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Numbers with an odd number of partitions.
15

%I #47 Jan 13 2020 13:54:37

%S 0,1,3,4,5,6,7,12,13,14,16,17,18,20,23,24,29,32,33,35,36,37,38,39,41,

%T 43,44,48,49,51,52,53,54,56,60,61,63,67,68,69,71,72,73,76,77,81,82,83,

%U 85,87,88,89,90,91,92,93,95,99,102,104,105,107,111,114,115,118,119,121

%N Numbers with an odd number of partitions.

%C A052003(n) = A000041(a(n+1)). - _Reinhard Zumkeller_, Nov 03 2015

%C Also, numbers having an odd number of partitions into distinct odd parts; that is, numbers m such that A000700(m) is odd. For example, 16 is in the list since 16 has 5 partitions into distinct odd parts, namely, 1 + 15, 3 + 13, 5 + 11, 7 + 9 and 1 + 3 + 5 + 7. See Formula section for a proof. - _Peter Bala_, Jan 22 2017

%H Clark Kimberling, <a href="/A052002/b052002.txt">Table of n, a(n) for n = 1..1000</a>

%H O. Kolberg, <a href="http://www.mscand.dk/article/view/10584/8605">Note on the parity of the partition function</a>, Math. Scand. 7 1959 377-378. MR0117213 (22 #7995).

%F From _Peter Bala_, Jan 22 2016: (Start)

%F Sum_{n>=0} x^a(n) = (1 + x)*(1 + x^3)*(1 + x^5)*... taken modulo 2. Proof: Product_{n>=1} 1 + x^(2*n-1) = Product_{n>=1} (1 - x^(4*n-2))/(1 - x^(2*n-1)) = Product_{n>=1} (1 - x^(2*n))*(1 - x^(4*n-2))/( (1 - x^(2*n)) * (1 - x^(2*n-1)) ) = ( 1 + 2*Sum_{n>=1} (-1)^n*x^(2*n^2) )/(Product_{n>=1} (1 - x^n)) == 1/( Product_{n>=1} (1 - x^n) ) (mod 2). (End)

%e From _Gus Wiseman_, Jan 13 2020: (Start)

%e The partitions of the initial terms are:

%e (1) (3) (4) (5) (6) (7)

%e (21) (22) (32) (33) (43)

%e (111) (31) (41) (42) (52)

%e (211) (221) (51) (61)

%e (1111) (311) (222) (322)

%e (2111) (321) (331)

%e (11111) (411) (421)

%e (2211) (511)

%e (3111) (2221)

%e (21111) (3211)

%e (111111) (4111)

%e (22111)

%e (31111)

%e (211111)

%e (1111111)

%e (End)

%p N:= 1000: # to get all terms <= N

%p V:= Vector(N+1):

%p V[1]:= 1:

%p for i from 1 to (N+1)/2 do

%p V[2*i..N+1]:= V[2*i..N+1] + V[1..N-2*i+2] mod 2

%p od:

%p select(t -> V[t+1]=1, [$1..N]); # _Robert Israel_, Jan 22 2017

%t f[n_, k_] := Select[Range[250], Mod[PartitionsP[#], n] == k &]

%t Table[f[2, k], {k, 0, 1}] (* _Clark Kimberling_, Jan 05 2014 *)

%o (PARI) for(n=0, 200, if(numbpart(n)%2==1, print1(n", "))) \\ _Altug Alkan_, Nov 02 2015

%o (Haskell)

%o import Data.List (findIndices)

%o a052002 n = a052002_list !! (n-1)

%o a052002_list = findIndices odd a000041_list

%o -- _Reinhard Zumkeller_, Nov 03 2015

%Y Cf. A000041, A000700, A001560, A052001, A052003.

%Y The strict version is A001318, with complement A090864.

%Y The version for prime instead of odd numbers is A046063.

%Y The version for squarefree instead of odd numbers is A038630.

%Y The version for set partitions appears to be A032766.

%Y The version for factorizations is A331050.

%Y The version for strict factorizations is A331230.

%K nonn,easy

%O 1,3

%A _Patrick De Geest_, Nov 15 1999

%E Offset corrected and b-file adjusted by _Reinhard Zumkeller_, Nov 03 2015