {k, 0, n/2}];
a[n_] := a[n] = 2/n * Sum[(2k + 1) M[k]^2, {k, 1, n}];
Table[a[n], {n, 1, 25}]

a(n) = 2*A005043(n+1)*((6+6/n)*A005043(n) + (2+1/n)*A005043(n+1)). - Mark van Hoeij, Nov 10 2022

A295371 a(n) = (1/(2n))Sum_{k=0..n-1} C(n-1, k)C(n+k, k)C(2k, k)(k+2)*(-3)^(n-1-k).

Original entry on oeis.org

1, 3, 19, 127, 921, 6921, 53523, 422199, 3382417, 27429043, 224636259, 1854761437, 15419579761, 128941830993, 1083686483259, 9147887134119, 77520233226537, 659167237928691, 5622149927918763, 48083938099637247

Offset: 1

Zhi-Wei Sun, Nov 20 2017

nonn

From Mark van Hoeij, Nov 10 2022: (Start)
The fact that a(n) is an integer follows from the formula for the generating function. It is not difficult to show that f(x) := (hypergeom([1/2, 1/2], [1], 16*x) - 1)/4 is an element of Z[[x]]. Substituting x -> x/(1 + 3*x)^2 then shows that the given g.f. is in Z[[x]] as well. The fact that this formula is indeed the g.f. follows from the recurrence.
One can also show that a(n) is odd, as follows. Reducing f(x) in Z[[x]] modulo 2 gives: x + x^2 + x^4 + x^8 + x^16 + ... Again substitute x -> x/(1 + 3*x)^2, which modulo 2 is: x + x^3 + x^5 + x^7 + ... Then use the fact that (a+b)^(2^i) is congruent to a^(2^i) + b^(2^i) modulo 2 to see that f(x/(1 + 3*x)^2) is congruent to x + x^2 + x^3 + x^4 + ... modulo 2, so every a(n) is congruent to 1 modulo 2.
The formula a(n) = (A002426(n)^2 + 3*A002426(n-1)^2)/4 gives a second proof that a(n) is an odd integer. The numbers A002426(n) are odd, and so their squares are congruent to 1 modulo 8. Hence A002426(n)^2 + 3*A002426(n-1)^2 is congruent to 1 + 3 * 1 modulo 8. Since a(n) is that number divided by 4, it follows that a(n) is an odd integer. (End)

			a(3) = 19 since (1/6)*Sum_{k=0,1,2} binomial(2,k)*binomial(3+k,k)*binomial(2k,k)*(k+2)*(-3)^(2-k) = (2*(-3)^2 + 2*4*2*3*(-3) + 10*6*4)/6 = 19.

Zhi-Wei Sun, Table of n, a(n) for n = 1..200
Heba Bou KaedBey, Mark van Hoeij, and Man Cheung Tsui, Solving Third Order Linear Difference Equations in Terms of Second Order Equations, arXiv:2402.11121 [math.AC], 2024. See p. 6.

Cf. A000984, A295113, A295132, A002426.

ogf := EllipticK((4*sqrt(x))/(3*x + 1))/(2*Pi) - (1/4); ser := series(ogf, x, 22): seq(coeff(ser, x, n), n = 1..20); # Peter Luschny, Nov 10 2022

f[n_,k_]:=f[n,k]=Binomial[n-1,k]Binomial[n+k,k]Binomial[2k,k](k+2)(-3)^(n-1-k);
s[n_]:=a[n]=Sum[f[n,k],{k,0,n-1}]/(2n);
Table[s[n],{n,1,20}]

Via the Zeilberger algorithm we find that the sequence the following recurrence: (2n + 1)*(n + 3)^2*a(n + 3) = (2n + 1)*(7n^2 + 38n + 52)*a(n + 2) + 3*(2n + 5)*(7n^2 + 4n + 1)*a(n + 1) - 27*(2n + 5)*n^2*a(n).
From Mark van Hoeij, Nov 10 2022: (Start)
G.f.: (hypergeom([1/2, 1/2], [1], 16*x/(1 + 3*x)^2) - 1)/4.
a(n) = (A002426(n)^2 + 3 * A002426(n-1)^2)/4. (End)
G.f.: EllipticK((4*sqrt(x))/(3*x + 1))/(2*Pi) - (1/4). - Peter Luschny, Nov 10 2022

Name simplified based on the proof of Mark van Hoeij by Peter Luschny, Nov 10 2022

cp's OEIS Frontend

A295112 a(n) = Sum_{k=0..n} binomial(n,2k)binomial(2k,k)/(2k-1).

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A295132 a(n) = (2/n)Sum_{k=1..n} (2k+1)M(k)^2 where M(k) is the Motzkin number A001006(k).

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A295371 a(n) = (1/(2n))Sum_{k=0..n-1} C(n-1, k)C(n+k, k)C(2k, k)(k+2)*(-3)^(n-1-k).

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cp's OEIS Frontend

A295112 a(n) = Sum_{k=0..n} binomial(n,2*k)*binomial(2*k,k)/(2*k-1).

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A295132 a(n) = (2/n)*Sum_{k=1..n} (2k+1)*M(k)^2 where M(k) is the Motzkin number A001006(k).

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A295371 a(n) = (1/(2n))*Sum_{k=0..n-1} C(n-1, k)*C(n+k, k)*C(2k, k)*(k+2)*(-3)^(n-1-k).

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A295112 a(n) = Sum_{k=0..n} binomial(n,2k)binomial(2k,k)/(2k-1).

A295132 a(n) = (2/n)Sum_{k=1..n} (2k+1)M(k)^2 where M(k) is the Motzkin number A001006(k).

A295371 a(n) = (1/(2n))Sum_{k=0..n-1} C(n-1, k)C(n+k, k)C(2k, k)(k+2)*(-3)^(n-1-k).