A358987 Omit the trailing 5 from double factorial of odd numbers (A001147(n)).

1, 1, 3, 1, 10, 94, 1039, 13513, 202702, 3445942, 65472907, 1374931057, 31623414322, 790585358062, 21345804667687, 619028335362937, 19189878396251062, 633265987076285062, 22164309547669977187, 820079453263789155937, 31983098677287777081562, 1311307045768798860344062

Offset: 0

Author

Stefano Spezia, Dec 10 2022

Keywords

• nonn
• base
• easy

Comments

A001147(n) has only a trailing five for n > 2.

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.

Links

• Stefano Spezia, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A001147.

Programs

Mathematica

Join[{1,1,3},Table[((2n-1)!!-5)/10,{n,3,21}]] (* or *)
CoefficientList[Series[(14-5Exp[x]+1/Sqrt[1-2x]+2x(7+8x))/10,{x,0,21}],x]Table[n!,{n,0,21}]

Formula

a(n) = (A001147(n) - 5)/10 for n > 2.

E.g.f.: (14 + 2*x*(7 + 8*x) - 5*exp(x) + 1/sqrt(1 - 2*x))/10.

D-finite with recurrence a(n) + (-2*n+1)*a(n-1) + (-n+1) = 0 for n > 3. - R. J. Mathar, Mar 25 2024