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Total number of digits in palindromes with n digits.
0

%I #19 Jan 14 2023 11:18:37

%S 10,18,270,360,4500,5400,63000,72000,810000,900000,9900000,10800000,

%T 117000000,126000000,1350000000,1440000000,15300000000,16200000000,

%U 171000000000,180000000000,1890000000000,1980000000000,20700000000000,21600000000000

%N Total number of digits in palindromes with n digits.

%C Let f(1) = g(1) = 10 and f(2) = 1; d(n) denotes the number of digits in f(n) and for n >= 3, f(n) = 10*f(n-1) + 5*10^(d(n-1)-1) if n is odd, otherwise f(n) = f(n-1) + 10^(d(n-1)-1)/2. Let g(n) = 18*f(n) for n > 1. It gives g(2) = 18, g(3) = 270, g(4) = 360, g(5) = 4500, .... In fact g(n) produces a different sequence than a(n).

%H <a href="/index/Pac#palindromes">Index entries for sequences related to palindromes</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,20,0,-100).

%F a(n) = n*A070252(n).

%F a(n) = 20*a(n-2)-100*a(n-4) for n>5. G.f.: 2*x*(50*x^4+35*x^2+9*x+5) / (10*x^2-1)^2. - _Colin Barker_, Mar 31 2014

%e There are nine 2-digit palindromes, so a(2) = 2*9 = 18.

%o (PARI) print1("10, 18, "); m=9; for(n=3, 24, if(bitand(n, 1), m=10*m); print1(m*n, ", "));

%o (PARI) Vec(2*x*(50*x^4+35*x^2+9*x+5)/(10*x^2-1)^2 + O(x^100)) \\ _Colin Barker_, Mar 31 2014

%Y Cf. A008600, A070252.

%K nonn,base,easy

%O 1,1

%A _Arkadiusz Wesolowski_, Mar 30 2014