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Let b(n) = b(n-1) + 5; then a(n) = b(n)*a(n-1). - Roger L. Bagula, Sep 17 2008
k=5; b[1]=2; b[n_]:= b[n] = b[n-1] +k; a[0]=1; a[1]=2; a[n_]:= a[n]= a[n-1]*b[n]; Table[a[n], {n, 0, 20}] (* Roger L. Bagula, Sep 17 2008 *)
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1, 2, 14, 168, 2856, 62832, 1696464, 54286848, 2008613376, 84361761792, 3965002804224, 206180145819648, 11752268311719936, 728640635326636032, 48818922566884614144, 3514962424815692218368, 270652106710808300814336, 22193472750286280666775552
Sum_{n>=0} 1/a(n) = 1 + (e/5^3)^(1/5)*(Gamma(2/5) - Gamma(2/5, 1/5)). - Amiram Eldar, Dec 19 2022
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