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.
Conjecture: This contains every nonnegative integer except those in A347298. - N. J. A. Sloane, Aug 26 2021
This is a theorem (proved in September 2021, see link for a new, more detailed proof). - Sebastian Karlsson, Nov 28 2024. See the other link for an alternative, shorter, proof. - N. J. A. Sloane, Nov 29 2024
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 the k-th meeting point in the present sequence, where the branches come together, is close to (c2*c1^k, 10^k).
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)
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.
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,changed
AUTHOR
Eric Angelini, Feb 26 2021
EXTENSIONS
Edited by N. J. A. Sloane, Nov 24 2024
STATUS
approved