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A109964
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a(n) = floor(sqrt(Sum_{i<n} a(i))), with a(0)=1.
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1
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1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 32, 32, 33, 33, 34, 34, 35, 35
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OFFSET
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0,5
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REFERENCES
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Related problem was offered at XXIX Moscow Mathematical Olympiad (1966).
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LINKS
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FORMULA
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a(n) = floor(sqrt(A109965(n)) = A109965(n+1)-A109965(n). Roughly (n-log_2(n))/2. 1 appears four times, other powers of 2 appear three times, other numbers appear twice.
For n>1, a(n)=2^j+k where j=floor(log_2(n))-1 and k=(n-2^(j+1)-j) mod 2.
a(2^(j+1)+j+2*k) = a(2^(j+1)+j+2*k+1) = 2^j+k; a(2^(j+1)+j-1) = 2^j for all j=0..infinity, k=0..(2^j-1).
(End)
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EXAMPLE
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a(5) = floor(sqrt(1+1+1+1+2)) = floor(sqrt(8)) = 2.
n=21; j=3; k=1; a(21)=2^3+1=9;
j=3; k=4; a(27)=a(28)=12.
(End)
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MAPLE
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sumr:=0: a(0):=1: for n from 1 to 1000 do sumr:=sumr+a(n-1): a(n):=floor(sqrt(sumr)): end do: # Paul Weisenhorn, Jun 22 2010
a(0..1)=1; for n from 2 to 100 do j:=floor(log[2](n))-1: k:=iquo(n-2^(j+1)-j, 2): a(n):=2^j+k: end do: # Paul Weisenhorn, Jun 26 2010
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MATHEMATICA
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lst={1}; Nest[AppendTo[lst, Floor[Sqrt[Total[lst]]]]&, 1, 85] (* Harvey P. Dale, May 24 2012 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Formula a(0..3)=1; a(n)=iquo(n+1-floor(log[2](n-2)),2); n=4..infinity; deleted and second Maple program changed Paul Weisenhorn, Aug 22 2010
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STATUS
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approved
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