A225367 Number of palindromes of length n in base 3 (A118594).

3, 2, 6, 6, 18, 18, 54, 54, 162, 162, 486, 486, 1458, 1458, 4374, 4374, 13122, 13122, 39366, 39366, 118098, 118098, 354294, 354294, 1062882, 1062882, 3188646, 3188646, 9565938, 9565938, 28697814, 28697814, 86093442, 86093442, 258280326, 258280326, 774840978

Offset: 1

Author

M. F. Hasler, May 05 2013

Keywords

• nonn
• base
• easy

Comments

Also: The number of n-digit terms in A006072. See there for further comments.

A palindrome of length L=2k-1 or of length L=2k is determined by the first k digits, which then determine the last k digits by symmetry. Since the first digit cannot be 0 (unless L=1), there are 2*3^(k-1) possibilities for L>1.

Except for the initial term, this is identical to A117855, which counts only nonzero palindromes.

Examples

			The a(1)=3 palindromes of length 1 are: 0, 1 and 2.
The a(2)=2 palindromes of length 2 are: 11 and 22.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,3).

Crossrefs

Cf. A050683 and A070252 for base 10 analogs.

Programs

Magma

[n eq 1 select 3 else 2*3^Floor((n-1)/2): n in [1..40]]; // Bruno Berselli, May 06 2013

Magma

I:=[3,2,6]; [n le 3 select I[n] else 3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, May 31 2017

Mathematica

Join[{3}, LinearRecurrence[{0, 3}, {2, 6}, 40]] (* Vincenzo Librandi, May 31 2017 *)

PARI

A225367(n)=2*3^((n-1)\2)+!n

Python

def A225367(n): return 3 if n==1 else 3**(n-1>>1)<<1 # Chai Wah Wu, Jul 30 2025

Formula

a(n) = 2*3^floor((n-1)/2) + [n=1].

a(n) = 3*a(n-2) for n>3.

G.f.: x*(3*x^2-2*x-3)/(3*x^2-1).

a(n) = (6-(1+(-1)^n)*(3-sqrt(3)))*sqrt(3)^(n-3) for n>1, a(1)=3. [Bruno Berselli, May 06 2013]