A024786 -id:A024786 - cp's OEIS frontend

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

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

Clark Kimberling

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

Alois P. Heinz, Table of n, a(n) for n = 1..1000
David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.

Cf. A000041, A066633, A024786, A024787, A024788, A024789, A024791, A024792, A024793, A024794.

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

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 *)

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

Clark Kimberling

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

Alois P. Heinz, Table of n, a(n) for n = 1..1000
David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.
Joseph Vandehey, Digital problems in the theory of partitions, Integers (2024) Vol. 24A, Art. No. A18. See p. 3.

Cf. A066633, A000070(n-1), A024786, A024787, A024788, A024789, A024790, A024791, A024792, A024793, A000041.

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

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 *)

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

A197126 Triangle T(n,k), n>=1, 1<=k<=n, read by rows: T(n,k) is the number of cliques of size k in all partitions of n.

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 4, 2, 0, 1, 8, 2, 1, 0, 1, 11, 4, 2, 1, 0, 1, 19, 5, 3, 1, 1, 0, 1, 26, 10, 3, 3, 1, 1, 0, 1, 41, 11, 7, 3, 2, 1, 1, 0, 1, 56, 20, 8, 5, 3, 2, 1, 1, 0, 1, 83, 25, 13, 6, 5, 2, 2, 1, 1, 0, 1, 112, 38, 17, 11, 5, 5, 2, 2, 1, 1, 0, 1, 160, 49, 25, 13, 9, 5, 4, 2, 2, 1, 1, 0, 1

Offset: 1

Alois P. Heinz, Oct 10 2011

All parts of a number partition with the same value form a clique. The size of a clique is the number of elements in the clique.

			T(4,1) = 4: [1,1,(2)], [(1),(3)], [(4)].
T(8,3) = 3: [1,1,(2,2,2)], [(1,1,1),2,3], [(1,1,1),5].
T(12,4) = 11: [(1,1,1,1),(2,2,2,2)], [1,(2,2,2,2),3], [(1,1,1,1),2,3,3], [(3,3,3,3)], [(1,1,1,1),2,2,4], [(2,2,2,2),4], [(1,1,1,1),4,4], [(1,1,1,1),3,5], [(1,1,1,1),2,6], [(1,1,1,1),8].  Here the first partition contains 2 cliques.
Triangle begins:
   1;
   1,  1;
   3,  0, 1;
   4,  2, 0, 1;
   8,  2, 1, 0, 1;
  11,  4, 2, 1, 0, 1;
  19,  5, 3, 1, 1, 0, 1;
  26, 10, 3, 3, 1, 1, 0, 1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Abdulaziz M. Alanazi and Augustine O. Munagi, On partition configurations of Andrews-Deutsch, Integers 17 (2017), #A7.

Columns k=1-10 give: A024786(n+1), A116646, A117524, A222704, A222705, A222706, A222707, A222708, A222709, A222710.
Row sums give: A000070(n-1). Diagonal gives: A000012.  Limit of reversed rows: T(2*n+1,n+1) = A002865(n).
Cf. A213180.

b:= proc(n, p, k) option remember; `if`(n=0, [1, 0], `if`(p<1, [0, 0],
      add((l->`if`(m=k, l+[0, l[1]], l))(b(n-p*m, p-1, k)), m=0..n/p)))
    end:
T:= (n, k)-> b(n, n, k)[2]:
seq(seq(T(n, k), k=1..n), n=1..20);

Table[CoefficientList[ 1/q* Tr[Flatten[q^Map[Length, Split /@ IntegerPartitions[n], {2}]]], q], {n, 24}] (* Wouter Meeussen, Apr 21 2012 *)
b[n_, p_, k_] := b[n, p, k] = If[n == 0, {1, 0}, If[p < 1, {0, 0}, Sum[ Function[l, If[m == k, l + {0, l[[1]]}, l]][b[n - p*m, p - 1, k]], {m, 0, n/p}]]]; T[n_, k_] := b[n, n, k][[2]]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 20}] // Flatten (* Jean-François Alcover, Aug 29 2016, after Alois P. Heinz *)

G.f. of column k: (x^k/(1-x^k)-x^(k+1)/(1-x^(k+1)))/Product_{j>0}(1-x^j).

Column k is asymptotic to exp(Pi*sqrt(2*n/3)) / (k*(k+1)*Pi*2^(3/2)*sqrt(n)). - Vaclav Kotesovec, May 24 2018

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

Clark Kimberling

nonn

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

Alois P. Heinz, Table of n, a(n) for n = 1..1000
David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.

