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.

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A206563 Triangle read by rows: T(n,k) = number of odd/even parts >= k in all partitions of n, if k is odd/even.

Original entry on oeis.org

1, 2, 1, 5, 1, 1, 8, 4, 1, 1, 15, 5, 3, 1, 1, 24, 11, 5, 3, 1, 1, 39, 15, 9, 4, 3, 1, 1, 58, 28, 13, 9, 4, 3, 1, 1, 90, 38, 23, 12, 8, 4, 3, 1, 1, 130, 62, 33, 21, 12, 8, 4, 3, 1, 1, 190, 85, 51, 29, 20, 11, 8, 4, 3, 1, 1, 268, 131, 73, 48, 28, 20, 11, 8, 4, 3, 1, 1
Offset: 1

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Author

Omar E. Pol, Feb 15 2012

Keywords

Comments

Let m and n be two positive integers such that m <= n. It appears that any set formed by m connected sections, or m disconnected sections, or a mixture of both, has the same properties described in the section example. (Cf. A135010, A207031, A207032, A212010). - Omar E. Pol, May 01 2012

Examples

			Calculation for n = 6. Write the partitions of 6 and below the sums of their columns:
.
.   6
.   3 + 3
.   4 + 2
.   2 + 2 + 2
.   5 + 1
.   3 + 2 + 1
.   4 + 1 + 1
.   2 + 2 + 1 + 1
.   3 + 1 + 1 + 1
.   2 + 1 + 1 + 1 + 1
.   1 + 1 + 1 + 1 + 1 + 1
. ------------------------
.  35, 16,  8,  4,  2,  1  --> Row 6 of triangle A181187.
.   |  /|  /|  /|  /|  /|
.   | / | / | / | / | / |
.   |/  |/  |/  |/  |/  |
.  19,  8,  4,  2,  1,  1  --> Row 6 of triangle A066633.
.
More generally, it appears that the sum of column k is also the total number of parts >= k in all partitions of n. It appears that the first differences of the column sums together with 1 give the number of occurrences of k in all partitions of n.
On the other hand we can see that the partitions of 6 contain:
24  odd parts >= 1 (the odd parts).
11 even parts >= 2 (the even parts).
5   odd parts >= 3.
3  even parts >= 4.
2   odd parts >= 5.
1  even part  >= 6.
Then, using the values of the column sums, it appears that:
T(6,1) = 35 - 16 + 8 - 4 + 2 - 1 = 24
T(6,2) =      16 - 8 + 4 - 2 + 1 = 11
T(6,3) =           8 - 4 + 2 - 1 = 5
T(6,4) =               4 - 2 + 1 = 3
T(6,5) =                   2 - 1 = 1
T(6,6) =                       1 = 1
So the 6th row of our triangle gives 24, 11, 5, 3, 1, 1.
Finally, for all partitions of 6, we can write:
The number of  odd parts      is equal to T(6,1) = 24.
The number of even parts      is equal to T(6,2) = 11.
The number of  odd parts >= 3 is equal to T(6,3) = 5.
The number of even parts >= 4 is equal to T(6,4) = 3.
The number of  odd parts >= 5 is equal to T(6,5) = 1.
The number of even parts >= 6 is equal to T(6,6) = 1.
More generally, we can write the same properties for any positive integer.
Triangle begins:
1;
2,    1;
5,    1,  1;
8,    4,  1,  1;
15,   5,  3,  1,  1;
24,  11,  5,  3,  1,  1;
39,  15,  9,  4,  3,  1,  1;
58,  28, 13,  9,  4,  3,  1,  1;
90,  38, 23, 12,  8,  4,  3,  1,  1;
130, 62, 33, 21, 12,  8,  4,  3,  1,  1;
		

Crossrefs

Formula

It appears that T(n,k) = abs(Sum_{j=k..n} (-1)^j*A181187(n,j)).
It appears that A066633(n,k) = T(n,k) - T(n,k+2). - Omar E. Pol, Feb 26 2012

Extensions

More terms from Alois P. Heinz, Feb 18 2012

A024788 Number of 4's in all partitions of n.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 6, 8, 13, 18, 28, 38, 55, 74, 105, 139, 190, 250, 336, 436, 575, 740, 963, 1228, 1577, 1995, 2538, 3186, 4013, 5005, 6256, 7751, 9617, 11847, 14605, 17894, 21927, 26730, 32582, 39531, 47942, 57915, 69920, 84114, 101116, 121176, 145095, 173248
Offset: 1

