A116646 -id:A116646 - cp's OEIS frontend

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.

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

A116644 Triangle read by rows: T(n,k) is the number of partitions of n having exactly k doubletons (n>=0, k>=0). By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] has two doubletons, shown between parentheses).

Original entry on oeis.org

1, 1, 1, 1, 3, 3, 2, 5, 2, 8, 2, 1, 10, 5, 13, 8, 1, 20, 9, 1, 26, 12, 4, 33, 21, 2, 46, 25, 5, 1, 58, 37, 6, 75, 48, 11, 1, 101, 59, 16, 125, 84, 19, 3, 157, 115, 23, 2, 206, 135, 39, 5, 253, 187, 46, 4, 317, 238, 63, 8, 1, 403, 292, 90, 7, 494, 382, 108, 17, 1, 608, 490, 139, 18

Offset: 0

Emeric Deutsch, Feb 20 2006

Apparently, rows n with p(p+1)<=n<(p+1)(p+2) have at most p+1 terms. Row sums are the partition numbers (A000041). T(n,0)=A116645(n). Sum(k*T(n,k),k>=0)=A116646(n).

			T(6,2) = 1 because [2,2,1,1] is the only partition of 6 with 2 doubletons.
Triangle starts:
1;
1;
1,  1;
3;
3,  2;
5,  2;
8,  2, 1;
10, 5;
13, 8, 1;

Alois P. Heinz, Rows n = 0..614, flattened

Cf. A000041, A116645, A116646.

g:=product(1+x^j+t*x^(2*j)+x^(3*j)/(1-x^j),j=1..35): gser:=simplify(series(g,x=0,35)): P[0]:=1: for n from 1 to 24 do P[n]:=coeff(gser,x^n) od: for n from 0 to 24 do seq(coeff(P[n],t,j),j=0..degree(P[n])) od; # sequence given in triangular form

G.f.: G(t,x) = product(1+x^j+tx^(2j)+x^(3j)/(1-x^j), j=1..infinity).

A117524 Total number of parts of multiplicity 3 in all partitions of n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 3, 3, 7, 8, 13, 17, 25, 32, 48, 59, 83, 108, 145, 183, 247, 310, 406, 512, 659, 824, 1055, 1307, 1651, 2047, 2558, 3146, 3913, 4788, 5904, 7202, 8821, 10707, 13054, 15770, 19118, 23027, 27775, 33312, 40029, 47835, 57231, 68182, 81261

Offset: 1

Vladeta Jovovic, Apr 26 2006

			a(9) = 7 because among the 30 (=A000041(9)) partitions of 9 only [6,(1,1,1)],[4,2,(1,1,1)],[(3,3,3)],[3,3,(1,1,1)],[3,(2,2,2)],[(2,2,2),(1,1,1)] contain parts of multiplicity 3 and their total number is 7 (shown between parentheses)

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A024786, A116646. Column k=3 of A197126.

g:=(x^3/(1-x^3)-x^4/(1-x^4))/product(1-x^i,i=1..65): gser:=series(g,x=0,62): seq(coeff(gser,x,n),n=1..58); # Emeric Deutsch, Apr 29 2006

G.f. for total number of parts of multiplicity m in all partitions of n is (x^m/(1-x^m)-x^(m+1)/(1-x^(m+1)))/Product(1-x^i,i=1..infinity).
a(n) = Sum(k*A118806(n,k), k>=0). - Emeric Deutsch, Apr 29 2006
a(n) ~ exp(Pi*sqrt(2*n/3)) / (24*Pi*sqrt(2*n)). - Vaclav Kotesovec, May 24 2018

A213180 Sum over all partitions lambda of n of Sum_{p:lambda} p^m(p,lambda), where m(p,lambda) is the multiplicity of part p in lambda.

Original entry on oeis.org

0, 1, 3, 7, 16, 28, 59, 91, 170, 269, 450, 655, 1162, 1602, 2527, 3793, 5805, 8034, 12660, 17131, 26484, 37384, 53738, 73504, 114683, 153613, 221225, 313339, 453769, 609179, 927968, 1223909, 1804710, 2522264, 3539835, 4855420, 7439870, 9765555, 14009545

Offset: 0

Alois P. Heinz, Feb 27 2013

nonn

			a(6) = 59: (1^6) + (2+1^4) + (2^2+1^2) + (2^3) + (3+1^3) + (3+2+1) + (3^2) + (4+1^2) + (4+2) + (5+1) + (6) = 1+3+5+8+4+6+9+5+6+6+6 = 59.

