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a(n) = Product_{i in row n of A245562} Fibonacci(i+2).
3

%I #39 Feb 04 2022 16:31:15

%S 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,

%T 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,

%U 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

%N a(n) = Product_{i in row n of A245562} Fibonacci(i+2).

%C This is the Run Length Transform of S(n) = Fibonacci(n+2).

%C 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).

%C a(n) = Sum_{k=0..n} ({binomial(3k,k)*binomial(n,k)} mod 2). - _Chai Wah Wu_, Oct 19 2016

%H Alois P. Heinz, <a href="/A245564/b245564.txt">Table of n, a(n) for n = 0..8191</a>

%H Chai Wah Wu, <a href="https://arxiv.org/abs/1610.06166">Sums of products of binomial coefficients mod 2 and run length transforms of sequences</a>, arXiv:1610.06166 [math.CO], 2016.

%p with(combinat); ans:=[];

%p for n from 0 to 100 do lis:=[]; t1:=convert(n,base,2); L1:=nops(t1); out1:=1; c:=0;

%p for i from 1 to L1 do

%p if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;

%p elif out1 = 0 and t1[i] = 1 then c:=c+1;

%p elif out1 = 1 and t1[i] = 0 then c:=c;

%p elif out1 = 0 and t1[i] = 0 then lis:=[c,op(lis)]; out1:=1; c:=0;

%p fi;

%p if i = L1 and c>0 then lis:=[c,op(lis)]; fi;

%p od:

%p a:=mul(fibonacci(i+2), i in lis);

%p ans:=[op(ans),a];

%p od:

%p ans;

%t a[n_] := Sum[Mod[Binomial[3k, k] Binomial[n, k], 2], {k, 0, n}];

%t a /@ Range[0, 100] (* _Jean-François Alcover_, Feb 29 2020, after _Chai Wah Wu_ *)

%o (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

%o (Python)

%o # use RLT function from A278159

%o from sympy import fibonacci

%o def A245564(n): return RLT(n,lambda m: fibonacci(m+2)) # _Chai Wah Wu_, Feb 04 2022

%Y Cf. A245562, A000045, A001045, A071053, A245565, A246028.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Aug 10 2014; revised Sep 05 2014