A295319 a(n) is the sum of all n-digit palindromes.

45, 495, 49500, 495000, 49500000, 495000000, 49500000000, 495000000000, 49500000000000, 495000000000000, 49500000000000000, 495000000000000000, 49500000000000000000, 495000000000000000000, 49500000000000000000000, 495000000000000000000000

Offset: 1

Author

Melvin Peralta, Nov 19 2017

Keywords

• base
• easy
• nonn

Comments

For n > 1, the sum of the digits of a(n) is always 18 (see AoPS link).

Conjecture: All terms after a(1) are of the form 495*10^x where x is any nonnegative integer k such that k != 1 (mod 3). - Harvey P. Dale, Mar 05 2023

Examples

			The sum of all nine two-digit palindromes is 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 = 495, and so a(2) = 495.
The sum of all three-digit palindromes is (101 + 999) + (111 + 989) + (121 + 979) + ... (545 + 565) + 555 = 49500, and so a(3) = 49500.
		

Links

• AoPS, problem 15 of the 2014 American Mathematics Competition 12
• Index entries for linear recurrences with constant coefficients, signature (0,1000).

Crossrefs

Cf. A002113, A050683, A261675.

Programs

Mathematica

palSum[n_] := 99/2*10^(n - 1) * 10^Floor[(n - 1)/2]; palSum[1] = 45; Array[ palSum, 16] (* Robert G. Wilson v, Nov 21 2017 *)

PARI

a(n) = if (n==1, 45, 9*10^floor((n-1)/2)*11*10^(n-1)/2); \\ Michel Marcus, Dec 26 2017

Formula

a(n) = A050683(n)*(5*10^(n-1) + (9/2)*10^(n-2) + ... + (9/2)*10 + 5) (calculates the sum by multiplying the expected value of a randomly selected n-digit palindrome with the number of n-digit palindromes).

For n > 1, a(n) = (A050683(n)/2)*11*10^(n-1).

For n > 3, a(n) = 1000 * a(n - 2). - David A. Corneth, Dec 26 2017

G.f.: x*(45 + 495*x + 4500*x^2)/(1 - 1000*x^2). - Chai Wah Wu, Jan 22 2018

E.g.f.: 9*(11*cosh(10*sqrt(10)*x) + 11*sqrt(10)*sinh(10*sqrt(10)*x) - 11 - 100*x)/200. - Stefano Spezia, Sep 19 2024