A386826 a(n) = Sum_{k=0..n} 3^(n-k)*C(2*n,n-k).

1, 7, 67, 694, 7459, 81922, 912718, 10273228, 116522275, 1329569290, 15244087642, 175472098996, 2026521318286, 23470106563924, 272476942589884, 3169997065488664, 36948020548661539, 431354994430077274, 5043279137171450914, 59041965004582271524, 692026745415822877594, 8119918150063503715324

Offset

0,2

Comments

Row sums of number triangle A386825.

Number of strings of length 2*n defined on {0,1,2,3} that have either the same number or more 0's than the combined number of 1's, 2's and 3's.

Links

Table of n, a(n) for n=0..21.

Formula

a(n) = 3^n*binomial(2*n, n)*hypergeom([1, -n], [1+n], -1/3). - Stefano Spezia, Aug 05 2025

a(n) ~ 2^(2*n-1) * 3^(n+1) / sqrt(Pi*n). - Vaclav Kotesovec, Aug 07 2025

D-finite with recurrence n*a(n) +(n+1)*a(n-1) +2*(-946*n+2017)*a(n-2) +144*(286*n-821)*a(n-3) +122112*(-2*n+7)*a(n-4)=0. - R. J. Mathar, Aug 26 2025

D-finite with recurrence n*(2*n-1)*a(n) +2*(-28*n^2+20*n+9)*a(n-1) +96*(2*n+1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 26 2025

Example

a(3)=694 counts the strings of length 6 as follows: 540 strings with three 0's, 135 with four 0's, 18 with five 0's, and 1 string with six 0's. Hence 694 = 540 + 135 + 18 + 1, where the summands come from the triangle in A386825.

Mathematica

Table[Sum[3^(n-k) Binomial[2n, n-k], {k, 0, n}], {n, 0, 21}]

Crossrefs

Cf. A128418, A386825.

Sequence in context: A036948 A385799 A020469 * A390590 A386812 A340973

Adjacent sequences: A386823 A386824 A386825 * A386827 A386828 A386829

Keyword

nonn

Author

Enrique Navarrete, Aug 04 2025

Status

approved