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a(n) is the sum of all n-digit palindromes.
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%I #45 Sep 20 2024 06:03:51

%S 45,495,49500,495000,49500000,495000000,49500000000,495000000000,

%T 49500000000000,495000000000000,49500000000000000,495000000000000000,

%U 49500000000000000000,495000000000000000000,49500000000000000000000,495000000000000000000000

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

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

%C 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

%H AoPS, <a href="http://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_15">problem 15 of the 2014 American Mathematics Competition 12</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,1000).

%F 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).

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

%F For n > 3, a(n) = 1000 * a(n - 2). - _David A. Corneth_, Dec 26 2017

%F G.f.: x*(45 + 495*x + 4500*x^2)/(1 - 1000*x^2). - _Chai Wah Wu_, Jan 22 2018

%F 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

%e The sum of all nine two-digit palindromes is 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 = 495, and so a(2) = 495.

%e The sum of all three-digit palindromes is (101 + 999) + (111 + 989) + (121 + 979) + ... (545 + 565) + 555 = 49500, and so a(3) = 49500.

%t 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 *)

%o (PARI) a(n) = if (n==1, 45, 9*10^floor((n-1)/2)*11*10^(n-1)/2); \\ _Michel Marcus_, Dec 26 2017

%Y Cf. A002113, A050683, A261675.

%K base,easy,nonn

%O 1,1

%A _Melvin Peralta_, Nov 19 2017