A055885
Euler transform applied twice to partition triangle A008284.
Original entry on oeis.org
1, 1, 3, 1, 3, 6, 1, 6, 9, 14, 1, 6, 18, 23, 27, 1, 9, 27, 54, 57, 58, 1, 9, 39, 87, 140, 131, 111, 1, 12, 51, 150, 259, 353, 295, 223, 1, 12, 69, 210, 470, 702, 832, 637, 424, 1, 15, 84, 314, 749, 1379, 1803, 1917, 1350, 817, 1, 15, 105, 416, 1176, 2352, 3730, 4403, 4245, 2789, 1527
Offset: 1
1;
1, 3;
1, 3, 6;
1, 6, 9, 14;
1, 6, 18, 23, 27;
...
A055886
Euler transform applied three times to partition triangle A008284.
Original entry on oeis.org
1, 1, 4, 1, 4, 10, 1, 8, 16, 30, 1, 8, 32, 54, 75, 1, 12, 48, 128, 176, 206, 1, 12, 70, 210, 443, 535, 518, 1, 16, 92, 362, 842, 1485, 1585, 1344, 1, 16, 124, 516, 1544, 3075, 4676, 4527, 3357, 1, 20, 152, 770, 2500, 6133, 10622, 14336, 12664, 8429, 1, 20, 190, 1030, 3952, 10718, 22524, 34918, 42426, 34631, 20759
Offset: 1
1;
1, 4;
1, 4, 10;
1, 8, 16, 30;
1, 8, 32, 54, 75;
...
A360764
Number T(n,k) of sets of nonempty strict integer partitions with a total of k parts and total sum of n; triangle T(n,k), n>=0, 0<=k<=max(i:T(n,i)>0), read by rows.
Original entry on oeis.org
1, 0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 0, 1, 4, 2, 0, 1, 4, 6, 1, 0, 1, 6, 8, 4, 0, 1, 6, 13, 9, 1, 0, 1, 8, 18, 16, 6, 0, 1, 8, 24, 29, 13, 2, 0, 1, 10, 30, 43, 29, 6, 0, 1, 10, 39, 64, 52, 19, 1, 0, 1, 12, 46, 89, 89, 42, 7, 0, 1, 12, 56, 122, 139, 85, 22, 1
Offset: 0
T(6,1) = 1: {[6]}.
T(6,2) = 4: {[1],[5]}, {[2],[4]}, {[1,5]}, {[2,4]}.
T(6,3) = 6: {[1,2,3]}, {[1],[1,4]}, {[1],[2,3]}, {[2],[1,3]}, {[3],[1,2]}, {[1],[2],[3]}.
T(6,4) = 1: {[1],[2],[1,2]}.
Triangle T(n,k) begins:
1;
0, 1;
0, 1;
0, 1, 2;
0, 1, 2, 1;
0, 1, 4, 2;
0, 1, 4, 6, 1;
0, 1, 6, 8, 4;
0, 1, 6, 13, 9, 1;
0, 1, 8, 18, 16, 6;
0, 1, 8, 24, 29, 13, 2;
0, 1, 10, 30, 43, 29, 6;
0, 1, 10, 39, 64, 52, 19, 1;
...
-
h:= proc(n, i) option remember; expand(`if`(n=0, 1,
`if`(i<1, 0, h(n, i-1)+x*h(n-i, min(n-i, i-1)))))
end:
g:= proc(n, i, j) option remember; expand(`if`(j=0, 1, `if`(i<0, 0, add(
g(n, i-1, j-k)*x^(i*k)*binomial(coeff(h(n$2), x, i), k), k=0..j))))
end:
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
`if`(i<1, 0, add(b(n-i*j, i-1)*g(i$2, j), 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);
-
h[n_, i_] := h[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, h[n, i - 1] + x*h[n - i, Min[n - i, i - 1]]]]];
g[n_, i_, j_] := g[n, i, j] = Expand[If[j == 0, 1, If[i<0, 0, Sum[g[n, i - 1, j - k]*x^(i*k)*Binomial[Coefficient[h[n, n], x, i], k], {k, 0, j}]]]];
b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, i - 1]*g[i, i, j], {j, 0, n/i}]]]] ;
T[n_] := CoefficientList[b[n, n], x];
Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Nov 17 2023, after Alois P. Heinz *)
A346632
Triangle read by rows giving the main diagonals of the matrices counting integer compositions by length and alternating sum (A345197).
