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A117940
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a(0)=1, thereafter a(3n) = a(3n+1)/3 = a(n), a(3n+2)=0.
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20
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1, 3, 0, 3, 9, 0, 0, 0, 0, 3, 9, 0, 9, 27, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 9, 0, 9, 27, 0, 0, 0, 0, 9, 27, 0, 27, 81, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 9, 0, 9, 27, 0, 0, 0, 0, 9, 27, 0, 27, 81, 0, 0, 0, 0, 0, 0
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OFFSET
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0,2
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COMMENTS
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a(n) = a(3n)/a(0) = a(3n+1)/a(1). a(n) mod 2 = A039966(n). Row sums of A117939.
Observation: if this is written as a triangle (see example) then at least the first five row sums coincide with A002001. - Omar E. Pol, Nov 28 2011
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LINKS
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FORMULA
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G.f.: Product{k>=0, 1+3x^(3^k)}; a(n)=sum{k=0..n, sum{j=0..n, L(C(n,j)/3)*L(C(n-j,k)/3)}} where L(j/p) is the Legendre symbol of j and p.
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EXAMPLE
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Contribution from Omar E. Pol, Nov 26 2011 (Start):
When written as a triangle this begins:
1,
3,0,
3,9,0,0,0,0,
3,9,0,9,27,0,0,0,0,0,0,0,0,0,0,0,0,0,
3,9,0,9,27,0,0,0,0,9,27,0,27,81,0,0,0,0,0,0,0,0,0,0,0,...
(End)
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CROSSREFS
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For generating functions Prod_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
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KEYWORD
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easy,nonn,tabf
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AUTHOR
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STATUS
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approved
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