A246028 -id:A246028 - cp's OEIS frontend

A246029 a(n) = Product_{i in row n of A245562} prime(i).

1, 2, 2, 3, 2, 4, 3, 5, 2, 4, 4, 6, 3, 6, 5, 7, 2, 4, 4, 6, 4, 8, 6, 10, 3, 6, 6, 9, 5, 10, 7, 11, 2, 4, 4, 6, 4, 8, 6, 10, 4, 8, 8, 12, 6, 12, 10, 14, 3, 6, 6, 9, 6, 12, 9, 15, 5, 10, 10, 15, 7, 14, 11, 13, 2, 4, 4, 6, 4, 8, 6, 10, 4, 8, 8, 12, 6, 12, 10, 14, 4, 8, 8, 12, 8, 16, 12, 20, 6, 12, 12, 18

Offset: 0

N. J. A. Sloane, Aug 15 2014; revised Sep 05 2014

nonn

This is the Run Length Transform of S(n) = {1,2,3,5,7,11,...} (1 followed by the primes).

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).

			From _Omar E. Pol_, Feb 12 2015: (Start)
Written as an irregular triangle in which row lengths is A011782:
1;
2;
2,3;
2,4,3,5;
2,4,4,6,3,6,5,7;
2,4,4,6,4,8,6,10,3,6,6,9,5,10,7,11;
2,4,4,6,4,8,6,10,4,8,8,12,6,12,10,14,3,6,6,9,6,12,9,15,5,10,10,15,7,14,11,13;
...
Right border gives the noncomposite numbers. This is simply a restatement of the theorem that this sequence is the Run Length Transform of A008578.
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..8191
Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, A Meta-Algorithm for Creating Fast Algorithms for Counting ON Cells in Odd-Rule Cellular Automata, arXiv:1503.01796 [math.CO], 2015; see also the Accompanying Maple Package.
Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, Odd-Rule Cellular Automata on the Square Grid, arXiv:1503.04249 [math.CO], 2015.
N. J. A. Sloane, On the No. of ON Cells in Cellular Automata, Video of talk in Doron Zeilberger's Experimental Math Seminar at Rutgers University, Feb. 05 2015: Part 1, Part 2
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for sequences related to cellular automata

Cf. A245562, A000045, A001045, A005940, A008578, A011782, A071053, A181819, A245565, A246028, A253081 (partial sums).

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(ithprime(i), i in lis);
ans:=[op(ans),a];
od:
ans;

f[n_, i_, x_] := f[n, i, x] = Which[n == 0, x, EvenQ[n], f[n/2, i + 1, x], True, f[(n - 1)/2, i, x*Prime[i]]];
a5940[n_] := f[n - 1, 1, 1];
a181819[n_] := Times @@ Prime[FactorInteger[n][[All, 2]]];
a[0] = 1; a[n_] := a181819[a5940[n + 1]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Aug 19 2018, after Antti Karttunen *)

from operator import mul
from functools import reduce
from re import split
from sympy import prime
def A246029(n):
    return reduce(mul,(prime(len(d)) for d in split('0+',bin(n)[2:]) if d != '')) if n > 0 else 1
# Chai Wah Wu, Sep 12 2014

a(n) = A181819(A005940(1+n)). - Antti Karttunen, Oct 15 2016

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

Original entry on oeis.org

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

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

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

			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)

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.

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.

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;

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 *)

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

# 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

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

cp's OEIS Frontend

A246029 a(n) = Product_{i in row n of A245562} prime(i).

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