A060656 a(n) = 2*a(n-1)*a(n-2)/a(n-3), with a(0)=a(1)=1.

1, 1, 2, 4, 16, 64, 512, 4096, 65536, 1048576, 33554432, 1073741824, 68719476736, 4398046511104, 562949953421312, 72057594037927936, 18446744073709551616, 4722366482869645213696, 2417851639229258349412352

Offset: 0

Author

Henry Bottomley, Apr 18 2001

Keywords

• nonn
• easy

Comments

a(n+1) is the Hankel transform of A135052. - Paul Barry, Nov 15 2007

a(n+1) is the Hankel transform of the aerated large Schroeder numbers. a(n) and a(n+1) both satisfy the trivial Somos-4 recurrence u(n)=4*u(n-2)^2/u(n-4). Associated with the elliptic curve y^2=1-6x^2+x^4 via Schroeder numbers. - Paul Barry, Dec 08 2009

Hankel transform of A089324. - Paul Barry, Mar 01 2010

a(n+1) is the number of n X n binary matrices that are symmetric about both diagonals (bisymmetric). For the derivation of this result, see the link below. - Dennis P. Walsh, Apr 03 2014

1 followed by {a(n-1)}A078495).%20-%20_Vladimir%20Shevelev">(n>=1) is the Somos-3 sequence: b(0)=b(1)=b(2)=1;for n>=3, b(n)=2*b(n-1)*b(n-2)/b(n-3) (cf. comment in A078495). - _Vladimir Shevelev, Apr 20 2016

If the Hankel transform is defined as in the link 'Sequence transformations' then a(n) is the Hankel transform of A151374. - Peter Luschny, Nov 30 2016

Examples

			a(6) = 2*64*16/4 = 512.
G.f. = 1 + x + 2*x^2 + 4*x^3 + 16*x^4 + 64*x^5 + 512*x^6 + 4096*x^7 + ...
		

Links

• Harry J. Smith, Table of n, a(n) for n = 0..100
• Peter Luschny, Sequence transformations
• Dennis P. Walsh, Notes on binary bisymmetric matrices

Crossrefs

Cf. A002416, A002620, A016116, A038754, A089324, A135052, A262666, A078495, A151374.

Programs

Maple

A060656:=n->2^floor(n^2/4); seq(A060656(n), n=0..20); # Wesley Ivan Hurt, Apr 30 2014

Mathematica

a[ n_] := 2^Quotient[n^2, 4]; (* Michael Somos, Jan 24 2014 *)
nxt[{a_,b_,c_}]:={b,c,(2c*b)/a}; NestList[nxt,{1,1,2},20][[All,1]] (* Harvey P. Dale, Nov 26 2017 *)

PARI

{ for (n=0, 100, write("b060656.txt", n, " ", 2^(n^2\4)); ) } \\ Harry J. Smith, Jul 09 2009

PARI

{a(n) = 2^(n^2\4)}; /* Michael Somos, Jan 24 2014 */

Formula

a(n) = 2^floor( n^2/4 ) = a(n - 1) * 2^floor( n/2 ) = a(n - 2) * 2^(n - 1) = a(n - 1) * A016116(n) = 2^A002620(n).

0 = a(n) * a(n+3) + a(n+1) * ( -2*a(n+2) ) for all n in Z. - Michael Somos, Jan 24 2014

0 = a(n) * a(n+4) + a(n+2) * ( -4*a(n+2) ) for all n in Z. - Michael Somos, Jan 24 2014