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A032531 An inventory sequence: triangle read by rows, where T(n, k), 0 <= k <= n, records the number of k's thus far in the flattened sequence.

%I A032531 #50 Nov 12 2024 17:23:16
%S A032531 0,1,1,1,3,0,2,3,1,2,2,4,3,3,1,2,5,4,4,3,1,2,6,5,5,3,3,1,2,7,6,7,3,3,
%T A032531 2,2,2,7,9,9,3,3,2,3,0,3,7,10,13,3,3,2,4,0,2,4,7,12,15,5,4,2,5,0,2,1,
%U A032531 5,8,14,15,6,6,4,5,1,2,1,0,6,10,15,15,7,7,5,7,1,2,2,0,1,7,12,17
%N A032531 An inventory sequence: triangle read by rows, where T(n, k), 0 <= k <= n, records the number of k's thus far in the flattened sequence.
%C A032531 Old name: a(n) = number of a(i) for 0<=i<n that are equal to A002262(n).
%C A032531 This sequence is a variation of the Inventory sequence A342585. The same rules apply except that in this variation each row ends after k terms, where k is the current row count which starts at 1. The behavior up to the first 1 million terms is similar to A342585 but beyond that the most common terms do not increase, likely due to the rows being cut off after k terms thus numbers such as 1 and 2 no longer make regular appearances. Larger number terms do increase and overtake the leading early terms, and it appears this pattern repeats as n increases. See the linked images. - _Scott R. Shannon_, Sep 13 2021
%C A032531 The complexity of this sequence derives from the totals being updated during the calculation of each row. If each row recorded an inventory of only the earlier rows, we would get the much simpler A025581. - _Peter Munn_, May 06 2023
%H A032531 Scott R. Shannon, <a href="/A032531/b032531.txt">Table of n, a(n) for n = 0..20000</a>.
%H A032531 Scott R. Shannon, <a href="/A032531/a032531.png">Image of the first 10^6 terms</a>.
%H A032531 Scott R. Shannon, <a href="/A032531/a032531_1.gif">Image of the first 10^7 terms</a>.
%H A032531 Scott R. Shannon, <a href="/A032531/a032531.gif">Image of the first 10^8 terms</a>.
%H A032531 <a href="/index/In#inventory">Index entries for sequences related to inventory sequences</a>
%e A032531 Triangle begins:
%e A032531   0;
%e A032531   1, 1;
%e A032531   1, 3, 0;
%e A032531   2, 3, 1, 2;
%e A032531   2, 4, 3, 3, 1;
%e A032531   2, 5, 4, 4, 3, 1;
%e A032531   2, 6, 5, 5, 3, 3, 1;
%e A032531   2, 7, 6, 7, 3, 3, 2, 2;
%e A032531   2, 7, 9, 9, 3, 3, 2, 3, 0;
%e A032531   ...
%p A032531 A002262 := proc(n)
%p A032531     n - binomial(floor(1/2+sqrt(2*(1+n))), 2);
%p A032531 end proc:
%p A032531 A032531 := proc(n)
%p A032531     option remember;
%p A032531     local a,piv,i ;
%p A032531     a := 0 ;
%p A032531     piv := A002262(n) ;
%p A032531     for i from 0 to n-1  do
%p A032531         if procname(i) = piv then
%p A032531             a := a+1 ;
%p A032531         end if;
%p A032531     end do:
%p A032531     a ;
%p A032531 end proc:
%p A032531 seq(A032531(n),n=0..100) ; # _R. J. Mathar_, May 08 2020
%t A032531 A002262[n_] :=  n - Binomial[Floor[1/2 + Sqrt[2*(1 + n)]], 2];
%t A032531 A032531[n_] := A032531[n] = Module[{a, piv, i}, a = 0; piv = A002262[n]; For[i = 0, i <= n-1, i++, If[A032531[i] == piv, a++]]; a];
%t A032531 Table[A032531[n], {n, 0, 100}] (* _Jean-François Alcover_, Mar 25 2024, after _R. J. Mathar_ *)
%o A032531 (Python)
%o A032531 from math import comb, isqrt
%o A032531 from collections import Counter
%o A032531 def idx(n): return n - comb((1+isqrt(8+8*n))//2, 2)
%o A032531 def aupton(nn):
%o A032531     num, alst, inventory = 0, [0], Counter([0])
%o A032531     for n in range(1, nn+1):
%o A032531         c = inventory[idx(n)]
%o A032531         alst.append(c)
%o A032531         inventory[c] += 1
%o A032531     return alst
%o A032531 print(aupton(93)) # _Michael S. Branicky_, May 07 2023
%Y A032531 Cf. A002262, A025581, A342585.
%K A032531 nonn,tabl,look
%O A032531 0,5
%A A032531 Dmitri Papichev (Dmitri.Papichev(AT)iname.com)
%E A032531 New name from _Peter Munn_, May 06 2023