Alternating Permutations
Richard P. Stanley
M.I.T.
Alternating Permutations – p. 1
Definitions
A sequence a1 , a2 , . . . , ak of distinct integers is
alternating if
a1 > a 2 < a 3 > a 4 < · · · ,
and reverse alternating if
a1 < a 2 > a 3 < a 4 > · · · .
Alternating Permutations – p. 2
Definitions
A sequence a1 , a2 , . . . , ak of distinct integers is
alternating if
a1 > a 2 < a 3 > a 4 < · · · ,
and reverse alternating if
a1 < a 2 > a 3 < a 4 > · · · .
Sn : symmetric group of all permutations of
1, 2, . . . , n
Alternating Permutations – p. 2
En
En = #{w ∈ Sn : w is alternating}
= #{w ∈ Sn : w is reverse alternating}
via
a1 a2 · · · an 7→ n + 1 − a1 , n + 1 − a2 , . . . , n + 1 − an
Alternating Permutations – p. 3
En
En = #{w ∈ Sn : w is alternating}
= #{w ∈ Sn : w is reverse alternating}
via
a1 a2 · · · an 7→ n + 1 − a1 , n + 1 − a2 , . . . , n + 1 − an
E.g., E4 = 5: 2143, 3142, 3241, 4132, 4231
Alternating Permutations – p. 3
André’s theorem
Theorem (Désiré André, 1879)
xn
En = sec x + tan x
y :=
n!
n≥0
X
Alternating Permutations – p. 4
André’s theorem
Theorem (Désiré André, 1879)
xn
En = sec x + tan x
y :=
n!
n≥0
X
En is an Euler number,
E2n a secant number,
E2n−1 a tangent number.
Alternating Permutations – p. 4
Naive proof
52439
| {z } 1826
|{z}
←−
−→
alternating alternating
Alternating Permutations – p. 5
Naive proof
52439
| {z } 1826
|{z}
←−
−→
alternating alternating
Obtain w that is either alternating or reverse
alternating.
Alternating Permutations – p. 5
Naive proof
52439
| {z } 1826
|{z}
←−
−→
alternating alternating
Obtain w that is either alternating or reverse
alternating.
⇒ 2En+1 =
n
X
k=0
n
Ek En−k , n ≥ 1
k
Alternating Permutations – p. 5
Completion of proof
2En+1 =
n X
n
k=0
k
Ek En−k , n ≥ 1
Alternating Permutations – p. 6
Completion of proof
2En+1 =
n X
n
k=0
k
Ek En−k , n ≥ 1
xn
2
En+1
= 2y 0
n!
n≥0
!
n
X X n
xn
Ek En−k
= 1 + y2
n!
k
n≥0
X
k=0
Alternating Permutations – p. 6
Completion of proof
2En+1 =
n X
n
k=0
k
Ek En−k , n ≥ 1
xn
2
En+1
= 2y 0
n!
n≥0
!
n
X X n
xn
Ek En−k
= 1 + y2
n!
k
n≥0
X
k=0
⇒ 2y 0 = 1 + y 2 , etc.
Alternating Permutations – p. 6
A generalization
∃ more sophisticated approaches.
E.g., let
fk (n) = #{w ∈ Sn : w(r) < w(r + 1) ⇔ k|r}
f2 (n) = En
Alternating Permutations – p. 7
A generalization
∃ more sophisticated approaches.
E.g., let
fk (n) = #{w ∈ Sn : w(r) < w(r + 1) ⇔ k|r}
f2 (n) = En
Theorem.
X
k≥0
xkn
fk (kn)
=
(kn)!
kn
x
(−1)n
(kn)!
n≥0
X
!−1
.
Alternating Permutations – p. 7
Combinatorial trigonometry
Define sec x and tan x by
xn
En .
sec x + tan x =
n!
n≥0
X
Alternating Permutations – p. 8
Combinatorial trigonometry
Define sec x and tan x by
xn
En .
sec x + tan x =
n!
n≥0
X
Exercise. sec2 x = 1 + tan2 x
Alternating Permutations – p. 8
Combinatorial trigonometry
Define sec x and tan x by
xn
En .
sec x + tan x =
n!
n≥0
X
Exercise. sec2 x = 1 + tan2 x
tan x + tan y
Exercise. tan(x + y) =
1 − (tan x)(tan y)
Alternating Permutations – p. 8
Some occurences of Euler numbers
(1) E2n−1 is the number of complete increasing
binary trees on the vertex set
[2n + 1] = {1, 2, . . . , 2n + 1}.