Views

Author

Keywords

Comments

The sums of four successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of 4th largest and the sum of 5th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012
a(n+4) is the number of n-vertex graphs that do not contain a triangle, P_4, or K_2,3 as induced subgraph. These are the K_2,3-free bipartite cographs. Bipartite cographs are graph that are disjoint unions of complete bipartite graphs [Babel et al. Corollary 2.2], and forbidding K_2,3 leaves one possible component for each size except size 4, where there are two. Thus, this number is A000041(n) + a(n) = a(n+4). - Falk Hüffner, Jan 11 2016
a(n) (n>=3) is the number of even singletons in all partitions of n-2 (by a singleton we mean a part that occurs exactly once). Example: a(7) = 3 because in the partitions [5], [4*,1], [3,2*], [3,1,1], [2,2,1], [2*,1,1,1], [1,1,1,1,1] we have 3 even singletons (marked by *). The statement of this comment can be obtained by setting k=2 in Theorem 2 of the Andrews et al. reference. - Emeric Deutsch, Sep 13 2016

Examples

			From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 7 we have:
--------------------------------------
.                             Number
Partitions of 7               of 4's
--------------------------------------
7 .............................. 0
4 + 3 .......................... 1
5 + 2 .......................... 0
3 + 2 + 2 ...................... 0
6 + 1 .......................... 0
3 + 3 + 1 ...................... 0
4 + 2 + 1 ...................... 1
2 + 2 + 2 + 1 .................. 0
5 + 1 + 1 ...................... 0
3 + 2 + 1 + 1 .................. 0
4 + 1 + 1 + 1 .................. 1
2 + 2 + 1 + 1 + 1 .............. 0
3 + 1 + 1 + 1 + 1 .............. 0
2 + 1 + 1 + 1 + 1 + 1 .......... 0
1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
------------------------------------
.           7 - 4 =              3
The difference between the sum of the fourth column and the sum of the fifth column of the set of partitions of 7 is 7 - 4 = 3 and equals the number of 4's in all partitions of 7, so a(7) = 3.
(End)
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local f, g;
          if n=0 or i=1 then [1, 0]
        else f:= b(n, i-1); g:= `if`(i>n, [0$2], b(n-i, i));
             [f[1]+g[1], f[2]+g[2]+`if`(i=4, g[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012
  • Mathematica
    Table[ Count[ Flatten[ IntegerPartitions[n]], 4], {n, 1, 50} ]
    (* second program: *)
    b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 4, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)

Formula

a(n) = A181187(n,4) - A181187(n,5). - Omar E. Pol, Oct 25 2012
From Peter Bala, Dec 26 2013: (Start)
a(n+4) - a(n) = A000041(n). a(n) + a(n+2) = A024786(n).
O.g.f.: x^4/(1 - x^4) * product {k >= 1} 1/(1 - x^k) = x^4 + x^5 + 2*x^6 + 3*x^7 + ....
Asymptotic result: log(a(n)) ~ 2*sqrt(Pi^2/6)*sqrt(n) as n -> inf. (End)
a(n) ~ exp(Pi*sqrt(2*n/3)) / (8*Pi*sqrt(2*n)) * (1 - 49*Pi/(24*sqrt(6*n)) + (49/48 + 1633*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016
G.f.: x^4/((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)) * Sum_{n >= 0} x^(4*n)/( Product_{k = 1..n} 1 - x^k ); that is, convolution of A026810 (partitions into 4 parts, or, modulo offset differences, partitions into parts <= 4) and A008484 (partitions into parts >= 4). - Peter Bala, Jan 17 2021

A024787 Number of 3's in all partitions of n.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 6, 9, 15, 21, 31, 45, 63, 87, 122, 164, 222, 298, 395, 519, 683, 885, 1146, 1475, 1887, 2401, 3050, 3845, 4837, 6060, 7563, 9402, 11664, 14405, 17751, 21807, 26715, 32634, 39784, 48352, 58649, 70969, 85690, 103232, 124143, 148951, 178407, 213277, 254509
Offset: 1

Views

Author

Keywords

Comments

Starting with the first 1 = row sums of triangle A173239. - Gary W. Adamson, Feb 13 2010
The sums of three successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of 3rd largest and the sum of 4th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012