Cf. A066633, A024786, A024787, A024788, A024789, A024790, A024792, A024793, A024794.

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

<< 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 *)

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

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

A024792 Number of 8's in all partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 15, 23, 31, 44, 59, 82, 108, 146, 191, 254, 328, 429, 549, 709, 900, 1148, 1446, 1829, 2286, 2865, 3559, 4427, 5465, 6752, 8288, 10178, 12429, 15175, 18442, 22404, 27102, 32767, 39473, 47516, 57012, 68349, 81703, 97579, 116236

Offset: 1

Clark Kimberling

The sums of eight successive terms give A000070. - Omar E. Pol, Jul 12 2012

a(n) is also the difference between the sum of 8th largest and the sum of 9th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012

Alois P. Heinz, Table of n, a(n) for n = 1..1000
David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.

Cf. A000041, A066633, A024786, A024787, A024788, A024789, A024790, A024791, A024793, A024794.

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=8, g[1], 0)]
      fi
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012

Table[ Count[ Flatten[ IntegerPartitions[n]], 8], {n, 1, 53} ]
(* 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 == 8, 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 *)

a(n) = A181187(n,8) - A181187(n,9). - Omar E. Pol, Oct 25 2012
From Peter Bala, Dec 26 2013: (Start)
a(n+8) - a(n) = A000041(n). a(n) + a(n+4) = A024788(n).
a(n) + a(n+2) + a(n+4) + a(n+6) = A024786(n).
O.g.f.: x^8/(1 - x^8) * product {k >= 1} 1/(1 - x^k) = x^8 + x^9 + 2*x^10 + 3*x^11 + ....
Asymptotic result: log(a(n)) ~ 2*sqrt(Pi^2/6)*sqrt(n) as n -> inf. (End)
a(n) ~ exp(Pi*sqrt(2*n/3)) / (16*Pi*sqrt(2*n)) * (1 - 97*Pi/(24*sqrt(6*n)) + (97/48 + 6337*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016

A024793 Number of 9's in all partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 15, 22, 31, 43, 58, 80, 106, 142, 187, 246, 319, 416, 533, 685, 872, 1108, 1397, 1762, 2204, 2755, 3426, 4251, 5250, 6476, 7950, 9746, 11905, 14514, 17638, 21403, 25888, 31265, 37661, 45288, 54329, 65079, 77775

Offset: 1

Clark Kimberling

The sums of nine successive terms give A000070. - Omar E. Pol, Jul 12 2012

a(n) is also the difference between the sum of 9th largest and the sum of 10th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012

Alois P. Heinz, Table of n, a(n) for n = 1..1000
David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.

Cf. A000041, A066633, A024786, A024787, A024788, A024789, A024790, A024791, A024792, A024794.

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=9, g[1], 0)]
      fi
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012

Table[ Count[ Flatten[ IntegerPartitions[n]], 9], {n, 1, 55} ]
(* 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 == 9, 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 *)

a(n) = A181187(n,9) - A181187(n,10). - Omar E. Pol, Oct 25 2012
From Peter Bala, Dec 26 2013: (Start)
a(n+9) - a(n) = A000041(n).
a(n) + a(n+3) + a(n+6) = A024787(n).
O.g.f.: x^9/(1 - x^9) * product {k >= 1} 1/(1 - x^k) = x^9 + x^10 + 2*x^11 + 3*x^12 + ....
Asymptotic result: log(a(n)) ~ 2*sqrt(Pi^2/6)*sqrt(n) as n -> inf. (End)
a(n) ~ exp(Pi*sqrt(2*n/3)) / (18*Pi*sqrt(2*n)) * (1 - 109*Pi/(24*sqrt(6*n)) + (109/48 + 7993*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016

A212551 Number of partitions T(n,k) of n containing at least one other part m-k if m is the largest part; triangle T(n,k), n>=0, 0<=k<=n.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 0, 0, 2, 3, 1, 1, 0, 0, 4, 3, 3, 1, 1, 0, 0, 4, 6, 4, 3, 1, 1, 0, 0, 7, 7, 7, 4, 3, 1, 1, 0, 0, 8, 11, 9, 8, 4, 3, 1, 1, 0, 0, 12, 13, 15, 10, 8, 4, 3, 1, 1, 0, 0, 14, 20, 18, 17, 11, 8, 4, 3, 1, 1, 0, 0

Offset: 0

Alois P. Heinz, May 20 2012

Reversed rows converge to A024786.

