A190840 a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.

1, 8, 288, 332928, 443365544448, 786292024016459316676608, 2473020588127600939387543243786675530709484249088

Offset: 0

Author

Alexander Zhukov, Aug 08 2011

Keywords

• nonn

Comments

For n>0, subsequence of A132592: both a(n)/2 and a(n)+1 are squares.

All terms (n > 0) are divisible by 8, yielding all terms of A185097, which is indexed from n=1, thus having the first term A185097(1) = 1.

The next term has 98 digits. - Harvey P. Dale, Jan 01 2014

For n>0, subsequence of A060355: both a(n) and a(n)+1 are powerful numbers. - Bernard Schott, Apr 24 2023

Crossrefs

Cf. A000217, A001601, A060355, A132592, A185097.

Programs

Mathematica

NestList[4#(#+1)&,1,7] (* Harvey P. Dale, Jan 01 2014 *)

Formula

a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.

a(n) = sinh(2^(n-2)*arccosh(17))^2. - Alexander R. Povolotsky, Aug 14 2011

a(n) = 8*A185097(n) for n > 0. - Alexander R. Povolotsky, Aug 14 2011

a(n) = (1 + sqrt(2))^(2^(n+1))/4 + (1 - sqrt(2))^(2^(n+1))/4 - 1/2. Therefore 2*a(n) + 1 = A001601(n+1). - Bruno Berselli, Feb 01 2017