OFFSET
0,3
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
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
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}]
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Stefano Spezia, Dec 10 2022
STATUS
approved