A306346 Start with the sequence S(1) = [0], and for n >= 1 define S(n+1) to equal the concatenation of S(n) with the RUNS transform of S(n) when read in reverse order. This sequence is the limit of that process as n goes to infinity.

0, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 7, 1, 1, 1, 1, 1, 5, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 1, 1, 1, 7, 1, 1, 1, 1, 1, 3, 1, 7, 1

Offset: 1

Author

Michel Lagneau, Feb 09 2019

Keywords

• nonn

Comments

Conjecture 1: a(n) <= 7.

Conjecture 2: a(n) is odd for n > 1 and when a(n) = 3, 5 or 7, n is odd.

Property of the sequence:

a(n) = 3 for the odd values n = 5, 9, 13, 15, 21, 23, 25, 37, 39, ...

a(n) = 5 for the odd values n = 33, 57, 75, 93, 111, 115, 129, 147, ...

a(n) = 7 for the odd values n = 51, 79, 87, 121, 133, 141, 185, 203, ...

Remark:

If row 1 = [1], the sequence a(n) becomes b(n) = 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 3, 1, 1, 1, ... and it is conjectured that b(n) <= 7.

Statistics for n <= 10^5:

+--------------+-------------------------+------------+

| a(n) | number of occurrences | percentage |

| | for n <= 10^5 | |

+--------------+-------------------------+------------+

| 1 | 72244 | 72.244% |

| 3 | 18030 | 18.030% |

| 5 | 6806 | 6.806% |

| 7 | 2919 | 2.919% |

+--------------+-------------------------+------------+

Examples

			We may consider this sequence to be the limit of the rows of the irregular triangle in which row n+1 equals the concatenation of row n with the RUNS transform of row n when read in reverse order, as illustrated below.
Start with row 1 = [0];
the RUNS of row 1 in reverse = [1], so row 2 = row 1 + [1] = [0, 1];
the RUNS of row 2 in reverse = [1, 1], so row 3 = row 2 + [1, 1] = [0, 1, 1, 1];
the RUNS of row 3 in reverse = [3, 1], so row 4 = row 3 + [3, 1] = [0, 1, 1, 1, 3, 1];
the RUNS of row 4 in reverse = [1, 1, 3, 1], so row 5 = row 4 + [1, 1, 3, 1] = [0, 1, 1, 1, 3, 1, 1, 1, 3, 1];
etc.
The irregular triangle starts:
  [0];
  [0, 1];
  [0, 1, 1, 1];
  [0, 1, 1, 1, 3, 1];
  [0, 1, 1, 1, 3, 1, 1, 1, 3, 1];
  [0, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1];
  [0, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1];
  ...
in which the limit of the rows yields this sequence.
The row lengths of the above triangle equal A381357 (offset 3).
		

Crossrefs

Cf. A000002, A381587, A381356, A381357.

Programs

Maple

n0:=1:T:=array(1..1000,[0$1000]):
for n from 1 to 50 do :
  it:=1:i:=n0:
  for k from n0 by -1 to 2 do:
    if T[k]=T[k-1]
     then
     it:=it+1:
     else
     i:=i+1:T[i]:=it:it:=1:
    fi:
   od:
  i:=i+1:T[i]:=1:n0:=i:
  od:
print(T):

PARI

\\ From Paul D. Hanna, Mar 04 2025: (Start)
\\ Print the first N rows of the irregular triangle.
\\ This sequence equals the limit of the rows.
\\ RUNS(V) Returns vector of run lengths in vector V:
{RUNS(V) = my(R=[], c=1); if(#V>1, for(n=2, #V, if(V[n]==V[n-1], c=c+1, R=concat(R, c); c=1))); R=concat(R, c)}
\\ REV(V) Reverses order of vector V:
{REV(V) = Vec(Polrev(Ser(V)))}
\\ Generates N rows as a vector A of row vectors
{N=12; A=vector(N); A[1]=[0];
for(n=1, #A-1, A[n+1] = concat(A[n], RUNS(REV(A[n]))); );}
for(n=1,N,print(A[n])) \\ (End)

Extensions

Name corrected and edited by Paul D. Hanna, Mar 05 2025