Examples

			From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 7 we have:
--------------------------------------
.                             Number
Partitions of 7               of 3's
--------------------------------------
7 .............................. 0
4 + 3 .......................... 1
5 + 2 .......................... 0
3 + 2 + 2 ...................... 1
6 + 1 .......................... 0
3 + 3 + 1 ...................... 2
4 + 2 + 1 ...................... 0
2 + 2 + 2 + 1 .................. 0
5 + 1 + 1 ...................... 0
3 + 2 + 1 + 1 .................. 1
4 + 1 + 1 + 1 .................. 0
2 + 2 + 1 + 1 + 1 .............. 0
3 + 1 + 1 + 1 + 1 .............. 1
2 + 1 + 1 + 1 + 1 + 1 .......... 0
1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
------------------------------------
.      13 - 7 =                  6
The difference between the sum of the third column and the sum of the fourth column of the set of partitions of 7 is 13 - 7 = 6 and equals the number of 3's in all partitions of 7, so a(7) = 6.
(End)
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local g;
          if n=0 or i=1 then [1, 0]
        else g:= `if`(i>n, [0$2], b(n-i, i));
             b(n, i-1) +g +[0, `if`(i=3, g[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012
  • Mathematica
    Table[ Count[ Flatten[ IntegerPartitions[n]], 3], {n, 1, 50} ]
    b[n_, i_] := b[n, i] = Module[{g}, If[n==0 || i==1, {1, 0}, g = If[i>n, {0, 0}, b[n-i, i]]; b[n, i-1] + g + {0, If[i==3, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
    Join[{0, 0}, (1/((1 - x^3) QPochhammer[x]) + O[x]^50)[[3]]] (* Vladimir Reshetnikov, Nov 22 2016 *)

Formula

a(n) = A181187(n,3) - A181187(n,4). - Omar E. Pol, Oct 25 2012
a(n) = Sum_{k=1..floor(n/3)} A263232(n,k). - Alois P. Heinz, Nov 01 2015
a(n) ~ exp(Pi*sqrt(2*n/3)) / (6*Pi*sqrt(2*n)) * (1 - 37*Pi/(24*sqrt(6*n)) + (37/48 + 937*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016
G.f.: x^3/((1 - x)*(1 - x^2)*(1 - x^3)) * Sum_{n >= 0} x^(3*n)/( Product_{k = 1..n} 1 - x^k ); that is, convolution of A069905 (partitions into 3 parts, or, modulo offset differences, partitions into parts <= 3) and A008483 (partitions into parts >= 3). - Peter Bala, Jan 17 2021

A302246 Irregular triangle read by rows in which row n lists all parts of all partitions of n, in nonincreasing order.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Apr 05 2018

Keywords

Comments

Also due to the correspondence divisor/part row n lists the terms of the n-th row of A338156 in nonincreasing order. In other words: row n lists in nonincreasing order the divisors of the terms of the n-th row of A176206. - Omar E. Pol, Jun 16 2022

Examples

			Triangle begins:
  1;
  2,1,1;
  3,2,1,1,1,1;
  4,3,2,2,2,1,1,1,1,1,1,1;
  5,4,3,3,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1;
  6,5,4,4,3,3,3,3,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
  ...
For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1]. There is only one 4, only one 3, three 2's and seven 1's, so the 4th row of this triangle is [4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1].
On the other hand for n = 4 the 4th row of A176206 is [4, 3, 2, 2, 1, 1, 1] and the divisors of these terms are [1, 2, 4], [1, 3], [1, 2], [1, 2], [1], [1], [1] the same as the 4th row of A338156. These divisors listed in nonincreasing order give the 4th row of this triangle. - _Omar E. Pol_, Jun 16 2022
		

Crossrefs

Both column 1 and 2 are A000027.
Row n has length A006128(n).
The sum of row n is A066186(n).
The number of parts k in row n is A066633(n,k).
The sum of all parts k in row n is A138785(n,k).
The number of parts >= k in row n is A181187(n,k).
The sum of all parts >= k in row n is A206561(n,k).
The number of parts <= k in row n is A210947(n,k).
The sum of all parts <= k in row n is A210948(n,k).
First differs from A036037, A080577, A181317, A237982 and A239512 at a(13) = T(4,3).
Cf. A302247 (mirror).