			T(4,0) = 2: [1,1,1,1], [2,2].
T(4,1) = 1: [2,1,1].
T(5,1) = 3: [2,1,1,1], [2,2,1], [3,2].
T(6,2) = 3: [3,1,1,1], [3,2,1], [4,2].
T(7,2) = 4: [3,1,1,1,1], [3,2,1,1], [3,3,1], [4,2,1].
T(8,4) = 3: [5,1,1,1], [5,2,1], [6,2].
Triangle T(n,k) begins:
1;
0, 0;
1, 0, 0;
1, 1, 0, 0;
2, 1, 1, 0, 0;
2, 3, 1, 1, 0, 0;
4, 3, 3, 1, 1, 0, 0;
4, 6, 4, 3, 1, 1, 0, 0;
7, 7, 7, 4, 3, 1, 1, 0, 0;

Alois P. Heinz, Rows n = 0..140, flattened
Wikipedia, Kronecker delta

Columns k=0-10 give: A002865, A083751(n+1), A119907, A212543, A212544, A212545, A212546, A212547, A212548, A212549, A212550.
Row sums give A000070(n-2) for n>1.
Cf. A024786.

b:= proc(n, i) option remember;
      `if`(n=0 or i=1, 1, b(n, i-1)+`if`(i>n, 0, b(n-i, i)))
    end:
T:= (n, k)-> `if`(n=0 and k=0, 1,
    add(b(n-2*m-k, min(n-2*m-k, m+k)), m=1..(n-k)/2)):
seq(seq(T(n, k), k=0..n), n=0..14);

b[n_, i_] := b[n, i] = If[n == 0 || i == 1, 1, b[n, i-1] + If[i > n, 0, b[n-i, i]]]; t[n_, k_] := If[n == 0 && k == 0, 1, Sum[b[n-2*m-k, Min[n-2*m-k, m+k]], {m, 1, (n-k)/2}]]; Table[Table[t[n, k], {k, 0, n}], {n, 0, 14}] // Flatten (* Jean-François Alcover, Dec 13 2013, translated from Maple *)

G.f. of column k: delta_{0,k} + Sum_{i>0} x^(2*i+k) / Product_{j=1..k+i} (1-x^j), where delta is the Kronecker delta.

A060244 Triangle a(n,k) of bipartite partitions of n objects into parts >1, k of which are black.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 2, 2, 3, 4, 4, 3, 2, 4, 5, 8, 8, 8, 5, 4, 4, 7, 11, 13, 13, 11, 7, 4, 7, 11, 19, 22, 26, 22, 19, 11, 7, 8, 15, 26, 35, 40, 40, 35, 26, 15, 8, 12, 22, 41, 54, 69, 70, 69, 54, 41, 22, 12, 14, 30, 56, 81, 104, 116, 116, 104, 81, 56, 30, 14, 21, 42

Offset: 0

N. J. A. Sloane, Mar 22 2001

			Series ends ... + 2*x^5 + 3*x^4*y + 4*x^3*y^2 + 4*x^2*y^3 + 3*x*y^4 + 2*y^5 + 2*x^4 + 2*x^3*y + 3*x^2*y^2 + 2*x*y^3 + 2*y^4 + x^3 + x^2*y + x*y^2 + y^3 + x^2 + x*y + y^2 + 1.
1;
0, 0;
1, 1, 1;
1, 1, 1, 1;
2, 2, 3, 2, 2;
...

P. A. MacMahon, Memoir on symmetric functions of the roots of systems of equations, Phil. Trans. Royal Soc. London, 181 (1890), 481-536; Coll. Papers II, 32-87.

Columns 0-6: A002865, A000041, A024786, A291553, A291589, A291590, A291596.
Row sums: A060285.
Cf. A005380, A054225.

read transforms; t1 := mul( mul( 1/(1-x^(i-j)*y^j), j=0..i), i=2..11): SERIES2(t1,x,y,7);

max = 12; gf = Product[1/(1 - x^(i - j)*y^j), {i, 2, max}, {j, 0, i}]; se = Series[gf, {x, 0, max}, {y, 0, max}] // Normal; t[n_, k_] := SeriesCoefficient[se, {x, 0, n}, {y, 0, k}]; Flatten[ Table[ t[n - k, k], {n, 0, max}, {k, 0, n}]] (* Jean-François Alcover, after Maple *)

G.f.: Product_{ i=2..infinity, j=0..i} 1/(1-x^(i-j)*y^j).

