A239830
Triangular array: T(n,k) = number of partitions of 2n that have alternating sum 2k, with T(0,0) = 1 for convenience.
Original entry on oeis.org
1, 1, 1, 2, 2, 1, 3, 5, 2, 1, 5, 9, 5, 2, 1, 7, 17, 10, 5, 2, 1, 11, 28, 20, 10, 5, 2, 1, 15, 47, 35, 20, 10, 5, 2, 1, 22, 73, 62, 36, 20, 10, 5, 2, 1, 30, 114, 102, 65, 36, 20, 10, 5, 2, 1, 42, 170, 167, 109, 65, 36, 20, 10, 5, 2, 1, 56, 253, 262, 182, 110
Offset: 0
First nine rows:
1
1 ... 1
2 ... 2 ... 1
3 ... 5 ... 2 ... 1
5 ... 9 ... 5 ... 2 ... 1
7 ... 17 .. 10 .. 5 ... 2 ... 1
11 .. 28 .. 20 .. 10 .. 5 ... 2 ... 1
15 .. 47 .. 35 .. 20 .. 10 .. 5 ... 2 ... 1
22 .. 73 .. 62 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
The partitions of 6 are 6, 51, 42, 411, 33, 321, 3111, 222, 2211, 21111, 111111, with respective alternating sums 6, 4, 2, 4, 0, 2, 2, 2, 0, 2, 0, so that row 3 (counting the top row as row 0) of the array is 3 .. 5 .. 2 .. 1.
-
b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(i<1, 0,
expand(b(n, i-1, t)+`if`(i>n, 0, b(n-i, i, -t)*x^((t*i)/2)))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(2*n$2, 1)):
seq(T(n), n=0..14); # Alois P. Heinz, Mar 30 2014
-
z = 16; s[w_] := s[w] = Total[Take[#, ;; ;; 2]] - Total[Take[Rest[#], ;; ;; 2]] &[w]; c[n_] := c[n] = Table[s[IntegerPartitions[n][[k]]], {k, 1, PartitionsP[n]}]; t[n_, k_] := Count[c[2 n], 2 k]; t[0, 0] = 1; u = Table[t[n, k], {n, 0, z}, {k, 0, n}]
TableForm[u] (* A239830, array *)
Flatten[u] (* A239830, sequence *)
(* Peter J. C. Moses, Mar 21 2014 *)
A240009
Number T(n,k) of partitions of n, where k is the difference between the number of odd parts and the number of even parts; triangle T(n,k), n>=0, -floor(n/2)+(n mod 2)<=k<=n, read by rows.
Original entry on oeis.org
1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 2, 1, 1, 1, 0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 0, 1, 1, 2, 3, 2, 2, 2, 1, 1, 0, 1, 1, 1, 2, 2, 2, 4, 3, 2, 2, 1, 1, 0, 1, 1, 2, 4, 5, 3, 4, 4, 2, 2, 1, 1, 0, 1, 1, 1, 2, 3, 3, 5, 7, 5, 4, 4, 2, 2, 1, 1, 0, 1, 1, 2, 4, 7, 7, 6, 8, 6, 4, 4, 2, 2, 1, 1, 0, 1
Offset: 0
T(5,-1) = 1: [2,2,1].
T(5,0) = 2: [4,1], [3,2].
T(5,1) = 1: [5].
T(5,2) = 1: [2,1,1,1].
T(5,3) = 1: [3,1,1].
T(5,5) = 1: [1,1,1,1,1].
Triangle T(n,k) begins:
: n\k : -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 ...
+-----+----------------------------------------------------
: 0 : 1;
: 1 : 1;
: 2 : 1, 0, 0, 1;
: 3 : 1, 1, 0, 1;
: 4 : 1, 1, 0, 1, 1, 0, 1;
: 5 : 1, 2, 1, 1, 1, 0, 1;
: 6 : 1, 1, 1, 1, 2, 2, 1, 1, 0, 1;
: 7 : 1, 2, 3, 2, 2, 2, 1, 1, 0, 1;
: 8 : 1, 1, 2, 2, 2, 4, 3, 2, 2, 1, 1, 0, 1;
: 9 : 1, 2, 4, 5, 3, 4, 4, 2, 2, 1, 1, 0, 1;
: 10 : 1, 1, 2, 3, 3, 5, 7, 5, 4, 4, 2, 2, 1, 1, 0, 1;
Columns k=(-1)-10 give:
A239832,
A045931,
A240010,
A240011,
A240012,
A240013,
A240014,
A240015,
A240016,
A240017,
A240018,
A240019.
Row lengths give
A016777(floor(n/2)).
Cf.
A240021 (the same for partitions into distinct parts),
A242618 (the same for parts counted without multiplicity).
-
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
expand(b(n, i-1)+`if`(i>n, 0, b(n-i, i)*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..14);
-
b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<1, 0, b[n, i-1] + If[i>n, 0, b[n-i, i]*x^(2*Mod[i, 2]-1)]]]; T[n_] := (degree = Exponent[b[n, n], x]; ldegree = -Exponent[b[n, n] /. x -> 1/x, x]; Table[Coefficient[b[n, n], x, i], {i, ldegree, degree}]); Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Jan 06 2015, translated from Maple *)
-
N=20; q='q+O('q^N);
e(n) = if(n%2!=0, u, 1/u);
gf = 1 / 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, Mar 31 2014 */
A239829
Triangular array: T(n,k) = number of partitions of 2n - 1 that have alternating sum 2k - 1.
