%I #53 Apr 02 2024 03:00:59
%S 1,1,3,1,10,94,1039,13513,202702,3445942,65472907,1374931057,
%T 31623414322,790585358062,21345804667687,619028335362937,
%U 19189878396251062,633265987076285062,22164309547669977187,820079453263789155937,31983098677287777081562,1311307045768798860344062
%N Omit the trailing 5 from double factorial of odd numbers (A001147(n)).
%C A001147(n) has only a trailing five for n > 2.
%C Proof: being trivial to prove that A001147(n) ends with at least a digit 5 for n > 2, it remains to prove that the tenth digit of A001147(n) is not equal to 5. Considering the product A001147(n) = A001147(n-1)*(2*n - 1) for n > 2, it is easy to verify that the tenth digit of A001147(n) is congruent to 2 modulo 5 if the tenth digit of A001147(n-1) is congruent to 2 modulo 5. Since for n <= 8 the tenth digit of A001147(n) is not equal to 5, and it is equal to 2 for n = 8, it follows that the tenth digit of A001147(n) for n > 8 is congruent to 2 modulo 5, and therefore, not equal to 5. QED.
%H Stefano Spezia, <a href="/A358987/b358987.txt">Table of n, a(n) for n = 0..400</a>
%F a(n) = (A001147(n) - 5)/10 for n > 2.
%F E.g.f.: (14 + 2*x*(7 + 8*x) - 5*exp(x) + 1/sqrt(1 - 2*x))/10.
%F D-finite with recurrence a(n) + (-2*n+1)*a(n-1) + (-n+1) = 0 for n > 3. - _R. J. Mathar_, Mar 25 2024
%t Join[{1,1,3},Table[((2n-1)!!-5)/10,{n,3,21}]] (* or *)
%t CoefficientList[Series[(14-5Exp[x]+1/Sqrt[1-2x]+2x(7+8x))/10,{x,0,21}],x]Table[n!,{n,0,21}]
%Y Cf. A001147.
%K nonn,base,easy
%O 0,3
%A _Stefano Spezia_, Dec 10 2022