A190840 - OEIS

%I #87 Dec 14 2025 01:44:44

%S 1,8,288,332928,443365544448,786292024016459316676608,

%T 2473020588127600939387543243786675530709484249088

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

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

%C 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.

%C The next term has 98 digits. - _Harvey P. Dale_, Jan 01 2014

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

%H Amiram Eldar, <a href="/A190840/b190840.txt">Table of n, a(n) for n = 0..10</a>

%H Mario Catalani, <a href="https://www.fq.math.ca/Papers1/42-1/February2004advanced.pdf">Problem H-608</a>, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 42, No. 1 (2004), p. 92; <a href="https://www.fq.math.ca/Problems/advanced43-1.pdf">Pell does it again</a>, Solution to Problem H-608 by Kenneth B. Davenport, ibid., Vol. 43, No. 1 (2005), pp. 95-96.

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

%F a(n) = sinh(2^(n-2)*arccosh(17))^2. - _Alexander R. Povolotsky_, Aug 14 2011

%F a(n) = 8*A185097(n) for n > 0. - _Alexander R. Povolotsky_, Aug 14 2011

%F 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

%F From _Amiram Eldar_, Dec 14 2025: (Start)

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

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

%t NestList[4#(#+1)&,1,7] (* _Harvey P. Dale_, Jan 01 2014 *)

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

%K nonn

%O 0,2

%A _Alexander Zhukov_, Aug 08 2011