A051120 -id:A051120 - cp's OEIS frontend

A055186 Cumulative counting sequence: method A (adjective-before-noun)-pairs with first term 0.

0, 1, 0, 2, 0, 1, 1, 3, 0, 3, 1, 1, 2, 4, 0, 5, 1, 2, 2, 2, 3, 5, 0, 6, 1, 5, 2, 3, 3, 1, 4, 1, 5, 6, 0, 9, 1, 6, 2, 5, 3, 2, 4, 4, 5, 1, 6, 7, 0, 11, 1, 8, 2, 6, 3, 4, 4, 6, 5, 4, 6, 0, 7, 0, 8, 1, 9, 10, 0, 13, 1, 9, 2, 7, 3, 7, 4, 7, 5, 7, 6, 2, 7, 2, 8, 2, 9, 0, 10, 1, 11, 12, 0, 15, 1, 13, 2, 8, 3, 8, 4

Offset: 1

Clark Kimberling, Apr 27 2000

Start with 0; at n-th step, write down what is in the sequence so far.
"Look and Say" how many times each integer (in increasing order), <= max {existing terms} appears in the sequence. Then concatenate.  Sequence's graph looks roughly like that of A080096.
For the original version, where "increasing order..." is "order of 1st appearance", see A217760.  The conjecture formerly placed here applies to A217760. - Clark Kimberling, Mar 24 2013

			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; ...

Zak Seidov, Table of n, a(n) for n = 1..1019 (the first 22 steps)

Cf. A005150. For other versions see A051120, A079668, A079686.
Cf. A055168-A055185 (method B) and A055187-A055191 (method A).
Cf. A217760.

s={0};Do[ta=Table[{Count[s, # ], # }&/@Range[0,Max[s]]]; s=Flatten[{s,ta}],{22}];s (* Zak Seidov, Oct 23 2009 *)

Conjectures: a(n) < 2*sqrt(n); limit as n goes to infinity Max( a(k) : 1<=k<=n)/sqrt(n) exist = 2. - Benoit Cloitre, Jan 28 2003

Edited by N. J. A. Sloane, Jan 17 2009 at the suggestion of R. J. Mathar
Removed a conjecture. - Clark Kimberling, Oct 24 2009
Entries changed to match b-file. - N. J. A. Sloane, Oct 04 2010

A079668 Start with 1; at n-th step, write down what is in the sequence so far.

Original entry on oeis.org

1, 1, 1, 3, 1, 4, 1, 0, 2, 1, 3, 1, 0, 6, 1, 1, 2, 2, 3, 1, 4, 2, 0, 10, 1, 3, 2, 3, 3, 2, 4, 0, 5, 1, 6, 4, 0, 12, 1, 6, 2, 6, 3, 3, 4, 1, 5, 2, 6, 0, 7, 0, 8, 0, 9, 1, 10, 8, 0, 15, 1, 8, 2, 8, 3, 5, 4, 2, 5, 5, 6, 1, 7, 1, 8, 1, 9, 2, 10, 0, 11, 1, 12, 10

Offset: 1

N. J. A. Sloane Jan 26 2003

After 1 1 1 3 1, we see "4 1's, 0 2's and 1 3", so next terms are 4 1 0 2 3 1.

			Row n lists all terms written at the n-th step:
  1;
  1, 1;
  3, 1;
  4, 1,  0, 2, 1, 3;
  1, 0,  6, 1, 1, 2, 2, 3, 1, 4;
  2, 0, 10, 1, 3, 2, 3, 3, 2, 4, 0, 5, 1, 6;
  4, 0, 12, 1, 6, 2, 6, 3, 3, 4, 1, 5, 2, 6, 0, 7, 0, 8, 0, 9, 1, 10;
  ...

Alois P. Heinz, Table of n, a(n) for n = 1..20377 (first 55 steps)

For other versions see A051120, A055186, A079686.

Cf. A030717, A333867.

b:= proc(n) option remember; `if`(n<1, 0, b(n-1)+add(x^i, i=T(n))) end:
T:= proc(n) option remember; `if`(n=1, 1, (p->
      seq([coeff(p, x, i), i][], i=ldegree(p)..degree(p)))(b(n-1)))
    end:
seq(T(n), n=1..10);  # Alois P. Heinz, Aug 24 2025

s={1}; Do[s=Flatten[{s,{Count[s,#],#}&/@Range[Min[s],Max[s]]}],{20}];s (* Peter J. C. Moses, Mar 21 2013 *)

from itertools import islice
def agen(): # generator of terms
    a = [1]
    yield 1
    while True:
        counts = []
        for d in range(min(a), max(a)+1):
            c = a.count(d)
            counts.extend([c, d])
        a += counts
        yield from counts
print(list(islice(agen(), 84))) # Michael S. Branicky, Aug 24 2025

More terms from Sean A. Irvine, Aug 24 2025

A079686 Start with 0; at n-th step, write down what is in the sequence so far.

Original entry on oeis.org

0, 1, 0, 1, 1, 2, 0, 1, 2, 3, 1, 3, 0, 2, 3, 2, 2, 5, 1, 4, 0, 1, 5, 1, 4, 3, 3, 5, 2, 6, 1, 5, 0, 1, 6, 4, 5, 2, 4, 5, 3, 6, 2, 9, 1, 6, 0, 1, 9, 0, 8, 0, 7, 4, 6, 6, 5, 4, 4, 6, 3, 8, 2, 11, 1, 7, 0, 1, 11, 0, 10, 2, 9, 2, 8, 2

Offset: 0

N. J. A. Sloane Jan 26 2003

Start with the highest value currently in the sequence and count down to zero.

			After the sequence is "0, 1, 0, 1, 1, 2, 0", we see one "2", three "1"s, and three "0"s, so we add 1, 2, 3, 1, 3, 0 to the sequence.

Nathaniel Johnston, Table of n, a(n) for n = 0..10000

Cf. A005150. For other versions see A051120, A079668, A055186.

a=[0];c=2;d=2; for j=1:7 for k=max(a):-1:0 a(c)=size(find(a(1:d-1)==k),2); a(c+1)=k; c=c+2; end; d=c; end; a' % Nathaniel Johnston, Apr 25 2011

s={0}; Do[s=Flatten[{s, {Count[s,#],#}&/@Range[Max[s],0,-1]}], {60}]; s (* Peter J. C. Moses, Mar 21 2013 *)

a(13) - a(75) from Nathaniel Johnston, Apr 25 2011

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A055186 Cumulative counting sequence: method A (adjective-before-noun)-pairs with first term 0.

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A079668 Start with 1; at n-th step, write down what is in the sequence so far.

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A079686 Start with 0; at n-th step, write down what is in the sequence so far.

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