The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A357237

(Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

A357237

Number of compositions (ordered partitions) of n into distinct parts of the form 2^j - 1.
(history; published version)

#11 by Peter Luschny at Sun Sep 25 11:03:13 EDT 2022

STATUS

reviewed

approved

#10 by Joerg Arndt at Sun Sep 25 10:54:29 EDT 2022

STATUS

proposed

reviewed

#9 by Ilya Gutkovskiy at Sun Sep 25 06:50:30 EDT 2022

STATUS

editing

proposed

Discussion

Sun Sep 25

10:54

Joerg Arndt: So we do want to have a sequence that has 0 where this is zero and k where this is k!; Name "Number of parts in the (unique) partition of n into parts of the form 2^j-1"; a(0) for that sequence should then be 0 (not one).

#8 by Ilya Gutkovskiy at Sun Sep 25 06:49:45 EDT 2022

COMMENTS

a(n) = Sum_{k=0..n} p(n,k) * k!, where p(n,k) = number of partitions of n into k distinct parts of the form 2^j - 1, or let Let b(n) the number of parts in partitions of n into distinct parts of the form 2^j-1, then a(n) = factorial(b(n)).

STATUS

proposed

editing

#7 by Ilya Gutkovskiy at Sun Sep 25 06:40:13 EDT 2022

STATUS

editing

proposed

#6 by Ilya Gutkovskiy at Sun Sep 25 06:39:46 EDT 2022

CROSSREFS

Cf. A000929, A079559, A093659, A104977.

STATUS

proposed

editing

#5 by Ilya Gutkovskiy at Fri Sep 23 05:02:36 EDT 2022

STATUS

editing

proposed

Discussion

Sat Sep 24

09:33

Joerg Arndt: For each n, that partition is unique, right?

13:47

Ilya Gutkovskiy: Yes.

#4 by Ilya Gutkovskiy at Fri Sep 23 05:01:38 EDT 2022

COMMENTS

a(n) = Sum_{k=0..n} p(n,k) * k!, where p(n,k) = number of partitions of n into k distinct parts of the form 2^j - 1, or let b(n) the number of parts in partitions of n into distinct parts of the form 2^j-1, then a(n) = factorial(b(n)).

STATUS

proposed

editing

Discussion

Fri Sep 23

05:02

Ilya Gutkovskiy: Done.

#3 by Ilya Gutkovskiy at Mon Sep 19 12:11:34 EDT 2022

STATUS

editing

proposed

Discussion

Tue Sep 20

02:55

Joerg Arndt: Let b(n) the number of partitions into k distinct such parts (A-number?), then a(n) = factorial(b(n)), right?

04:30

Ilya Gutkovskiy: a(n) = Sum_{k=0..n} p(n,k) * k!, where p(n,k) = number of partitions of n into k distinct parts of the form 2^j - 1, or let b(n) the number of parts in partitions of n into distinct parts of the form 2^j-1, then a(n) = factorial(b(n)) (see A079559).

Fri Sep 23

03:20

N. J. A. Sloane: Maybe you could add those two remarks in the Comments section?

#2 by Ilya Gutkovskiy at Mon Sep 19 08:06:56 EDT 2022

NAME

allocated for Ilya GutkovskiyNumber of compositions (ordered partitions) of n into distinct parts of the form 2^j - 1.

DATA

1, 1, 0, 1, 2, 0, 0, 1, 2, 0, 2, 6, 0, 0, 0, 1, 2, 0, 2, 6, 0, 0, 2, 6, 0, 6, 24, 0, 0, 0, 0, 1, 2, 0, 2, 6, 0, 0, 2, 6, 0, 6, 24, 0, 0, 0, 2, 6, 0, 6, 24, 0, 0, 6, 24, 0, 24, 120, 0, 0, 0, 0, 0, 1, 2, 0, 2, 6, 0, 0, 2, 6, 0, 6, 24, 0, 0, 0, 2, 6, 0, 6, 24, 0, 0, 6, 24

OFFSET

0,5

LINKS

<a href="/index/Com#comp">Index entries for sequences related to compositions</a>

CROSSREFS

Cf. A000929, A079559, A104977.

KEYWORD

allocated

nonn

AUTHOR

Ilya Gutkovskiy, Sep 19 2022

STATUS

approved

editing