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 1-6 of 6 results.

A138121 Triangle read by rows in which row n lists the partitions of n that do not contain 1 as a part in juxtaposed reverse-lexicographical order followed by A000041(n-1) 1's.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 4, 2, 2, 1, 1, 1, 5, 3, 2, 1, 1, 1, 1, 1, 6, 3, 3, 4, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 7, 4, 3, 5, 2, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 4, 4, 5, 3, 6, 2, 3, 3, 2, 4, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 5, 4, 6, 3, 3, 3, 3, 7, 2, 4, 3, 2, 5, 2, 2, 3, 2, 2
Offset: 1

Views

Author

Omar E. Pol, Mar 21 2008

Keywords

Comments

Mirror of triangle A135010.

Examples

			Triangle begins:
[1];
[2],[1];
[3],[1],[1];
[4],[2,2],[1],[1],[1];
[5],[3,2],[1],[1],[1],[1],[1];
[6],[3,3],[4,2],[2,2,2],[1],[1],[1],[1],[1],[1],[1];
[7],[4,3],[5,2],[3,2,2],[1],[1],[1],[1],[1],[1],[1],[1],[1],[1],[1];
...
The illustration of the three views of the section model of partitions (version "tree" with seven sections) shows the connection between several sequences.
---------------------------------------------------------
Partitions                A194805            Table 1.0
.  of 7       p(n)        A194551             A135010
---------------------------------------------------------
7              15                    7     7 . . . . . .
4+3                                4       4 . . . 3 . .
5+2                              5         5 . . . . 2 .
3+2+2                          3           3 . . 2 . 2 .
6+1            11    6       1             6 . . . . . 1
3+3+1                  3     1             3 . . 3 . . 1
4+2+1                    4   1             4 . . . 2 . 1
2+2+2+1                    2 1             2 . 2 . 2 . 1
5+1+1           7            1   5         5 . . . . 1 1
3+2+1+1                      1 3           3 . . 2 . 1 1
4+1+1+1         5        4   1             4 . . . 1 1 1
2+2+1+1+1                  2 1             2 . 2 . 1 1 1
3+1+1+1+1       3            1 3           3 . . 1 1 1 1
2+1+1+1+1+1     2          2 1             2 . 1 1 1 1 1
1+1+1+1+1+1+1   1            1             1 1 1 1 1 1 1
.               1                         ---------------
.               *<------- A000041 -------> 1 1 2 3 5 7 11
.                         A182712 ------->   1 0 2 1 4 3
.                         A182713 ------->     1 0 1 2 2
.                         A182714 ------->       1 0 1 1
.                                                  1 0 1
.                         A141285           A182703  1 0
.                    A182730   A182731                 1
---------------------------------------------------------
.                              A138137 --> 1 2 3 6 9 15..
---------------------------------------------------------
.       A182746 <--- 4 . 2 1 0 1 2 . 4 ---> A182747
---------------------------------------------------------
.
.       A182732 <--- 6 3 4 2 1 3 5 4 7 ---> A182733
.                    . . . . 1 . . . .
.                    . . . 2 1 . . . .
.                    . 3 . . 1 2 . . .
.      Table 2.0     . . 2 2 1 . . 3 .     Table 2.1
.                    . . . . 1 2 2 . .
.                            1 . . . .
.
.  A182982  A182742       A194803       A182983  A182743
.  A182992  A182994       A194804       A182993  A182995
---------------------------------------------------------
.
From _Omar E. Pol_, Sep 03 2013: (Start)
Illustration of initial terms (n = 1..6). The table shows the six sections of the set of partitions of 6. Note that before the dissection the set of partitions was in the ordering mentioned in A026792. More generally, the six sections of the set of partitions of 6 also can be interpreted as the first six sections of the set of partitions of any integer >= 6.
Illustration of initial terms:
---------------------------------------
n  j     Diagram          Parts
---------------------------------------
.         _
1  1     |_|              1;
.         _ _
2  1     |_  |            2,
2  2       |_|            .  1;
.         _ _ _
3  1     |_ _  |          3,
3  2         | |          .  1,
3  3         |_|          .  .  1;
.         _ _ _ _
4  1     |_ _    |        4,
4  2     |_ _|_  |        2, 2,
4  3           | |        .  1,
4  4           | |        .  .  1,
4  5           |_|        .  .  .  1;
.         _ _ _ _ _
5  1     |_ _ _    |      5,
5  2     |_ _ _|_  |      3, 2,
5  3             | |      .  1,
5  4             | |      .  .  1,
5  5             | |      .  .  1,
5  6             | |      .  .  .  1,
5  7             |_|      .  .  .  .  1;
.         _ _ _ _ _ _
6  1     |_ _ _      |    6,
6  2     |_ _ _|_    |    3, 3,
6  3     |_ _    |   |    4, 2,
6  4     |_ _|_ _|_  |    2, 2, 2,
6  5               | |    .  1,
6  6               | |    .  .  1,
6  7               | |    .  .  1,
6  8               | |    .  .  .  1,
6  9               | |    .  .  .  1,
6  10              | |    .  .  .  .  1,
6  11              |_|    .  .  .  .  .  1;
...
(End)
		

