This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
Write 0, thus having 1 0, thus having 2 0's and 1 1, thus having 3 0's and 3 1's and 1 2, etc. 0; 1,0; 2,0,1,1; 3,0,3,1,1,2; ...
s={0};Do[ta=Table[{Count[s, # ], # }&/@Range[0,Max[s]]]; s=Flatten[{s,ta}],{22}];s (* Zak Seidov, Oct 23 2009 *)
Writing pairs vertically, the initial segments are 1..1..3..4 1..6 2 1..8 3 2 1 1..11 5 3 2 3 1 ...1..1..1 3..1 3 4..1 3 4 6 2...1 3 4 6 2 8. The 5th segment is read "6 1's and 2 3's and 1 4," this being a count of all the previously written numbers. The numbers 6,2,1 are used as adjectives, whereas 1,3,4 are used as nouns. Here, the nouns are kept in order of first appearance; in A055187, they are in increasing order. - _Clark Kimberling_, Mar 25 2013
s = {1}; Do[s = Flatten[{s, {Count[s, #], #} & /@ DeleteDuplicates[s]}], {14}]; s (* Peter J. C. Moses, Mar 21 2013 *)
At stage n >= 1 we only look at the numbers 1 up to n, and ignore numbers bigger than n. Stage 0: start with a(1) = 1, a(2) = 1. Stage 1: we see 1's at 1,2, so we adjoin 1,2, getting 1,1, 1,2. 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. 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. 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. We obtain an irregular triangle by writing the results of the stages as separate rows: 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, 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, 18, 30, 8, 13, 21, 34, 12, 19, 31, 14, 22, 35, ... (_N. J. A. Sloane_, Nov 07 2022)
terms = [1, 1]
for i in range(1,11):
new_terms = []
for j in range(1, i+1):
for k in range(len(terms)):
if terms[k] == j: new_terms.append(k+1)
terms.extend(new_terms)
print(terms) # Gleb Ivanov, Nov 01 2022
Comments