A035363 Number of partitions of n into even parts.
1, 0, 1, 0, 2, 0, 3, 0, 5, 0, 7, 0, 11, 0, 15, 0, 22, 0, 30, 0, 42, 0, 56, 0, 77, 0, 101, 0, 135, 0, 176, 0, 231, 0, 297, 0, 385, 0, 490, 0, 627, 0, 792, 0, 1002, 0, 1255, 0, 1575, 0, 1958, 0, 2436, 0, 3010, 0, 3718, 0, 4565, 0, 5604, 0, 6842, 0, 8349, 0, 10143, 0, 12310, 0
Offset: 0
Examples
From _Joerg Arndt_, Jun 11 2013: (Start)
There are a(12)=11 symmetric unimodal compositions of 12+2=14 where the maximal part appears twice:
01: [ 1 1 1 1 1 2 2 1 1 1 1 1 ]
02: [ 1 1 1 1 3 3 1 1 1 1 ]
03: [ 1 1 1 4 4 1 1 1 ]
04: [ 1 1 2 3 3 2 1 1 ]
05: [ 1 1 5 5 1 1 ]
06: [ 1 2 4 4 2 1 ]
07: [ 1 6 6 1 ]
08: [ 2 2 3 3 2 2 ]
09: [ 2 5 5 2 ]
10: [ 3 4 4 3 ]
11: [ 7 7 ]
There are a(14)=15 symmetric unimodal compositions of 14 where the maximal part appears an even number of times:
01: [ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ]
02: [ 1 1 1 1 1 2 2 1 1 1 1 1 ]
03: [ 1 1 1 1 3 3 1 1 1 1 ]
04: [ 1 1 1 2 2 2 2 1 1 1 ]
05: [ 1 1 1 4 4 1 1 1 ]
06: [ 1 1 2 3 3 2 1 1 ]
07: [ 1 1 5 5 1 1 ]
08: [ 1 2 2 2 2 2 2 1 ]
09: [ 1 2 4 4 2 1 ]
10: [ 1 3 3 3 3 1 ]
11: [ 1 6 6 1 ]
12: [ 2 2 3 3 2 2 ]
13: [ 2 5 5 2 ]
14: [ 3 4 4 3 ]
15: [ 7 7 ]
(End)
a(8)=5 because we have [8], [6,2], [4,4], [4,2,2], and [2,2,2,2]. - _Emeric Deutsch_, Jan 27 2016
From _Gus Wiseman_, May 22 2021: (Start)
The a(0) = 1 through a(12) = 11 partitions into even parts are the following (empty columns shown as dots, A = 10, C = 12). The Heinz numbers of these partitions are given by A066207.
() . (2) . (4) . (6) . (8) . (A) . (C)
(22) (42) (44) (64) (66)
(222) (62) (82) (84)
(422) (442) (A2)
(2222) (622) (444)
(4222) (642)
(22222) (822)
(4422)
(6222)
(42222)
(222222)
(End)
References
- Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education, Vol. 31, No. 1, pp. 24-28, Winter 1997. MathEduc Database (Zentralblatt MATH, 1997c.01891).
- Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem II, Missouri Journal of Mathematical Sciences, Vol. 16, No. 1, Winter 2004, pp. 12-17. Zentralblatt MATH, Zbl 1071.05501.
Links
- Robert Price, Table of n, a(n) for n = 0..2001
Crossrefs
Bisection (even part) gives the partition numbers A000041.
Note: A-numbers of ranking sequences are in parentheses below.
The version for parts divisible by 3 instead of 2 is A035377.
The strict case is A035457.
The Heinz numbers of these partitions are given by A066207.
The ordered version (compositions) is A077957 prepended by (1,0).
This is column k = 2 of A168021.
The multiplicative version (factorizations) is A340785.
The following count partitions of even length:
Programs
-
Maple
ZL:= [S, {C = Cycle(B), S = Set(C), E = Set(B), B = Prod(Z,Z)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=0..69); # Zerinvary Lajos, Mar 26 2008 g := 1/mul(1-x^(2*k), k = 1 .. 100): gser := series(g, x = 0, 80): seq(coeff(gser, x, n), n = 0 .. 78); # Emeric Deutsch, Jan 27 2016 # Using the function EULER from Transforms (see link at the bottom of the page). [1,op(EULER([0,1,seq(irem(n,2),n=0..66)]))]; # Peter Luschny, Aug 19 2020 # next Maple program: a:= n-> `if`(n::odd, 0, combinat[numbpart](n/2)): seq(a(n), n=0..84); # Alois P. Heinz, Jun 22 2021 -
Mathematica
nmax = 50; s = Range[2, nmax, 2]; Table[Count[IntegerPartitions@n, x_ /; SubsetQ[s, x]], {n, 0, nmax}] (* Robert Price, Aug 05 2020 *) -
Python
from sympy import npartitions def A035363(n): return 0 if n&1 else npartitions(n>>1) # Chai Wah Wu, Sep 23 2023
Formula
G.f.: Product_{k even} 1/(1 - x^k).
Convolution with the number of partitions into distinct parts (A000009, which is also number of partitions into odd parts) gives the number of partitions (A000041). - Franklin T. Adams-Watters, Jan 06 2006
If n is even then a(n)=A000041(n/2) otherwise a(n)=0. - Omar E. Pol, Nov 20 2009
G.f.: 1 + x^2*(1 - G(0))/(1-x^2) where G(k) = 1 - 1/(1-x^(2*k+2))/(1-x^2/(x^2-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 23 2013
G.f.: exp(Sum_{k>=1} x^(2*k)/(k*(1 - x^(2*k)))). - Ilya Gutkovskiy, Aug 13 2018
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