A001255
Squares of partition numbers.
Original entry on oeis.org
1, 1, 4, 9, 25, 49, 121, 225, 484, 900, 1764, 3136, 5929, 10201, 18225, 30976, 53361, 88209, 148225, 240100, 393129, 627264, 1004004, 1575025, 2480625, 3833764, 5934096, 9060100, 13823524, 20839225, 31404816, 46812964, 69705801, 102880449, 151536100
Offset: 0
A260664
Number of ordered triples of partitions of n with no common parts.
Original entry on oeis.org
1, 0, 6, 18, 90, 192, 864, 1710, 5970, 13110, 36810, 75984, 210546, 410130, 1003908, 2045808, 4616730, 8950176, 19746720, 37297710, 78247344, 147410640, 294299424, 543058032, 1067679540, 1925323308, 3653769792, 6555529158, 12129597486, 21348640230
Offset: 0
a(3) = 18 because of the 18 triples of partitions of 3: (3,3,21), (3,3,111), (3,21,3), (3,21,21), (3,21,111), (3,111,3), (3,111,21), (3,111,111), (21,3,3), (21,3,21), (21,3,111), (21,21,3), (21,111,3), (111,3,3), (111,3,21), (111,3,111), (111,21,3) and (111,111,3);
a(3) = A000041(3-A001318(0))^3 - A000041(3-A001318(1))^3 - A000041(3-A001318(2))^3 = 3^3 - 2^3 - 1^3 = 27 - 8 - 1 = 18.
- Reinhard Zumkeller, Table of n, a(n) for n = 0..5000
- Sylvie Corteel, Carla D. Savage, Herbert S. Wilf, Doron Zeilberger, A pentagonal number sieve, J. Combin. Theory Ser. A 82 (1998), no. 2, 186-192.
- Eric Weisstein's World of Mathematics, Pentagonal Number Theorem
- Wikipedia, Pentagonal number theorem
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a260664 = sum . zipWith (*) a087960_list . map a133042 . a260672_row
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Table[Sum[(Cos[Pi*j/2] - Sin[Pi*j/2]) * PartitionsP[n - ((6*j^2 + 6*j + 1)/16 - (2*j + 1)*(-1)^j/16)]^3, {j, 0, Floor[Sqrt[8*n/3]]}], {n, 0, 30}] (* Vaclav Kotesovec, Jul 04 2016 *)
nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^3*x^k, {k, 0, nmax}] / Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 04 2016 *)
A304873
G.f.: Sum_{k>=0} p(k)^4 * x^k / Sum_{k>=0} p(k)*x^k, where p(n) is the partition function A000041(n).
Original entry on oeis.org
1, 0, 14, 64, 528, 1696, 11616, 33600, 169072, 525760, 2069922, 5928066, 22259874, 59321760, 193797792, 526647420, 1566376990, 4012181104, 11456306798, 28263784110, 75995086336, 184440427360, 468750673616, 1104027571108, 2730165482640, 6239956155696
Offset: 0
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nmax = 25; CoefficientList[Series[Sum[PartitionsP[k]^4*x^k, {k, 0, nmax}] / Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x]
A129668
Number of different ways to divide an n X n X n cube into subcubes, considering only the list of parts.
Original entry on oeis.org
1, 2, 3, 11, 19, 121, 291, 1656
Offset: 1
a(3) = 3 because the 3 X 3 X 3 cube can be divided into subcubes in 3 different ways: a single 3 X 3 X 3 cube, a 2 X 2 X 2 plus 19 1 X 1 X 1 cubes, or 27 1 X 1 X 1 cubes.
a(4) = 11 because the 4 X 4 X 4 cube can be divided into 11 different combinations of subcubes. The table below lists each of the 11 combinations and gives the number of ways those subcubes can be arranged:
(1) 64 1 X 1 X 1 cubes in 1 way
(2) 56 1 X 1 X 1 cubes and 1 2 X 2 X 2 cube in 27 ways
(3) 48 1 X 1 X 1 cubes and 2 2 X 2 X 2 cubes in 193 ways
(4) 40 1 X 1 X 1 cubes and 3 2 X 2 X 2 cubes in 544 ways
(5) 32 1 X 1 X 1 cubes and 4 2 X 2 X 2 cubes in 707 ways
(6) 24 1 X 1 X 1 cubes and 5 2 X 2 X 2 cubes in 454 ways
(7) 16 1 X 1 X 1 cubes and 6 2 X 2 X 2 cubes in 142 ways
(8) 8 1 X 1 X 1 cubes and 7 2 X 2 X 2 cubes in 20 ways
(9) 8 2 X 2 X 2 cubes in 1 way
(10) 37 1 X 1 X 1 cubes and 1 3 X 3 X 3 cube in 8 ways
(11) 1 4 X 4 X 4 cube in 1 way
The total number of arrangements is 2098 = A228267(4,4,4).
Cf.
A034295 (same problem in 2 dimensions rather than 3).
A304992
G.f.: Sum_{k>=0} A000041(k)^3 * x^k / Sum_{k>=0} A000009(k) * x^k.
Original entry on oeis.org
1, 0, 7, 18, 98, 210, 969, 1938, 7037, 15258, 44815, 93180, 262391, 518550, 1311015, 2657328, 6189160, 12124098, 27239760, 52063668, 111630480, 211503288, 432900236, 806091180, 1610854427, 2940167268, 5691072911, 10289144976, 19402974147, 34523231688
Offset: 0
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nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^3*x^k, {k, 0, nmax}] / Sum[PartitionsQ[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x]
Showing 1-5 of 5 results.
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