Gregory L. Simay - cp's OEIS frontend

A308654 The overpartition triangle: T(n,k) is the number of overpartitions of n with exactly k positive integer parts, 0 <= k <= n.

1, 0, 2, 0, 2, 2, 0, 2, 4, 2, 0, 2, 6, 4, 2, 0, 2, 8, 8, 4, 2, 0, 2, 10, 14, 8, 4, 2, 0, 2, 12, 20, 16, 8, 4, 2, 0, 2, 14, 28, 26, 16, 8, 4, 2, 0, 2, 16, 38, 40, 28, 16, 8, 4, 2, 0, 2, 18, 48, 60, 46, 28, 16, 8, 4, 2, 0, 2, 20, 60, 84, 72, 48, 28, 16, 8, 4, 2

Offset: 0

Gregory L. Simay, Jun 14 2019

T(n,0) = A000007(n).
T(n,1) = A040000(n) for n > 0.
T(n,2) = A005843(n-1).
T(n,3) = 2*A007980(n-3).
T(n,4) = 2*A061866(n-1).
T(n,5) = 2*A091773(n-5).
Conjecture: T(n,k) = 2*(the associated Poincaré series). If T(n,1) were 1 for n>0, then T(n, k>1) would be a Poincaré series.

			T(5,3) = 8 and counts the overpartitions 3,1,1; 3',1,1; 3,1',1; 3',1',1; 2,2,1; 2',2,1; 2,2,1' and 2',2,1'.
T(16,5) = 404 = T(11,5) + 2*( T(11,4) + T(11,3) + T(11,2) + T(11,1)) = 72 + 2*(84 + 60 + 20 + 2) = 404.
T(16, 5) = T(15,4) + T(11,4) + T(10,4) + T(6,4) + T(5,4) = 248 + 84 + 60 + 8 + 4 = 404.
T(9,1) + T(8,2) + T(7,3) + T(6,4) + T(6,5)= 2 + 14 + 20 + 8 + 2 = 46 =A300415(10).
Triangle: T(n,k) begins:
  1;
  0, 2;
  0, 2,  2;
  0, 2,  4,  2;
  0, 2,  6,  4,  2;
  0, 2,  8,  8,  4,  2;
  0, 2, 10, 14,  8,  4,  2;
  0, 2, 12, 20, 16,  8,  4,  2;
  0, 2, 14, 28, 26, 16,  8,  4, 2;
  0, 2, 16, 38, 40, 28, 16,  8, 4, 2;
  0, 2, 18, 48, 60, 46, 28, 16, 8, 4, 2;
  ...

Row sums give A015128.
Main diagonal T(n,n) gives A040000.
Cf. A005843, A007980, A061866, A091773, A235792, A300415.

b:= proc(n,i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(
      expand(`if`(j>0, 2*x^j, 1)*b(n-i*j, i-1)), j=0..n/i)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
seq(T(n), n=0..14);  # Alois P. Heinz, Jun 15 2019

b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[Expand[If[j > 0, 2*x^j, 1]*b[n - i*j, i - 1]], {j, 0, n/i}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][ b[n, n]];
T /@ Range[0, 14] // Flatten (* Jean-François Alcover, Dec 10 2019, after Alois P. Heinz *)

Sum_{k=0..n} T(n,k) = A015128(n), the number of overpartitions of n.
If k > n, T(n,k) = 0.
If n >= k > n/2, T(n,k) = 2*A015128(n-k).
Conjecture: T(n,k) = T(n-k, k) + 2*(T(n-k, k-1) + ... + T(n-k, 1)).
Conjecture: T(n,k) = T(n-1, k-1) + T(n-k, k-1) + T(n-k-1, k-1) + T(n-2k, k-1) + T(n-2k-1) + ...
Conjecture: T(n,1) + T(n-1,2) + ... + T(n-floor(n/2),floor(n/2)) = A300415(n+1).
T(n,2) = 2n - 2.
Conjecture: g.f. T(n,k) = 2*(1+x)(1+x^2)...(1+x^(k-1))/((1-x)...(1-x^k)).
Sum_{k=1..n} k * T(n,k) = A235792(n). - Alois P. Heinz, Jun 15 2019

A297120 Number of compositions derived from the overpartitions of n.

