A042974 -id:A042974 - cp's OEIS frontend

A237591 Irregular triangle read by rows: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts (n>=1, 1<=k<=A003056(n)).

1, 2, 2, 1, 3, 1, 3, 2, 4, 1, 1, 4, 2, 1, 5, 2, 1, 5, 2, 2, 6, 2, 1, 1, 6, 3, 1, 1, 7, 2, 2, 1, 7, 3, 2, 1, 8, 3, 1, 2, 8, 3, 2, 1, 1, 9, 3, 2, 1, 1, 9, 4, 2, 1, 1, 10, 3, 2, 2, 1, 10, 4, 2, 2, 1, 11, 4, 2, 1, 2, 11, 4, 3, 1, 1, 1, 12, 4, 2, 2, 1, 1, 12, 5, 2, 2, 1, 1, 13, 4, 3, 2, 1, 1, 13, 5, 3, 1, 2, 1, 14, 5, 2, 2, 2, 1

Offset: 1

Omar E. Pol, Feb 22 2014

The original name was: Triangle read by rows: T(n,k) = A235791(n,k) - A235791(n,k+1), assuming that the virtual right border of triangle A235791 is A000004.
T(n,k) is also the length of the k-th segment in a zig-zag path on the first quadrant of the square grid, connecting the point (n, 0) with the point (m, m), starting with a segment in vertical direction, where m <= n.
Conjecture: the area of the polygon defined by the x-axis, this zig-zag path and the diagonal [(0, 0), (m, m)], is equal to A024916(n)/2, one half of the sum of all divisors of all positive integers <= n. Therefore the reflected polygon, which is adjacent to the y-axis, with the zig-zag path connecting the point (0, n) with the point (m, m), has the same property. And so on for each octant in the four quadrants.
For the representation of A024916 and A000203 we use two octants, for example: the first octant and the second octant, or the 6th octant and the 7th octant, etc., see A237593.
At least up to n = 128, two zig-zag paths never cross (checked by hand).
The finite sequence formed by the n-th row of triangle together with its mirror row gives the n-th row of triangle A237593.
The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> this sequence --> A237593 --> A239660 --> A237270 --> A237271.
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014. (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
The row sum is A235791(n,1) - A235791(n,floor((sqrt(8n+1)-1)/2)+1) = n - 0.
Mathematica function has been written to check the conjecture as well as non-crossing zig-zag paths (Dyck paths rotated by 90 degrees) up through n=30000 (same applies to A237593). (End)
The n-th zig-zag path ending at the point (m, m), where m = A240542(n). - Omar E. Pol, Apr 16 2014
From Omar E. Pol, Aug 23 2015: (Start)
n is an odd prime if and only if T(n,2) = 1 + T(n-1,2) and T(n,k) = T(n-1,k) for the rest of the values of k.
The elements of the n-th row of triangle together with the elements of the n-th row of triangle A261350 give the n-th row of triangle A237593.
T(n,k) is also the area (or the number of cells) of the k-th vertical side at the n-th level (starting from the top) in the left hand part of the front view of the stepped pyramid described in A245092, see Example section.
(End)
From Omar E. Pol, Nov 19 2015: (Start)
T(n,k) is also the number of cells between the k-th and the (k+1)st line segments (from left to right) in the n-th row of the diagram as shown in Example section.
Note that the number of horizontal line segments in the n-th row of the diagram equals A001227(n), the number of odd divisors of n. (End)
Conjecture: the values f(n,k) in the n-th row of the triangle are either 1 or 2 for all k with ceiling((sqrt(4*n+1)-1)/2) <= k <= floor((sqrt(8*n+1)-1)/2) = r(n), the length of the n-th row, though the lower bound need not be minimal; tested through 2500000. See also A285356. - Hartmut F. W. Hoft, Apr 17 2017
Conjecture: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts. - Omar E. Pol, Apr 30 2017
From Omar E. Pol, Aug 31 2021: (Start)
It appears that T(n,2)/T(n,1) converges to 1/3.
It appears that T(n,3)/T(n,2) converges to 1/2.
It appears that T(n,4)/T(n,3) converges to 3/5.
It appears that T(n,5)/T(n,4) converges to 2/3. (End)
In other words: T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(n). - Omar E. Pol, Sep 08 2021

