# Revision history [back]

Note that in the answers of tmonteil you do not have the condition $a_1 \leq a_2 \leq \ldots \leq a_m$. To do so, you may use the very flexible IntegerListsLex as in

sage: V = IntegerListsLex(5, floor=lambda x: 1, min_slope=0)
sage: list(V)
[[5], [2, 3], [1, 4], [1, 2, 2], [1, 1, 3], [1, 1, 1, 2], [1, 1, 1, 1, 1]]


the argument floor specifies the lower bound for the element of the list, and the slope is the difference between two consecutive elemtns of the list. In particular, for increasing sequence you may use

sage: V = IntegerListsLex(7, floor=lambda x: 1, min_slope=1)
sage: list(V)
[[7], [3, 4], [2, 5], [1, 6], [1, 2, 4]]


Note that in the answers of tmonteil you do not have the condition $a_1 \leq a_2 \leq \ldots \leq a_m$. To do so, you may use the very flexible IntegerListsLex as in

sage: V = IntegerListsLex(5, floor=lambda x: 1, min_slope=0)
sage: list(V)
[[5], [2, 3], [1, 4], [1, 2, 2], [1, 1, 3], [1, 1, 1, 2], [1, 1, 1, 1, 1]]


the argument floor specifies the lower bound for the element of the list, and the slope is the difference between two consecutive elemtns of the list. In particular, for increasing sequence you may use

sage: V = IntegerListsLex(7, floor=lambda x: 1, min_slope=1)
sage: list(V)
[[7], [3, 4], [2, 5], [1, 6], [1, 2, 4]]


But note that with that solution you need to specify the sum of the elements.