|
%I #29 Sep 24 2023 08:35:48
%S 1,1,3,15,165,3465,148995,12664575,2165642325,738484032825,
%T 504384594419475,688484971382583375,1880252456845835197125,
%U 10268058666835106011499625,112158004817839862963610403875
%N Product of first n nonzero Jacobsthal numbers (A001045).
%C Inspired by A015013.
%F a(n) = abs(A015013(n)).
%F a(n) ~ c * 2^(n*(n+1)/2) / 3^n, where c = QPochhammer(-2, 1/4)*QPochhammer(1/4)/3 = 1.21072413030105918013617285610590504636804163112313764347615924554000... - _Vaclav Kotesovec_, Mar 04 2021, updated Jul 19 2021
%F Equivalently, c = QPochhammer(-1/2). - _Vaclav Kotesovec_, Sep 24 2023
%e a(4) = 15 because a(4) = 1*1*3*5 = 15.
%t Table[Abs@QFactorial[n, -2], {n, 20}] (* _Vladimir Reshetnikov_, Sep 16 2016 *)
%t FoldList[Times,LinearRecurrence[{1,2},{1,1},20]] (* _Harvey P. Dale_, Apr 22 2019 *)
%t Table[(-1)^Floor[n/2] * QPochhammer[-2, 4, 1 + Floor[(n-1)/2]] * QPochhammer[4, 4, Floor[n/2]]/3^n, {n, 1, 20}] (* _Vaclav Kotesovec_, Mar 04 2021 *)
%o (PARI) a001045(n) = (2^n - (-1)^n) / 3;
%o a(n) = prod(i=1, n, a001045(i));
%Y Cf. A001045, A015013.
%K nonn,easy
%O 1,3
%A _Altug Alkan_, Apr 05 2016
|