More terms from Vladeta Jovovic, Mar 23 2001

Edited by Christian G. Bower, Jan 08 2004

A103628 Total sum of parts of multiplicity 1 in all partitions of n.

Original entry on oeis.org

0, 1, 2, 6, 10, 21, 33, 59, 89, 145, 212, 325, 463, 680, 948, 1348, 1845, 2558, 3446, 4681, 6219, 8306, 10901, 14352, 18632, 24230, 31151, 40077, 51074, 65088, 82290, 103986, 130517, 163679, 204078, 254174, 314975, 389839, 480369, 591133, 724600, 886965

Offset: 0

Vladeta Jovovic, Mar 25 2005

Total number of parts of multiplicity 1 in all partitions of n is A024786(n+1).

Equals A000041 convolved with A026741. - Gary W. Adamson, Jun 11 2009

			Partitions of 4 are [1,1,1,1], [1,1,2], [2,2], [1,3], [4] and a(4) = 0 + 2 + 0 + (1+3) + 4 = 10.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A026741. - Gary W. Adamson, Jun 11 2009

Column k=1 of A222730. - Alois P. Heinz, Mar 03 2013

gf:=x*(1+x+x^2)/(1-x^2)^2/product((1-x^k), k=1..500): s:=series(gf, x, 100): for n from 0 to 60 do printf(`%d,`,coeff(s, x, n)) od: # James Sellers, Apr 22 2005
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, [1, 0],
      `if`(i<1, [0, 0], add((l->`if`(j=1, [l[1],
       l[2]+l[1]*i], l))(b(n-i*j, i-1)), j=0..n/i)))
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=0..50);  # Alois P. Heinz, Feb 03 2013

b[n_, p_] := b[n, p] = If[n == 0 && p == 0, {1, 0}, If[p == 0, Array[0&, n+2], Sum[Function[l, ReplacePart[l, m+2 -> p*l[[1]] + l[[m+2]]]][Join[b[n-p*m, p-1], Array[0&, p*m]]], {m, 0, n/p}]]]; a[n_] := b[n, n][[3]]; a[0] = 0; Table[a[n], {n, 0, 50}]  (* Jean-François Alcover, Jan 24 2014, after Alois P. Heinz *)

G.f.: x*(1+x+x^2)/(1-x^2)^2 /Product_{k>0}(1-x^k).
a(n) = A066186(n) - A194544(n). - Omar E. Pol, Nov 20 2011
a(n) = 3*A014153(n)/4 - 3*A000070(n)/4 - A270143(n+1)/4 + A087787(n)/4. - Vaclav Kotesovec, Nov 05 2016
a(n) ~ 3^(3/2) * exp(Pi*sqrt(2*n/3)) / (8*Pi^2) * (1 - Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Nov 05 2016

More terms from James Sellers, Apr 22 2005

A173238 Triangle by columns, A000041 in every column shifted down twice for columns > 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 5, 2, 1, 7, 3, 1, 11, 5, 2, 1, 15, 7, 3, 1, 22, 11, 5, 2, 1, 30, 15, 7, 3, 1, 42, 22, 11, 5, 2, 1, 56, 30, 15, 7, 3, 1, 77, 42, 22, 11, 5, 2, 1, 101, 56, 30, 15, 7, 3, 1, 135, 77, 42, 22, 11, 5, 2, 1, 176, 101, 56, 30, 15, 7, 3, 1

