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%I #12 Jun 24 2013 23:32:00
%S 1,6,23,40,273,1870,7343,12816,87841,602070,2364359,4126648,28284465,
%T 193864606,761316191,1328767776,9107509825,62423800998,245141449079,
%U 427859097160,2932589879121,20100270056686,78934785287183
%N Numbers n such that n^2= (1/5)*(n+floor(sqrt(5)*n*floor(sqrt(5)*n))).
%C Conjecture: if m is an integer and sqrt(m) is irrational, the sequence of n such that n^2 = (1/m)*(n + floor(sqrt(m)*n*floor(sqrt(m)*n))) always satisfies a recurrence of order m. For example: if m=6, the sequence n=b(k) satisfies: b(6k)=4*b(6k-1)+4*b(6k-2)-b(6k-3)-1; b(6k+1)=.... etc.
%F a(1)=1; a(2)=6; a(3)=23; a(4)=40; a(4n)=2*a(4n-1)-a(4n-2); a(4n+1)=7*a(4n)-a(4n-2)-1; a(4n+2)=7*a(4n+1)-a(4n-1)-1; a(4n+3)=4*a(4n+2)-a(4n+1)/2-1/2.
%F Empirical g.f.: x*(x^7+x^6+13*x^5+89*x^4-17*x^3-17*x^2-5*x-1) / ((x-1)*(x^2-4*x-1)*(x^2+4*x-1)*(x^4+18*x^2+1)). - _Colin Barker_, Jun 24 2013
%o (PARI) x=1; y=6; z=23; u=40; for(n=5,50,v=if((n%4-1)*(n%4-2),if(n%4,4*u-z/2-1/2,2*u-z),if(n%4-1,7*u-z-1,7*u-y-1)); x=y; y=z; z=u; u=v; print1(v,","))
%Y Cf. A046090.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Apr 15 2003
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