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.

A240021 Number T(n,k) of partitions of n into distinct parts, where k is the difference between the number of odd parts and the number of even parts; triangle T(n,k), n>=0, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 1, 1, 1, 0, 1, 1, 1, 3, 1, 1, 1, 0, 2, 2, 2, 4, 1, 0, 1, 2, 1, 1, 4, 2, 4, 5, 1, 1, 1, 1, 2, 1, 2, 6, 3, 1, 6, 6, 1, 2, 2, 1, 3, 1, 5, 9, 3, 2, 9, 7, 2, 4, 3, 2, 3, 2, 8, 12, 4, 0, 1, 4, 12, 8, 3, 7, 4, 3, 4, 3, 14, 16, 4, 1, 1, 7, 16, 9, 6, 11, 5, 1, 4, 4, 6, 20, 20, 5, 2, 2
Offset: 0

Views

Author

Alois P. Heinz, Mar 31 2014

Keywords

Comments

T(n,k) is defined for all n >= 0, k in A001057. Row n contains all terms from the leftmost to the rightmost nonzero term. All other terms (not in the triangle) are equal to 0. First nonzero term of column k>=0 is at n = k^2, first nonzero term of column k<=0 is at n = k*(k+1).
T(n,k) = T(n+k,-k).
T(2n*(2n+1),2n) = A000041(n).
T(4n^2+14n+11,2n+2) = A000070(n).
T(n^2,n) = 1.
T(n^2,n-1) = 0.
T(n^2,n-2) = A209815(n+1).
T(n^2+1,n-1) = A000065(n).
T(n,0) = A239241(n).
Sum_{k<=-1} T(n,k) = A239239(n).
Sum_{k<=0} T(n,k) = A239240(n).
Sum_{k>=1} T(n,k) = A239242(n).
Sum_{k>=0} T(n,k) = A239243(n).
Sum_{k=-1..1} T(n,k) = A239881(n).
T(n,-1) + T(n,1) = A239880(n).
Sum_{k=-n..n} T(n,k) = A000009 (row sums).

Examples

			T(12,-3) = 1: [6,4,2].
T(12,-2) = 2: [10,2], [8,4].
T(12,-1) = 1: [12].
T(12,0) = 2: [6,3,2,1], [5,4,2,1].
T(12,1) = 6: [9,2,1], [8,3,1], [7,4,1], [7,3,2], [6,5,1], [5,4,3].
T(12,2) = 3: [11,1], [9,3], [7,5].
T(13,-1) = 6: [10,2,1], [8,4,1], [8,3,2], [7,4,2], [6,5,2], [6,4,3].
T(14,-2) = 3: [12,2], [10,4], [8,6].
Triangle T(n,k) begins:
: n\k : -3 -2 -1  0  1  2  3  ...
+-----+--------------------------
:  0  :           1
:  1  :              1
:  2  :        1
:  3  :           1, 1
:  4  :        1, 0, 0, 1
:  5  :           2, 1
:  6  :     1, 1, 0, 1, 1
:  7  :        1, 3, 1
:  8  :     1, 1, 0, 2, 2
:  9  :        2, 4, 1, 0, 1
: 10  :     2, 1, 1, 4, 2
: 11  :        4, 5, 1, 1, 1
: 12  :  1, 2, 1, 2, 6, 3
: 13  :     1, 6, 6, 1, 2, 2
: 14  :  1, 3, 1, 5, 9, 3
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n>i*(i+1)/2, 0, `if`(n=0, 1,
          expand(b(n, i-1)+`if`(i>n, 0, b(n-i, i-1)*x^(2*irem(i, 2)-1)))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=ldegree(p)..degree(p)))(b(n$2)):
    seq(T(n), n=0..20);
  • Mathematica
    b[n_, i_] := b[n, i] = If[n>i*(i+1)/2, 0, If[n == 0, 1, Expand[b[n, i-1] + If[i>n, 0, b[n-i, i-1]*x^(2*Mod[i, 2]-1)]]]]; T[n_] := Function[{p}, Table[ Coefficient[p, x, i], {i, Exponent[p, x, Min], Exponent[p, x]}]][b[n, n]]; Table[ T[n], {n, 0, 20}] // Flatten (* Jean-François Alcover, Feb 11 2015, after Alois P. Heinz *)
  • PARI
    N=20; q='q+O('q^N);
    e(n) = if(n%2!=0, u, 1/u);
    gf = prod(n=1,N, 1 + e(n)*q^n );
    V = Vec( gf );
    { for (j=1, #V,  \\ print triangle, including leading zeros
        for (i=0, N-j, print1("   "));  \\ padding
        for (i=-j+1, j-1, print1(polcoeff(V[j], i, u),", "));
        print();
    ); }
    /* Joerg Arndt, Apr 01 2014 */

