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

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

Offset

1,1

Comments

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.

Links

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.

Formula

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

Example

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.

Maple

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

Mathematica

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}]

Crossrefs

Cf. A001006, A005043, A295112, A295113.

Sequence in context: A241394 A054491 A282710 * A013261 A013265 A228703

Adjacent sequences: A295129 A295130 A295131 * A295133 A295134 A295135

Keyword

nonn

Author

Zhi-Wei Sun, Nov 15 2017

Status

approved