# A general term of this sequence [closed]

My question is about the set theory.
I have c[0]={{1}} and i want to define c[n] as follows:

c[n] = Prepend 1 to each element of c[n-1] UNION add 1 to the first element of each element of c[n-1].


For example:

c[1]={{1,1},{2}}

c[2]={{1,1,1},{2,1},{1,2},{3}}

c[3]={{1,1,1,1},{2,1,1},{1,2,1},{3,1},{1,1,2},{2,2},{1,3},{4}}


• Welcome to Mma.SE! Your question needs more from your side. Here it's considered helpful and polite to show your own efforts and share your data and code attempts in a well formatted form, so we can quickly see the problem you are facing. What have you tried? Please help us to help you and edit your question accordingly. Also, please take the tour, it will help you understand the site. If you write an excellent question it will inspire great answers. May 6 at 14:15
• Did you search the site for similar questions? Did you find 190310, 126150, 224738? Why do those questions' answers don't answer yours? May 6 at 14:19

Prepend has an curried form:

Prepend[1] /@ {{a, b}, {c, d, e}}
(* outputs {{1, a, b}, {1, c, d, e}} *)


MapAt can be used to apply a function to parts of a list:

MapAt[1 + # &, {{a, b}, {c, d, e}}, {All, 1}]
(* {{1 + a, b}, {1 + c, d, e}} *)


The function Union does unions. So, we have a pretty straightforward implementation:

c[0] = {{1}};
c[n_] := Union[Prepend[1] /@ c[n - 1], MapAt[1 + # &, c[n - 1], {All, 1}]]


It might be worth adding checks to avoid calculating c[n] with negative or non-integer n. It might also be worth using some memoization. These choices would depend on your context.

• Great Thanks for your help Dear @lericr. May 6 at 17:01