A364027 -id:A364027 - cp's OEIS frontend

A364036 a(0) = 0, a(1) = 0; for n > 1, a(n) is the number of pairs of consecutive terms prior to a(n-1) that sum to the same value as a(n-2) + a(n-1).

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

Offset: 0

Scott R. Shannon, Jul 02 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(3620001) = a(3620002) = a(3620003) = 1159. It is likely such triplets occur infinitely often although this is unknown.

			a(2) = 0 as there are no previous pairs prior to a(1).
a(3) = 1 as a(1) + a(2) = 0 + 0 = 0, and there has been one previous pair that also sums to 0, namely a(0) + a(1).
a(6) = 2 as a(4) + a(5) = 0 + 1 = 1, and there has been two previous pairs that also sums to 1, 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. A364027 (include previous pair), A342585, A347062.

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

def A364036_gen(): # generator of terms
    a, b, ndict = 0, 0, {0:1}
    while True:
        a, b = b, ndict[a+b]
        yield b-1
        ndict[a+b] = ndict.get(a+b,0)+1 # Chai Wah Wu, Jul 02 2023

A373902 a(1) = 2; for n > 1, a(n) is the smallest positive number that does not equal a(n-1), shares a factor with a(n-1), and has not appeared as an adjacent term to a(n-1) previously in the sequence.

Original entry on oeis.org

2, 4, 6, 2, 8, 4, 10, 2, 12, 3, 6, 8, 10, 5, 15, 3, 9, 6, 10, 12, 4, 14, 2, 16, 4, 18, 2, 20, 4, 22, 2, 24, 3, 18, 6, 12, 8, 14, 6, 15, 9, 12, 14, 7, 21, 3, 27, 6, 16, 8, 18, 9, 21, 6, 20, 5, 25, 10, 14, 16, 10, 15, 12, 16, 18, 10, 20, 8, 22, 6, 24, 4, 26, 2, 28, 4, 30, 2, 32, 4, 34, 2, 36, 3

Offset: 1

Scott R. Shannon, Jun 22 2024

See A371618 for the indices where the primes first appear.

			a(7) = 10 as a(6) = 4 and, although 2 and 6 share factors with 4, 2 and 4 form a previous adjacent pair, as do 4 and 6. This leaves 10 as the smallest number that shares a factor with 4 while 4 and 10 have not previously appeared as adjacent terms.

Scott R. Shannon, Table of n, a(n) for n = 1..10000
Scott R. Shannon, Image of the first 250000 terms. Prime terms are shown in red.

Cf. A371618, A064413, A027748, A364027.

nn = 120; c[_] := {}; a[1] = j = 2; c[2] = {2};
Do[If[PrimePowerQ[j],
  (k = 1;
     While[Or[j == #  k, CoprimeQ[j, #  k], ! FreeQ[c[j], #  k]], k++];
     k *= #) &[FactorInteger[j][[1, 1]]],
   k = FactorInteger[j][[1, 1]];
     While[Or[j == k, CoprimeQ[j, k], ! FreeQ[c[j], k]], k++] ];
  Set[{a[n], c[j], c[k], j},
      {k, Union[c[j], {k}], Union[c[k], {j}], k}], {n, 2, nn}];
Array[a, nn] (* Michael De Vlieger, Jun 22 2024 *)

from math import gcd
from itertools import count, islice
def agen(): # generator of terms
    an, adjacent = 2, {2: set()}
    while True:
        yield an
        A = adjacent[an]
        m = next(k for k in count(2) if k!=an and gcd(k, an)>1 and k not in A)
        adjacent[an].add(m)
        if m not in adjacent: adjacent[m] = set()
        adjacent[m].add(an)
        an = m
print(list(islice(agen(), 84))) # Michael S. Branicky, Jun 22 2024

A362947 a(0) = 0, a(1) = 0; for n > 1, a(n) is the number of pairs of consecutive terms whose product has same value as a(n-2) * a(n-1).

Original entry on oeis.org

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

Offset: 0

Scott R. Shannon, Jul 05 2023

nonn

Similarly to A364027 the same number cannot occur four times in a row. In the first 10 million terms three consecutive equal numbers occurs twenty-three times, the last such triplet being a(8247993)..a(8247995) = 59. It is likely such triplets occur infinitely often although this is unknown.

			a(2) = 1 as there is one pair whose product equals a(0) * a(1) = 0, namely a(0) * a(1).
a(3) = 2 as a(1) * a(2) = 0 * 1 = 0, and there has been two previous pairs whose product is 0, namely a(0) * a(1) and a(1) * a(2).
a(11) = 4 as a(9) * a(10) = 1 * 2 = 2, and there has been four previous pairs whose product is 2, namely a(2) * a(3), a(3) * a(4), a(4) * a(5) and a(9) * a(10).

Scott R. Shannon, Table of n, a(n) for n = 0..10000.
Scott R. Shannon, Image of the first 10 million terms.

Cf. A364027, A364036, A342585, A347062.

cp's OEIS Frontend

A364036 a(0) = 0, a(1) = 0; for n > 1, a(n) is the number of pairs of consecutive terms prior to a(n-1) that sum to the same value as a(n-2) + a(n-1).

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A373902 a(1) = 2; for n > 1, a(n) is the smallest positive number that does not equal a(n-1), shares a factor with a(n-1), and has not appeared as an adjacent term to a(n-1) previously in the sequence.

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A362947 a(0) = 0, a(1) = 0; for n > 1, a(n) is the number of pairs of consecutive terms whose product has same value as a(n-2) * a(n-1).

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