533697

Appears in sequences

• Smallest number k such that 2^n divides A066796(k) = Sum(i=1,k,binomial(2*i,i)).at n=17A126806
• Smallest number k such that 2^n divides A066796(k) = Sum(i=1,k,binomial(2*i,i)).at n=18A126806
• Smallest number k such that 2^n divides A066796(k) = Sum(i=1,k,binomial(2*i,i)).at n=19A126806