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a(n) = Product_{k=1..n} (k^2 + 21).
0

%I #15 Oct 13 2017 21:39:54

%S 1,22,550,16500,610500,28083000,1600731000,112051170000,9524349450000,

%T 971483643900000,117549520911900000,16692031969489800000,

%U 2754185274965817000000,523295202243505230000000,113555058886840634910000000,27934544486162796187860000000,7737868822667094544037220000000

%N a(n) = Product_{k=1..n} (k^2 + 21).

%C Yin et al. prove that a(n) is never a square for n > 0.

%H Q. Yin, Q. Tan and Y. Luo, <a href="http://www.seams-bull-math.ynu.edu.cn/downloadfile.jsp?filemenu=_201505&amp;filename=11_39(5).pdf">p-Adic Valuation of (12+21)...(n2+21) and Applications</a>, Southeast Asian Bulletin of Mathematics, Vol. 39(5) (2015) page: 747-754.

%t Table[Product[k^2 + 21, {k, n}], {n, 0, 16}] (* _Michael De Vlieger_, Apr 03 2016 *)

%o (PARI) a(n) = prod(k=1, n, (k^2+21));

%Y Cf. A101686, A241851.

%K nonn

%O 0,2

%A _Michel Marcus_, Apr 03 2016