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A082389
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a(n) = floor((n+2)*phi) - floor((n+1)*phi) where phi=(1+sqrt(5))/2.
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6
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1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2
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OFFSET
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1,2
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COMMENTS
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Alternative descriptions (1): unique positive integer sequence taking values in {1,2} satisfying a(1)=1, a(2)=2 and a(a(1)+...+a(n))=a(n) for n >= 3.
(2) Start with 1,2; then for any k>=1, a(a(1)+...+a(k))=a(k), fill in any undefined terms by the rule that a(t) = 1 if a(t-1) = 2 and a(t) = 2 if a(t-1) = 1.
(3) a(1)= 1, a(2)=2, a(a(1)+a(2)+...+a(n))=a(n); a(a(1)+a(2)+...+a(n)+1)=3-a(n).
More generally, the sequence a(n)=floor(r*(n+2))-floor(r*(n+1)), r= (1/2) *(z+sqrt(z^2+4)), z integer >=1, is defined by a(1), a(2) and a(a(1)+a(2)+...+a(n)+f(z))=a(n); a(a(1)+a(2)+...+a(n)+f(z)+1)=(2z+1)-a(n) where f(1)=0, f(z)=z-2 for z>=2.
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LINKS
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FORMULA
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EXAMPLE
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a(1)+a(2)=3 and a(a(1)+a(2)) must be a(2) so a(3)=2. Therefore a(a(1)+a(2)+a(3))=a(5)=2 and from the rule the "hole" a(4) is 1. Hence sequence begins 1,2,2,1,2,...
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MAPLE
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MATHEMATICA
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Rest@Nest[ Flatten[ # /. {1 -> 2, 2 -> {2, 1}}] &, {1}, 11] (* Robert G. Wilson v, Jan 26 2006 *)
#[[2]]-#[[1]]&/@Partition[Table[Floor[GoldenRatio n], {n, 0, 110}], 2, 1] (* Harvey P. Dale, Sep 04 2019 *)
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PROG
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(Python)
from math import isqrt
def A082389(n): return (n+2+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 29 2022
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CROSSREFS
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Same as A014675 without the first term.
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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