A371652 a(n) = (Sum_{k=0..n-1}(145*k^2+104*k+18)*C(2k,k)*C(3k,k)^2/(2k+1))/(6n*(2n-1)*C(3n,n)).

1, 3, 29, 399, 6514, 117711, 2275251, 46139015, 969837866, 20962468086, 463305649245, 10428205097283, 238311987683964, 5516455670448105, 129108299508906255, 3050631709840606455, 72685647150198891642, 1744632999762729504318, 42150287092525653156282, 1024327224110685261526062

Offset

1,2

Comments

Conjecture: All the terms are integers.

This is motivated by Conjecture 4.13 and Remark 4.13 in the linked 2023 paper of Z.-W. Sun.

Links

Table of n, a(n) for n=1..20.

Zhi-Wei Sun, New congruences involving harmonic numbers, Nanjing Univ. J. Math. Biquarterly 40 (2023), 1-33.

Formula

a(n) ~ 3^(3*n - 1/2) / (20*Pi*n^2). - Vaclav Kotesovec, Apr 01 2024

Example

a(2) = 3 since (145*0^2+104*0+18)*C(2*0,0)*C(3*0,0)^2/(2*0+1) + (145*1^2+104*1+18)*C(2*1,1)*C(3*1,1)^2/(2*1+1) divided by 6*2*(2*2-1)*C(3*2,2) coincides with (18+267*2*3^2/3)/(36*15) = 3.

Mathematica

a[n_]:=a[n]=Sum[(145k^2+104k+18)Binomial[2k, k]Binomial[3k, k]^2/(2k+1), {k, 0, n-1}]/(6n*(2n-1)Binomial[3n, n]); Table[a[n], {n, 1, 20}]

Crossrefs

Cf. A000984, A001764, A005809.

Sequence in context: A262640 A302582 A335867 * A302923 A370922 A366005

Adjacent sequences: A371649 A371650 A371651 * A371653 A371654 A371655

Keyword

nonn

Author

Zhi-Wei Sun, Apr 01 2024

Status

approved