Alternating Permutations – p. 9
Five vertices
1
2
2
3
4
5
4
3
5
5
2
4
2 4
1
3
5
4
2
5
4
5
3
5
4
4
3
1
3
2
4
5
4
3
1
2
3
5
2
1
4
3
1
5
3
5
2
2
1
3
4
5
4
1
3
1
2
2
5
2
4
1
3
5
1
2
3
1
3
1
2
1
4
1
1
2
5
3
5
4
Alternating Permutations – p. 10
Five vertices
1
2
2
3
4
5
4
3
5
5
2
4
2 4
1
3
5
4
2
5
4
5
3
5
4
4
3
1
3
2
4
5
4
3
1
2
3
5
2
1
4
3
1
5
3
5
2
2
1
3
4
5
4
1
3
1
2
2
5
2
4
1
3
5
1
2
3
1
3
1
2
1
4
1
1
2
5
3
5
4
Slightly more complicated for E2n
Alternating Permutations – p. 10
Simsun permutations
Definition. A permutation w = a1 · · · an ∈ Sn is
simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk
of w to 1, 2, . . . , k has no double descents
bi−1 > bi > bi+1 .
Alternating Permutations – p. 11
Simsun permutations
Definition. A permutation w = a1 · · · an ∈ Sn is
simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk
of w to 1, 2, . . . , k has no double descents
bi−1 > bi > bi+1 .
Example. 346251 is not simsun: restriction to
1, 2, 3, 4 is 3421.
Alternating Permutations – p. 11
Simsun permutations
Definition. A permutation w = a1 · · · an ∈ Sn is
simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk
of w to 1, 2, . . . , k has no double descents
bi−1 > bi > bi+1 .
Example. 346251 is not simsun: restriction to
1, 2, 3, 4 is 3421.
Theorem (R. Simion and S. Sundaram) The
number of simsun permutations w ∈ Sn is En .
Alternating Permutations – p. 11
Chains of partitions
(2) Start with n one-element sets {1}, . . . , {n}.
Alternating Permutations – p. 12
Chains of partitions
(2) Start with n one-element sets {1}, . . . , {n}.
Merge together two at a time until reaching
{1, 2, . . . , n}.
Alternating Permutations – p. 12
Chains of partitions
(2) Start with n one-element sets {1}, . . . , {n}.
Merge together two at a time until reaching
{1, 2, . . . , n}.
1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6
125−34−6, 125−346, 123456
Alternating Permutations – p. 12
Chains of partitions
(2) Start with n one-element sets {1}, . . . , {n}.
Merge together two at a time until reaching
{1, 2, . . . , n}.
1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6
125−34−6, 125−346, 123456
Sn acts on these chains.
Alternating Permutations – p. 12
Chains of partitions
(2) Start with n one-element sets {1}, . . . , {n}.
Merge together two at a time until reaching
{1, 2, . . . , n}.
1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6
125−34−6, 125−346, 123456
Sn acts on these chains.
Theorem. The number of Sn -orbits is En−1 .
Alternating Permutations – p. 12
Orbit representatives for n = 5
12−3−4−5
123−4−5
1234−5
12−3−4−5
123−4−5
123−45
12−3−4−5
12−34−5
125−34
12−3−4−5
12−34−5
12−345
12−3−4−5
12−34−5
1234−5
Alternating Permutations – p. 13
A polytope volume
Let Pn be the convex polytope in Rn with
equations
xi ≥ 0, 1 ≤ i ≤ n
xi + xi+1 ≤ 1, 1 ≤ i ≤ n − 1.
Alternating Permutations – p. 14
A polytope volume
Let Pn be the convex polytope in Rn with
equations
xi ≥ 0, 1 ≤ i ≤ n
xi + xi+1 ≤ 1, 1 ≤ i ≤ n − 1.
1
Theorem. vol(Pn ) = En
n!
Alternating Permutations – p. 14
Ulam’s problem
Let E(n) be the expected length is(w) of the
longest increasing subsequence of w.
Alternating Permutations – p. 15
Ulam’s problem
Let E(n) be the expected length is(w) of the
longest increasing subsequence of w.
w = 5241736 ⇒ is(w) = 3
Alternating Permutations – p. 15
Two highlights
Long story . . .
Alternating Permutations – p. 16
Two highlights
Long story . . .
Logan-Shepp, Vershik-Kerov:
√
E(n) ∼ 2 n
Alternating Permutations – p. 16
Two highlights
Long story . . .
Logan-Shepp, Vershik-Kerov:
√
E(n) ∼ 2 n
Write isn (w) for is(w) when w ∈ Sn .
