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

1,3

Comments

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

Links

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

Example

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;
  ...

Prog

(PARI) \\ 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)

Crossrefs

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

Cf. A306211.

Sequence in context: A003647 A353352 A084217 * A245548 A025903 A175327

Adjacent sequences: A381584 A381585 A381586 * A381588 A381589 A381590

Keyword

tabf,nonn,new

Author

Neal Gersh Tolunsky, Feb 27 2025

Status

approved