			Triangle begins:
   1;
   2;
   2, 1;
   3, 1;
   3, 2;
   4, 1, 1;
   4, 2, 1;
   5, 2, 1;
   5, 2, 2;
   6, 2, 1, 1;
   6, 3, 1, 1;
   7, 2, 2, 1;
   7, 3, 2, 1;
   8, 3, 1, 2;
   8, 3, 2, 1, 1;
   9, 3, 2, 1, 1;
   9, 4, 2, 1, 1;
  10, 3, 2, 2, 1;
  10, 4, 2, 2, 1;
  11, 4, 2, 1, 2;
  11, 4, 3, 1, 1, 1;
  12, 4, 2, 2, 1, 1;
  12, 5, 2, 2, 1, 1;
  13, 4, 3, 2, 1, 1;
  13, 5, 3, 1, 2, 1;
  14, 5, 2, 2, 2, 1;
  14, 5, 3, 2, 1, 2;
  15, 5, 3, 2, 1, 1, 1;
  ...
For n = 10 the 10th row of triangle A235791 is [10, 4, 2, 1] so row 10 is [6, 2, 1, 1].
From _Omar E. Pol_, Aug 23 2015: (Start)
Illustration of initial terms:
  Row                                                         _
   1                                                        _|1|
   2                                                      _|2 _|
   3                                                    _|2  |1|
   4                                                  _|3   _|1|
   5                                                _|3    |2 _|
   6                                              _|4     _|1|1|
   7                                            _|4      |2  |1|
   8                                          _|5       _|2 _|1|
   9                                        _|5        |2  |2 _|
  10                                      _|6         _|2  |1|1|
  11                                    _|6          |3   _|1|1|
  12                                  _|7           _|2  |2  |1|
  13                                _|7            |3    |2 _|1|
  14                              _|8             _|3   _|1|2 _|
  15                            _|8              |3    |2  |1|1|
  16                          _|9               _|3    |2  |1|1|
  17                        _|9                |4     _|2 _|1|1|
  18                      _|10                _|3    |2  |2  |1|
  19                    _|10                 |4      |2  |2 _|1|
  20                  _|11                  _|4     _|2  |1|2 _|
  21                _|11                   |4      |3   _|1|1|1|
  22              _|12                    _|4      |2  |2  |1|1|
  23            _|12                     |5       _|2  |2  |1|1|
  24          _|13                      _|4      |3    |2 _|1|1|
  25        _|13                       |5        |3   _|1|2  |1|
  26      _|14                        _|5       _|2  |2  |2 _|1|
  27    _|14                         |5        |3    |2  |1|2 _|
  28   |15                           |5        |3    |2  |1|1|1|
  ...
Also the diagram represents the left part of the front view of the pyramid described in A245092. For the other half front view see A261350. For more information about the pyramid and the symmetric representation of sigma see A237593. (End)
From _Omar E. Pol_, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below:
.                           _
                           | |
                           | |
                           | |
                           | |
                           | |
                      _ _ _| |
                    _|    _ _|
                  _|     |
                 |      _|
                 |  _ _|1
      _ _ _ _ _ _| |  2
     |_ _ _ _ _ _ _|2
            7
.
The lengths of the successive line segments from the first vertex to the central vertex of the largest Dyck path are [7, 2, 2, 1] respectively, the same as the 12th row of triangle. (End)

G. C. Greubel, Table of n, a(n) for the first 150 rows

Row n has length A003056(n) hence column k starts in row A000217(k).
Row sums give A000027.
Column 1 is A008619, n >= 1.
Right border gives A042974.
Cf. A000203, A001227, A024916, A196020, A235791, A236104, A237048, A237270, A237271, A237593, A239660, A239931-A239934, A240542, A244580, A245092, A249351, A259176, A259177, A261350, A261699, A262626, A285356, A286000, A286001.