Crossrefs

Row n has length A138137(n).
Rows sums give A138879.

Programs

  • Mathematica
    less[run1_, run2_] := (lg1 = run1 // Length; lg2 = run2 // Length; lg = Max[lg1, lg2]; r1 = If[lg1 == lg, run1, PadRight[run1, lg, 0]]; r2 = If[lg2 == lg, run2, PadRight[run2, lg, 0]]; Order[r1, r2] != -1); row[n_] := Join[Array[1 &, {PartitionsP[n - 1]}], Sort[Reverse /@ Select[IntegerPartitions[n], FreeQ[#, 1] &], less]] // Flatten // Reverse; Table[row[n], {n, 1, 9}] // Flatten (* Jean-François Alcover, Jan 15 2013 *)
    Table[Reverse/@Reverse@DeleteCases[Sort@PadRight[Reverse/@Cases[IntegerPartitions[n], x_ /; Last[x]!=1]], x_ /; x==0, 2]~Join~ConstantArray[{1}, PartitionsP[n - 1]], {n, 1, 9}]  // Flatten (* Robert Price, May 11 2020 *)

A182712 Number of 2's in the last section of the set of partitions of n.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 4, 3, 8, 7, 15, 15, 27, 29, 48, 53, 82, 94, 137, 160, 225, 265, 362, 430, 572, 683, 892, 1066, 1370, 1640, 2078, 2487, 3117, 3725, 4624, 5519, 6791, 8092, 9885, 11752, 14263, 16922, 20416, 24167, 29007, 34254, 40921, 48213, 57345, 67409
Offset: 0

Views

Author

Omar E. Pol, Nov 28 2010

Keywords

Comments

Essentially the same as A087787 but here a(n) is the number of 2's in the last section of n, not n-2. See also A100818.
Note that a(1)..a(11) coincide with a(2)..a(12) of A005291.
Also number of 2's in all partitions of n that do not contain 1's as a part, if n >= 1. Also 0 together with the first differences of A024786. - Omar E. Pol, Nov 13 2011
Also number of 2's in the n-th section of the set of partitions of any integer >= n. For the definition of "section" see A135010. - Omar E. Pol, Dec 01 2013

Examples

			a(6) = 4 counts the 2's in 6 = 4+2 = 2+2+2. The 2's in 6 = 3+2+1 = 2+2+1+1 = 2+1+1+1+1 do not count. - _Omar E. Pol_, Nov 13 2011
From _Omar E. Pol_, Oct 27 2012: (Start)
----------------------------------
Last section               Number
of the set of                of
partitions of 6             2's
----------------------------------
6 .......................... 0
3 + 3 ...................... 0
4 + 2 ...................... 1
2 + 2 + 2 .................. 3
.   1 ...................... 0
.       1 .................. 0
.       1 .................. 0
.           1 .............. 0
.           1 .............. 0
.               1 .......... 0
.                   1 ...... 0
---------------------------------
.   8 - 4 =                  4
.
In the last section of the set of partitions of 6 the difference between the sum of the second column and the sum of the third column is 8 - 4 = 4, the same as the number of 2's, so a(6) = 4 (see also A024786).
(End)
		

