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

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

Offset

0,2

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

Links

Amiram Eldar, Table of n, a(n) for n = 0..10

Mario Catalani, Problem H-608, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 42, No. 1 (2004), p. 92; Pell does it again, Solution to Problem H-608 by Kenneth B. Davenport, ibid., Vol. 43, No. 1 (2005), pp. 95-96.

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

From Amiram Eldar, Dec 14 2025: (Start)

a(n) = 2*A051009(n+1)^2 = 2*A000129(2^(n+1))^2 for n >= 1.

Product_{n>=0} (1 + 1/sqrt(a(n)+1)) = 1 + sqrt(2) (A014176) (Catalani, 2004). (End)

Mathematica

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

Crossrefs

Cf. A000129, A000217, A001601, A014176, A051009, A060355, A132592, A185097.

Sequence in context: A034977 A384088 A065141 * A000842 A202559 A232372

Adjacent sequences: A190837 A190838 A190839 * A190841 A190842 A190843

Keyword

nonn

Author

Alexander Zhukov, Aug 08 2011

Status

approved