A380037 For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) as 1 more than the number of positive terms between a(m+1) and a(n-1). Otherwise, a(n+1) = -1. Start with a(1) = 0.

0, -1, -1, 1, -1, 2, -1, 2, 1, 3, -1, 4, -1, 2, 4, 2, 2, 1, 7, -1, 7, 1, 3, 10, -1, 5, -1, 2, 8, -1, 3, 5, 4, 14, -1, 5, 3, 5, 2, 9, -1, 6, -1, 2, 3, 6, 3, 2, 4, 13, -1, 8, 18, -1, 3, 6, 8, 4, 7, 30, -1, 7, 2, 12, -1, 4, 6, 9, 22, -1, 5, 26

Offset: 1

Author

Connor Criss, Jan 10 2025

Keywords

• sign

Comments

This sequence was meant to be a version of A181391 that is more agnostic to individual natural numbers. When a new number is encountered, -1 is used as a placeholder instead of 0, and the -1 terms are not counted when determining the next term in the sequence. So, all natural numbers are in a sense treated equally.

No zeros appear beyond the first term in the sequence. By the definition of the sequence, beyond the first term, all terms must be equal to -1 or an integer of value at least 1.

a(n) < n for all n. a(1) < 1, and for n > 1 either a(n) = -1, which is less than n for all n, or there exists m < n - 1 such that a(m) = a(n - 1) and a(n) <= n - 1 - m which is less than n.

Links

• Connor Criss, Table of n, a(n) for n = 1..10000

Crossrefs

Similar to the Van Eck sequence A181391.

Programs

PARI

a(max_n) = {my(v = [0], r = [], n = 1); while(#v < max_n, r = select(x->x==v[n], v, 1); if(#r>1, v=concat(v, [1+sum(m=r[#r-1]+1, r[#r]-1, v[m]>0)]), v=concat(v,[-1])); n++); v} \\ Thomas Scheuerle, Jan 11 2025

Python

k = [0]
i = 0
j = 0
n = 0
negatives = 0
total_vals = 0
MAX = 10000
#change value of MAX to modify number of terms generated
index = 0
while(index < MAX):
    n = k[-1]
    i = len(k) - 2
    j = 1
    m = 0
    while(i > -1):
        if(k[i] == k[-1]):
            k.append(j)
            m = 1
            break
        if(k[i] != -1 or k[-1] == -1):
            j = j + 1
        i = i - 1
    if m == 0:
        k.append(-1)
        negatives = negatives + 1
    index = index + 1
for x in range(MAX):
    print(str(k[x]))