A347299 Lengths of successive rows when the Inventory Sequence A342585 is written as a triangle.

1, 3, 4, 6, 6, 8, 9, 9, 11, 12, 13, 13, 15, 15, 17, 17, 18, 19, 20, 22, 22, 25, 25, 25, 27, 28, 28, 30, 30, 33, 33, 34, 35, 35, 37, 37, 39, 39, 42, 42, 43, 44, 47, 47, 47, 49, 49, 49, 50, 52, 52, 54, 54

Offset

1,2

Comments

First differences of A343880, preceded by 1.

Links

Michael De Vlieger, Table of n, a(n) for n = 1..16384 (first 10000 terms from Hugo Pfoertner)

Index entries for sequences related to the inventory sequence

Mathematica

Prepend[Differences[#], First[#]] &@ Position[Block[{c, k, m}, c[0] = 1; {0}~Join~Reap[Do[k = 0; While[IntegerQ[c[k]], Set[m, c[k]]; Sow[m]; If[IntegerQ@ c[m], c[m]++, c[m] = 1]; k++]; Sow[0]; c[0]++, 52]][[-1, -1]]], 0][[All, 1]] (* Michael De Vlieger, Oct 12 2021 *)
(* Faster code: *)
Block[{c, m, k}, c[_] = 0; Reap[Do[k = 0; While[c[k] > 0, Set[m, c[k]]; c[m]++; k++]; Set[m, c[k]]; Sow[k + 1]; c[m]++, {i, 53}]][[-1, -1]]] (* Michael De Vlieger, Nov 13 2021 *)

Prog

(Python)
from collections import Counter
def aupton(terms):
    num, A342585lst, inventory, alst, rowlen = 0, [0], Counter([0]), [1], 0
    while len(alst) < terms:
        rowlen += 1
        c = inventory[num]
        if c == 0:
            num = 0
            alst.append(rowlen)
            rowlen = 0
        else:
            num += 1
        A342585lst.append(c)
        inventory.update([c])
    return alst
print(aupton(53)) # Michael S. Branicky, Aug 27 2021

Crossrefs

Cf. A342585, A343878, A343880.

Sequence in context: A023836 A064800 A078574 * A230300 A347318 A275883

Adjacent sequences: A347296 A347297 A347298 * A347300 A347301 A347302

Keyword

nonn

Author

N. J. A. Sloane, Aug 27 2021

Status

approved