Programs

  • Mathematica
    nrows=10;Array[ReverseSort[Flatten[IntegerPartitions[#]]]&,nrows] (* Paolo Xausa, Jun 16 2022 *)
  • PARI
    row(n) = my(list = List()); forpart(p=n, for (k=1, #p, listput(list, p[k]));); vecsort(Vec(list), , 4); \\ Michel Marcus, Jun 16 2022

A302247 Irregular triangle read by rows in which row n lists all parts of all partitions of n, in nondecreasing order.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Omar E. Pol, Apr 05 2018

Keywords

Comments

Also due to the correspondence divisor/part row n lists the terms of the n-th row of A338156 in nondecreasing order. In other words: row n lists in nondecreasing order the divisors of the terms of the n-th row of A176206. - Omar E. Pol, Jun 16 2022

Examples

			Triangle begins:
  1;
  1,1,2;
  1,1,1,1,2,3;
  1,1,1,1,1,1,1,2,2,2,3,4;
  1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5;
  1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,4,4,5,6;
  ...
For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1]. There are seven 1's, three 2's, only one 3 and only one 4, so the 4th row of this triangle is [1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4].
On the other hand for n = 4 the 4th row of A176206 is [4, 3, 2, 2, 1, 1, 1] and the divisors of these terms are [1, 2, 4], [1, 3], [1, 2], [1, 2], [1], [1], [1] the same as the 4th row of A338156. These divisors listed in nondecreasing order give the 4th row of this triangle. - _Omar E. Pol_, Jun 16 2022
		

Crossrefs

Mirror of A302246.
Row n has length A006128(n).
The sum of row n is A066186(n).
The number of parts k in row n is A066633(n,k).
The sum of all parts k in row n is A138785(n,k).
The number of parts >= k in row n is A181187(n,k).
The sum of all parts >= k in row n is A206561(n,k).
The number of parts <= k in row n is A210947(n,k).
The sum of all parts <= k in row n is A210948(n,k).
First differs from both A026791 and A080576 at a(17) = T(4,7).

Programs

  • Mathematica
    nrows=10; Array[Sort[Flatten[IntegerPartitions[#]]]&,nrows] (* Paolo Xausa, Jun 16 2022 *)
  • PARI
    row(n) = my(list = List()); forpart(p=n, for (k=1, #p, listput(list, p[k]));); vecsort(Vec(list)); \\ Michel Marcus, Jun 16 2022

A024789 Number of 5's in all partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 3, 5, 8, 12, 17, 25, 35, 50, 68, 94, 126, 170, 226, 299, 391, 511, 660, 853, 1091, 1393, 1766, 2235, 2811, 3527, 4403, 5484, 6800, 8415, 10369, 12752, 15627, 19110, 23298, 28346, 34389, 41642, 50295, 60636, 72929, 87563, 104903, 125470
Offset: 1

Views

Author

Keywords

Comments

The sums of five successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of 5th largest and the sum of 6th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012

Examples

			From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 8 we have:
--------------------------------------
.                             Number
Partitions of 8               of 5's
--------------------------------------
8 .............................. 0
4 + 4 .......................... 0
5 + 3 .......................... 1
6 + 2 .......................... 0
3 + 3 + 2 ...................... 0
4 + 2 + 2 ...................... 0
2 + 2 + 2 + 2 .................. 0
7 + 1 .......................... 0
4 + 3 + 1 ...................... 0
5 + 2 + 1 ...................... 1
3 + 2 + 2 + 1 .................. 0
6 + 1 + 1 ...................... 0
3 + 3 + 1 + 1 .................. 0
4 + 2 + 1 + 1 .................. 0
2 + 2 + 2 + 1 + 1 .............. 0
5 + 1 + 1 + 1 .................. 1
3 + 2 + 1 + 1 + 1 .............. 0
4 + 1 + 1 + 1 + 1 .............. 0
2 + 2 + 1 + 1 + 1 + 1 .......... 0
3 + 1 + 1 + 1 + 1 + 1 .......... 0
2 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 .. 0
------------------------------------
.               7 - 4 =          3
The difference between the sum of the fifth column and the sum of the sixth column of the set of partitions of 8 is 7 - 4 = 3 and equals the number of 5's in all partitions of 8, so a(8) = 3.
(End)
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local g;
          if n=0 or i=1 then [1, 0]
        else g:= `if`(i>n, [0$2], b(n-i, i));
             b(n, i-1) +g +[0, `if`(i=5, g[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012
  • Mathematica
    Table[ Count[ Flatten[ IntegerPartitions[n]], 5], {n, 1, 50} ]
    (* second program: *)
    b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 5, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
  • PARI
    x='x+O('x^50); concat([0, 0, 0, 0], Vec(x^5/(1 - x^5) * prod(k=1, 50, 1/(1 - x^k)))) \\ Indranil Ghosh, Apr 06 2017

