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A116528
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Starting a(0)=0, a(1)=1, recursively a(2n)=a(n), a(2n+1)= 2*a(n)+a(n+1).
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2
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0, 1, 1, 3, 1, 5, 3, 7, 1, 7, 5, 13, 3, 13, 7, 15, 1, 9, 7, 19, 5, 23, 13, 29, 3, 19, 13, 33, 7, 29, 15, 31, 1, 11, 9, 25, 7, 33, 19, 43, 5, 33, 23, 59, 13, 55, 29, 61, 3, 25, 19, 51, 13, 59, 33, 73, 7, 43, 29, 73, 15, 61, 31, 63, 1, 13
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OFFSET
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0,4
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COMMENTS
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Equals row 2 of the array in A178239, an infinite set of sequences of the form a(n) = a(2n), a(2n+1) = r*a(n) + a(n+1). - Gary W. Adamson, May 23 2010
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REFERENCES
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H. Harborth, Number of Odd Binomial Coefficients, Proc. Amer. Math. Soc. 62, 19-22, 1977
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LINKS
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Table of n, a(n) for n=0..65.
Eric Weisstein's World of Mathematics, Stolarsky-Harborth Constant
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FORMULA
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Given an infinite lower triangular matrix M with (1, 1, 2, 0, 0, 0,...) in every column, shifted down twice for columns k>1; Lim_{n->inf} M^n = A116528, the left-shifted vector considered as a sequence with offset 1. - Gary W. Adamson, May 05 2010
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MAPLE
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A116528 := proc(n)
option remember;
if n <= 1 then
n;
elif type(n, 'even') then
procname(n/2) ;
else
2* procname((n-1)/2)+procname((n+1)/2) ;
end if;
end proc:
seq(A116528(n), n=0..80) ; # R. J. Mathar, Nov 16 2011
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MATHEMATICA
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b[0] := 0 b[1] := 1 b[n_?EvenQ] := b[n] = b[n/2] b[n_?OddQ] := b[n] = 2*b[(n - 1)/2] + b[(n + 1)/2] a = Table[b[n], {n, 1, 50}]
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CROSSREFS
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Cf. A006046.
Sequence in context: A161398 A204455 A161820 * A081431 A053641 A161825
Adjacent sequences: A116525 A116526 A116527 * A116529 A116530 A116531
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KEYWORD
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nonn,easy
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AUTHOR
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Roger Bagula, Mar 15 2006
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STATUS
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approved
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