A306251 -id:A306251 - cp's OEIS frontend

A306246 a(1) = 1, a(2) = 2, and for any n > 2, a(n) = o(n-1) + o(n-2) where o(k) is the number of occurrences of a(k) among a(1), ..., a(k).

1, 2, 2, 3, 3, 3, 5, 4, 2, 4, 5, 4, 5, 6, 4, 5, 8, 5, 6, 7, 3, 5, 10, 7, 3, 7, 8, 5, 9, 8, 4, 8, 9, 6, 5, 11, 9, 4, 9, 10, 6, 6, 9, 10, 8, 8, 11, 8, 9, 13, 7, 5, 13, 11, 5, 13, 13, 7, 9, 12, 8, 9, 16, 9, 10, 13, 9, 15, 11, 5, 15, 13, 8, 15, 12, 5, 14, 13, 8

Offset: 1

Rémy Sigrist, Jan 31 2019

nonn

The sequence o corresponds to the ordinal transform of the sequence a.
This sequence has a taste of Fibonacci, as each pair of consecutive terms gives rise to the next term.
This sequence is unbounded:
- if the sequence was bounded, say a(n) <= M for any n > 0, then some value in the range 1..M, say t, would appear at least M times among the first M^2 terms, and we would have a value > M after the M-th occurrence of t, a contradiction.
This sequence has interesting graphical features (see Links section).

			The first terms, alongside o, are:
  n   a(n)  o(n)
  --  ----  ----
   1     1     1
   2     2     1
   3     2     2
   4     3     1
   5     3     2
   6     3     3
   7     5     1
   8     4     1
   9     2     3
  10     4     2
  11     5     2
  12     4     3
  13     5     3
  14     6     1
  15     4     4
  16     5     4
  17     8     1
  18     5     5
  19     6     2
  20     7     1

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Density plot of the first 10000000 terms

See A306251 for the ordinal transform of this sequence.

o = vector(18); for (n=1, 79, if (n<=2, v=n, v=pp+p); print1 (v ", "); [pp,p]=[p,o[1+v]++])

A364027 a(0) = 0, a(1) = 0; for n > 1, a(n) is the number of pairs of consecutive terms that sum to the same value as a(n-2) + a(n-1).

Original entry on oeis.org

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

Offset: 0

Scott R. Shannon, Jul 01 2023

The same number cannot occur four times in a row as the second pair in a triplet of the same numbers increments the appearance count of the first pair by one, so the fourth number is always one more than the previous three numbers.

The occurrences of three consecutive equal numbers is quite rare, only occurring thirteen times in the first 20 million terms. The last such triplet is a(3641208) = a(3641209) = a(3641210) = 1177. It is likely such triplets occur infinitely often although this is unknown.

			a(2) = 1 as there is one pair that sums to a(0) + a(1) = 0, namely a(0) + a(1).
a(4) = 1 as a(2) + a(3) = 1 + 1 = 2, and there has been one previous pair that also sums to 2, namely a(2) + a(3).
a(5) = 2 as a(3) + a(4) = 1 + 1 = 2, and there has been two previous pairs that also sums to 2, namely a(2) + a(3) and a(3) + a(4).

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Image of the first 20 million terms.

Cf. A306251, A364036 (do not include previous pair), A342585, A347062.

nn = 120; c[] := 0; a[0] = a[1] = i = j = 0; Do[Set[k, 1 + c[i + j]++]; i = j; j = a[n] = k, {n, 2, nn}]; Array[a, nn + 1, 0] (* _Michael De Vlieger, Jul 02 2023 *)

def A364027_gen(): # generator of terms
    a, b, ndict = 0, 0, {0:1}
    yield from (0,0)
    while True:
        a, b = b, ndict[a+b]
        yield b
        ndict[a+b] = ndict.get(a+b,0)+1 # Chai Wah Wu, Jul 02 2023

a(n) = A306251(n-2) for any n >= 3. - Rémy Sigrist, Apr 03 2025

cp's OEIS Frontend

A306246 a(1) = 1, a(2) = 2, and for any n > 2, a(n) = o(n-1) + o(n-2) where o(k) is the number of occurrences of a(k) among a(1), ..., a(k).

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