Original entry on oeis.org
1, 2, 1, 4, 2, 1, 7, 5, 2, 1, 12, 10, 5, 2, 1, 19, 19, 10, 5, 2, 1, 30, 33, 20, 10, 5, 2, 1, 45, 57, 36, 20, 10, 5, 2, 1, 67, 92, 64, 36, 20, 10, 5, 2, 1, 97, 147, 107, 65, 36, 20, 10, 5, 2, 1, 139, 227, 177, 110, 65, 36, 20, 10, 5, 2, 1, 195, 345, 282, 184
Offset: 1
First nine rows:
1
2 ... 1
4 ... 2 ... 1
7 ... 5 ... 2 ... 1
12 .. 10 .. 5 ... 2 ... 1
19 .. 19 .. 10 .. 5 ... 2 ... 1
30 .. 33 .. 20 .. 10 .. 5 ... 2 ... 1
45 .. 57 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
67 .. 92 .. 64 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
The partitions of 5 are 5, 41, 32, 311, 221, 2111, 11111, with respective alternating sums 5, 3, 1, 3, 1, 1, 1, so that row 2 of the array is 4 .. 2 .. 1.
-
b:= proc(n, i, t) option remember; `if`(n=0, x^(1/2), `if`(i<1, 0,
expand(b(n, i-1, t)+`if`(i>n, 0, b(n-i, i, -t)*x^((t*i)/2)))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(2*n-1$2, 1)):
seq(T(n), n=1..14); # Alois P. Heinz, Mar 30 2014
-
z = 15; s[w_] := s[w] = Total[Take[#, ;; ;; 2]] - Total[Take[Rest[#], ;; ;; 2]] &[w]; c[n_] := c[n] = Table[s[IntegerPartitions[n][[k]]], {k, 1, PartitionsP[n]}]; t[n_, k_] := Count[c[2 n - 1], 2 k - 1]; u = Table[t[n, k], {n, 1, z}, {k, 1, n}]
TableForm[u] (* A239829, array *)
Flatten[u] (* A239829, sequence *)
(* Peter J. C. Moses, Mar 21 2014 *)
b[n_, i_, t_] := b[n, i, t] = If[n == 0, x^(1/2), If[i<1, 0, Expand[b[n, i-1, t] + If[i>n, 0, b[n-i, i, -t]*x^((t*i)/2)]]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[2n-1, 2n-1, 1]]; Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, Aug 27 2016, after Alois P. Heinz *)
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
The three partitions counted by a(10) are [10], [4,1,2,1,2], and [2,3,2,1,2].
-
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 *)
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
a(8) counts these 8 partitions: 8, 161, 521, 341, 4121, 323, 3212, 21212.
-
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
-
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 *)
A239881
Number of strict partitions of n having an ordering in which no parts of equal parity are juxtaposed.
Original entry on oeis.org
1, 1, 1, 2, 1, 3, 2, 5, 3, 7, 6, 10, 9, 13, 15, 18, 22, 23, 33, 31, 46, 41, 65, 55, 87, 73, 117, 99, 153, 132, 199, 177, 254, 236, 324, 313, 408, 412, 512, 540, 639, 701, 795, 904, 986, 1159, 1221, 1473, 1509, 1861, 1862, 2336, 2298, 2915, 2830, 3615, 3485
Offset: 0
a(12) counts these 9 partitions: [12], [9,2,1], [3,8,1], [7,4,1], [7,2,3], [5,6,1], [6,3,2,1], [5,4,3], [5,4,1,2].
-
d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];
p[n_] := p[n] = Select[d[n], Abs[Count[#, ?OddQ] - Count[#, ?EvenQ]] <= 1 &]; t = Table[p[n], {n, 0, 12}]
TableForm[t] (* shows the partitions *)
u = Table[Length[p[n]], {n, 0, 60}] (* A239880 *)
(* Peter J. C. Moses, Mar 10 2014 *)
A239880
Number of strict partitions of n having an ordering in which no parts of equal parity are juxtaposed and the first and last terms have the same parity.
Original entry on oeis.org
0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 5, 5, 7, 7, 10, 11, 14, 15, 19, 22, 26, 30, 35, 42, 47, 56, 62, 76, 83, 100, 108, 132, 142, 171, 184, 222, 239, 284, 306, 363, 394, 460, 500, 581, 636, 730, 802, 914, 1010, 1139, 1262, 1415, 1577, 1753, 1956, 2163, 2423, 2663
Offset: 0
a(12) counts these 7 partitions: [12], [9,2,1], [3,8,1], [7,4,1], [7,2,3], [5,6,1], [5,4,3].
-
b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2 or
abs(t)-n>1, 0, `if`(n=0, abs(t), b(n, i-1, t)+
`if`(i>n, 0, b(n-i, i-1, t+(2*irem(i, 2)-1)))))
end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..80); # Alois P. Heinz, Apr 02 2014
-
d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];
p[n_] := p[n] = Select[d[n], Abs[Count[#, ?OddQ] - Count[#, ?EvenQ]] == 1 &]; t = Table[p[n], {n, 0, 12}]
TableForm[t] (* shows the partitions *)
u = Table[Length[p[n]], {n, 0, 60}] (* A239880 *)
(* Peter J. C. Moses, Mar 10 2014 *)
b[n_, i_, t_] := b[n, i, t] = If[n > i*(i+1)/2 || Abs[t]-n>1, 0, If[n==0, Abs[t], b[n, i-1, t] + If[i>n, 0, b[n-i, i-1, t + (2*Mod[i, 2]-1)]]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Aug 29 2016, after Alois P. Heinz *)
Showing 1-7 of 7 results.
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