cp's OEIS Frontend

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

%I A347315 #24 Sep 18 2022 11:05:38
%S A347315 0,2,6,11,17,24,32,40,51,63,76,89,102,116,132,149,169,188,208,228,249,
%T A347315 272,297,322,349,377,404,432,461,494,528,562,597,632,667,703,740,778,
%U A347315 820,862,903,945,991,1038,1085,1132,1181,1229,1277,1328,1380,1434,1487
%N A347315 a(n) = sum of row beginning with n when inventory sequence A342585 is written as an irregular triangle.
%C A347315 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.
%C A347315 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
%H A347315 Michael De Vlieger, <a href="/A347315/b347315.txt">Table of n, a(n) for n = 0..1000</a>
%e A347315 As an irregular triangle A342585 begins:
%e A347315    0;
%e A347315    1,  1,  0;
%e A347315    2,  2,  2,  0;
%e A347315    3,  2,  4,  1,  1,  0;
%e A347315    4,  4,  4,  1,  4,  0;
%e A347315 ...
%e A347315 and the row sums are 0, 2, 6, 11, 17, ...
%t A347315 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 *)
%o A347315 (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
%o A347315 (Python)
%o A347315 from collections import Counter
%o A347315 def aupton(nn):
%o A347315     num, inventory, rowsum, alst = 0, Counter([0]), 0, [0]
%o A347315     while len(alst) <= nn:
%o A347315         c = inventory[num]
%o A347315         num += 1
%o A347315         rowsum += c
%o A347315         inventory.update([c])
%o A347315         if c == 0:
%o A347315             alst.append(rowsum)
%o A347315             num = rowsum = 0
%o A347315     return alst
%o A347315 print(aupton(52)) # _Michael S. Branicky_, Nov 14 2021
%Y A347315 Cf. A342585.
%K A347315 nonn
%O A347315 0,2
%A A347315 _N. J. A. Sloane_, Sep 09 2021
%E A347315 More terms from _Alois P. Heinz_, Sep 09 2021