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%I #46 Nov 08 2022 09:44:13
%S 1,1,1,2,1,2,3,4,1,2,3,5,4,6,7,1,2,3,5,9,4,6,10,7,11,8,13,1,2,3,5,9,
%T 16,4,6,10,17,7,11,18,8,13,21,12,19,1,2,3,5,9,16,28,4,6,10,17,29,7,11,
%U 18,30,8,13,21,34,12,19,31,14,22,35,1,2,3,5,9,16,28,46,4,6,10,17,29,47,7,11,18,30,48
%N Inventory sequence: record where the 1's, 2's, etc. are located starting with a(1) = 1, a(2) = 1 (see example).
%e At stage n >= 1 we only look at the numbers 1 up to n, and ignore numbers bigger than n.
%e Stage 0: start with a(1) = 1, a(2) = 1.
%e Stage 1: we see 1's at 1,2, so we adjoin 1,2, getting 1,1, 1,2.
%e Stage 2: we see 1's at 1,2,3, and 2's at 4, so we adjoin 1,2,3,4, getting 1,1,1,2, 1,2,3,4.
%e Stage 3: we see 1's at 1,2,3,5, 2's at 4,6 and 3's at 7, so we adjoin 1,2,3,5,4,6,7, getting 1,1,1,2,1,2,3,4, 1,2,3,5,4,6,7.
%e Stage 4: we see 1's at 1,2,3,5,9, 2's at 4,6,10, 3's at 7,11, 4's at 8,13, so we adjoin 1,2, ..., 8,13 and so on.
%e We obtain an irregular triangle by writing the results of the stages as separate rows:
%e 1, 1,
%e 1, 2,
%e 1, 2, 3, 4,
%e 1, 2, 3, 5, 4, 6, 7,
%e 1, 2, 3, 5, 9, 4, 6, 10, 7, 11, 8, 13,
%e 1, 2, 3, 5, 9, 16, 4, 6, 10, 17, 7, 11, 18, 8, 13, 21, 12, 19,
%e 1, 2, 3, 5, 9, 16, 28, 4, 6, 10, 17, 29, 7, 11, 18, 30, 8, 13, 21, 34, 12, 19, 31, 14, 22, 35,
%e ... (_N. J. A. Sloane_, Nov 07 2022)
%o (Python)
%o terms = [1, 1]
%o for i in range(1,11):
%o new_terms = []
%o for j in range(1, i+1):
%o for k in range(len(terms)):
%o if terms[k] == j: new_terms.append(k+1)
%o terms.extend(new_terms)
%o print(terms) # _Gleb Ivanov_, Nov 01 2022
%Y See A357443 for another version.
%Y Cf. A030717, A055187, A217780, A342585, A357317, A356784.
%K nonn,tabf
%O 1,4
%A _Ctibor O. Zizka_, Oct 29 2022
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