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%I #56 Nov 23 2016 11:30:06
%S 0,4,12,24,48,80,140,216,344,512,768,1100,1596,2224,3120,4272,5852,
%T 7860,10576,13992,18520,24208,31596,40824,52696,67404,86088,109176,
%U 138180,173812,218252,272540,339708,421464,521848,643504,792056,971248,1188804,1450348,1766184,2144416,2599164,3141748,3791248,4563780
%N a(n) = sum of the perimeters of the Ferrers boards of the partitions of n. Also, sum of the perimeters of the diagrams of the regions of the set of partitions of n.
%C a(n) is also 4 times the total number of parts in all partitions of n.
%C Hence a(n) is also 4 times the sum of largest parts of all partitions of n.
%C Hence a(n) is also twice the total number of parts in all partitions of n plus twice the sum of largest parts of all partitions of n.
%C a(n) is also the sum of the perimeters of the first n polygons constructed with the Dyck path (and its mirror) that arises from the minimalist diagram of the regions of the set of partitions of n. The n-th odd-indexed segment of the diagram has A141285(n) up-steps and the n-th even-indexed segment has A194446(n) down-steps. The k-th polygon of the diagram is associated to the k-th section of the set of partitions of n, with 1<=k<=n. See the bottom of Example section. For the definition of "section" see A135010. For the definition of "region" see A206437.
%F a(n) = 4*A006128(n) = 2*A211978(n).
%F a(n) = 2*A225600(2*A000041(n)) = 2*A225600(A139582(n)), n >= 1.
%F a(n) = 2*(Sum_{m=1..p(n)} A194446(m)) + (Sum_{m=1..p(n)} A141285(m))) = 4*Sum_{m=1..p(n)} A194446(m) = 4*Sum_{m=1..p(n)} A141285(m), where p(n) = A000041(n), n >= 1.
%e For n = 5 consider the partitions of 5 in colexicographic order (as shown in the 5th row of the triangle A211992) and its associated diagram of regions as shown below:
%e . Regions Minimalist
%e . Partitions of 5 diagram version
%e . _ _ _ _ _
%e . 1, 1, 1, 1, 1 |_| | | | | _| | | | |
%e . 2, 1, 1, 1 |_ _| | | | _ _| | | |
%e . 3, 1, 1 |_ _ _| | | _ _ _| | |
%e . 2, 2, 1 |_ _| | | _ _| | |
%e . 4, 1 |_ _ _ _| | _ _ _ _| |
%e . 3, 2 |_ _ _| | _ _| |
%e . 5 |_ _ _ _ _| _ _ _ _ _|
%e .
%e Then consider the following table which contains the Ferrers boards of the partitions of 5 and the diagram of every region of the set of partitions of 5:
%e -------------------------------------------------------------------------
%e | Partitions | | | Regions | | |
%e | of 5 | Ferrers | Peri- | of 5 | Region | Peri- |
%e |(See A211992)| board | meter |(see A220482)| diagram | meter |
%e -------------------------------------------------------------------------
%e | _ | _ |
%e | 1 |_| | 1 |_| 4 |
%e | 1 |_| | _ |
%e | 1 |_| | 1 _|_| |
%e | 1 |_| | 2 |_|_| 8 |
%e | 1 |_| 12 | _ |
%e | _ _ | 1 |_| |
%e | 2 |_|_| | 1 _ _|_| |
%e | 1 |_| | 3 |_|_|_| 12 |
%e | 1 |_| | _ _ |
%e | 1 |_| 12 | 2 |_|_| 6 |
%e | _ _ _ | _ |
%e | 3 |_|_|_| | 1 |_| |
%e | 1 |_| | 1 |_| |
%e | 1 |_| 12 | 1 _|_| |
%e | _ _ | 2 _ _|_|_| |
%e | 2 |_|_| | 4 |_|_|_|_| 18 |
%e | 2 |_|_| | _ _ _ |
%e | 1 |_| 10 | 3 |_|_|_| 8 |
%e | _ _ _ _ | _ |
%e | 4 |_|_|_|_| | 1 |_| |
%e | 1 |_| 12 | 1 |_| |
%e | _ _ _ | 1 |_| |
%e | 3 |_|_|_| | 1 |_| |
%e | 2 |_|_| 10 | 1 _|_| |
%e | _ _ _ _ _ | 2 _ _ _|_|_| |
%e | 6 |_|_|_|_|_| 12 | 5 |_|_|_|_|_| 24 |
%e | | |
%e -------------------------------------------------------------------------
%e | Sum of perimeters: 80 <-- equals --> 80 |
%e -------------------------------------------------------------------------
%e The sum of the perimeters of the Ferrers boards is 12 + 12 + 12 + 10 + 12 + 10 + 12 = 80, so a(5) = 80.
%e On the other hand, the sum of the perimeters of the diagrams of regions is 4 + 8 + 12 + 6 + 18 + 8 + 24 = 80, equaling the sum of the perimeters of the Ferrers boards.
%e .
%e Illustration of first six polygons of an infinite diagram constructed with the boundary segments of the minimalist diagram of regions and its mirror (note that the diagram looks like reflections on a mountain lake):
%e 11............................................................
%e . /\
%e . / \
%e . / \
%e 7................................... / \
%e . /\ / \
%e 5..................... / \ /\/ \
%e . /\ / \ /\ / \
%e 3........... / \ / \ / \/ \
%e 2....... /\ / \ /\/ \ / \
%e 1... /\ / \ /\/ \ / \ /\/ \
%e 0 /\/ \/ \/ \/ \/ \
%e . \/\ /\ /\ /\ /\ /
%e . \/ \ / \/\ / \ / \/\ /
%e . \/ \ / \/\ / \ /
%e . \ / \ / \ /\ /
%e . \/ \ / \/ \ /
%e . \ / \/\ /
%e . \/ \ /
%e . \ /
%e . \ /
%e . \ /
%e . \/
%e n:
%e . 0 1 2 3 4 5 6
%e Perimeter of the n-th polygon:
%e . 0 4 8 12 24 32 60
%e a(n) is the sum of the perimeters of the first n polygons:
%e . 0 4 12 24 48 80 140
%e .
%e For n = 5, the sum of the perimeters of the first five polygons is 4 + 8 + 12 + 24 + 32 = 80, so a(5) = 80.
%e For n = 6, the sum of the perimeters of the first six polygons is 4 + 8 + 12 + 24 + 32 + 60 = 140, so a(6) = 140.
%e For another version of the above diagram see A228109.
%Y Cf. A000041, A006128, A135010, A138137, A139582, A141285, A194446, A211992, A220482, A225600, A211978, A233968, A244968.
%K nonn
%O 0,2
%A _Omar E. Pol_, Nov 19 2016