Formula

G.f.: prod(n>=1, 1 + e(n)*q^n ) = 1 + sum(n>=1, e(n)*q^n * prod(k=1..n-1, 1+e(k)*q^k) ) where e(n) = u if n odd, otherwise 1/u; see Pari program. [Joerg Arndt, Apr 01 2014]

A239832 Number of partitions of n having 1 more even part than odd, so that there is an ordering of parts for which the even and odd parts alternate and the first and last terms are even.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 2, 2, 4, 3, 7, 6, 11, 11, 17, 19, 27, 31, 41, 51, 62, 79, 95, 121, 142, 182, 212, 269, 314, 393, 459, 570, 665, 816, 958, 1160, 1364, 1639, 1928, 2297, 2706, 3200, 3768, 4434, 5212, 6105, 7170, 8361, 9799, 11396, 13322, 15450, 18022
Offset: 0

Views

Author

Clark Kimberling, Mar 29 2014

Keywords

Comments

Let c(n) be the number of partitions of n having 1 more odd part than even, so that there is an ordering of parts for which the even and odd parts alternate and the first and last terms are odd. Then c(n) = a(n+1) for n >= 0.

Examples

			The three partitions counted by a(10) are [10], [4,1,2,1,2], and [2,3,2,1,2].
		

Crossrefs

Column k=-1 of A240009.

Programs

  • Mathematica
    p[n_] := p[n] = Select[IntegerPartitions[n], Count[#, ?OddQ] == -1 + Count[#, ?EvenQ] &]; t = Table[p[n], {n, 0, 10}]
    TableForm[t] (* shows the partitions *)
    Table[Length[p[n]], {n, 0, 30}]  (* A239832 *)
    (* Peter J. C. Moses, Mar 10 2014 *)

A239833 Number of partitions of n having an ordering of parts in which no parts of equal parity are adjacent and the first and last terms have the same parity.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 3, 4, 6, 7, 10, 13, 17, 22, 28, 36, 46, 58, 72, 92, 113, 141, 174, 216, 263, 324, 394, 481, 583, 707, 852, 1029, 1235, 1481, 1774, 2118, 2524, 3003, 3567, 4225, 5003, 5906, 6968, 8202, 9646, 11317, 13275, 15531, 18160, 21195, 24718, 28772
Offset: 0