Crossrefs

Programs

  • Mathematica
    Table[Count[Flatten@Cases[IntegerPartitions[n], x_ /; Last[x] != 1], 2], {n, 0, 49}] (* Robert Price, May 15 2020 *)
  • Sage
    A182712 = lambda n: sum(list(p).count(2) for p in Partitions(n) if 1 not in p) # Omar E. Pol, Nov 13 2011

Formula

It appears that A000041(n) = a(n+1) + a(n+2), n >= 0. - Omar E. Pol, Feb 04 2012
G.f.: (x^2/(1 + x))*Product_{k>=1} 1/(1 - x^k). - Ilya Gutkovskiy, Jan 03 2017
a(n) ~ exp(Pi*sqrt(2*n/3)) / (8*sqrt(3)*n). - Vaclav Kotesovec, Jun 02 2018

A194812 Square array read by antidiagonals: T(n,k) = number of parts of size k in the last section of the set of partitions of n.

Original entry on oeis.org

1, 1, 0, 2, 1, 0, 3, 0, 0, 0, 5, 2, 1, 0, 0, 7, 1, 0, 0, 0, 0, 11, 4, 1, 1, 0, 0, 0, 15, 3, 2, 0, 0, 0, 0, 0, 22, 8, 2, 1, 1, 0, 0, 0, 0, 30, 7, 3, 1, 0, 0, 0, 0, 0, 0, 42, 15, 6, 3, 1, 1, 0, 0, 0, 0, 0, 56, 15, 6, 2, 1, 0, 0, 0, 0, 0, 0, 0, 77, 27, 10
Offset: 1

Views

Author

Omar E. Pol, Feb 04 2012

Keywords

Comments

It appears that in the column k, starting in row n, the sum of k successive terms is equal to A000041(n-1).

Examples

			Array begins:
.  1,  0,  0,  0, 0, 0, 0, 0, 0, 0, 0, 0,...
.  1,  1,  0,  0, 0, 0, 0, 0, 0, 0, 0, 0,...
.  2,  0,  1,  0, 0, 0, 0, 0, 0, 0, 0, 0,...
.  3,  2,  0,  1, 0, 0, 0, 0, 0, 0, 0, 0,...
.  5,  1,  1,  0, 1, 0, 0, 0, 0, 0, 0, 0,...
.  7,  4,  2,  1, 0, 1, 0, 0, 0, 0, 0, 0,...
. 11,  3,  2,  1, 1, 0, 1, 0, 0, 0, 0, 0,...
. 15,  8,  3,  3, 1, 1, 0, 1, 0, 0, 0, 0,...
. 22,  7,  6,  2, 2, 1, 1, 0, 1, 0, 0, 0,...
. 30, 15,  6,  5, 3, 2, 1, 1, 0, 1, 0, 0,...
. 42, 15, 10,  5, 4, 2, 2, 1, 1, 0, 1, 0,...
. 56, 27, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1,...
...
For n = 7, from the conjecture we have that p(n-1) = p(6) = 11 = 3+8 = 2+3+6 = 1+3+2+5 = 1+1+2+3+4 = 0+1+1+2+2+5, etc. where p(n) = A000041(n).
		

Crossrefs

Columns 1-4: A000041, A182712, A182713, A182714. Main triangle: A182703.

Formula

It appears that A000041(n) = Sum_{j=1..k} T(n+j,k), n >= 0, k >= 1.

A182714 Number of 4's in the last section of the set of partitions of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 1, 3, 2, 5, 5, 10, 10, 17, 19, 31, 34, 51, 60, 86, 100, 139, 165, 223, 265, 349, 418, 543, 648, 827, 992, 1251, 1495, 1866, 2230, 2758, 3289, 4033, 4803, 5852, 6949, 8411, 9973, 12005, 14194, 17002, 20060, 23919, 28153, 33426, 39256, 46438
Offset: 1