Original entry on oeis.org
1, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 1, 2, 0, 0, 0, 1, 2, 3, 0, 0, 0, 1, 2, 6, 6, 0, 0, 0, 1, 2, 9, 12, 0, 0, 0, 0, 1, 2, 12, 18, 10, 0, 0, 0, 0, 1, 2, 15, 24, 30, 20, 0, 0, 0, 0, 1, 2, 18, 30, 60, 60, 0, 0, 0, 0, 0, 1, 2, 21, 36, 100, 120, 35, 0, 0, 0, 0
Offset: 0
Triangle begins:
1
0 0
0 1 0
0 1 2 0
0 1 2 0 0
0 1 2 3 0 0
0 1 2 6 6 0 0
0 1 2 9 12 0 0 0
0 1 2 12 18 10 0 0 0
0 1 2 15 24 30 20 0 0 0
0 1 2 18 30 60 60 0 0 0 0
0 1 2 21 36 100 120 35 0 0 0 0
0 1 2 24 42 150 200 140 70 0 0 0 0
0 1 2 27 48 210 300 350 280 0 0 0 0 0
0 1 2 30 54 280 420 700 700 126 0 0 0 0 0
The first nonzero element in each column appears to be
A001405.
These are the diagonals of the matrices given by
A345197.
Antidiagonals of the same matrices are
A345907.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse:
A344612).
A316524 gives the alternating sum of prime indices (reverse:
A344616).
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
-
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Table[Length[Select[Join@@Permutations/@IntegerPartitions[n,{k}],k==(n+ats[#])/2&]],{k,n}],{n,0,15}]
A360468
Number of multisets of nonempty integer partitions with a total of n parts and total sum of 2n.
Original entry on oeis.org
1, 1, 4, 12, 43, 134, 448, 1387, 4347, 13128, 39350, 115285, 334179, 952512, 2684714, 7468402, 20556838, 55963935, 150896053, 402999801, 1066962557, 2801089402, 7295920768, 18859954024, 48404773852, 123381167011, 312438704848, 786231143489, 1966628476977
Offset: 0
a(3) = 12: {[1,1,4]}, {[1,2,3]}, {[2,2,2]}, {[1],[1,4]}, {[1],[2,3]}, {[2],[1,3]}, {[2],[2,2]}, {[3],[1,2]}, {[4],[1,1]}, {[1],[1],[4]}, {[1],[2],[3]}, {[2],[2],[2]}.
A360072
Number of pairs of positive integers (k,i) such that k >= i and there exists an integer partition of n of length k with i distinct parts.
Original entry on oeis.org
0, 1, 2, 3, 5, 5, 9, 9, 13, 14, 18, 19, 26, 25, 30, 34, 39, 40, 48, 48, 56, 59, 64, 67, 78, 78, 84, 89, 97, 99, 111, 111, 121, 125, 131, 137, 149, 149, 158, 165, 176, 177, 190, 191, 202, 210, 216, 222, 238, 239, 250, 256, 266, 270, 284, 289, 302, 307, 316, 323
Offset: 0
The a(5) = 5 pairs are: (1,1), (2,2), (3,2), (4,2), (5,1). The pair (3,3) is absent because it is not possible to partition 5 into 3 parts, all 3 of which are distinct.
The a(6) = 9 pairs are: (1,1), (2,1), (2,2), (3,1), (3,2), (3,3), (4,2), (5,2), (6,1). The pair (3,3) is present because (3,2,1) is a partition of 6 into 3 parts, all 3 of which are distinct.
A116608 counts partitions by number of distinct parts.
-
Table[Count[Flatten[Sign[Table[Length[Select[IntegerPartitions[n], Length[#]==k&&Length[Union[#]]==i&]],{k,1,n},{i,1,k}]]],1],{n,0,30}]
-
a(n) = if(n < 1, 0, numdiv(n) + sum(k=2, (sqrtint(8*n+1)-1)\2, n-binomial(k+1,2)+1)) \\ Andrew Howroyd, Jan 30 2023
A179314
An irregular table of values refining Table A061260.
Original entry on oeis.org
1, 2, 1, 3, 2, 1, 5, 3, 3, 2, 1, 7, 5, 6, 3, 3, 2, 1, 11, 7, 10, 6, 5, 6, 4, 3, 3, 2, 1, 15, 11, 14, 15, 7, 10, 6, 9, 5, 6, 4, 3, 3, 2, 1
Offset: 1
Row six Col three of A061260 can be derived from partitions 4+1+1, 3+2+1, and 2+2+2 yielding 5+6+4 = 15 cases.
Partition 2+2+2 yields four cases because we can write 2*3*4/3!=4.
1;
2, 1;
3, 2, 1;
5, 3, 3, 2, 1;
7, 5, 6, 3, 3, 2, 1;
11, 7, 10, 6, 5, 6, 4, 3, 3, 2, 1;
15, 11, 14, 15, 7, 10, 6, 9, 5, 6, 4, 3, 3, 2, 1;
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