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proposed

The a(9) = 20 wayssplit partitions:

Cf. A001970, A063834, A316245, A317508, A317546, A317715, A318434, A318683, A319794.

Cf. A001970, A063834, A316245, A317546, A317715, A318434, A318683, A319794.

allocated for Gus WisemanNumber of ways to split a strict integer partition of n into consecutive subsequences with strictly decreasing sums.

1, 1, 1, 3, 3, 5, 8, 11, 14, 20, 28, 35, 48, 61, 79, 105, 129, 162, 208, 257, 318, 404, 489, 600, 732, 896, 1075, 1315, 1576, 1895, 2272, 2715, 3217, 3851, 4537, 5377, 6353, 7484, 8765, 10314, 12044, 14079, 16420, 19114, 22184, 25818, 29840, 34528, 39903, 46030

0,4

The a(9) = 20 ways:

(9)

(81) (8)(1)

(72) (7)(2)

(63) (6)(3)

(54) (5)(4)

(432) (43)(2) (4)(3)(2)

(621) (62)(1) (6)(2)(1) (6)(21)

(531) (53)(1) (5)(3)(1) (5)(31)

comps[q_]:=Table[Table[Take[q, {Total[Take[c, i-1]]+1, Total[Take[c, i]]}], {i, Length[c]}], {c, Join@@Permutations/@IntegerPartitions[Length[q]]}];

Table[Sum[Length[Select[comps[y], OrderedQ[Total/@#, Greater]&]], {y, Select[IntegerPartitions[n], UnsameQ@@#&]}], {n, 30}]

Cf. A001970, A063834, A316245, A317546, A317715, A318434, A318683.

allocated

nonn

Gus Wiseman, Sep 29 2018

approved

editing

allocated for Gus Wiseman

recycled

allocated

editing

approved

Triangle read by rows: T(0,0) = 1; T(n,k) = - T(n-1,k) + 2 T(n-5,k-1) for k = 0..floor(n/5); T(n,k)=0 for n or k < 0.

1, -1, 1, -1, 1, -1, 2, 1, -4, -1, 6, 1, -8, -1, 10, 1, -12, 4, -1, 14, -12, 1, -16, 24, -1, 18, -40, 1, -20, 60, -1, 22, -84, 8, 1, -24, 112, -32, -1, 26, -144, 80, 1, -28, 180, -160, -1, 30, -220, 280, 1, -32, 264, -448, 16

0,7

The numbers in rows of the triangle are along "fourth layer" skew diagonals pointing top-left in center-justified triangle given in A065109 ((2-x)^n) and along "fourth layer" skew diagonals pointing top-right in center-justified triangle given in A303872 ((-1+2x)^n), see links. (Note: First layer skew diagonals in center-justified triangles of coefficients in expansions of (2-x)^n and (-1+2x)^n are given in A133156 (coefficients of Chebyshev polynomials of the second kind) and A305098 respectively.) The coefficients in the expansion of 1/(1+x-2x^5) are given by the sequence generated by the row sums. The row sums give A317509.

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Shara Lalo, <a href="/A318684/a318684.pdf">Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (2 - x)^n</a>

Shara Lalo, <a href="/A318684/a318684_1.pdf">Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (-1 + 2x)^n</a>

T(n,k) = (-1)^(n - 5*k) * 2^k / ((n - 5*k)! k!) * (n - 4*k)! where n >= 0 and 0 <= k <= floor(n/5).

Triangle begins:

1;

-1;

1;

-1;

1;

-1, 2;

1, -4;

-1, 6;

1, -8;

-1, 10;

1, -12, 4;

-1, 14, -12;

1, -16, 24;

-1, 18, -40;

1, -20, 60;

-1, 22, -84, 8;

1, -24, 112, -32;

-1, 26, -144, 80;

1, -28, 180, -160;

-1, 30, -220, 280;

t[n_, k_] := t[n, k] = (-1)^(n - 5 k) * (2)^k/((n - 5 k)! k!) * (n - 4 k)!; Table[t[n, k], {n, 0, 20}, {k, 0, Floor[n/5]} ] // Flatten.

t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, - t[n - 1, k] + 2 t[n - 5, k - 1]]; Table[t[n, k], {n, 0, 20}, {k, 0, Floor[n/5]}] // Flatten.

Row sums give A317509.

Cf. A065109, A303872.

Cf. A133156, A305098.

sign

recycled

Shara Lalo, Sep 04 2018

proposed

editing

editing

proposed

Triangle read by rows: T(0,0) = 1; T(n,k) = - T(n-1,k) + 2 T(n-5,k-1) for k = 0,1..floor(n/5); T(n,k)=0 for n or k < 0.

The numbers in rows of the triangle are along "fourth layer" skew diagonals pointing top-left in center-justified triangle given in A065109 ((2-x)^n) and along "fourth layer" skew diagonals pointing top-right in center-justified triangle given in A303872 ((-1+2x)^n), see links. (Note: First layer skew diagonals in center-justified triangles of coefficients in expansions of (2-x)^n and (-1+2x)^n are given in A133156 (coefficients of Chebyshev polynomials of the second kind) and A305098 respectively). ) The coefficients in the expansion of 1/(1+x-2x^5) are given by the sequence generated by the row sums. The row sums give A317509.

T(n,k) = (-1)^(n - 5 *k) * (2)^k / ((n - 5 *k)! k!) * (n - 4 *k)! where n is a nonnegative integer >= 0 and 0 <= k <= 0,1..floor(n/5).

1;

-1;

1;

-1;

1;

-1, 2;

1, -4;

-1, 6;

1, -8;

-1, 10;

1, -12, 4;

-1, 14, -12;

1, -16, 24;

-1, 18, -40;

1, -20, 60;

-1, 22, -84, 8;

1, -24, 112, -32;

-1, 26, -144, 80;

1, -28, 180, -160;

-1, 30, -220, 280;

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