Views

Author

Omar E. Pol, Nov 13 2011

Keywords

Comments

Zero together with the first differences of A024788.
Also number of 4's in all partitions of n that do not contain 1 as a part.
a(n) is the number of partitions of n such that m(1) < m(3), where m = multiplicity; e.g., a(7) counts these 3 partitions: [4, 3], [3, 3, 1], [3, 2, 2]. - Clark Kimberling, Apr 01 2014
The last section of the set of partitions of n is also the n-th section of the set of partitions of any integer >= n. - Omar E. Pol, Apr 07 2014

Examples

			a(8) = 3 counts the 4's in 8 = 4+4 = 4+2+2. The 4's in 8 = 4+3+1 = 4+2+1+1 = 4+1+1+1+1 do not count.
From _Omar E. Pol_, Oct 25 2012: (Start)
--------------------------------------
Last section                   Number
of the set of                    of
partitions of 8                 4's
--------------------------------------
8 .............................. 0
4 + 4 .......................... 2
5 + 3 .......................... 0
6 + 2 .......................... 0
3 + 3 + 2 ...................... 0
4 + 2 + 2 ...................... 1
2 + 2 + 2 + 2 .................. 0
.   1 .......................... 0
.       1 ...................... 0
.       1 ...................... 0
.           1 .................. 0
.       1 ...................... 0
.           1 .................. 0
.           1 .................. 0
.               1 .............. 0
.           1 .................. 0
.               1 .............. 0
.               1 .............. 0
.                   1 .......... 0
.                   1 .......... 0
.                       1 ...... 0
.                           1 .. 0
------------------------------------
.           6 - 3 =              3
.
In the last section of the set of partitions of 8 the difference between the sum of the fourth column and the sum of the fifth column is 6 - 3 = 3 equaling the number of 4's, so a(8) = 3 (see also A024788).
(End)
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local g, h;
          if n=0 then [1, 0]
        elif i<2 then [0, 0]
        else g:= b(n, i-1); h:= `if`(i>n, [0, 0], b(n-i, i));
             [g[1]+h[1], g[2]+h[2]+`if`(i=4, h[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq (a(n), n=1..70);  # Alois P. Heinz, Mar 19 2012
  • Mathematica
    z = 60; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Count[p, 1] < Count[p, 3]], {n, 0, z}] (* Clark Kimberling, Apr 01 2014 *)
    b[n_, i_] := b[n, i] = Module[{g, h}, If[n==0, {1, 0}, If[i<2, {0, 0}, g = b[n, i-1]; h = If[i>n, {0, 0}, b[n-i, i]]; {g[[1]] + h[[1]], g[[2]] + h[[2]] + If[i==4, h[[1]], 0]}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 70}] (* Jean-François Alcover, Sep 21 2015, after Alois P. Heinz *)
    Table[Count[Flatten@Cases[IntegerPartitions[n], x_ /; Last[x] != 1], 4], {n, 52}] (* Robert Price, May 15 2020 *)
  • Sage
    A182714 = lambda n: sum(list(p).count(4) for p in Partitions(n) if 1 not in p)

Formula

It appears that A000041(n) = a(n+1) + a(n+2) + a(n+3) + a(n+4), n >= 0. - Omar E. Pol, Feb 04 2012

A174455 Number of partitions where the number of 1's and 2's are equal.

Original entry on oeis.org

1, 0, 0, 2, 1, 1, 4, 3, 4, 8, 8, 10, 17, 18, 23, 34, 39, 48, 67, 78, 97, 127, 151, 185, 237, 281, 343, 428, 511, 616, 759, 902, 1084, 1315, 1562, 1863, 2242, 2649, 3147, 3752, 4424, 5222, 6190, 7266, 8545, 10062, 11776, 13782, 16157, 18832, 21964, 25622, 29777, 34589, 40200, 46556, 53912
Offset: 0

Views

Author

Joerg Arndt, Nov 28 2010

Keywords

Comments

From Omar E. Pol, Jan 19 2013: (Start)
Column 3 of triangle A220504.
With offset 3, a(n) is also the number of appearances of 3 as the smallest part in all partitions of n.
Also consider the sequence formed by [0, 0] together with this sequence, with offset 1, then it appears that A027336(n) = Sum_{j=1..3} a(n+j), n >= 0.
(End)

