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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Alexander Zhukov

Alexander Zhukov's wiki page.

Alexander Zhukov has authored 1 sequences.

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

Original entry on oeis.org

1, 8, 288, 332928, 443365544448, 786292024016459316676608, 2473020588127600939387543243786675530709484249088
Offset: 0

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Author

Alexander Zhukov, Aug 08 2011

Keywords

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

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