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Revision History for A332275 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

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Number of totally co-strong integer partitions of n.
(history; published version)
#8 by Susanna Cuyler at Fri Feb 14 08:02:38 EST 2020
STATUS

proposed

approved

#7 by Gus Wiseman at Fri Feb 14 03:40:21 EST 2020
STATUS

editing

proposed

#6 by Gus Wiseman at Fri Feb 14 03:38:37 EST 2020
COMMENTS

Also the number of totally strong reversed partitions of n. A sequence is totally co-strong if it is empty, equal to (1), or weakly decreasing (strong) with totally strong run-lengths.

Also the number of totally strong reversed integer partitions of n.

#5 by Gus Wiseman at Thu Feb 13 20:57:00 EST 2020
#4 by Gus Wiseman at Thu Feb 13 20:56:11 EST 2020
NAME

Number of recursively totally co-strong integer partitions of n.

COMMENTS

A sequence is recursively totally co-strong if it is empty, equal to (1), or its run-lengths are weakly increasing (co-strong) and are themselves a recursively totally co-strong sequence.

Also the number of totally co-strong compositions reversed partitions of n. A sequence is totally co-strong if it is empty, equal to (1), or weakly increasing decreasing (co-strong) with totally co-strong run-lengths.

EXAMPLE

For example, the partition y = (5,4,4,4,3,3,3,2,2,2,2,2,2,1,1,1,1,1,1) has run-lengths (1,3,3,6,6), with run-lengths (1,2,2), with run-lengths (1,2), with run-lengths (1,1), with run-lengths (2), with run-lengths (1). All of these (except y) are having weakly increasing, run-lengths, and the last is a singleton, (1), so y is counted under a(44).

CROSSREFS

The total strong version is A316496.

The strong version is (also) A316496.

#3 by Gus Wiseman at Wed Feb 12 03:09:26 EST 2020
COMMENTS

A sequence is recursively co-strong if it is empty, equal to (1), or its run-lengths are weakly increasing (co-strong) and are themselves a recursively co-strong sequence.

Also the number of totally co-strong compositions of n. A sequence is totally co-strong if it is empty, equal to (1), or weakly increasing (co-strong) with totally co-strong run-lengths.

#2 by Gus Wiseman at Wed Feb 12 03:07:45 EST 2020
NAME

allocated for Gus WisemanNumber of recursively co-strong integer partitions of n.

DATA

1, 1, 2, 3, 5, 6, 11, 12, 17, 22, 30, 32, 49, 53, 70, 82, 108, 119, 156, 171, 219, 250, 305, 336, 424, 468, 562, 637, 754, 835, 1011, 1108, 1304, 1461, 1692, 1873, 2212, 2417, 2787, 3109, 3562, 3911, 4536, 4947, 5653, 6265, 7076, 7758, 8883, 9669, 10945, 12040

OFFSET

0,3

COMMENTS

A sequence is recursively co-strong if it is empty, equal to (1), or its run-lengths are weakly increasing and are themselves a recursively co-strong sequence.

Also the number of totally co-strong compositions of n. A sequence is totally co-strong if it is empty, equal to (1), or weakly increasing with totally co-strong run-lengths.

EXAMPLE

The a(1) = 1 through a(7) = 12 partitions:

(1) (2) (3) (4) (5) (6) (7)

(11) (21) (22) (32) (33) (43)

(111) (31) (41) (42) (52)

(211) (311) (51) (61)

(1111) (2111) (222) (322)

(11111) (321) (421)

(411) (511)

(2211) (4111)

(3111) (22111)

(21111) (31111)

(111111) (211111)

(1111111)

For example, the partition y = (5,4,4,4,3,3,3,2,2,2,2,2,2,1,1,1,1,1,1) has run-lengths (1,3,3,6,6), with run-lengths (1,2,2), with run-lengths (1,2), with run-lengths (1,1), with run-lengths (2). All of these (except y) are weakly increasing, and the last is a singleton, so y is counted under a(44).

MATHEMATICA

totincQ[q_]:=Or[q=={}, q=={1}, And[LessEqual@@Length/@Split[q], totincQ[Length/@Split[q]]]];

Table[Length[Select[IntegerPartitions[n], totincQ]], {n, 0, 30}]

CROSSREFS

The total version is A316496.

The strong version is (also) A316496.

The version for reversed partitions is (also) A316496.

The alternating version is A317256.

The generalization to compositions is A332274.

Cf. A001462, A100883, A181819, A182850, A317491, A329746, A332289, A332297, A332336, A332337, A332338, A332339.

KEYWORD

allocated

nonn

AUTHOR

Gus Wiseman, Feb 12 2020

STATUS

approved

editing

#1 by Gus Wiseman at Sat Feb 08 23:13:02 EST 2020
NAME

allocated for Gus Wiseman

KEYWORD

allocated

STATUS

approved