Original entry on oeis.org

1, 2, 5, 14, 36, 92, 234, 586, 1452, 3562, 8674, 20956, 50290, 119922, 284308, 670458, 1573250, 3674700, 8546282, 19796234, 45681908, 105041402, 240723618, 549919604, 1252492674, 2844551866, 6442833156, 14555300218, 32801922154, 73749649900, 165443000338

Offset: 0

Gregory L. Simay, Dec 25 2017

nonn

Start by enumerating the overpartitions of n, then allow the parts to vary their arrangements.

			The A015128(4) = 14 overpartitions of 4 are: 4; 4'; 3,1; 3,1'; 3'1; 3',1', 2,2; 2',2; 2,1,1; 2,1',1; 2',1,1; 2',1',1; 1,1,1,1; and 1',1,1,1.  The corresponding 36 compositions are 4; 4'; 3,1; 1,3; 3,1'; 1',3; 3',1; 1,3'; 3',1'; 1',3'; 2,2; 2,2'; 2',2; 2,1,1; 1,2,1; 1,1,2; 2,1,1'; 2,1',1; 1,2,1'; 1,1',2'; 1',1,2; 1',2,1; 2',1,1; 1,2',2; 1,1,2'; 2',1,1'; 2',1',1; 1,2',1'; 1,1',2'; 1',2',1; 1',1,2'; 1,1,1,1; 1,1,1,1'; 1,1,1',1; 1,1',1,1; and 1',1,1,1. Note: For a sequence of like parts p,p,...p, an overcomposition of n will only recognize p,p...p and p',p...,p; the p' is not allowed to be other than the initial p term.

Vaclav Kotesovec, Table of n, a(n) for n = 0..3000 (terms 0..1000 from Alois P. Heinz)

Cf. A015128, A236002.

b:= proc(n, i, p) option remember; `if`(n=0 or i=1, (p+n)!*
      (1+n)/n!, add(b(n-i*j, i-1, p+j)*(1+j)/j!, j=0..n/i))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..35);  # Alois P. Heinz, Dec 26 2017

b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1, (p + n)!*(n + 1)/n!, Sum[b[n - i*j, i - 1, p + j]*(j + 1)/j!, {j, 0, n/i}]];
a[n_] := b[n, n, 0];
Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Dec 27 2017, after Alois P. Heinz *)

{my(n=30); apply(p->subst(serlaplace(p), y, 1), Vec(prod(k=1, n, (1+y*x^k)*exp(y*x^k + O(x*x^n)))))} \\ Andrew Howroyd, Dec 26 2017

from sympy.core.cache import cacheit
from sympy import factorial
@cacheit
def b(n, i, p):  return factorial(p + n)*(n + 1)//factorial(n) if n==0 or i==1 else sum(b(n - i*j, i - 1, p + j)*(j + 1)//factorial(j) for j in range(n//i + 1))
def a(n): return b(n, n, 0)
print([a(n) for n in range(41)]) # Indranil Ghosh, Dec 29 2017, after Maple code

More terms from Alois P. Heinz, Dec 26 2017

A289249 Number of compositions of n if only the order of parts 1 and 2 matters.

Original entry on oeis.org

1, 1, 2, 4, 7, 12, 21, 35, 59, 98, 162, 266, 437, 713, 1163, 1893, 3077, 4995, 8105, 13139, 21293, 34492, 55858, 90438, 146406, 236974, 383538, 620703, 1004471, 1625447, 2630249, 4256087, 6886804, 11143447, 18030911, 29175137, 47206975, 76383199, 123591458

Offset: 0

Gregory L. Simay, Jun 29 2017

nonn

If only the order of parts 1 and 2 matters, then the remaining parts can be frozen "[]" in a partition subsequence; e.g., a(15) would count the sequence 5,4,3,2,1 twice: [5,4,3]2,1 and [5,4,3]1,2. (Also see example.)

			For n=6, the 21 sequences counted are [6]; [5],1; [4],2; [3,3], [4],1,1; [3],2,1; [3],1,2; 2,2,2; [3],1,1,1; 2,2,1,1; 2,1,2,1; 1,2,2,1; 1,2,1,2; 1,1,2,2; 2,1,1,2; 2,1,1,1,1; 1,2,1,1,1; 1,1,2,1,1; 1,1,1,2,1; 1,1,1,1,2; and 1,1,1,1,1,1.

Cf. A000041, A000045, A275388.