Formula

a(n) = A181187(n,5) - A181187(n,6). - Omar E. Pol, Oct 25 2012
a(n) ~ exp(Pi*sqrt(2*n/3)) / (10*Pi*sqrt(2*n)) * (1 - 61*Pi/(24*sqrt(6*n)) + (61/48 + 2521*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016
G.f.: x^5/(1 - x^5) * Product_{k>=1} 1/(1 - x^k). - Ilya Gutkovskiy, Apr 06 2017

A024790 Number of 6's in all partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 12, 16, 24, 33, 47, 63, 89, 117, 159, 209, 278, 360, 474, 607, 786, 1001, 1280, 1615, 2049, 2565, 3222, 4011, 4998, 6180, 7653, 9407, 11571, 14154, 17308, 21063, 25630, 31044, 37586, 45339, 54646, 65646, 78804, 94305, 112761, 134473
Offset: 1

Views

Author

Keywords

Comments

The sums of six successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of 6th largest and the sum of 7th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local g;
          if n=0 or i=1 then [1, 0]
        else g:= `if`(i>n, [0$2], b(n-i, i));
             b(n, i-1) +g +[0, `if`(i=6, g[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012
  • Mathematica
    Table[ Count[ Flatten[ IntegerPartitions[n]], 6], {n, 1, 52} ]
    b[n_, i_] := b[n, i] = Module[{g}, If [n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 6, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)

Formula

a(n) = A181187(n,6) - A181187(n,7). - Omar E. Pol, Oct 25 2012
From Peter Bala, Dec 26 2013: (Start)
a(n+6) - a(n) = A000041(n). a(n) + a(n+3) = A024787(n).
a(n) + a(n+2) + a(n+4) = A024786(n).
O.g.f.: x^6/(1 - x^6) * product {k >= 1} 1/(1 - x^k) = x^6 + x^7 + 2*x^8 + 3*x^9 + ....
Asymptotic result: log(a(n)) ~ 2*sqrt(Pi^2/6)*sqrt(n) as n -> inf. (End)
a(n) ~ exp(Pi*sqrt(2*n/3)) / (12*Pi*sqrt(2*n)) * (1 - 73*Pi/(24*sqrt(6*n)) + (73/48 + 3601*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016

A024794 Number of 10's in all partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 43, 57, 79, 104, 140, 183, 242, 312, 407, 520, 670, 849, 1081, 1359, 1715, 2141, 2678, 3322, 4125, 5085, 6274, 7691, 9430, 11502, 14025, 17024, 20655, 24959, 30140, 36270, 43612, 52274, 62604, 74763
Offset: 1

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Author

Keywords

Comments

The sums of ten successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of 10th largest and the sum of 11th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012
In general, if m>0 and a(n+m)-a(n) = A000041(n), then a(n) ~ exp(sqrt(2*n/3)*Pi) / (2*Pi*m*sqrt(2*n)) * (1 - Pi*(1/24 + m/2)/sqrt(6*n) + (1/48 + Pi^2/6912 + m/4 + m*Pi^2/288 + m^2*Pi^2/72)/n). - Vaclav Kotesovec, Nov 05 2016

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local g;
          if n=0 or i=1 then [1, 0]
        else g:= `if`(i>n, [0$2], b(n-i, i));
             b(n, i-1) +g +[0, `if`(i=10, g[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012
  • Mathematica
    Table[ Count[ Flatten[ IntegerPartitions[n]], 10], {n, 1, 55} ]
    b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 10, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)

