A084594 -id:A084594 - cp's OEIS frontend

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A084595 For n > 0: a(n) = Sum_{r=0..2^(n-1)-1} binomial(2^n, 2r+1)*3^r.

Original entry on oeis.org

1, 2, 16, 896, 2781184, 26794772135936, 2487085750646543836443049984, 21427531469765285263614058238314319540132878612321796096

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), May 31 2003

A084594(n)/a(n) converges to sqrt(3). Related to Newton's iteration.

a(n) is divisible by 2^n.

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Eric Weisstein's World of Mathematics, Newton's Iteration.

Cf. A001146, A002605, A084594.

For n>0: Table[Sum[Binomial[2^n, 2 r + 1]3^r, {r, 0, 2^(n - 1) - 1}], {n, 1, 8}]

a(n) = if (n==0, 1, sum(r=0, 2^(n-1)-1, binomial(2^n, 2*r+1)*3^r)); \\ Michel Marcus, Sep 09 2019; corrected Jun 13 2022

a(n) = ((1+sqrt(3))^(2^n) - (1-sqrt(3))^(2^n))/(2*sqrt(3)).
For n > 1:
a(n) = 2*a(n-1)*sqrt(3*a(n-1)^2 + A001146(n-1)).
a(n) = 2*a(n-1)*A084594(n-1).
a(n) = A002605(2^n).

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