Table[PartitionsP[n] + Sum[Fibonacci[k] PartitionsP[n - 2 - k], {k, n - 2}], {n, 0, 50}] (* Indranil Ghosh, Jun 29 2017 *)

a275388(n)=sum(k=1, n, fibonacci(k)*numbpart(n - k));
a(n)=numbpart(n)+a275388(n - 2); \\ Indranil Ghosh, Jun 29 2017

from sympy import fibonacci, npartitions
def a(n): return npartitions(n) + sum([fibonacci(k)*npartitions(n - 2 - k) for k in range(1, n - 1)])
print([a(n) for n in range(51)]) # Indranil Ghosh, Jun 29 2017

a(n) = A000041(n) + A275388(n-2), the sum of the n-th partition number and the (n-2)th convolution of partition numbers with Fibonacci numbers. E.g., a(8) = 59 = A000041(8) + A275388(6) = 22 + 37 = 59.
a(n) = A275388(n+1) - A275388(n) - A275388(n-1) + A275388(n-2).
G.f.: (1/x)*(1-x)*(1-x^2)*(g.f. of A275388) =(1/x)*(1-x)*(1-x^2)*Sum_{k=1..n} A000045(k)*A000041(n-k).

More terms from Indranil Ghosh, Jun 29 2017

A285981 Number of Carlitz compositions having at least two identical parts.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 3, 10, 20, 44, 67, 149, 277, 528, 959, 1673, 3107, 5572, 9992, 17801, 31647, 55379, 98445, 173288, 305355, 536709, 943353, 1655316, 2900221, 5088098, 8916905, 15624332, 27368888, 47935241, 83939143, 146974040, 257277523, 450432510, 788487147

Offset: 0

Gregory L. Simay, Apr 29 2017

nonn

			For n=7, there are a(7) = 10 Carlitz compositions with at least two identical parts: 1,5,1; 2,3,2; 3,1,3; 1,2,1,3; 1,2,3,1; 1,3,1,2; 1,3,2,1; 2,1,3,1; 3,1,2,1 and 1,2,1,2,1.

Alois P. Heinz, Table of n, a(n) for n = 0..4100

Cf. A003242, A032020.

a(n) = A003242(n) - A032020(n).

More terms from Alois P. Heinz, Apr 29 2017

A276129 a(n) is the number of ordered ways to tile a strip of length n+2 with white tiles of odd lengths summing to length n and two red squares.

Original entry on oeis.org

1, 3, 6, 13, 27, 54, 106, 204, 387, 725, 1344, 2469, 4500, 8145, 14652, 26213, 46665, 82704, 145982, 256722, 449937, 786109, 1369494, 2379447, 4123944, 7130895, 12303714, 21186013, 36411399, 62466906, 106987282, 182946888, 312367887, 532587461, 906840060

Offset: 0

Gregory L. Simay, Aug 21 2016

nonn

a(n) is a specific case of b(r,n), the number of ordered ways to rearrange a tiling of length n + r, with odd(1,3,5...) white tiles summing to n and r red squares.
Define the following summation: b(0,r,n) = b(r,n); b(1,r,n) = b(r, n-2) + b(r, n-4) + b(r, n-6) + ..; b(s, r, n) = b(s-1, r, n-2) + b(s-1, r, n-4) + b(r-1, s, n-6) + ...
The number of compositions of n with exactly r even numbers is b(r, r, n-2r).
Except for the initial 1, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = (1 - S)^3. See A291219. - Clark Kimberling, Sep 04 2017

			Let 1,3 be the lengths of the odd tiles summing to 3 and let r,r be the two odd squares. Then the resulting number of compositions is a(3) = 13. The 6 compositions are 3,r,r; r,3,r; r,r,3; 1,1,1,r,r; 1,1,r,r,1; 1,r,r,1,1; r,r,1,1,1; 1,1,r,1,r; 1,r,1,r,1; r,1,r,1,1; r,1,1,r,1; 1,r,1,1,r; r,1,1,1,r.