Formula

a(n) = A181187(n,10) - A181187(n,11). - Omar E. Pol, Oct 25 2012
From Peter Bala, Dec 26 2013: (Start)
a(n+10) - a(n) = A000041(n). a(n) + a(n+5) = A024789(n).
a(n) + a(n+2) + a(n+4) + a(n+6) + a(n+8) = A024786(n).
O.g.f.: x^10/(1 - x^10) * product {k >= 1} 1/(1 - x^k) = x^10 + x^11 + 2*x^12 + 3*x^13 + ....
Asymptotic result: log(a(n)) ~ 2*sqrt(Pi^2/6)*sqrt(n) as n -> inf. (End)
a(n) ~ exp(Pi*sqrt(2*n/3)) / (20*Pi*sqrt(2*n)) * (1 - 121*Pi/(24*sqrt(6*n)) + (121/48 + 9841*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016

A207032 Triangle read by rows: T(n,k) = number of odd/even parts >= k in the last section of the set of partitions of n, if k is odd/even.

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 3, 3, 0, 1, 7, 1, 2, 0, 1, 9, 6, 2, 2, 0, 1, 15, 4, 4, 1, 2, 0, 1, 19, 13, 4, 5, 1, 2, 0, 1, 32, 10, 10, 3, 4, 1, 2, 0, 1, 40, 24, 10, 9, 4, 4, 1, 2, 0, 1, 60, 23, 18, 8, 8, 3, 4, 1, 2, 0, 1, 78, 46, 22, 19, 8, 9, 3, 4, 1, 2, 0, 1
Offset: 1

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Author

Omar E. Pol, Feb 17 2012

Keywords

Comments

For the calculation of row n, the number of odd/even parts, etc, take the row n from the triangle A207031 and then follow the same rules of A206563.

Examples

			Triangle begins:
  1;
  1,   1;
  3,   0,  1;
  3,   3,  0,  1;
  7,   1,  2,  0, 1;
  9,   6,  2,  2, 0, 1;
  15,  4,  4,  1, 2, 0, 1;
  19, 13,  4,  5, 1, 2, 0, 1;
  32, 10, 10,  3, 4, 1, 2, 0, 1;
  40, 24, 10,  9, 4, 4, 1, 2, 0, 1;
  60, 23, 18,  8, 8, 3, 4, 1, 2, 0, 1;
  78, 46, 22, 19, 8, 9, 3, 4, 1, 2, 0, 1;
		

Crossrefs

Formula

It appears that T(n,k) = abs(Sum_{j=k..n} (-1)^j*A207031(n,j)).
It appears that A182703(n,k) = T(n,k) - T(n,k+2). - Omar E. Pol, Feb 26 2012

A024791 Number of 7's in all partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 16, 23, 32, 45, 61, 84, 112, 151, 199, 263, 342, 446, 574, 739, 943, 1201, 1518, 1917, 2404, 3010, 3749, 4661, 5766, 7122, 8759, 10753, 13153, 16059, 19544, 23743, 28759, 34774, 41938, 50491, 60642, 72718, 87004, 103934, 123908
Offset: 1

Views

Author

Keywords

Comments

The sums of seven successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of 7th largest and the sum of 8th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local g;
          if n=0 or i=1 then [1, 0]
        else g:= `if`(i>n, [0$2], b(n-i, i));
             b(n, i-1) +g +[0, `if`(i=7, g[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012
  • Mathematica
    << DiscreteMath`Combinatorica`; Table[ Count[ Flatten[ Partitions[n]], 7], {n, 1, 52} ]
    Table[Count[Flatten[IntegerPartitions[n]],7],{n,55}] (* Harvey P. Dale, Feb 26 2015 *)
    b[n_, i_] := b[n, i] = Module[{g}, If[n == 0 || i == 1, {1, 0}, g = If[i > n, {0, 0}, b[n - i, i]]; b[n, i - 1] + g + {0, If[i == 7, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
  • PARI
    x='x+O('x^50); concat([0, 0, 0, 0, 0, 0], Vec(x^7/(1 - x^7) * prod(k=1, 50, 1/(1 - x^k)))) \\ Indranil Ghosh, Apr 06 2017

Formula

a(n) = A181187(n,7) - A181187(n,8). - Omar E. Pol, Oct 25 2012
a(n) ~ exp(Pi*sqrt(2*n/3)) / (14*Pi*sqrt(2*n)) * (1 - 85*Pi/(24*sqrt(6*n)) + (85/48 + 4873*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016
G.f.: x^7/(1 - x^7) * Product_{k>=1} 1/(1 - x^k). - Ilya Gutkovskiy, Apr 06 2017
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