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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). 2
6, 23, 90, 432, 2286, 13176, 80418, 513764, 3400518, 23167311, 161640554, 1150633512, 8332048638, 61232315553, 455830692210, 3432015694314, 26101221114582, 200295455169015, 1549473966622602, 12074304397434552, 94713783502786686, 747454269790900728 (list; graph; refs; listen; history; text; internal format)
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, 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
Sequence in context: A241394 A054491 A282710 * A013261 A013265 A228703
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 15 2017
STATUS
approved

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Last modified May 27 21:52 EDT 2024. Contains 372882 sequences. (Running on oeis4.)