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A245564
a(n) = Product_{i in row n of A245562} Fibonacci(i+2).
3
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
OFFSET
0,2
COMMENTS
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
MAPLE
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;
MATHEMATICA
a[n_] := Sum[Mod[Binomial[3k, k] Binomial[n, k], 2], {k, 0, n}];
a /@ Range[0, 100] (* Jean-François Alcover, Feb 29 2020, after Chai Wah Wu *)
PROG
(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)
# use RLT function from A278159
from sympy import fibonacci
def A245564(n): return RLT(n, lambda m: fibonacci(m+2)) # Chai Wah Wu, Feb 04 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Aug 10 2014; revised Sep 05 2014
STATUS
approved