A232730 - OEIS

%I #34 Feb 29 2024 18:02:48

%S 5,45,495,4905,49500,494550,4950000,49495500,495000000,4949955000,

%T 49500000000,494999550000,4950000000000,49499995500000,

%U 495000000000000,4949999955000000,49500000000000000,494999999550000000,4950000000000000000,49499999995500000000

%N Number of n-digit numbers that yield an (n+1)-digit number after Reverse and Add.

%C A232729(n) + a(n) = 9*10^(n-1).

%H Robert Israel, <a href="/A232730/b232730.txt">Table of n, a(n) for n = 1..990</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (10,10,-100).

%F a(1) = 5, a(3) = 495, a(2*k+1) = 100*a(2*k-1), k > 1.

%F a(2) = 45, a(4) = 4905, a(2*k) = 110*a(2*k-2) - 1000*a(2*k-4), k > 2.

%F G.f. = 5*x*(1+x)*(1-x)^2 / ((1-10*x)*(1-10*x^2)). - _M. F. Hasler_, Nov 30 2013

%F From _Colin Barker_, Mar 20 2017: (Start)

%F a(n) = -45*(10^(n/2-2) - 11*10^(n-3)) for n>2 even.

%F a(n) = 99*2^(n-3)*5^(n-2) for n>2 odd.

%F a(n) = 10*a(n-1) + 10*a(n-2) - 100*a(n-3) for n>4. (End)

%F E.g.f.: (99*cosh(10*x) - 90*cosh(sqrt(10)*x) + 99*sinh(10*x) + 10*x - 9)/200. - _Stefano Spezia_, Oct 27 2022

%e There are 5 1-digit numbers (5,6,7,8,9) that yield a 2-digit number (10,12,14,16,18), so a(1)=5.

%p a[1]:=5: t[0]:= 0: t[1]:= 5:

%p for n from 2 to 50 do

%p a[n]:= 45*10^(n-2) + 9*t[n-2];

%p t[n]:= a[n] + t[n-2];

%p od:

%p seq(a[n],n=1..50); # _Robert Israel_, Apr 21 2016

%t LinearRecurrence[{10,10,-100},{5,45,495,4905},20] (* _Harvey P. Dale_, Feb 29 2024 *)

%o (PARI) Vec(5*x*(1+x)*(1-x)^2 / ((1-10*x)*(1-10*x^2)) + O(x^30)) \\ _Colin Barker_, Mar 20 2017

%Y Cf. A232729, A232731.

%K nonn,base,easy

%O 1,1

%A _Lars Blomberg_, Nov 29 2013

%E G.f. corrected and empirical formulas proved by _Robert Israel_, Apr 21 2016