Cf. A000045, A029907, A239242.

b:= proc(n, m) option remember;
      `if`(n+m=0, 1, `if`(m>0, b(n, m-1), 0)+
      add(`if`(j::odd, b(n-j, m), 0), j=1..n))
    end:
a:= n-> b(n, 2):
seq(a(n), n=0..50);  # Alois P. Heinz, Aug 29 2016

a[0] = 1; a[n_] := Sum[Binomial[n - 2*k + 2, 2]*Binomial[n - k - 1, k], {k, 0, (n - 1)/2}];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Dec 27 2017, after Andrew Howroyd *)

a(n)={if(n<=0, n==0, sum(k=0, (n-1)\2, binomial(n-2*k+2, 2)*binomial(n-k-1, k)))} \\ Andrew Howroyd, Dec 26 2017

a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-2*k+2, 2)*binomial(n-k-1, k) for n > 0. - Andrew Howroyd, Dec 26 2017
b(0,0)=1. For n>=1, b(0,n) = b(0,0,n) = the n-th Fibonacci number, A000045(n).
b(1,0)=1. For n>=1, b(1,n) = A239342(n+1).
b(2,n) = a(n) = a(n-1) + a(n-2) + A239342(n+1) + A239342(n-1).
G.f. for b(2,n): ((1-x^2)/(1-x-x^2))^3.
G.f. for b(r,n): ((1-x^2)/(1-x-x^2))^(r+1).
b(1,1,n) = A029907(n+1).
b(r,n) = b(r, n-1) + b(r, n-2) + b(r-1, n) - b(r-1, n-2).
b(r,r,n) = b(r-1, r-1, n) + b(r, r, n-1) + b(r, r, n-2).
G.f. for b(r,r,n): (1-x^2)/((1-x-x^2)^(r+1)).

More terms from Alois P. Heinz, Aug 29 2016

A275548 Number of compositions of n if only the order of the odd numbers matter.

Original entry on oeis.org

1, 1, 2, 3, 6, 9, 16, 25, 43, 68, 113, 181, 298, 479, 781, 1260, 2048, 3308, 5364, 8672, 14048, 22720, 36782, 59502, 96305, 155807, 252136, 407943, 660113, 1068056, 1728210, 2796266, 4524531, 7320797, 11845394, 19166191, 31011673, 50177864, 81189642, 131367506

Offset: 0

Gregory L. Simay, Aug 01 2016

nonn

The number of partitions of n = 2k with only even numbers is p(k) = A000041(k). The number of compositions of n with only odd numbers is F(n) = the n-th Fibonacci number = A000045(n). Enumerating a(n) is therefore a sum of products of partition numbers and Fibonacci numbers.

			The compositions enumerated by a(6) = 16 are (6),(5,1),(1,5),(4,2)=(2,4), (3,3), (4,1,1)=(1,4,1)=(1,1,4), (2,3,1)=(3,2,1)=(3,1,2), (2,1,3)=(1,2,3)=(1,3,2), (2,2,2), (3,1,1,1),(1,3,1,1),(1,1,3,1),(1,1,1,3), (2,2,1,1)=(2,1,2,1)=(2,1,1,2)=(1,2,1,2)=(1,1,2,2)=(1,2,2,1), (2,1,1,1,1)=(1,2,1,1,1)=(1,1,2,1,1)=(1,1,1,2,1)=(1,1,1,1,2), (1,1,1,1,1,1).
The compositions enumerated by a(5) = 9 are (5), (4,1)=(1,4), (3,2)=(2,3), (3,1,1), (1,3,1), (1,1,3), (2,2,1)=(2,1,2)=(1,2,2), (2,1,1,1)=(1,2,1,1)=(1,1,2,1)=(1,1,1,2), (1,1,1,1,1).

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

Cf. A000041, A000045, A275592.

b:= proc(n, i, p) option remember; (t-> `if`(n=0, p!,
      `if`(i<1, 0, add(b(n-i*j, i-1, p+`if`(t, j, 0))/
      `if`(t, j, 0)!, j=0..n/i))))(i::odd)
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..50);  # Alois P. Heinz, Aug 03 2016

b[n_, i_, p_] := b[n, i, p] = If[n == 0, p!, If[i < 1, 0, Sum[b[n - i*j, i - 1, p + If[#, j, 0]]/If[#, j, 0]!, {j, 0, n/i}]]]&[OddQ[i]];
a[n_] := b[n, n, 0];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, May 21 2018, after Alois P. Heinz *)
nmax = 40; CoefficientList[Series[1/(1 - x - x^2) * Product[1/(1 - x^(2*k)), {k, 2, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 02 2018 *)

