A105395 -id:A105395 - cp's OEIS frontend

A054966 Numbers that are congruent to {0, 1, 8} mod 9.

0, 1, 8, 9, 10, 17, 18, 19, 26, 27, 28, 35, 36, 37, 44, 45, 46, 53, 54, 55, 62, 63, 64, 71, 72, 73, 80, 81, 82, 89, 90, 91, 98, 99, 100, 107, 108, 109, 116, 117, 118, 125, 126, 127, 134, 135, 136, 143, 144, 145, 152, 153, 154, 161, 162, 163, 170, 171, 172, 179, 180

Offset: 1

Henry Bottomley, May 24 2000

n == n^3 mod 9, so the iterated sum of the decimal digits of n and n^3 are equal.

H. I. Okagbue, M.O.Adamu, S.A. Bishop and A.A. Opanuga, Properties of Sequences Generated by Summing the Digits of Cubed Positive Integers, Indian Journal Of Natural Sciences, Vol. 6 / Issue 32 / October 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A047523. Complement of A275910.

[n : n in [0..200] | n mod 9 in [0, 1, 8]]; // Wesley Ivan Hurt, Jun 14 2016

A054966:=n->3*n-3+2*cos(2*n*Pi/3)+2*sin(2*n*Pi/3)/sqrt(3): seq(A054966(n), n=1..100); # Wesley Ivan Hurt, Jun 14 2016

Select[Range[0, 200], MemberQ[{0, 1, 8}, Mod[#, 9]] &] (* Wesley Ivan Hurt, Jun 14 2016 *)
LinearRecurrence[{1, 0, 1, -1}, {0, 1, 8, 9}, 100] (* Vincenzo Librandi, Jun 15 2016 *)

G.f.: x^2*(1+7*x+x^2) / ((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 14 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = 3*n-3+2*cos(2*n*Pi/3)+2*sin(2*n*Pi/3)/sqrt(3).
a(3k) = 9k-1, a(3k-1) = 9k-8, a(3k-2) = 9k-9. (End)
A008591 UNION A056020. - R. J. Mathar, Jul 19 2024
a(n) -a(n-1) = A105395(n+1), n>1. - R. J. Mathar, Jul 19 2024

A281899 a(n) = n + 6*floor(n/3).

Original entry on oeis.org

0, 1, 2, 9, 10, 11, 18, 19, 20, 27, 28, 29, 36, 37, 38, 45, 46, 47, 54, 55, 56, 63, 64, 65, 72, 73, 74, 81, 82, 83, 90, 91, 92, 99, 100, 101, 108, 109, 110, 117, 118, 119, 126, 127, 128, 135, 136, 137, 144, 145, 146, 153, 154, 155, 162, 163, 164, 171, 172, 173, 180, 181, 182, 189

Offset: 0

Bruno Berselli, Feb 06 2017

Equivalently, numbers that are congruent to {0, 1, 2} mod 9.
Also numbers m such that floor(m/3) = 3*floor(m/9).
The n-th term is 3*n, 3*n-2 or 3*n-4.
For n > 0, numbers k such that 3 | floor(k/3). - Wesley Ivan Hurt, Dec 01 2020

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A002264.
Subsequence of A060464 and A248375.
The first differences are in A105395.
Cf. similar sequences with formula n+i*floor(n/3): A004773 (i=1), A047217 (i=2), A047240 (i=3), A047354 (i=4), A047469 (i=5), this sequence (i=6).
Cf. numbers that are congruent to {0, 1, 2} mod j: the sequences are listed in the previous row for j = 4..9, respectively.

