A337277 Stern's triangle read by rows.

1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 2, 1, 1, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 4, 3, 5, 2, 5, 3, 4, 1, 3, 2, 3, 1, 2, 1, 1, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 4, 3, 5, 2, 5, 3, 4, 1, 3, 2, 3, 1, 2, 1, 1

Offset

0,7

Comments

The first two rows are 1, then 1,1,1. To get row n, copy row n-1, and insert c+d between every pair of adjacent terms c,d, and finally insert a 1 at the beginning and end of the row.

The maximum value in row n is A000045(n+1). - Alois P. Heinz, Sep 09 2020

References

Stanley, Richard P. "Some Linear Recurrences Motivated by Stern’s Diatomic Array." The American Mathematical Monthly 127.2 (2020): 99-111.

Links

Alois P. Heinz, Rows n = 0..14, flattened

Richard P. Stanley, Some Linear Recurrences Motivated by Stern's Diatomic Array, arXiv:1901.04647 [math.CO], 2019.

Formula

T(n,n) = A002487(n+1). - Alois P. Heinz, Sep 09 2020

Example

Triangle begins:
  1;
  1, 1, 1;
  1, 1, 2, 1, 2, 1, 1;
  1, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 2, 1, 1;
  1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 4, 3, 5, 2, 5, 3, 4, 1, 3, 2, 3, 1, 2, 1, 1;
  ...

Maple

T:= proc(n) option remember; `if`(n=0, 1, (L-> [1, L[1], seq(
      [L[i-1]+L[i], L[i]][], i=2..nops(L)), 1][])([T(n-1)]))
    end:
seq(T(n), n=0..6);  # Alois P. Heinz, Sep 09 2020

Mathematica

Nest[Append[#, Flatten@ Join[{1}, If[Length@ # > 1, Map[{#1, #1 + #2} & @@ # &, Partition[#[[-1]], 2, 1] ], {}], {#[[-1, -1]]}, {1}]] &, {{1}}, 5] // Flatten (* Michael De Vlieger, Sep 09 2020 *)

Crossrefs

Cf. A000045, A002487, A049456, A334627.

Row sums give A000244.

Row lengths give A126646.

Sequence in context: A103414 A092400 A303598 * A269974 A269975 A308069

Adjacent sequences: A337274 A337275 A337276 * A337278 A337279 A337280

Keyword

nonn,tabf

Author

N. J. A. Sloane, Sep 09 2020

Status

approved