row[n_]:= Floor[(Sqrt[8*n+1] -1)/2];  f[n_,k_]:= Ceiling[(n+1)/k-(k+1)/2] - Ceiling[(n+1)/(k+1)-(k+2)/2];
Table[f[n,k],{n,1,50},{k,1,row[n]}]//Flatten
(* Hartmut F. W. Hoft, Apr 08 2014 *)

row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
row(n) = {my(orow = concat(row235791(n), 0)); vector(#orow -1, i, orow[i] - orow[i+1]);} \\ Michel Marcus, Mar 27 2014

from sympy import sqrt
import math
def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))
for n in range(1, 29): print([T(n, k) for k in range(1, int((sqrt(8*n + 1) - 1)/2) + 1)]) # Indranil Ghosh, Apr 30 2017

T(n,k) = ceiling((n+1)/k - (k+1)/2) - ceiling((n+1)/(k+1) - (k+2)/2), for 1 <= n and 1 <= k <= floor((sqrt(8n+1)-1)/2). - Hartmut F. W. Hoft, Apr 07 2014

3 more rows added by Omar E. Pol, Aug 23 2015

New name from a comment dated Apr 30 2017. - Omar E. Pol, Jun 18 2023

A235791 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 2, 3, 1, 4, 1, 5, 2, 6, 2, 1, 7, 3, 1, 8, 3, 1, 9, 4, 2, 10, 4, 2, 1, 11, 5, 2, 1, 12, 5, 3, 1, 13, 6, 3, 1, 14, 6, 3, 2, 15, 7, 4, 2, 1, 16, 7, 4, 2, 1, 17, 8, 4, 2, 1, 18, 8, 5, 3, 1, 19, 9, 5, 3, 1, 20, 9, 5, 3, 2, 21, 10, 6, 3, 2, 1, 22, 10, 6, 4, 2, 1, 23, 11, 6, 4, 2, 1, 24, 11, 7, 4, 2, 1

Offset: 1

Omar E. Pol, Jan 23 2014

The alternating sum of the squares of the elements of the n-th row equals the sum of all divisors of all positive integers <= n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*(T(n,k))^2 = A024916(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
For more information see A236104.
The sum of row n gives A060831(n), the sum of the number of odd divisors of all positive integers <= n. - Omar E. Pol, Mar 01 2014. [An equivalent assertion is that the sum of row n of A237048 is the number of odd divisors of n, and this was proved by Hartmut F. W. Hoft in a comment in A237048. - N. J. A. Sloane, Dec 07 2020]
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
Mathematica function has been written to check the first property up to n = 20000.
T(n,(sqrt(8n+1)-1)/2+1) = 0 for all n >= 1, which is useful for formulas for A237591 and A237593. (End)
Alternating row sums give A240542. - Omar E. Pol, Apr 16 2014
Conjecture: T(n,k) is also the total number of partitions of all positive integers <= n into exactly k consecutive parts, i.e., the partial column sum of A285898, or in accordance with the triangles of the same family: the partial column sum of A237048. - Omar E. Pol, Apr 28 2017, Nov 24 2020
The above conjecture is true. The proof will be added soon (it uses the generating function for the columns). - N. J. A. Sloane, Nov 24 2020
T(n,k) is also the total length of all line segments between the k-th vertex and the central vertex of the largest Dyck path of the symmetric representation of sigma(n). In other words: T(n,k) is the sum of the last (A003056(n)-k+1) terms of the n-th row of A237591. - Omar E. Pol, Sep 07 2021
T(n,k) is also the Manhattan distance between the k-th vertex and the central vertex of the Dyck path described in the n-th row of the triangle A237593. - Omar E. Pol, Jan 11 2023