a(2k+1) = Sum_{j=0..k} p(j)*F(2k + 1 - 2j), where  p(j) = A000041(j), the number of partitions of j, and F(j) = A000045(j), the j-th Fibonacci number.
a(2k) = p(k) + Sum_{j=0..(k-1)} p(j)*F(2k - 2j).
a(2k+1) = a(2k) + a(2k-1).
a(2k) = a(2k-1) + a(2k-2) + p(k) - p(k-1).
G.f.: 1/(1 - x - x^2) * Product_{n>=2} 1/(1 - x^(2*n)). - Peter Bala, Aug 03 2016 [corrected by Vaclav Kotesovec, Jun 02 2018]
a(n) ~ c * phi^n, where c = 1 / (sqrt(5) * QPochhammer[1/phi^2]) = 0.92890318501026782066... and phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 02 2018

A275592 Number of compositions of n if only the order of the even numbers matter.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 12, 16, 26, 35, 56, 74, 117, 154, 241, 317, 492, 645, 998, 1306, 2014, 2634, 4053, 5296, 8139, 10630, 16321, 21310, 32699, 42684, 65472, 85452, 131038, 171012, 262198, 342161, 524552, 684497, 1049300, 1369216, 2098849, 2738710, 4198011

Offset: 0

Gregory L. Simay, Aug 02 2016

nonn

The number of compositions of n = 2k with only even numbers is c(k) = A011782(k). The number of partitions of n with only odd numbers is the strict partition q(n) = A000009(n). Enumerating a(n) is therefore a sum of products of composition numbers and strict partition numbers. (See formulas.)

			The compositions enumerated by a(6) = 12 are (6), (5,1)=(1,5), (4,2), (2,4), (3,3), (4,1,1)=(1,4,1)=(1,1,4), (3,2,1)=(1,2,3)=(2,3,1)=(2,1,3)=(3,1,2)=(1,3,2), (2,2,2), (3,1,1,1)=(1,3,1,1)=(1,1,3,1)=(1,1,1,3), (2,2,1,1)=(2,1,2,1)=(2,1,1,2)=(1,2,1,2)=(1,1,2,2)=(1,2,2,1), (2,1,1,1,1)=(1,2,1,1,1)=(1,1,2,1,1,)=(1,1,1,2,1)=(1,1,1,1,2), (1,1,1,1,1,1).
The compositions enumerated by a(5) = 7 are (5), (4,1)=(1,4), (3,2)=(2,3), (3,1,1)=(1,3,1)=(1,1,3), (2,2,1)=(2,1,2)=(1,2,2), (2,1,1,1)=(1,2,1,1)=(1,1,2,1)=(1,1,1,2), (1,1,1,1,1).

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

Cf. A000009, A011782, A275548.

b:= proc(n, i, p) option remember; (t-> `if`(n=0, p!,
      `if`(i<1, 0, add(b(n-i*j, i-1, p+`if`(t, j, 0))/
      `if`(t, j, 0)!, j=0..n/i))))(i::even)
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..50);  # Alois P. Heinz, Aug 03 2016

nmax = 40; CoefficientList[Series[(1 - x^2)/(1 - 2*x^2)*Product[(1 + x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 02 2018 *)

a(2k+1) = Sum_{j=0..k} c(j)*q(2k+1-2j), where c(j) = A011782(j), the number of compositions of j, and q(j) = A000009(j), the number of strict partitions of j.
a(2k) = Sum_{j=0..k} c(j)*q(2k - 2j).
a(n) = 2*a(n-2) + q(n) - q(n-2).
G.f.: (1 - x^2)/(1 - 2*x^2) * Product_{n>=1} (1 + x^n). - Peter Bala, Aug 03 2016
a(n) ~ c * 2^(n/2), where c = (QPochhammer[-1, 1/sqrt(2)] + (-1)^n*QPochhammer[-1, -1/sqrt(2)])/8, c = 2.002012668882683075956932277149607919866122388... if n is even and c = 1.8471591618236152130512812517147483461076894... if n is odd. - Vaclav Kotesovec, Jun 02 2018

A274174 Number of compositions of n if all summand runs are kept together.

