A245564 a(n) = Product_{i in row n of A245562} Fibonacci(i+2).

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

Author

N. J. A. Sloane, Aug 10 2014; revised Sep 05 2014

Keywords

• nonn
• easy

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).

Also the number of sparse subsets of the binary indices of n, where a set is sparse iff 1 is not a first difference. The maximal case is A384883. For prime instead of binary indices we have A166469. - Gus Wiseman, Jul 05 2025

Examples

			From _Gus Wiseman_, Jul 05 2025: (Start)
The binary indices of 11 are {1,2,4}, with sparse subsets {{},{1},{2},{4},{1,4},{2,4}}, so a(11) = 6.
The maximal runs of binary indices of 11 are ((1,2),(4)), with lengths (2,1), so a(11) = F(2+2)*F(1+2) = 6.
The a(0) = 1 through a(12) = 3 sparse subsets are:
  0    1    2    3    4    5    6    7    8    9    10    11    12
  ------------------------------------------------------------------
  {}   {}   {}   {}   {}   {}   {}   {}   {}   {}    {}    {}    {}
       {1}  {2}  {1}  {3}  {1}  {2}  {1}  {4}  {1}   {2}   {1}   {3}
                 {2}       {3}  {3}  {2}       {4}   {4}   {2}   {4}
                           {1,3}     {3}       {1,4} {2,4} {4}
                                     {1,3}                 {1,4}
                                                           {2,4}
The greatest number whose set of binary indices is a member of column n above is A374356(n).
(End)
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..8191
• Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.

Crossrefs

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

A034839 counts subsets by number of maximal runs, strict partitions A116674.

A384877 gives lengths of maximal anti-runs of binary indices, firsts A384878.

A384893 counts subsets by number of maximal anti-runs, for partitions A268193, A384905.

Cf. A000071, A001629, A010049, A053538, A166469, A202023, A202064, A208342, A374356, A384883.

Programs

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 *)
spars[S_]:=Select[Subsets[S],FreeQ[Differences[#],1]&];
bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length[spars[bpe[n]]],{n,0,30}] (* Gus Wiseman, Jul 05 2025 *)

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

Formula

a(n) = Sum_{k=0..n} ({binomial(3k,k)*binomial(n,k)} mod 2). - Chai Wah Wu, Oct 19 2016