Magma
```
[n+6*(n div 3): n in [0..70]];
```

A281899:=n->n+6*floor(n/3): seq(A281899(n), n=0..100); # Wesley Ivan Hurt, Feb 09 2017

Table[n + 6 Floor[n/3], {n, 0, 70}]
LinearRecurrence[{1,0,1,-1},{0,1,2,9},90] (* Harvey P. Dale, Feb 25 2018 *)

Maxima
```
makelist(n+6*floor(n/3), n, 0, 70);
```

a(n)=n\3*6 + n \\ Charles R Greathouse IV, Feb 07 2017

Python
```
[n+6*int(n/3) for n in range(70)]
```
Sage
```
[n+6*floor(n/3) for n in range(70)]
```

G.f.: x*(1 + x + 7*x^2)/((1 - x)^2*(1 + x + x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4).
a(n) = 3*n - 2*(n mod 3). In general, n + 3*h*floor(n/3) = (h+1)*n - h*(n mod 3).
a(n) + a(n+s) = a(2*n+s-1) + 1, where s is nonnegative and not divisible by 3. Example: for s=14, a(n) + a(n+14) = a(2*n+13) + 1; for n=3, a(3) + a(17) = a(19) + 1 = 9 + 47 = 55 + 1 = 56.
a(6*k+r) = 18*k + a(r), where 0 <= r <= 5.
a(n) = 7*A002264(n) + A002264(n+1) + A002264(n+2).

A105647 Lexicographically earliest sequence of increasing numbers whose digits satisfy the "Fractal Jump" rule using only the digits 2 and 5: keep the first digit "d" of the sequence, then jump over the next "d" digits and keep the digit "e" on which you have landed. Jump now over the next "e" digits and keep the digit "f" on which you have landed, etc. The succession "def..." of kept digits is the sequence itself.

Original entry on oeis.org

2, 5, 25, 52, 55, 222, 252, 255, 552, 555, 2222, 5222, 5252, 5255, 22222, 22252, 22522, 22525, 22555, 25222, 52222, 55222, 55522, 55525, 55552, 55555, 222522, 225222, 225225, 225522, 252222, 252225, 522222, 2225222, 2225252, 2225255, 5222225, 5222522, 5225222

Offset: 1

Eric Angelini and Alexandre Wajnberg, May 03 2005

			The sequence and the "kept" digits begin
  2, 5, 25, 52, 55, 222, 252, 255, 552, ...
  ^      ^           ^    ^         ^
  2      5           2    5         5

Tyler Busby, Table of n, a(n) for n = 1..200

Cf. A105395, A105396, A105397, A105398.

Corrected and extended by Tyler Busby, Jan 08 2023

A188656 Decimal expansion of (1+sqrt(65))/8.

Original entry on oeis.org

1, 1, 3, 2, 7, 8, 2, 2, 1, 8, 5, 3, 7, 3, 1, 8, 7, 0, 6, 5, 4, 5, 8, 2, 6, 6, 5, 3, 7, 8, 7, 9, 7, 1, 3, 9, 1, 3, 9, 1, 7, 9, 9, 5, 3, 8, 2, 0, 1, 0, 7, 1, 6, 7, 3, 4, 9, 2, 0, 7, 4, 0, 4, 8, 6, 5, 7, 9, 8, 4, 3, 6, 8, 8, 7, 8, 2, 1, 1, 0, 2, 5, 3, 7, 0, 0, 1, 9, 2, 8, 3, 3, 3, 9, 6, 5, 3, 8, 3, 0, 4, 5, 4, 6, 8, 0, 3, 0, 8, 2, 6, 7, 4, 9, 3, 2, 3, 9, 0, 2, 6, 7, 1, 8, 5, 8, 1, 5, 1, 5

Offset: 1

Clark Kimberling, Apr 09 2011

Apart from the second digit the same as A177707.
Decimal expansion of the shape of a (1/4)-extension rectangle.
See A188640 for definitions of shape and r-extension rectangle.
A (1/4)-extension rectangle matches the continued fraction [1,7,1,1,7,1,1,7,1,1,7,1,1,7,...] for the shape L/W= (1+sqrt(65))/8.  This is analogous to the matching of a golden rectangle to the continued fraction [1,1,1,1,1,1,1,...].  Specifically, for the (4/3)-extension rectangle, 1 square is removed first, then 7 squares, then 1 square, then 1 square, then 7 squares,..., so that the original rectangle is partitioned into an infinite collection of squares.

			length/width = 1.13278221853731870654582665....