Offset: 0

Gary W. Adamson, Feb 13 2010

Row sums = A024786 starting with A024786(2): (1, 1, 3, 4, 8, 11, 19, 26, ...) = number of 2's in all partitions of n.
A173238 as an infinite lower triangular matrix * [1, 2, 3, ...] = A103650.
Let the triangle = M. Then Lim_{n->inf} M^n = (1, 1, 3, 4, 10, 13, 26, ...), the left-shifted vector considered as a sequence = A092119, the Euler transform of the ruler sequence, A001511.
Given P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), then P(x) = A(x)/A(x^2), where A(x) = polcoeff A092119: (1 + x + 3x^2 + 4x^3 + ...).
Conjectures: Given the infinite set of triangles with A000041 in every column shifted down 0, 1, 2, ... n times, row sums of n-th triangle (where A173238 = 2nd in the set) = the numbers of n's in all partitions of n. E.g., row sums = of A173238 = A024786, the numbers of 2's in all partitions of n.
Similarly, row sums of triangle A173239 with columns >0 shifted down thrice = numbers of 3's in all partitions of n, and so on. Refer to comments in A000041 regarding the numbers of 1's in partitions of n.
...
Second conjecture: Given the infinite set of analogous triangles with columns shifted down 2, 3, 4, ..., k times, we let such triangles = T(k) and perform lim_{n->inf} T^n(k), obtaining the left-shifted vectors considered as sequences. The conjecture states that the infinite set of such left-shifted vectors = the Euler transform of the infinite set of Ruler functions starting with the ruler function for k=2 = A001511: (1, 2, 1, 3, 1, 2, 1, ...)
To obtain the k-th ruler functions, begin with the natural numbers, 1,...2,...3,...4,...5,...6,...7,...8,...9,...; then for k = 2 we get A001511: 1,...2,...1,...3,...1,...2,...1,...4,...1,...; by finding the highest exponent of k dividing n, then adding 1. Similarly, for k = 2, we obtain A051064: 1,...1,...2,...1,...1,...2,...1,...1,...3,...
Next, we obtain the Euler transforms of the ruler functions (e.g., Euler transform of A001511 = A092119: (1, 1, 3, 4, 10, 13, 26, ...), noting that A092119 is lim_{n->inf} A173238^n, the left-shifted vector.
...
Third conjecture: Let P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + ...), and A(k)(x) = the Euler transform of k-th ruler sequence, (k=2,3,...). Then P(x) = A(k)(x) / A(k)(x^k).
Examples: for k=2, A(x) = A092119: (1, 1, 3, 4, 10, 13, ...), then P(x) = (1 + x + 2x^2 + 3x^3 + ...) = (1 + x + 3x^2 + 4x^3 + ...) / (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...). For k=3 relating to triangle A173238, the left-shifted vector = the Euler transform of A051064 = A(x) for k=3, then P(x) = A(x) / A(x^3).
The conjecture extends the analogous conclusions to all k.
From Gary W. Adamson, Feb 25 2010: (Start)
Proof of second conjecture received from Helmut Prodinger 02/28/10 with a summary by R. J. Mathar:
Consider Product_{n>=0} z^(t^n)/(1-z^(t^n)) = Sum_{k>=1} (1+v_t(k))z^k where v_t(n) is the number of trailing zeros in the t-ary expansion of n, and its Euler transform A(z) = product_{k >= 1} 1/(1-z^k)^{1+v_t(k)}, then A(z)/A(z^t) = product_{k >= 1} 1/(1-z^k) is the partition generating function.
Here is the proof: A(z)/A(z^t) = Product_{k>=1} (1-z^(tk))^(1+v_t(k))/(1-z^k)^(1+v_t(k))
= Product_{k>=1} (1-z^(tk))^(v_t(tk))/(1-z^k)^(1+v_t(k))
= Product_{k>=1} (1-z^k)^(v_t(k))/(1-z^k)^(1+v_t(k)) (*)
= Product_{k>=1} 1/(1-z^k)
as desired. Notice that for (*), that v_t(n)=0 if n is not divisible by t. [Helmut Prodinger, hproding(AT)sun.ac.za, Feb 28 2010] (End)

			First few rows of the triangle:
    1;
    1;
    2,   1;
    3,   1;
    5,   2,  1;
    7,   3,  1;
   11,   5,  2,  1;
   15,   7,  3,  1;
   22,  11,  5,  2,  1;
   30,  15,  7,  3,  1;
   42,  22, 11,  5,  2, 1;
   56,  30, 15,  7,  3, 1;
   77,  42, 22, 11,  5, 2, 1;
  101,  56, 30, 15,  7, 3, 1;
  135,  77, 42, 22, 11, 5, 2, 1;
  176, 101, 56, 30, 15, 7, 3, 1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..9999 (rows 0 <= n <= 199, flattened)

Cf. A000041, A001511, A092119, A173239.

Table[PartitionsP[n - 2 k], {n, 17}, {k, Floor[n/2]}] // Flatten (* Michael De Vlieger, Nov 23 2021 *)

T(n,k) = A000041(n-2*k) for k=0..floor(n/2).

a(24), a(25) corrected by Georg Fischer, Nov 23 2021

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