Views

Author

Clark Kimberling, Mar 29 2014

Keywords

Examples

			a(10) counts these 10 partitions:  [10], [1,8,1], [7,2,1], [3,6,1], [5,4,1], [5,3,2], [3,4,3], [4,1,2,1,2], [2,3,2,1,2], [1,2,1,2,1,2,1].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(abs(t)>n, 0,
          `if`(n=0, 1, `if`(i<1, 0, b(n, i-1, t)+
          `if`(i>n, 0, b(n-i, i, t+(2*irem(i, 2)-1))))))
        end:
    a:= n-> b(n$2, -1) +b(n$2, 1):
    seq(a(n), n=0..80);  # Alois P. Heinz, Apr 02 2014
  • Mathematica
    p[n_] := p[n] = Select[IntegerPartitions[n], Abs[Count[#, ?OddQ] - Count[#, ?EvenQ]] == 1 &]; t = Table[p[n], {n, 0, 10}]
    TableForm[t] (* shows the partitions*)
    t = Table[Length[p[n]], {n, 0, 60}] (* A239833 *)
    (* Peter J. C. Moses, Mar 10 2014 *)
    b[n_, i_, t_] := b[n, i, t] = If[Abs[t]>n, 0, If[n==0, 1, If[i<1, 0, b[n, i-1, t] + If[i>n, 0, b[n-i, i, t+(2*Mod[i, 2]-1)]]]]]; a[n_] := b[n, n, -1] + b[n, n, 1]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Oct 12 2015, after Alois P. Heinz *)

Formula

a(n) = A239832(n) + A239832(n+1) for n >= 0.
a(n) = A240009(n,-1) + A240009(n,1). - Alois P. Heinz, Apr 02 2014

A239835 Number of partitions of n such that the absolute value of the difference between the number of odd parts and the number of even parts is <=1.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 4, 7, 8, 12, 15, 20, 26, 33, 44, 54, 71, 86, 113, 136, 175, 211, 268, 323, 403, 487, 601, 726, 885, 1068, 1292, 1556, 1867, 2244, 2678, 3208, 3809, 4547, 5379, 6398, 7542, 8937, 10506, 12404, 14542, 17110, 20011, 23465, 27381, 32006, 37267
Offset: 0

Views

Author

Clark Kimberling, Mar 29 2014

Keywords

Comments

Number of partitions of n having an ordering of parts in which no parts of equal parity are adjacent, as in Example.

Examples

			a(8) counts these 8 partitions:  8, 161, 521, 341, 4121, 323, 3212, 21212.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(abs(t)-n>1, 0,
          `if`(n=0, 1, `if`(i<1, 0, b(n, i-1, t)+
          `if`(i>n, 0, b(n-i, i, t+(2*irem(i, 2)-1))))))
        end:
    a:= n-> b(n$2, 0):
    seq(a(n), n=0..80);  # Alois P. Heinz, Apr 01 2014
  • Mathematica
    p[n_] := p[n] = Select[IntegerPartitions[n], Abs[Count[#, ?OddQ] - Count[#, ?EvenQ]] <= 1 &]; t = Table[p[n], {n, 0, 10}]
    TableForm[t] (* shows the partitions *)
    Table[Length[p[n]], {n, 0, 60}] (* A239835 *)
    (* Peter J. C. Moses, Mar 10 2014 *)
    b[n_, i_, t_] := b[n, i, t] = If[Abs[t]-n>1, 0, If[n==0, 1, If[i<1, 0, b[n, i-1, t] + If[i>n, 0, b[n-i, i, t+(2*Mod[i, 2]-1)]]]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)

Formula

a(n) = A045931(n) + A239833(n) for n >= 0.
a(n) = Sum_{k=-1..1} A240009(n,k). - Alois P. Heinz, Apr 01 2014

A239872 Number of strict partitions of 2n having 1 more even part than odd, so that there is at least one ordering of the parts in which the even and odd parts alternate, and the first and last terms are even.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 2, 3, 6, 10, 17, 26, 40, 57, 81, 110, 148, 193, 250, 316, 397, 491, 603, 732, 885, 1061, 1268, 1508, 1790, 2120, 2510, 2970, 3517, 4170, 4950, 5887, 7013, 8371, 10005, 11979, 14353, 17217, 20654, 24785, 29725, 35637, 42672, 51046, 60962
Offset: 0

Views

Author

Clark Kimberling, Mar 29 2014

Keywords

Comments

Let c(n) be the number of strict partitions (that is, every part has multiplicity 1) of 2n having 1 more even part than odd, so that there is an ordering of parts for which the even and odd parts alternate and the first and last terms are even. This sequence is nondecreasing, unlike A239871, of which it is a bisection; the other bisection is A239873.