Daniel Starodubtsev, Table of n, a(n) for n = 1..10000
Clark Kimberling, A Visual Euclidean Algorithm, The Mathematics Teacher 76 (1983) 108-109.

Cf. A188640, A105395.

RealDigits[(1 + Sqrt[65])/8, 10, 111][[1]] (* Robert G. Wilson v, Aug 19 2011 *)

A359385 The lexicographically earliest "Increasing Term Fractal Jump Sequence" that does not use the digit 0 in any terms.

Original entry on oeis.org

1, 2, 21, 22, 23, 112, 122, 132, 133, 134, 141, 221, 311, 2112, 2113, 3111, 21111, 31113, 31114, 31124, 31131, 34111, 41121, 42111, 43111, 111121, 111122, 112111, 112311, 131111, 211112, 211113, 1111311, 1111312, 3111311, 3111312, 4111131, 4111132, 4141111

Offset: 1

Tyler Busby, Dec 29 2022

The rules of an "Increasing Term Fractal Jump Sequence" are described in A105647.

The digit zero is omitted as it can create situations where the next term can require a 0 in the first digit.

			The sequence and the "kept" digits begin
  1, 2, 21, 22, 23, 112, 122, 132, 133, ...
  ^     ^    ^      ^ ^    ^    ^    ^
  1     2    2      1 2    2    2    3

Tyler Busby, Table of n, a(n) for n = 1..5000

Cf. A105647, A105395, A105396, A105397, A105398.

A359611 The lexicographically earliest "Increasing Term Fractal Jump Sequence".

Original entry on oeis.org

1, 2, 20, 22, 100, 200, 201, 1000, 20000, 20001, 110000, 2000000, 2000001, 110100000, 200000000, 200000001, 1101001000000, 2000000000020, 2000000010101, 10100010000000, 20000000000002, 20020000000001, 101001010010000, 100000000200000000000000

Offset: 1

Tyler Busby, Jan 06 2023

The rules of an "Increasing Term Fractal Jump Sequence" are described in A105647.
We define a "forced" digit in Fractal Jump Sequences as a digit that is required to be a specific value by a digit that occurred previously in the sequence. This is in opposition to digits that could have any value selected for them without breaking the Fractal Jump Sequence rules. In the diagram below, the digits with carets below them are the forced digits.
To find a(n), increment a(n-1) until all of the forced digits that will positionally occur in a(n) satisfy their forced values. Then, to avoid leading zeros in a(n+1), if there are forced zeros immediately following the candidate a(n), continue to increment until it is the same number of digits longer as there are consecutive forced zeros, and continue to increment until the candidate a(n) once again satisfies all forcing criteria (including the new zeros).
The only digits that appear in this sequence are 0, 1, and 2, even though no numerals are arbitrarily restricted from appearing.

			The sequence and the "kept"/"forced" digits begin
  1, 2, 20, 22, 100, 200, 201, 1000, 20000, ...
  ^     ^    ^    ^  ^    ^    ^ ^^  ^  ^^
  1     2    2    0  2    2    1 00  2  00
In the case of computing a(5), we have a 22 for a(4), so we would normally increment to 23, as there is nothing forcing the next two digits. However, since there is a 0 forcing the following digit, we must increment to the smallest number that satisfies this forced 0 (as we can't have leading zeros in a(6)).

Cf. A359385 (no zeros), A105647, A105395, A105396, A105397, A105398.

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A054966 Numbers that are congruent to {0, 1, 8} mod 9.

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A281899 a(n) = n + 6*floor(n/3).

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A188656 Decimal expansion of (1+sqrt(65))/8.

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A359385 The lexicographically earliest "Increasing Term Fractal Jump Sequence" that does not use the digit 0 in any terms.

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A359611 The lexicographically earliest "Increasing Term Fractal Jump Sequence".

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