Original entry on oeis.org

1, 1, 2, 4, 7, 12, 22, 36, 60, 97, 162, 254, 406, 628, 974, 1514, 2305, 3492, 5254, 7842, 11598, 17292, 25294, 37090, 53866, 78113, 112224, 161092, 230788, 328352, 466040, 658846, 928132, 1302290, 1821770, 2537156, 3536445, 4897310, 6777806, 9341456, 12858960, 17625970, 24133832, 32910898, 44813228, 60922160, 82569722

Offset: 0

Gregory L. Simay, Jun 12 2016

nonn

a(n^2) is odd. - Gregory L. Simay, Jun 23 2019

Also the number of compositions of n avoiding the patterns (1,2,1) and (2,1,2). - Gus Wiseman, Jul 07 2020

			If the summand runs are blocked together, there are 22 compositions of a(6): 6; 5+1, 1+5, 4+2, 2+4, (3+3), 4+(1+1), (1+1)+4, 1+2+3, 1+3+2, 2+1+3, 2+3+1, 3+1+2, 3+2+1, (2+2+2), 3+(1+1+1), (1+1+1)+3, (2+2)+(1+1), (1+1)+(2+2), 2+(1+1+1+1), (1+1+1+1)+2, (1+1+1+1+1+1).
a(0)=1; a(1)= 1; a(4) = 7; a(9) = 97; a(16) = 2305; a(25) = 78113 and a(36) = 3536445. - _Gregory L. Simay_, Jun 23 2019

Alois P. Heinz, Table of n, a(n) for n = 0..5000
Gus Wiseman, Sequences counting and ranking compositions by the patterns they match or avoid.

Cf. A000070, A101880, A116608.
The version for patterns is A001339.
The version for prime indices is A333175.
The complement (i.e., the matching version) is A335548.
Anti-run compositions are A003242.
(1,2,1)- and (2,1,2)-matching permutations of prime indices are A335462.
(1,2,1)-matching compositions are A335470.
(1,2,1)-avoiding compositions are A335471.
(2,1,2)-matching compositions are A335472.
(2,1,2)-avoiding compositions are A335473.
Cf. A000670, A056986, A181796, A335451, A335452, A335460, A335463.

b:= proc(n, i, p) option remember; `if`(n=0, p!, `if`(i<1, 0,
       add(b(n-i*j, i-1, p+`if`(j=0, 0, 1)), j=0..n/i)))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..50);  # Alois P. Heinz, Jun 12 2016

Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[Split[#]]==Length[Union[#]]&]],{n,0,10}] (* Gus Wiseman, Jul 07 2020 *)
b[n_, i_, p_] := b[n, i, p] = If[n == 0, p!, If[i < 1, 0,
    Sum[b[n - i*j, i - 1, p + If[j == 0, 0, 1]], {j, 0, n/i}]]];
a[n_] := b[n, n, 0];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Jul 11 2021, after Alois P. Heinz *)

a(n) = Sum_{k>=0} k! * A116608(n,k). - Joerg Arndt, Jun 12 2016

Terms a(9) and beyond from Joerg Arndt, Jun 12 2016

A265317 The number of partitions of 2n having segment structure symmetry.

Original entry on oeis.org

1, 1, 3, 5, 10, 17, 33, 53

Offset: 0

Gregory L. Simay, Dec 06 2015

Define a segmented partition a(n, k, ) to be a partition of n with exactly k parts, with s(j) parts t(j) identical to each other and distinct from all other parts. Note that n>=k, j<=k, 0<=s(j)<=k, s(1)t(1) + ... + s(j)t(j) = n and s(1) + ... + s(j) = k.
A partition of 2n has segment structure symmetry if there is at least one way it can be arranged into two consecutive sequences (no central summand), with each sequence exhibiting the same segment structure.
If the total number of parts is an odd number, then a partition of 2n cannot exhibit segment structure symmetry.
Every partition of n into exactly 2 parts has structure symmetry <1><1>.
Except for the case of partitions with a segment structure of <3,1>, every partition of n into exactly 4 parts has segment structure symmetry.
Except for the case of partitions with a segment structure of <5,1> or <3,1,1>, every partition of 2n into exactly 6 integers has segment structure symmetry.

