login
a(1)=1, then a(n) = smallest number whose square is larger than 2*(a(n-1))^2.
1

%I #19 May 01 2024 17:02:13

%S 1,2,3,5,8,12,17,25,36,51,73,104,148,210,297,421,596,843,1193,1688,

%T 2388,3378,4778,6758,9558,13518,19118,27037,38237,54076,76476,108154,

%U 152953,216309,305908,432620,611818,865242,1223637,1730485,2447276,3460971

%N a(1)=1, then a(n) = smallest number whose square is larger than 2*(a(n-1))^2.

%C The sequence satisfies an almost recurrence relation, that is, there are 4 sequences b_0, b_1, b_2, b_3 taking values in {-2,-1,1,2} such that 2b_0(n)a(n) + 2b_1(n)a(n+1) + b_2(n)a(n+2) + b_3(n)a(n+3) = 0. For instance, we have a(103) - a(102) - 2a(101) + 2a(100) = 0, 2a(106) - a(105) - 4a(104) + 2a(103) = 0. - _Benoit Cloitre_, Oct 16 2012

%H Harvey P. Dale, <a href="/A175539/b175539.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = ceiling(sqrt(2)*a(n-1)) with a(1)=1. - _Benoit Cloitre_, Oct 16 2012

%t NestList[Floor[Sqrt[2#^2]]+1&,1,50] (* _Harvey P. Dale_, Oct 19 2014 *)

%o (PARI) a=1;s=[a];for(i=2,100,a=1+sqrtint(2*a^2);s=concat(s,a));s

%o (PARI) a(n)=if(n<2,1,floor(sqrt(2)*a(n-1))) \\ _Benoit Cloitre_, Oct 16 2012

%Y Cf. A087057 (smallest number whose square is larger than 2*n^2).

%K nonn

%O 1,2

%A _Zak Seidov_, Jun 14 2010