%I #36 Jan 14 2024 11:48:56
%S 1,1,1,1,2,2,3,3,5,5,7,7,11,11,15,15,22,22,30,30,42,42,56,56,77,77,
%T 101,101,135,135,176,176,231,231,297,297,385,385,490,490,627,627,792,
%U 792,1002,1002,1255,1255,1575,1575,1958,1958,2436,2436,3010,3010,3718,3718
%N Number of partitions of floor(3n/2) into n parts each from {1,2,...,n}.
%C The bisection {1,1,2,3,5,7,11,15,22,...} agrees with the initial terms of A008641, Number of partitions of n into at most 12 parts and also A008635, Molien series for A_12.
%C a(2n+1)=a(2n) for all n>0. If the partition {...,1} is a member of a(2n) then the partition {...,1,1} is a member of a(2n+1). - _Robert G. Wilson v_, Jun 09 2006
%C Number of partitions of n where all parts (except for possibly the first part) are even; see example. [_Joerg Arndt_, Apr 22 2013]
%C For n >= 2, a(n) = number of partitions p of n such that floor(n/2) is a part of p. For n >= 1, a(n) = number of partitions p of n such that ceiling(n/2) is a part of p. - _Clark Kimberling_, Feb 28 2014
%C From _Gus Wiseman_, Oct 28 2021: (Start)
%C If we insert zeros every three terms, this counts partitions of n such that n = floor(3*k/2), where k is the number of parts. This counts by sum rather than length. These partitions are ranked by A347452.
%C Also the number of integer partitions of n with alternating product 1, where the alternating product of a sequence (y_1,...,y_k) is Product_i y_i^((-1)^(i-1)). These are the conjugates of the partitions (ranked by A336119) described in Arndt's comment above. For example, the a(2) = 1 through a(10) = 7 partitions are:
%C 11 111 22 221 33 331 44 441 55
%C 1111 11111 2211 22111 2222 22221 3322
%C 111111 1111111 3311 33111 4411
%C 221111 2211111 222211
%C 11111111 111111111 331111
%C 22111111
%C 1111111111
%C These partitions are ranked by A028982. The odd-length case is A035363 (shifted), which is also the version for sum instead of product. The multiplicative version (factorizations) is A347438.
%C (End)
%F a(n) = A000041(floor(n/2)). - _Vladeta Jovovic_, Jun 10 2006
%F G.f.: ( Sum_{n>=0, x^(2*n) / prod(k=1..n, 1-x^k ) ) / (1 - x). - _Michael Somos_, Mar 01 2014
%e For n=8, Floor[3n/2] is 12 and there are five partitions of 12 into 8 parts each in the range 1-8 inclusive, namely: {5,1,1,1,1,1,1,1}, {4,2,1,1,1,1,1,1}, {3,3,1,1,1,1,1,1}, {3,2,2,1,1,1,1,1} and {2,2,2,2,1,1,1,1}. Thus a(8)=5.
%e From _Joerg Arndt_, Apr 22 2013: (Start)
%e a(8) = a(9) = 5, counting the following partitions where all parts (except for possibly the first part) are even:
%e 01: [ 2 2 2 2 ]
%e 02: [ 4 2 2 ]
%e 03: [ 4 4 ]
%e 04: [ 6 2 ]
%e 05: [ 8 ]
%e and
%e 01: [ 3 2 2 2 ]
%e 02: [ 5 2 2 ]
%e 03: [ 5 4 ]
%e 04: [ 7 2 ]
%e 05: [ 9 ]
%e (End)
%e G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 3*x^7 + 5*x^8 + 5*x^9 + 7*x^10 + ...
%p # Using the function EULER from Transforms (see link at the bottom of the page).
%p [1, op(EULER([1,0,seq(irem(n,2),n=2..55)]))]; # _Peter Luschny_, Aug 19 2020
%t (* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) f[n_] := f[n] = Length@ Select[ Partitions[ Floor[3n/2], n], Length@# == n &]; Table[ If[n > 1, f[2Floor[n/2]], f[n]], {n, 57}] (* _Robert G. Wilson v_, Jun 09 2006 *)
%t Table[ PartitionsP[ Floor[n/2]], {n, 57}] (* _Robert G. Wilson v_, Jun 09 2006 *)
%t Table[Count[IntegerPartitions[n], p_ /; MemberQ[p, Ceiling[n/2]]], {n, 50}] (* _Clark Kimberling_, Feb 28 2014 *)
%t a[ n_] := SeriesCoefficient[ (1 + x) / QPochhammer[x^2], {x, 0, n}]; (* _Michael Somos_, Mar 01 2014 *)
%o (PARI) a(n)=numbpart(n\2); \\ _Joerg Arndt_, Apr 22 2013
%Y Cf. A008641, A008635.
%Y Both bisections are A000041.
%Y An adjoint version is A108711.
%Y A027187 counts partitions of even length.
%Y A027193 counts partitions of odd length.
%Y A325534 counts separable partitions.
%Y A325535 counts inseparable partitions.
%Y Cf. A000070, A001700, A028983, A086543, A182616, A236559, A236914, A344654, A347444, A347452.
%K nonn
%O 0,5
%A _John W. Layman_, Jun 07 2006
%E More terms from _Robert G. Wilson v_, Jun 09 2006
%E Added a(0)=1. - _Michael Somos_, Mar 01 2014