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A245564
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a(n) = Product_{i in row n of A245562} Fibonacci(i+2).
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3
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1, 2, 2, 3, 2, 4, 3, 5, 2, 4, 4, 6, 3, 6, 5, 8, 2, 4, 4, 6, 4, 8, 6, 10, 3, 6, 6, 9, 5, 10, 8, 13, 2, 4, 4, 6, 4, 8, 6, 10, 4, 8, 8, 12, 6, 12, 10, 16, 3, 6, 6, 9, 6, 12, 9, 15, 5, 10, 10, 15, 8, 16, 13, 21, 2, 4, 4, 6, 4, 8, 6, 10, 4, 8, 8, 12, 6, 12, 10, 16, 4, 8, 8, 12, 8, 16, 12, 20, 6, 12, 12, 18
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OFFSET
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0,2
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COMMENTS
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This is the Run Length Transform of S(n) = Fibonacci(n+2).
The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).
a(n) = Sum_{k=0..n} ({binomial(3k,k)*binomial(n,k)} mod 2). - Chai Wah Wu, Oct 19 2016
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LINKS
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MAPLE
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with(combinat); ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
od:
a:=mul(fibonacci(i+2), i in lis);
ans:=[op(ans), a];
od:
ans;
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MATHEMATICA
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a[n_] := Sum[Mod[Binomial[3k, k] Binomial[n, k], 2], {k, 0, n}];
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PROG
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(PARI) a(n)=my(s=1, k); while(n, n>>=valuation(n, 2); k=valuation(n+1, 2); s*=fibonacci(k+2); n>>=k); s \\ Charles R Greathouse IV, Oct 21 2016
(Python)
from sympy import fibonacci
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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