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A203317 a(n) = A203315(n)/A000178(n) where A000178=(superfactorials).

%I A203317 #12 May 27 2025 11:11:32
%S A203317 1,2,8,256,10240,3440640,1233125376,4058744094720,286551971651911680,
%T A203317 13351029463205868994560,18128348229848045861669437440,
%U A203317 80945830355202461675325011924090880,223346912509970707926726595810215906508800
%N A203317 a(n) = A203315(n)/A000178(n) where A000178=(superfactorials).
%H A203317 R. Chapman, <a href="https://web.archive.org/web/20230227162446/https://www.maa.org/sites/default/files/Robin_Chapman27238.pdf">A polynomial taking integer values</a>, Mathematics Magazine, 29 (1996), 121.
%t A203317 f[j_] := Prime[j + 1]; z = 17;
%t A203317 v[n_] := Product[Product[f[k] - f[j], {j, 1, k - 1}], {k, 2, n}]
%t A203317 d[n_] := Product[(i - 1)!, {i, 1, n}]
%t A203317 Table[v[n], {n, 1, z}]                  (* A203315 *)
%t A203317 Table[v[n + 1]/(2 v[n]), {n, 1, z - 1}] (* A203316 *)
%t A203317 Table[v[n]/d[n], {n, 1, 20}]            (* A203317 *)
%Y A203317 Cf. A203315, A203316, A000040.
%K A203317 nonn
%O A203317 1,2
%A A203317 _Clark Kimberling_, Jan 01 2012