OFFSET
1,3
COMMENTS
The definition refers to the digit-stream in the sequence (ignoring the commas), which starts 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 3, 0, ...
The sequence is always extended with the smallest nonnegative integer not yet present that doesn't lead to a contradiction.
Theorem: The sequence contains every nonnegative integer except those in A347298.
Proved in September 2021. See S.K. link for a new, more detailed proof. - Sebastian Karlsson, Nov 28 2024. See N.J.A.S. link for an alternative, shorter, proof. - N. J. A. Sloane, Nov 29 2024
Comments added by N. J. A. Sloane, Dec 04 2024 (Start):
Let S = present sequence, P = A377912. By definition the terms in P appear in their natural order. There are A377917(k) terms in P of decimal length k >= 1. They form a consecutive block in P, starting at P(i1) and ending at P(i2), where i1 = A377918(k), i2 = A377918(k+1)-1.
We know S contains exactly the same terms as P, but in a different order.
Conjecture 1. For k >= 1, the terms of length k in S form a consecutive block with the same starting and ending points as in P. In both P and S, the block begins with 10101... (1's and 0's alternate, length is k) and end with 99...9 (k 9's).
Conjecture 2. We know every number appears in S. Suppose x = S(m) = 899...9 (with k-1 9's). Then x is the last term of length k in S that begins with a digit <= 8. The remaining terms of length k have leading digit 9 and appear in order, ending with 99...9 (k 9's).
(Some k-digit numbers beginning with 9 may appear before x.)
(End)
Comment from N. J. A. Sloane, Dec 01 2024 (Start)
Let c1 = 7.422574840... and c2 = 1.3824387... be the constants defined in A377918. Then assuming Conjecture 1, the index of the last term of length k in the present sequence is close to (c2*c1^k, 10^k). [Thanks to Sebastian Karlsson for pointing out that Conjecture 1 is required and is as yet unproved.]
Let x = c2*c1^k, and express k in terms of x.
Then this point has coordinates (x,y) where y = (x/c2)^c3, with c3 = (log 10)/(log c1) = 1.14869... This defines a curve that is a good approximation to the lower envelope of the present sequence.
For example, the fifth meeting point has coordinates (31148, 101010) (see A377918) and the formula here gives (x,y) = (31148, 100003.0039).
(End)
Comment from Sebastian Karlsson, Dec 12 2024: (Start)
Theorem: Let d be in {1, 2, ..., 8}. For every positive integer k, the k-digit number d99...9 appears in the sequence before the k-digit number (d+1)99...9.
A proof can be found in the links. Since all k-digit numbers starting with 9 appears before any (k+1)-digit number, we get that terms of a certain length form a consecutive block. In particular, this proves Conjectures 1 and 2 above.
(End)
LINKS
N. J. A. Sloane, Table of n, a(n) for n = 1..20000 [Computed using Rémy Sigrist's PARI program. The first 10000 terms were computed by Peter Kagey]
Sebastian Karlsson, Proof of Theorem.
Sebastian Karlsson, Proof of the second Theorem.
Rémy Sigrist, Colored scatterplot of the first 20000 terms (where the color is function of the leading digit of a(n))
Rémy Sigrist, PARI program for A342042
N. J. A. Sloane, A Nasty Surprise in a Sequence and Other OEIS Stories, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; Slides [Mentions this sequence]
N. J. A. Sloane, An Alternative Proof of the Theorem.
PROG
(PARI) \\ See Links section.
(Python)
def cond(s, minfirst):
return all(s[i+1] > s[i] for i in range(len(s)-1) if s[i] in "02468")
def aupton(terms):
alst, seen = [0], {0}
while len(alst) < terms:
d = alst[-1]%10
an = minfirst = (1 - d%2)*(d+1)
stran = str(an)
while an in seen or not cond(stran, minfirst):
an += 1
stran = str(an)
if int(stran[0]) < minfirst:
an = minfirst*10**(len(stran)-1)
alst.append(an); seen.add(an)
return alst
print(aupton(77)) # Michael S. Branicky, Sep 07 2021
CROSSREFS
KEYWORD
base,nonn,nice
AUTHOR
Eric Angelini, Feb 26 2021
EXTENSIONS
Edited by N. J. A. Sloane, Nov 24 2024
STATUS
approved