A381358 -id:A381358 - cp's OEIS frontend

A381587 a(1) = 1; thereafter the sequence is extended by iteratively appending the run length transform of the reverse of the sequence thus far.

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, 3, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 3, 1, 3, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3

Offset: 1

Neal Gersh Tolunsky, Feb 27 2025

The run length transform replaces each run of consecutive equal values with a single value representing the length of that run.
Is 2 the greatest even number in the sequence?
It appears that the limit of the rows read in reverse equals A306346 (ignoring the initial terms). - Paul D. Hanna, Mar 03 2025

			Irregular triangle begins:
  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;
  1,3,1,3,1,1,1,3,1,1,1,3,1,1,1,2;
  1,3,1,3,1,3,1,1,1,1,1,3,1,3,1,1,1,3,1,1,1,3,1,1,1,2;
  ...

Paul D. Hanna, Table of n, a(n) for n = 1..9259 (first 20 rows)

Cf. A381357 (row lengths), A381358 (row sums), A381356 (limit of rows), A306346.

Cf. A306211.

\\ From Paul D. Hanna, Mar 03 2025: (Start)
\\ 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=15; A=vector(N);A[1]=[1];A[2]=[1];A[3]=[2];
for(n=3,#A-1, A[n+1] = concat(RUNS(REV(A[n])),A[n]);); A}
for(n=1,N,print(A[n])) \\ Print N rows of this triangle (End)

A381357 Row lengths of irregular triangle A381587.

Original entry on oeis.org

1, 1, 1, 2, 4, 6, 10, 16, 26, 42, 66, 102, 156, 238, 364, 560, 868, 1354, 2120, 3322, 5198, 8112, 12624, 19602, 30400, 47138, 73138, 113598, 176630, 274858

Offset: 1

Paul D. Hanna, Mar 03 2025

If it exists, what is the limit of a(n)^(1/n) as n increases?

			Row n+1 of irregular triangle A381587 equals the run lengths of the first n rows of the triangle (flattened) when read in reverse order, starting with
  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;
  1,3,1,3,1,1,1,3,1,1,1,3,1,1,1,2; ...
This sequence gives the row lengths [1, 1, 1, 2, 4, 6, 10, 16, ...].

Cf. A381587, A381358.

\\ Print the row lengths of irregular triangle A381587
\\ 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=25; A=vector(N); A[1]=[1]; A[2]=[1]; A[3]=[2];
for(n=3, #A-1, A[n+1] = concat(RUNS(REV(A[n])), A[n]); );}
\\ Print the row lengths of the first N rows
for(n=1, N, print1(#A[n],", "))

cp's OEIS Frontend

A381587 a(1) = 1; thereafter the sequence is extended by iteratively appending the run length transform of the reverse of the sequence thus far.

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