login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A237048 Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists 1's interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2. 77
1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET

1

COMMENTS

The sum of row n gives A001227(n), the number of odd divisors of n.

Row n has length A003056(n) hence column k starts in row A000217(k).

If n = 2^j then the only positive integer in row n is T(n,1) = 1.

If n is an odd prime then the only two positive integers in row n are T(n,1) = 1 and T(n,2) = 1.

The partial sums of column k give the column k of A235791.

The connection with A196020 is as follow: A235791 --> A236104 --> A196020.

The connection with the symmetric representation of sigma is as follows: A235791 --> A237591 --> A237593 --> A239660 --> A237270.

From Hartmut F. W. Hoft, Oct 23 2014: (Start)

Property: Let n = 2^m*s*t with m >= 0 and 1 <= s, t odd, and r(n) = floor(sqrt(8*n+1) - 1)/2) = A003056(n). T(n, k) = 1 precisely when k is odd and k|n, or k = 2^(m+1)*s when 1 <= s < 2^(m+1)*s <= r(n) < t. Thus each odd divisor greater than r(n) is matched by a unique even index less than or equal to r(n).

For further connections with the symmetric representation of sigma see also A249223. (End)

From Omar E. Pol, Jan 21 2017: (Start)

Conjecture 1: alternating sum of row n gives A067742(n), the number of middle divisors of n.

The sum of row n also gives the number of subparts in the symmetric representation of sigma(n), equaling A001227(n), the number of odd divisors of n. For more information see A279387. (End)

From Omar E. Pol, Feb 08 2017, Feb 22 2017: (Start)

Conjectures:

2. Alternating sum of row n also gives the number of central subparts in the symmetric representation of sigma(n), equaling the width of the terrace at the n-th level in the main diagonal of the pyramid described in A245092.

3. The sum of the odd indexed terms in row n gives A082647(n): the number of odd divisors of n less than sqrt(2*n), also the number of partitions of n into an odd number of consecutive parts.

4. The sum of the even indexed terms in row n gives A131576(n): the number of odd divisors of n greater than sqrt(2*n), also the number of partitions of n into an even number of consecutive parts.

5. The sum of the even indexed terms in row n also gives the number of pairs of equidistant subparts in the symmetric representation of sigma(n). (End)

Conjecture 6: T(n,k) is also the number of partitions of n into exactly k consecutive parts. - Omar E. Pol, Apr 28 2017

LINKS

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

FORMULA

For n >= 1 and k = 1, ..., A003056(n) : if k is odd then T(n, k) = 1 if k|n else 0, and if k is even then T(n, k) = 1 if k|(n-k/2) else 0. - Hartmut F. W. Hoft, Oct 23 2014

a(n) = A057427(A196020(n)) = A057427(A261699(n)). - Omar E. Pol, Nov 14 2016

EXAMPLE

Triangle begins (rows 1..28):

1;

1;

1,  1;

1,  0;

1,  1;

1,  0,  1;

1,  1,  0;

1,  0,  0;

1,  1,  1;

1,  0,  0,  1;

1,  1,  0,  0;

1,  0,  1,  0;

1,  1,  0,  0;

1,  0,  0,  1;

1,  1,  1,  0,  1;

1,  0,  0,  0,  0;

1,  1,  0,  0,  0;

1,  0,  1,  1,  0;

1,  1,  0,  0,  0;

1,  0,  0,  0,  1;

1,  1,  1,  0,  0,  1;

1,  0,  0,  1,  0,  0;

1,  1,  0,  0,  0,  0;

1,  0,  1,  0,  0,  0;

1,  1,  0,  0,  1,  0;

1,  0,  0,  1,  0,  0;

1,  1,  1,  0,  0,  1;

1,  0,  0,  0,  0,  0,  1;

...

For n = 20 the divisors of 20 are 1, 2, 4, 5, 10, 20.

There are two odd divisors: 1 and 5. On the other hand the 20th row of triangle is [1, 0, 0, 0, 1] and the row sum is 2, equaling the number of odd divisors of 20.

From Hartmut F. W. Hoft, Oct 23 2014: (Start)

For n = 18 the divisors are 1, 2, 3, 6, 9, 18.

There are three odd divisors: 1 and 3 are in their respective columns, but 9 is accounted for in column 4 = 2^2*1 since 18 = 2^1*1*9 and 9>5, the number of columns in row 18. (End)

From Omar E. Pol, Dec 17 2016: (Start)

Illustration of initial terms:

Row                                                         _

1                                                         _|1|

2                                                       _|1 _|

3                                                     _|1  |1|

4                                                   _|1   _|0|

5                                                 _|1    |1 _|

6                                               _|1     _|0|1|

7                                             _|1      |1  |0|

8                                           _|1       _|0 _|0|

9                                         _|1        |1  |1 _|

10                                      _|1         _|0  |0|1|

11                                    _|1          |1   _|0|0|

12                                  _|1           _|0  |1  |0|

13                                _|1            |1    |0 _|0|

14                              _|1             _|0   _|0|1 _|

15                            _|1              |1    |1  |0|1|

16                          _|1               _|0    |0  |0|0|

17                        _|1                |1     _|0 _|0|0|

18                      _|1                 _|0    |1  |1  |0|

19                    _|1                  |1      |0  |0 _|0|

20                  _|1                   _|0     _|0  |0|1 _|

21                _|1                    |1      |1   _|0|0|1|

22              _|1                     _|0      |0  |1  |0|0|

23            _|1                      |1       _|0  |0  |0|0|

24          _|1                       _|0      |1    |0 _|0|0|

25        _|1                        |1        |0   _|0|1  |0|

26      _|1                         _|0       _|0  |1  |0 _|0|

27    _|1                          |1        |1    |0  |0|1 _|

28   |1                            |0        |0    |0  |0|0|1|

...

Note that the 1's are placed exactly below the horizontal line segments.

Also the above structure represents the left hand part of the front view of the pyramid described in A245092. For more information about the pyramid and the symmetric representation of sigma see A237593. (End)

MATHEMATICA

cd[n_, k_] := If[Divisible[n, k], 1, 0]

row[n_] := Floor[(Sqrt[8n+1] - 1)/2]

a237048[n_, k_] := If[OddQ[k], cd[n, k], cd[n - k/2, k]]

a237048[n_] := Map[a237048[n, #]&, Range[row[n]]]

Flatten[Map[a237048, Range[24]]] (* data: 24 rows of triangle *)

(* Hartmut F. W. Hoft, Oct 23 2014 *)

PROG

(PARI) it(n, k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0);

tabf(nn) = {for (n=1, nn, for (k=1, floor((sqrt(1+8*n)-1)/2), print1(t(n, k), ", "); ); print(); ); } \\ Michel Marcus, Sep 20 2015

(Python)

from sympy import sqrt

import math

def T(n, k): return (n%k == 0)*1 if k%2 == 1 else (((n - k/2)%k) == 0)*1

for n in xrange(1, 21): print [T(n, k) for k in xrange(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)] # Indranil Ghosh, Apr 21 2017

CROSSREFS

Cf. A000203, A000217, A001227, A003056, A057427, A067742, A082647, A131576, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239657, A245092, A249351, A261699, A262611, A262626, A279387, A279693.

Sequence in context: A114592 A221641 A196308 * A236862 A294934 A204441

Adjacent sequences:  A237045 A237046 A237047 * A237049 A237050 A237051

KEYWORD

nonn,easy,tabf

AUTHOR

Omar E. Pol, Mar 01 2014

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy .

Last modified February 24 22:35 EST 2018. Contains 299627 sequences. (Running on oeis4.)