A052002 Numbers with an odd number of partitions.
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, 43, 44, 48, 49, 51, 52, 53, 54, 56, 60, 61, 63, 67, 68, 69, 71, 72, 73, 76, 77, 81, 82, 83, 85, 87, 88, 89, 90, 91, 92, 93, 95, 99, 102, 104, 105, 107, 111, 114, 115, 118, 119, 121
Offset: 1
Examples
From _Gus Wiseman_, Jan 13 2020: (Start) The partitions of the initial terms are: (1) (3) (4) (5) (6) (7) (21) (22) (32) (33) (43) (111) (31) (41) (42) (52) (211) (221) (51) (61) (1111) (311) (222) (322) (2111) (321) (331) (11111) (411) (421) (2211) (511) (3111) (2221) (21111) (3211) (111111) (4111) (22111) (31111) (211111) (1111111) (End)
Links
- Clark Kimberling, Table of n, a(n) for n = 1..1000
- O. Kolberg, Note on the parity of the partition function, Math. Scand. 7 1959 377-378. MR0117213 (22 #7995).
Crossrefs
Programs
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Haskell
import Data.List (findIndices) a052002 n = a052002_list !! (n-1) a052002_list = findIndices odd a000041_list -- Reinhard Zumkeller, Nov 03 2015
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Maple
N:= 1000: # to get all terms <= N V:= Vector(N+1): V[1]:= 1: for i from 1 to (N+1)/2 do V[2*i..N+1]:= V[2*i..N+1] + V[1..N-2*i+2] mod 2 od: select(t -> V[t+1]=1, [$1..N]); # Robert Israel, Jan 22 2017
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Mathematica
f[n_, k_] := Select[Range[250], Mod[PartitionsP[#], n] == k &] Table[f[2, k], {k, 0, 1}] (* Clark Kimberling, Jan 05 2014 *)
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PARI
for(n=0, 200, if(numbpart(n)%2==1, print1(n", "))) \\ Altug Alkan, Nov 02 2015
Formula
From Peter Bala, Jan 22 2016: (Start)
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)
Extensions
Offset corrected and b-file adjusted by Reinhard Zumkeller, Nov 03 2015
Comments