Alois P. Heinz, Table of n, a(n) for n = 0..5000

Cf. A000070 (Sum 1), A006128 (Sum m), A014153 (Sum p), A024786 (Sum floor(1/m)), A066183 (Sum p^2*m), A066186 (Sum p*m), A073336 (Sum floor(m/p)), A116646 (Sum delta(m,2)), A117524 (Sum delta(m,3)), A103628 (Sum delta(m,1)*p), A117525 (Sum delta(m,2)*p), A197126, A213191.

b:= proc(n, p) option remember; `if`(n=0, [1, 0], `if`(p<1, [0, 0],
      add((l->`if`(m=0, l, l+[0, l[1]*p^m]))(b(n-p*m, p-1)), m=0..n/p)))
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=0..40);

b[n_, p_] := b[n, p] = If[n==0, {1, 0}, If[p<1, {0, 0}, Sum[Function[l, If[m==0, l, l+{0, l[[1]]*p^m}]][b[n-p*m, p-1]], {m, 0, n/p}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 15 2017, translated from Maple *)

From Vaclav Kotesovec, May 24 2018: (Start)
a(n) ~ c * 3^(n/3), where
c = 5.0144820680945600131204662934686439430547... if mod(n,3)=0
c = 4.6144523178014379613985400559486878971522... if mod(n,3)=1
c = 4.5237761454818383598444208605033385016299... if mod(n,3)=2
(End)

A316861 a(n) = Sum_{p in P} y(1)*y(2), where P is the set of partitions of n, and y(k) is the number of parts with multiplicity at least k in p.

Original entry on oeis.org

0, 0, 1, 1, 4, 7, 13, 22, 38, 58, 93, 139, 208, 302, 438, 616, 869, 1200, 1650, 2239, 3026, 4038, 5374, 7081, 9292, 12103, 15704, 20236, 25992, 33191, 42237, 53490, 67524, 84860, 106341, 132736, 165212, 204928, 253518, 312629, 384585, 471734, 577276, 704584, 858078

Offset: 0

Emily Anible, Jul 15 2018

nonn

Also (1/2)*Sum_{p in P} H(1)*H(2), where P is the set of partitions of n, and H(k) is the number of k-hooks in p.

			For n=6, we sum over the partitions of 6. For each partition, we count the parts with multiplicity at least one, and those of at least two.
6............y(1)*y(2) = 1*0 = 0
5,1..........y(1)*y(2) = 2*0 = 0
4,2..........y(1)*y(2) = 2*0 = 0
4,1,1........y(1)*y(2) = 2*1 = 2
3,3..........y(1)*y(2) = 1*1 = 1
3,2,1........y(1)*y(2) = 3*0 = 0
3,1,1,1......y(1)*y(2) = 2*1 = 2
2,2,2........y(1)*y(2) = 1*1 = 1
2,2,1,1......y(1)*y(2) = 2*2 = 4
2,1,1,1,1....y(1)*y(2) = 2*1 = 2
1,1,1,1,1,1..y(1)*y(2) = 1*1 = 1
--------------------------------
Total.........................13

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

Cf. A000097, A116646, A000070, A024786.

b:= proc(n, i, x, y) option remember;
     `if`(n=0, x*y, `if`(i<1, 0, add(b(n-i*j, i-1,
     `if`(j>0, 1, 0)+x, `if`(j>1, 1, 0)+y), j=0..n/i)))
    end:
a:= n-> b(n$2, 0$2):
seq(a(n), n=0..55);  # Alois P. Heinz, Jul 30 2018

Array[Total[
   Count[Split@#, (_?(Length@# >= 1 &))] Count[
       Split@#, (_?(Length@# >= 2 &))] & /@
    IntegerPartitions[#]] &, 50]
(* Second program: *)
b[n_, i_, x_, y_] := b[n, i, x, y] = If[n == 0, x*y, If[i < 1, 0, Sum[b[n - i*j, i - 1, If[j > 0, 1, 0] + x, If[j > 1, 1, 0] + y], {j, 0, n/i}]]];
a[n_] := b[n, n, 0, 0];
a /@ Range[0, 55] (* Jean-François Alcover, Sep 16 2019, after Alois P. Heinz *)

seq(n)={Vec(x*(1 + x^2 + x^3)/((1 - x)^2*(1 + x)*(1 + x + x^2)*prod(i=1, n-1, 1 - x^i + O(x^n))) + O(x^n), -n)} \\ Andrew Howroyd, Jul 15 2018

G.f.: (q^3/((1-q)(1-q^2)) + q^2/(1-q^2) - q^3/(1-q^3))*Product_{j>=1} 1/(1-q^j).
a(n) = A000097(n+3) + A116646(n).
In general, Sum_{n>=0} q^n Sum_{p in P} y(s)*y(t) for s < t is given by (q^(s+t)/((1-q^s)(1-q^t)) + q^t/(1-q^t) - q^(s+t)/(1-q^(s+t))) * Product_{j>=1} 1/(1-q^j).

cp's OEIS Frontend

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.

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A116644 Triangle read by rows: T(n,k) is the number of partitions of n having exactly k doubletons (n>=0, k>=0). By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] has two doubletons, shown between parentheses).

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A117524 Total number of parts of multiplicity 3 in all partitions of n.

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A213180 Sum over all partitions lambda of n of Sum_{p:lambda} p^m(p,lambda), where m(p,lambda) is the multiplicity of part p in lambda.

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A316861 a(n) = Sum_{p in P} y(1)*y(2), where P is the set of partitions of n, and y(k) is the number of parts with multiplicity at least k in p.

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