			The partition 5,3,3,1,1,1 can be rearranged into two consecutive sequences (separated by a / for clarity) 1,1,5/3,3,1 which exhibits the segment structure symmetry <2,1><2,1>, and so counts as one of the 53 partitions of 14 exhibiting this symmetry.
The partitions of 14 with exactly 2 parts are 13,1; 12,2; 11,3; 10,4; 9,5; 8,6; and 7,7. All of them, including 7,7, exhibit segment structure symmetry in the form of <1><1>.
The partition 5,5,3,1 is an example of a partition of 14 with exactly 4 parts with segment structure <2,1,1>. This partition can be arranged into two consecutive sequences 5,3/5,1, which exhibits the segment structure symmetry, <1,1><1,1>.
The partition 3,3,3,2,2,1 is an example of a partition of 14 with exactly 6 parts with segment structure <3,2,1>. This partition can be arranged into two consecutive sequences 3,3,1/2,2,3, which exhibits the segment structure symmetry <2,1><2,1>.

Cf. A000041.

A263982 Number of partitions of n with a palindromicity of 3.

Original entry on oeis.org

1, 1, 3, 4, 8, 10, 18, 22, 36, 44, 67, 81, 119, 142

Offset: 6

Gregory L. Simay, Nov 01 2015

A non-strict partition of n can be arranged into two consecutive sequences that are mirror images of each other, separated by a central sequence of k distinct summands, k>=0. Strict partitions (ref. A000009) only have the central sequence. In both cases, the palindromicity of the partition is k.

Palindromic partitions (ref. A025065) have palindromicity 0 (no central summand) or 1 (central summand). Non-palindromic partitions have palindromicities >=2.

			If considered unordered rather than a nonincreasing sequence, the partition 9,7,7,5,4,4,2,2,1,1,1 can be arranged as 7,4,2,1 [central sequence of 9,5,1] 1,2,4,7. Therefore the palindromicity of this particular partition is 3.
w(14,3) = w(11,2) + w(11,3) = A014153(4) + 10 = 26 + 10 = 36.
w(15,3) = w(2*8-1,3) = A014153(4) + A014153(3) + A014153(1) + A014153(0) = 26 + 14 + 3 + 1 = 44.

Cf. A000070, A014153, A025065.

p(n,k) = number of partitions of n with palindromicity k.
If k*(k+1)/2 <= p(n) < (k+1)*(k+2)/2, then p(n) = p(n,0) + .. + p(n,k)
Let q(n,k)= number of strict partitions of n (ref. A000009) with exactly k parts. Then p(n,k) = Sum_{j>=0} q(n-2j,k)*p(j), which affords another way to demonstrate that the convolution of q(2n-j) with p(j) equals p(2n).
p(2n,0) = p(n) and p(2n+1,0) = 0 (ref. A025065).
p(2n,1) = p(2n+1,1) = A000070(n-1), the first partial sum of A000041 (ref. A025065).
p(2n,2) = p(2n-1,2) = A014153(n-2), the second partial sum of A000041.
p(2n,3) = A014153(n-3) + A014153(n-5) + A014153(n-6) + A014153(n-8) + A014153(n-9) + A014153(n-11) + A014153(n-12) + ...
G.f. for p(2n,3): p(x)* x^3*(1+x+x^2+x^3)/(1-x)*(1-x^2)*(1-x^3) where p(x) is the g.f. for A000041.
p(2n-1,3) = A014153(n-4) + A014153(n-5) + A014153(n-7) + A014153(n-8) + A014153(n-10) + A014153(n-11) + A014153(n-13) + A014153(n-14) + ...
G.f. for p(2n-1,3): p(x)* x^3*(1+2x+x^2)/(1-x)*(1-x^2)*(1-x^3) where p(x) is the g.f. for A000041.
More generally, p(n,k>=3) = p(n-k,k-1) + p(n-2k, k-1) + p(n-3k,n-1) + ... for k>=3 = p(n-k, k-1) + p(n-k,k).

cp's OEIS Frontend

User: Gregory L. Simay

A308654 The overpartition triangle: T(n,k) is the number of overpartitions of n with exactly k positive integer parts, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A297120 Number of compositions derived from the overpartitions of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Extensions

A289249 Number of compositions of n if only the order of parts 1 and 2 matters.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Python

Formula

Extensions

A285981 Number of Carlitz compositions having at least two identical parts.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

Extensions

A276129 a(n) is the number of ordered ways to tile a strip of length n+2 with white tiles of odd lengths summing to length n and two red squares.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A275548 Number of compositions of n if only the order of the odd numbers matter.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A275592 Number of compositions of n if only the order of the even numbers matter.

Views

Author

Keywords

Comments

Examples