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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 1-3 of 3 results.

A239516 Number of partitions p of n that are separable by the 2*min(p); see Comments.

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 2, 4, 6, 6, 8, 12, 14, 16, 22, 27, 32, 41, 49, 60, 73, 88, 106, 130, 154, 184, 220, 262, 313, 373, 440, 520, 616, 723, 849, 1002, 1173, 1373, 1606, 1873, 2182, 2543, 2955, 3431, 3979, 4608, 5327, 6160, 7105, 8190, 9435, 10851, 12469, 14317
Offset: 1

Views

Author

Clark Kimberling, Mar 26 2014

Keywords

Comments

Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)-separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)-separable if there is an ordering x, h, x, h, ..., x, h, where the number of h's on the ends is 1; next, p is (h,2)-separable if there is an ordering h, x, h, ..., x, h. Finally, p is h-separable if it is (h,i)-separable for i = 0,1, or 2.

Examples

			Let h = 2*min(p). The (h,0)-separable partition of 8 is 521; the (h,1)-separable partition is 3212; the (h,2)-separable partitions are 242, 21212. So, there are 1 + 1 + 2 = 4 h-separable partitions of 8.

Crossrefs

Cf. A239515, A239517, A239518, A239482.

Programs

  • Mathematica
    z = 75; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239515 *)
t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2*Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239516 *)
t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Max[p]] <= Length[p] + 1], {n, 1, z}]  (* A239517 *)
t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Length[p]] <= Length[p] + 1], {n, 1, z}]  (* A239518 *)

A239517 Number of partitions of n that are separable by the greatest part; see Comments.

Original entry on oeis.org

0, 0, 1, 2, 4, 5, 8, 10, 13, 16, 20, 24, 29, 33, 39, 46, 53, 59, 67, 77, 87, 97, 107, 120, 134, 147, 163, 180, 196, 216, 236, 259, 281, 305, 332, 363, 393, 423, 456, 496, 534, 577, 619, 667, 718, 770, 823, 887, 949, 1016, 1087, 1165, 1240, 1325, 1414, 1512
Offset: 1

Views

Author

Clark Kimberling, Mar 26 2014

Keywords

Comments

Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)-separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)-separable if there is an ordering x, h, x, h, ..., x, h, where the number of h's on the ends is 1; next, p is (h,2)-separable if there is an ordering h, x, h, ..., x, h. Finally, p is h-separable if it is (h,i)-separable for i = 0, 1, or 2.

Examples

			Let h = max(p). The (h,0)-separable partition of 8 are 161, 251, 341, 242; the (h,1)-separable partitions are 71, 62, 323, 1313; the (h,2)-separable partitions are 323, 21212. So, there are 4 + 4 + 2 = 10 h-separable partitions of 8.

Crossrefs

Cf. A239515, A239516, A239518, A239482.

Programs

  • Mathematica
    z = 75; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239515 *)
t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2*Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239516 *)
t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Max[p]] <= Length[p] + 1], {n, 1, z}]  (* A239517 *)
t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Length[p]] <= Length[p] + 1], {n, 1, z}]  (* A239518 *)

A239518 Number of partitions p of n that are separable by the number of parts of p; see Comments.

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 3, 3, 3, 5, 6, 7, 8, 9, 11, 13, 15, 17, 20, 22, 25, 28, 32, 36, 42, 45, 52, 57, 65, 71, 81, 88, 100, 109, 122, 134, 149, 162, 180, 197, 218, 238, 262, 286, 315, 343, 376, 410, 449, 488, 534, 580, 633, 687, 749, 812, 883, 956, 1038, 1123
Offset: 1

Views

Author

Clark Kimberling, Mar 26 2014

Keywords

Comments

Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)-separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)-separable if there is an ordering x, h, x, h, ..., x, h, where the number of h's on the ends is 1; next, p is (h,2)-separable if there is an ordering h, x, h, ..., x, h. Finally, p is h-separable if it is (h,i)-separable for i = 0, 1, or 2.

Examples

			Let h = number of parts of p. The (h,0)-separable partition of 11 are 92, 731, 632, 434; the (h,1)-separable partition is 2414; the (h,2)-partition is 353. So, there are 4 + 1 + 1 = 6 h-separable partitions of 11.

Crossrefs

Cf. A239515, A239516, A239517, A239482.

Programs

  • Mathematica
    z = 75; t1 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239515 *)
t2 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, 2*Min[p]] <= Length[p] + 1], {n, 1, z}]  (* A239516 *)
t3 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Max[p]] <= Length[p] + 1], {n, 1, z}]  (* A239517 *)
t4 = -1 + Table[Count[IntegerPartitions[n], p_ /; Length[p] - 1 <= 2 Count[p, Length[p]] <= Length[p] + 1], {n, 1, z}]  (* A239518 *)
Showing 1-3 of 3 results.

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