A394611 a(1) = 0. For n > 1, a(n) is the number of ways to write a(n-1) as the sum of two previous terms with distinct positions in the sequence.

0, 0, 1, 2, 2, 4, 3, 4, 6, 6, 8, 7, 6, 10, 11, 6, 12, 13, 10, 15, 8, 14, 18, 14, 20, 16, 19, 13, 13, 15, 17, 21, 26, 18, 22, 22, 24, 27, 22, 26, 30, 28, 37, 20, 26, 36, 27, 27, 29, 29, 31, 30, 40, 36, 35, 38, 31, 33, 44, 43, 37, 47, 33, 46, 44, 48, 46, 48, 50, 51, 43, 45, 38, 35, 46, 55, 45, 43, 49, 61, 49, 63, 52

Offset

1,4

Comments

Conjecture: Many numbers, beginning with 5 and 9, never occur in this sequence (see A397471).

Links

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Sean A. Irvine, Table of n, a(n) for n = 1..1000000 [4.3M, gzip]

Example

a(2) = 0, since we only have one previous term.
a(3) = 1, since a(1) + a(2) = a(2).
a(4) = 2, since a(1) + a(3) = a(3) and a(2) + a(3) = a(3).

Prog

(Python)
def sequence(N=100):
    a = [0]  # a(1) = 0
    for n in range(1, N):
        target = a[n - 1]  # a(n)
        count = 0
        # count pairs (i < j) with i, j in [0..n-1]
        for i in range(n):
            for j in range(i + 1, n):
                if a[i] + a[j] == target:
                    count += 1
        a.append(count)
    return a
seq = sequence(100)
print(seq)
(Python)
from collections import Counter
from itertools import islice
def A394611_gen(): # generator of terms
    alst, an, c = [], 0, Counter()
    while True:
        an = c[an]
        yield an
        c.update(an+ai for ai in alst)
        alst.append(an)
print(list(islice(A394611_gen(), 83))) # Michael S. Branicky, Jun 26 2026

Crossrefs

Cf. A395907, A397173, A397174, A397471.

Sequence in context: A233511 A205793 A178431 * A233574 A353925 A157927

Adjacent sequences: A394608 A394609 A394610 * A394612 A394613 A394614

Keyword

nonn,new

Author

Joshua B. Weinstein, Jun 18 2026

Status

approved