A216149 Numbers k such that Sum_{j=1..k-1} (3*j)!/5^j is an integer.
1, 3, 16
Offset: 1
Programs
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Mathematica
seq={}; sum=0; fak=1; k=0; While[k<10000, sum+=fak; If[Denominator[sum]==1, AppendTo[seq,k+1]]; k++; fak*=3*k*(3*k-1)*(3*k-2)/5;]; seq Select[Range[20],IntegerQ[Sum[(3k)!/5^k,{k,#-1}]]&] (* Harvey P. Dale, Apr 07 2019 *)
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