			Triangle begins:
   1;
   2;
   3,  1;
   4,  1;
   5,  2;
   6,  2,  1;
   7,  3,  1;
   8,  3,  1;
   9,  4,  2;
  10,  4,  2,  1;
  11,  5,  2,  1;
  12,  5,  3,  1;
  13,  6,  3,  1;
  14,  6,  3,  2;
  15,  7,  4,  2,  1;
  16,  7,  4,  2,  1;
  17,  8,  4,  2,  1;
  18,  8,  5,  3,  1;
  19,  9,  5,  3,  1;
  20,  9,  5,  3,  2;
  21, 10,  6,  3,  2,  1;
  22, 10,  6,  4,  2,  1;
  23, 11,  6,  4,  2,  1;
  24, 11,  7,  4,  2,  1;
  25, 12,  7,  4,  3,  1;
  26, 12,  7,  5,  3,  1;
  27, 13,  8,  5,  3,  2;
  28, 13,  8,  5,  3,  2,  1;
  ...
For n = 10 the 10th row of triangle is 10, 4, 2, 1, so we have that 10^2 - 4^2 + 2^2 - 1^2 = 100 - 16 + 4 - 1 = 87, the same as A024916(10) = 87, the sum of all divisors of all positive integers <= 10.
From _Omar E. Pol_, Nov 19 2015: (Start)
Illustration of initial terms in the third quadrant:
.                                                            y
Row                                                         _|
1                                                         _|1|
2                                                       _|2 _|
3                                                     _|3  |1|
4                                                   _|4   _|1|
5                                                 _|5    |2 _|
6                                               _|6     _|2|1|
7                                             _|7      |3  |1|
8                                           _|8       _|3 _|1|
9                                         _|9        |4  |2 _|
10                                      _|10        _|4  |2|1|
11                                    _|11         |5   _|2|1|
12                                  _|12          _|5  |3  |1|
13                                _|13           |6    |3 _|1|
14                              _|14            _|6   _|3|2 _|
15                            _|15             |7    |4  |2|1|
16                          _|16              _|7    |4  |2|1|
17                        _|17               |8     _|4 _|2|1|
18                      _|18                _|8    |5  |3  |1|
19                    _|19                 |9      |5  |3 _|1|
20                  _|20                  _|9     _|5  |3|2 _|
21                _|21                   |10     |6   _|3|2|1|
22              _|22                    _|10     |6  |4  |2|1|
23            _|23                     |11      _|6  |4  |2|1|
24          _|24                      _|11     |7    |4 _|2|1|
25        _|25                       |12       |7   _|4|3  |1|
26      _|26                        _|12      _|7  |5  |3 _|1|
27    _|27                         |13       |8    |5  |3|2 _|
28   |28                           |13       |8    |5  |3|2|1|
...
T(n,k) is also the number of cells between the k-th vertical line segment (from left to right) and the y-axis in the n-th row of the structure.
Note that the number of horizontal line segments in the n-th row of the structure equals A001227(n), the number of odd divisors of n.
Also the diagram represents the left part of the front view of the pyramid described in A245092. (End)
For more information about the diagram see A286001. - _Omar E. Pol_, Dec 19 2020
From _Omar E. Pol_, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below:
                            _
                           | |
                           | |
                           | |
                           | |
                           | |
                      _ _ _| |
                    _|    _ _|
                  _|     |
                 |      _|
                 |  _ _|
      _ _ _ _ _ _| |3   1
     |_ _ _ _ _ _ _|
    12              5
.
For n = 12 and k = 1 the total length of all line segments between the first vertex and the central vertex of the largest Dyck path is equal to 12, so T(12,1) = 12.
For n = 12 and k = 2 the total length of all line segments between the second vertex and the central vertex of the largest Dyck path is equal to 5, so T(12,2) = 5.
For n = 12 and k = 3 the total length of all line segments between the third vertex and the central vertex of the largest Dyck path is equal to 3, so T(12,3) = 3.
For n = 12 and k = 4 the total length of all line segments between the fourth vertex and the central vertex of the largest Dyck path is equal to 1, so T(12,4) = 1.
Hence the 12th row of triangle is [12, 5, 3, 1]. (End)

G. C. Greubel, Table of n, a(n) for the first 150 rows, flattened

Columns 1..3: A000027, A008619, A008620.
Operations on rows: A003056 (number of terms), A237591 (differences between terms), A060831 (sums), A339577 (products), A240542 (alternating sums), A236104 (squares), A339576 (sums of squares), A024916 (alternating sums of squares), A237048 (differences between rows), A042974 (right border).
Cf. A000203, A000217, A001227, A196020, A211343, A228813, A231345, A231347, A235794, A236106, A236112, A237270, A237271, A237593, A239660, A245092, A261699, A262626, A286000, A286001, A280850, A280851, A296508, A335616.

