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A086450
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a(0) = 1, a(2n+1) = a(n), a(2n) = a(n) + a(n-1) +...+ a(n-m) +... where a(n<0) = 0.
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2
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1, 1, 2, 1, 4, 2, 5, 1, 9, 4, 11, 2, 16, 5, 17, 1, 26, 9, 30, 4, 41, 11, 43, 2, 59, 16, 64, 5, 81, 17, 82, 1, 108, 26, 117, 9, 147, 30, 151, 4, 192, 41, 203, 11, 246, 43, 248, 2, 307, 59, 323, 16, 387, 64, 392, 5, 473, 81, 490, 17, 572, 82, 573, 1, 681, 108, 707
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OFFSET
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0,3
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COMMENTS
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Sequence has itself and its partial sums as bisections.
Setting m=1 gives Stern-Brocot sequence (A002487).
Conjecture: a(n) mod 2 repeats the 7-pattern 1,1,0,1,0,0,1 (A011657).
The conjecture is easily proved by induction: a(0) to a(14)=1, 1, 2, 1, 4, 2, 5, 1, 9, 4, 11, 2, 16, 5 read mod 2 gives 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1. Assume the conjecture is true up to n = 14k. Then the 7 next odd entries a(14k+1), a(14k+3), ..., a(14k+13) are read from a(7k) to a(7k+6), which follow the correct mod 2 pattern by assuption. For the even entries a(14k), a(14k+10)... a(14k+12), the sum over the first 7k-1 addends is even, simply because of each consecutive 7 addends exactly 4 are odd. So again a(7k) to a(7k+6) determines the outcome and again gives the desired pattern. a(14k) is odd, since a(7k) is odd, a(14k+2) is even, since a(7k) and a(7k+1) are odd and so on ... - Lambert Herrgesell (zero815(AT)googlemail.com), May 08 2007
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LINKS
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Table of n, a(n) for n=0..66.
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PROG
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(PARI) a(n)=if(n<2, n>=0, if(n%2==0, sum(k=0, n/2, a(n/2-k)), a((n-1)/2)))
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CROSSREFS
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Cf. A086449.
Partial sums are in A085765.
Sequence in context: A120988 A095979 A054269 * A106044 A124896 A008742
Adjacent sequences: A086447 A086448 A086449 * A086451 A086452 A086453
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KEYWORD
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nonn,easy
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AUTHOR
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Ralf Stephan, Jul 20 2003
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STATUS
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approved
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