Examples

			a(8)=9, there are 8 such partitions of 9, they are
  #1:    9 =  3* 1 + 3* 2 + 0    + 0    + 0    + 0    + 0    + 0    + 0
  #2:    9 =  2* 1 + 2* 2 + 1* 3 + 0    + 0    + 0    + 0    + 0    + 0
  #3:    9 =  1* 1 + 1* 2 + 2* 3 + 0    + 0    + 0    + 0    + 0    + 0
  #4:    9 =  0    + 0    + 3* 3 + 0    + 0    + 0    + 0    + 0    + 0
  #5:    9 =  0    + 0    + 0    + 1* 4 + 1* 5 + 0    + 0    + 0    + 0
  #6:    9 =  1* 1 + 1* 2 + 0    + 0    + 0    + 1* 6 + 0    + 0    + 0
  #7:    9 =  0    + 0    + 1* 3 + 0    + 0    + 1* 6 + 0    + 0    + 0
  #8:    9 =  0    + 0    + 0    + 0    + 0    + 0    + 0    + 0    + 1* 9
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local j, r; if n=0 or i<1 then 0
          else `if`(i=3 and irem(n, 3, 'r')=0, r, 0); for j from 0
          to n/i do %+b(n-i*j, i-1) od; % fi
        end:
    a:= n-> b(n+3, n+3):
    seq(a(n), n=0..60);  # Alois P. Heinz, Jan 20 2013
  • Mathematica
    (* See A240056. - Clark Kimberling, Mar 31 2014 *)
    m = 66; gf = 1/((1-x^3)*Product[1-x^n, {n, 3, m}]) + O[x]^m; CoefficientList[gf, x] (* Jean-François Alcover, Jul 02 2015, after Joerg Arndt *)
  • PARI
    N=66;  x='x+O('x^N);
    gf=1/( (1-x^3) * prod(n=3,N, 1-x^n) );
    Vec(gf)
    /* Joerg Arndt, Jul 07 2012 */

Formula

G.f.: 1/( (1-x^3) * Product_{n>=3} (1-x^n) ). - Joerg Arndt, Jul 07 2012
a(n) = A182713(n+2) - A182713(n) = A240056(n+1) - A240056(n) for n >= 0. - Clark Kimberling, Mar 31 2014
a(n) ~ Pi * exp(sqrt(2*n/3)*Pi) / (9 * 2^(3/2) * n^(3/2)). - Vaclav Kotesovec, Jan 15 2022

A240056 Number of partitions of n such that m(1) > m(2), where m = multiplicity.

Original entry on oeis.org

0, 1, 1, 1, 3, 4, 5, 9, 12, 16, 24, 32, 42, 59, 77, 100, 134, 173, 221, 288, 366, 463, 590, 741, 926, 1163, 1444, 1787, 2215, 2726, 3342, 4101, 5003, 6087, 7402, 8964, 10827, 13069, 15718, 18865, 22617, 27041, 32263, 38453, 45719, 54264, 64326, 76102, 89884
Offset: 0

Views

Author

Clark Kimberling, Mar 31 2014

Keywords

Examples

			a(7) counts these 9 partitions: 61, 511, 4111, 331, 3211, 31111, 22111, 211111, 1111111.
		

Crossrefs

Programs

  • Mathematica
    z = 60; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Count[p, 1] < Count[p, 2]], {n, 0, z}]  (* A182713 *)
    t2 = Table[Count[f[n], p_ /; Count[p, 1] <= Count[p, 2]], {n, 0, z}] (* A182713(n+2) *)
    t3 = Table[Count[f[n], p_ /; Count[p, 1] == Count[p, 2]], {n, 0, z}] (* A174455 *)
    t4 = Table[Count[f[n], p_ /; Count[p, 1] > Count[p, 2]], {n, 0, z}]  (* A240056 *)
    t5 = Table[Count[f[n], p_ /; Count[p, 1] >= Count[p, 2]], {n, 0, z}] (* A240056(n+1) *)

Formula

a(n) = A000041(n) - A182713(n+2) = a(n+1) - A174455(n) for n >= 0.
a(n) ~ exp(sqrt(2*n/3)*Pi) / (2 * 3^(3/2) * n). - Vaclav Kotesovec, Jan 15 2022
Showing 1-6 of 6 results.