Examples

			a(9) counts these 3 partitions of 18:  [18], [8,3,4,1,2], [6,5,4,1,2].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2 or
          abs(t)-n>0, 0, `if`(n=0, 1, b(n, i-1, t)+
          `if`(i>n, 0, b(n-i, i-1, t+(2*irem(i, 2)-1)))))
        end:
    a:= n-> b(2*n$2, 1):
    seq(a(n), n=0..60);  # Alois P. Heinz, Apr 01 2014
  • Mathematica
    d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &]; p[n_] := p[n] = Select[d[n], Count[#, ?OddQ] == -1 + Count[#, ?EvenQ] &]; t = Table[p[n], {n, 0, 20}]
    TableForm[t] (* shows the partitions *)
    u = Table[Length[p[2 n]], {n, 0, 40}]  (* A239872 *)
    (* Peter J. C. Moses, Mar 10 2014 *)
    b[n_, i_, t_] := b[n, i, t] = If[n > i*(i+1)/2 || Abs[t]-n > 0, 0, If[n == 0, 1, b[n, i-1, t] + If[i>n, 0, b[n-i, i-1, t + (2*Mod[i, 2] - 1)]]]]; a[n_] := b[2*n, 2*n, 1]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Oct 28 2015, after Alois P. Heinz *)

A239873 Number of strict partitions of 2n + 1 having 1 more even part than odd, so that there is at least one ordering of the parts in which the even and odd parts alternate, and the first and last terms are even.

Original entry on oeis.org

0, 0, 0, 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 43, 51, 61, 74, 91, 113, 144, 184, 239, 311, 407, 530, 692, 895, 1155, 1478, 1882, 2375, 2983, 3715, 4602, 5660, 6925, 8418, 10187, 12257, 14686, 17514, 20809, 24624, 29049, 34154, 40051, 46842, 54668, 63667
Offset: 0

Views

Author

Clark Kimberling, Mar 29 2014

Keywords

Comments

Let c(n) be the number of strict partitions (that is, every part has multiplicity 1) of 2n having 1 more even part than odd, so that there is an ordering of parts for which the even and odd parts alternate and the first and last terms are even. This sequence is nondecreasing, unlike A239871, of which it is a bisection; the other bisection is A239872.

Examples

			a(7) counts these 9 partitions of 15:  [12,1,2], [10,1,4], [10,3,2], [4,9,2], [8,1,6], [8,5,2], [8,3,4], [6,7,2], [6,5,4].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2 or
          abs(t)>n, 0, `if`(n=0, 1, b(n, i-1, t)+
          `if`(i>n, 0, b(n-i, i-1, t+(2*irem(i, 2)-1)))))
        end:
    a:= n-> b(2*n+1$2, 1):
    seq(a(n), n=0..80);  # Alois P. Heinz, Apr 02 2014
  • Mathematica
    d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &]; p[n_] := p[n] = Select[d[n], Count[#, ?OddQ] == -1 + Count[#, ?EvenQ] &]; t = Table[p[n], {n, 0, 20}]
    TableForm[t] (* shows the partitions *)
    u = Table[Length[p[2 n + 1]], {n, 0, 38}]  (* A239873 *)
    (* Peter J. C. Moses, Mar 10 2014 *)
    b[n_, i_, t_] := b[n, i, t] = If[n > i (i + 1)/2 || Abs[t] > n, 0, If[n == 0, 1, b[n, i-1, t] + If[i>n, 0, b[n-i, i-1, t + (2 Mod[i, 2] - 1)]]]]; a[n_] := b[2n+1, 2n+1, 1]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Aug 29 2016, after Alois P. Heinz *)
Showing 1-6 of 6 results.