cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A342042 When a digit d in the digit-stream of this sequence is even, the next digit is > d.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 30, 13, 14, 50, 15, 16, 70, 17, 18, 90, 19, 23, 24, 51, 25, 26, 71, 27, 28, 91, 29, 31, 32, 33, 34, 52, 35, 36, 72, 37, 38, 92, 39, 45, 46, 73, 47, 48, 93, 49, 53, 54, 55, 56, 74, 57, 58, 94, 59, 67, 68, 95, 69, 75, 76, 77, 78, 96, 79, 89, 97, 98, 99, 101
Offset: 1

Views

Author

Eric Angelini, Feb 26 2021

Keywords

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)

Crossrefs

Cf. A342043, A342044, A342045, A342046 and A342047 (variations on the same idea).
See A377913 and A377914 for records.
See also A347298.

Programs

  • 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

Extensions

Edited by N. J. A. Sloane, Nov 24 2024

A377912 Numbers such that every even digit except the last is immediately followed by a strictly larger digit.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117
Offset: 1

Views

Author

N. J. A. Sloane, Nov 27 2024, with thanks to Paolo Xausa for correcting the initial definition

Keywords

Comments

In other words, numbers that do not contain a pair of successive digits i, j where i is even and j <= i.

Crossrefs

The complement of A347298. Cf. A342042, A377914, A377917, A377918.

Programs

  • Mathematica
    A377912Q[k_] := FreeQ[Partition[IntegerDigits[k], 2, 1], {i_?EvenQ, j_} /; j <= i];
    Select[Range[0, 200], A377912Q] (* Paolo Xausa, Mar 17 2025 *)
  • Python
    def ok(n):
        s = str(n)
        return not any(s[i] in "2468" and s[i+1] <= s[i] for i in range(len(s)-1))
    print([k for k in range(0, 118) if ok(k)]) # Michael S. Branicky, Nov 28 2024

Extensions

Added 0 to match A342042, and replaced negative definition by a positive one. - N. J. A. Sloane, Nov 29 2024

A377918 a(n) = index in A377912 (or, equally, in A342042) of the first n-digit term.

Original entry on oeis.org

1, 11, 77, 566, 4197, 31148, 231193, 1716043, 12737453, 94544693, 701765055, 5208903636, 38663477066, 286982552081, 2130149470506, 15811193864583, 117359769764941, 871111674250772, 6465891595866732, 47993564275737877, 356235822660837879, 2644187054283807954, 19626676300599636003
Offset: 1

Views

Author

Keywords

Comments

These are the points in the graph of A342042 where the separate paths come together.
The first differences are in A377917, which is the more fundamental sequence. To get this sequence from A377917, add an initial zero, take partial sums, and add 1 to each term.

Crossrefs

Programs

  • Maple
    A377918 := proc(n) local S; option remember;
    S:=[1, 11, 77, 566, 4197, 31148, 231193, 1716043];
    if n <= 8 then S[n] else
    6*A377918(n-1)+10*A377918(n-2)+5*A377918(n-3)-5*A377918(n-4)-9*A377918(n-5)-5*A377918(n-6)-A377918(n-7); fi;
    end;
    [seq(A377918(i),i=1..20)];
  • Mathematica
    LinearRecurrence[{6, 10, 5, -5, -9, -5, -1}, {1, 11, 77, 566, 4197, 31148, 231193, 1716043}, 25] (* Paolo Xausa, Dec 02 2024  *)

Formula

G.f. = (x^7+6*x^6+15*x^5+19*x^4+11*x^3-x^2-5*x-1)/((1-x)*(x^6+6*x^5+15*x^4+20*x^3+15*x^2+5*x-1)) (From g.f. for A377917).
Recurrence: See Maple code.
The smallest root of the denominator of the g.f. is 0.134724138401519... whose reciprocal is (say) c1 = 7.422574840... Then a(n) is asymptotically c2*c1^n, for n >= 0, where c2 = 1.3824387... This is an excellent approximation. It gives a(22) = 0.1962667617*10^20, compared with a(22) = 19626676300599636003.
This also enables us to give a formula for the lower envelope of A342042 - see that entry for details.

Extensions

More terms added based on A377917. - N. J. A. Sloane, Dec 01 2024
Showing 1-3 of 3 results.