row[n_] := Floor[(Sqrt[8*n + 1] - 1)/2]; f[n_, k_] := Ceiling[(n + 1)/k - (k + 1)/2]; Table[f[n, k], {n, 1, 150}, {k, 1, row[n]}] // Flatten (* Hartmut F. W. Hoft, Apr 07 2014 *)

row(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i); \\ Michel Marcus, Mar 27 2014

from sympy import sqrt
import math
def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))
for n in range(1, 21): print([T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 25 2017

T(n,k) = ceiling((n+1)/k - (k+1)/2) for 1 <= n, 1 <= k <= floor((sqrt(8n+1)-1)/2) = A003056(n). - Hartmut F. W. Hoft, Apr 07 2014
G.f. for column k (k >= 1): x^(k*(k+1)/2)/( (1-x)*(1-x^k) ). - N. J. A. Sloane, Nov 24 2020
T(n,k) = Sum_{j=1..n} A237048(j,k). - Omar E. Pol, May 18 2017
T(n,k) = sqrt(A236104(n,k)). - Omar E. Pol, Feb 14 2018
Sigma(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k)^2 - T(n-1,k)^2), assuming that T(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 10 2018
a(s(n,k)) = T(n,k), n >= 1, 1 <= k <= r = floor((sqrt(8*n + 1) - 1)/2), where s(n,k) = r*n - r*(r+1)*(r+2)/6 + k translates position (row n, column k) in the triangle of this sequence to its position in the sequence. - Hartmut F. W. Hoft, Feb 24 2021

A042975 Decimal expansion of sqrt(0.121121112...).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
C. V. Eynden, Problem 10439. An irrational mimic of 1/9, Amer. Math. Monthly, 104 (1997), 873.

Cf. A042974.

With[{c=FromDigits[Flatten[Most[Riffle[Table[PadRight[{},n,1],{n,50}], 2]]]]},RealDigits[Sqrt[c/10^IntegerLength[c]],10,120][[1]]] (* Harvey P. Dale, Oct 12 2013 *)

A056030 Continued fraction for 0.12112111211112... .

Original entry on oeis.org

0, 8, 3, 1, 9, 3, 110, 1, 7, 1, 10, 4, 2, 10, 1, 3, 1, 1, 10, 2, 5, 8, 1, 9, 8, 1, 997, 1, 109, 8, 1, 111, 4, 9, 2, 2, 1, 1, 1, 1, 2, 1, 53, 1, 3, 1, 6, 5, 1, 1, 3, 1, 4, 3, 1, 1, 6, 1, 1, 59, 2, 1, 1, 1, 8, 2, 9, 11, 5, 6, 33, 1, 1, 3, 2, 6, 9, 3, 2, 20, 1, 7, 1, 27, 1, 20, 5, 1, 229, 1, 2, 12, 6

Offset: 1

Robert G. Wilson v, Jul 24 2000

			0.121121112111121111121111112... = 0 + 1/(8 + 1/(3 + 1/(1 + 1/(9 + ...)))). - _Harry J. Smith_, May 08 2009

Harry J. Smith, Table of n, a(n) for n = 1..20000 [corrected by _Kevin Ryde_, Aug 04 2025]
G. Xiao, Contfrac
Index entries for continued fractions for constants

Cf. A042974.

ContinuedFraction[N[1/9,1000]+N[Sum[10^-(n(n+3)/2),{n,1,50}],1000],100]

A181133 a(n) = n + A003056(n).

Original entry on oeis.org

2, 3, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 77, 78, 79, 80, 81, 82, 83, 84, 85

Offset: 1

Craig Michoski (michoski(AT)google.com), Oct 05 2010

Obtained starting with a triangle with 1's and a trailing 2, and accumulating a partial sum along rows and columns:
2; # 2
1,2; # 3,5
1,1,2; # 6,7,9
1,1,1,2; # 10,11,12,14
1,1,1,1,2; # 15,16,17,18,20
1,1,1,1,1,2;

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A003056, A042974.

