A382502 Lexicographically earliest sequence of positive integers such that no two subsequences {a(j), a(j+k), a(j+2k)} and {a(i), a(i+m), a(i+2m)} with different k and m values are the same.

1, 1, 1, 2, 3, 1, 2, 3, 4, 5, 6, 7, 8, 1, 9, 2, 3, 4, 5, 6, 1, 10, 7, 8, 11, 9, 12, 7, 10, 5, 13, 12, 14, 4, 6, 15, 16, 11, 17, 8, 18, 2, 3, 9, 5, 18, 1, 19, 14, 5, 15, 4, 20, 21, 13, 12, 22, 23, 24, 2, 21, 11, 25, 8, 26, 16, 20, 3, 27, 17, 12, 28, 29, 30, 31

Offset: 1

Author

Neal Gersh Tolunsky, Mar 29 2025

Keywords

• nonn

Comments

In other words, each unique subsequence of the form {a(j), a(j+k), a(j+2k)} (j, k >= 1) occurs with only one k value (or index spacing).

Note that a candidate term is sometimes denied because it would create a scenario in which a future term is inevitably the third term in two identical subsequences with different k values. See example.

Examples

			To find a(4), we first try 1. If we allowed a(4) = 1, then the subsequences at i = 1,3,5 and i = 3,4,5 would be the same since they both begin 1,1 and their final index is i=5. Since these two subsequences have distinct k values, they cannot be the same, so a(4) cannot be 1. a(4) = 2 as this creates only one new subsequence, and does not create a scenario where a future value will necessarily contradict the definition.
		

Links

• Pontus von Brömssen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A364057, A382501.

Programs

Python

from itertools import count
def A382502_generator():
    a_list = []
    spacings = {} # spacings[t] is the spacing (k) used by the triple t.
    pairs = {} # pairs[i] is a set of pairs (x,y) such that there exist j and k>0 with a(j)=x, a(j+k)=y, and i=j+2k.
    for n in count():
        for a in count(1):
            ok = True
            spacings_new = {}
            for k in range(1,n//2+1):
                t = a_list[n-2*k],a_list[n-k],a
                if t in spacings and k != spacings[t] or t in spacings_new:
                    ok = False
                    break
                spacings_new[t] = k
            if not ok: continue
            pairs_new = []
            for i,a0 in enumerate(reversed(a_list),n+1):
                p = (a0,a)
                if i <= 2*(n-1) and p in pairs[i]:
                    ok = False
                    break
                pairs_new.append(p)
            if ok: break
        yield a
        a_list.append(a)
        spacings.update(spacings_new)
        if n >= 3: del pairs[n+1]
        for i,p in enumerate(pairs_new[1:],n+2):
            if i <= 2*(n-1): pairs[i].add(p)
            else: pairs[i] = {p} # Pontus von Brömssen, Apr 01 2025

Extensions

More terms from Pontus von Brömssen, Mar 30 2025