A347315 a(n) = sum of row beginning with n when inventory sequence A342585 is written as an irregular triangle.

0, 2, 6, 11, 17, 24, 32, 40, 51, 63, 76, 89, 102, 116, 132, 149, 169, 188, 208, 228, 249, 272, 297, 322, 349, 377, 404, 432, 461, 494, 528, 562, 597, 632, 667, 703, 740, 778, 820, 862, 903, 945, 991, 1038, 1085, 1132, 1181, 1229, 1277, 1328, 1380, 1434, 1487

Offset: 0

Author

N. J. A. Sloane, Sep 09 2021

Keywords

• nonn

Comments

The inventory sequence A342585 counts, for k = 0, 1, 2, ..., the k's that have occurred so far, and if zero, restarts with k = 0. The rows end where the zeros occur.

The sequence appears to grow approximately quadratically. More precisely, b(n) = sqrt(a(n)) is roughly a straight line over increasingly large intervals, but the slope is slightly larger at the beginning and then decreasing towards the end of these intervals. For example, on [1..80] the slope is almost exactly 0.72; on [150..250] the slope is roughly 1.0, over [320..420] the slope is again 0.8, over [430..520] it is again 1.0, over [530..620] it is again 0.8; then the slope increases: b(780..1000) is again a nearly straight line with slope 1.67, etc. - M. F. Hasler, Nov 14 2021

Examples

			As an irregular triangle A342585 begins:
   0;
   1,  1,  0;
   2,  2,  2,  0;
   3,  2,  4,  1,  1,  0;
   4,  4,  4,  1,  4,  0;
...
and the row sums are 0, 2, 6, 11, 17, ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A342585.

Programs

Mathematica

Join[{0}, Total /@ SplitBy[Block[{c, k, m, nn = 52}, c[0] = 1; 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]++, nn]][[-1, -1]]], # == 0 &][[1 ;; -1 ;; 2]]] (* Michael De Vlieger, Oct 12 2021 *)

PARI

A347315_vec(N, c=[], i, s)=vector(N, j, until(c[1+c[i]]++&&!c[i]||j==1, while(#c<=i||#c<=c[i+1], c=concat(c, 0)); s+=c[i+=1]); s+s=i=0) \\ M. F. Hasler, Nov 14 2021

Python

from collections import Counter
def aupton(nn):
    num, inventory, rowsum, alst = 0, Counter([0]), 0, [0]
    while len(alst) <= nn:
        c = inventory[num]
        num += 1
        rowsum += c
        inventory.update([c])
        if c == 0:
            alst.append(rowsum)
            num = rowsum = 0
    return alst
print(aupton(52)) # Michael S. Branicky, Nov 14 2021

Extensions

More terms from Alois P. Heinz, Sep 09 2021