A003056:= [seq(n$(n+1),n=1..20)]:
A003056+[$1..nops(A003056)]; # Robert Israel, Dec 24 2017

Array[# + Floor[(Sqrt[1 + 8 #] - 1)/2] &, 74] (* Michael De Vlieger, Dec 24 2017 *)
Accumulate[Flatten[Table[Join[PadRight[{},n,1],{2}],{n,0,15}]]] (* Harvey P. Dale, Aug 14 2022 *)

a(n) = n + floor((sqrt(1+8*n)-1)/2) \\ Iain Fox, Dec 25 2017

from math import isqrt
def A181133(n): return n+(isqrt((n<<3)+1)-1>>1) # Chai Wah Wu, Feb 10 2023

a(n) = 2 + Sum_{k=1..n-1} A042974(k). - R. J. Mathar, Oct 08 2010

G.f.: (2*x-1)/(1-x)^2 + Theta_2(0,sqrt(x))/(x^(1/8)*(2-2*x)) where Theta_2 is a Jacobi theta function. - Robert Israel, Dec 24 2017

Definition re-fitted to something precise, sequence extended beyond a(15), and comment added by R. J. Mathar, Oct 24 2010

A042976 Decimal expansion of 0.121121112...^2.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

C. V. Eynden, Problem 10439. An irrational mimic of 1/9, Amer. Math. Monthly, 104 (1997), 873.

Cf. A042974.

A230517 An irrational x such that the decimal representation of neither x nor sqrt(x) contains the digit 0.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 1, 3, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Michel Marcus, Oct 22 2013

The rational number 1/9 is an example of a number in [0, 1] such that the decimal representation of neither x nor sqrt(x) contains the digit 0. The object of Problem 10439 of the Amer. Math. Monthly was to find an irrational with the same property (see link).
The solution proposed by Jerrold Grossman defines a sequence of irrationals starting with c1= 0.121121112... (A042974). Moving from left to right, the 0's in the decimal expansion of sqrt(cn) are eliminated by increasing the corresponding digit in the decimal expansion of cn by 2. The limit of cn is a number with the desired property.
The indices of the decimals that are successively changed are 4, 8, 29, 38, 40, 54, 62, 70, 72, 96, 118, ... (see print(ndeci) in PARI script).
The decimal expansion of sqrt(x) begins with 0.3483118317127931144162557719319698175373163374567....

			0.12132113211112111112111111213111112113131112111111111411111113...

C. V. Eynden, Problem 10439. An irrational mimic of 1/9, Amer. Math. Monthly, 104 (1997), 873.

Cf. A042974, A042975.

pdeci(x, nb) = {x = x * 10; for (n=1, nb, d = floor(x); x = (x-d)*10; print1(d, ", ");); print();}
finddeci(x) = {x = x * 10; found = 0; nd = 1; while (! found, d = floor(x); x = (x-d)*10; if (d == 0, found = 1, nd++);); nd;}
changedeci(x, ndeci) = {deci = floor(x * 10^ndeci) - 10*floor(x * 10^(ndeci-1)); x += 2/10^ndeci; x;}
lista(nn) = {prec = 2*nn; default(realprecision, prec); x = 0; for (n=1, prec, x = 10*x + 1 + issquare(9+8*n);); x /= 10^prec; ok = 0; while (! ok, y = sqrt(x); ndeci = finddeci(y); print1(ndeci, ", "); x = changedeci(x, ndeci); if (ndeci > nn, ok =1);); print(); pdeci(x, nn); print("sqrt(x)=", sqrt(x));} \\ Michel Marcus, Oct 22 2013

cp's OEIS Frontend

A237591 Irregular triangle read by rows: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts (n>=1, 1<=k<=A003056(n)).

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A235791 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

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A042975 Decimal expansion of sqrt(0.121121112...).

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A056030 Continued fraction for 0.12112111211112... .

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A181133 a(n) = n + A003056(n).

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A042976 Decimal expansion of 0.121121112...^2.

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A230517 An irrational x such that the decimal representation of neither x nor sqrt(x) contains the digit 0.

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PARI