Baik-Deift-Johansson:
√
isn (w) − 2 n
lim Prob
≤ t = F (t),
1/6
n→∞
n
the Tracy-Widom distribution.
Alternating Permutations – p. 16
Alternating subsequences?
as(w) = length of longest alternating subseq. of w
w = 56218347 ⇒ as(w) = 5
Alternating Permutations – p. 17
The main lemma
MAIN LEMMA. ∀ w ∈ Sn ∃ alternating
subsequence of maximal length that contains n.
Alternating Permutations – p. 18
The main lemma
MAIN LEMMA. ∀ w ∈ Sn ∃ alternating
subsequence of maximal length that contains n.
ak (n) = #{w ∈ Sn : as(w) = k}
bk (n) = a1 (n) + a2 (n) + · · · + ak (n)
= #{w ∈ Sn : as(w) ≤ k}.
Alternating Permutations – p. 18
Recurrence for ak (n)
⇒ ak (n) =
X
2r+s=k−1
n X
n−1
j=1
j−1
(a2r (j − 1) + a2r+1 (j − 1)) as (n − j)
Alternating Permutations – p. 19
The main generating function
Define B(x, t) =
X
k,n≥0
n
x
bk (n)tk
n!
Theorem.
2/ρ
1
B(x, t) =
1−ρ ρx − ρ ,
1− t e
√
where ρ= 1 − t2 .
Alternating Permutations – p. 20
Formulas for bk (n)
Corollary.
⇒ b1 (n) = 1
b2 (n) = n
b3 (n) = 14 (3n − 2n + 3)
b4 (n) =
..
.
1 n
8 (4
− (2n − 4)2n )
Alternating Permutations – p. 21
Formulas for bk (n)
Corollary.
⇒ b1 (n) = 1
b2 (n) = n
b3 (n) = 14 (3n − 2n + 3)
b4 (n) =
..
.
1 n
8 (4
− (2n − 4)2n )
no such formulas for longest increasing
subsequences
Alternating Permutations – p. 21
Mean (expectation) of as(w)
1 X
D(n) =
as(w),
n!
w∈Sn
the expectation of as(n) for w ∈ Sn
Alternating Permutations – p. 22
Mean (expectation) of as(w)
1 X
D(n) =
as(w),
n!
w∈Sn
the expectation of as(n) for w ∈ Sn
Let
n
x
A(x, t) =
ak (n)tk
= (1 − t)B(x, t)
n!
k,n≥0
!
1
2/ρ
.
= (1 − t)
1−ρ ρx −
ρ
1− t e
X
Alternating Permutations – p. 22
Formula for D(n)
X
n≥0
D(n)x
n
∂
=
A(x, 1)
∂t
6x − 3x2 + x3
=
6(1 − x)2
X 4n + 1
= x+
xn .
6
n≥2
Alternating Permutations – p. 23
Formula for D(n)
X
n≥0
D(n)x
n
∂
=
A(x, 1)
∂t
6x − 3x2 + x3
=
6(1 − x)2
X 4n + 1
= x+
xn .
6
n≥2
4n + 1
, n≥2
⇒ A(n) =
6
Alternating Permutations – p. 23
Formula for D(n)
X
n≥0
D(n)x
n
∂
=
A(x, 1)
∂t
6x − 3x2 + x3
=
6(1 − x)2
X 4n + 1
= x+
xn .
6
n≥2
4n + 1
, n≥2
⇒ A(n) =
6
√
Recall E(n) ∼ 2 n.
Alternating Permutations – p. 23
Variance of as(w)
2
X
4n + 1
1
as(w) −
V (n) =
, n≥2
n!
6
w∈Sn
the variance of as(n) for w ∈ Sn
Alternating Permutations – p. 24
Variance of as(w)
2
X
4n + 1
1
as(w) −
V (n) =
, n≥2
n!
6
w∈Sn
the variance of as(n) for w ∈ Sn
Corollary.
8
13
V (n) = n −
, n≥4
45
180
Alternating Permutations – p. 24
Variance of as(w)
2
X
4n + 1
1
as(w) −
V (n) =
, n≥2
n!
6
w∈Sn
the variance of as(n) for w ∈ Sn
Corollary.
8
13
V (n) = n −
, n≥4
45
180
similar results for higher moments
Alternating Permutations – p. 24
A new distribution?
P (t) = lim Probw∈Sn
n→∞
as(w) − 2n/3
√
≤t
n
Alternating Permutations – p. 25
A new distribution?
P (t) = lim Probw∈Sn
n→∞
as(w) − 2n/3
√
≤t
n
Stanley distribution?
Alternating Permutations – p. 25
Limiting distribution
Theorem (Pemantle, Widom, (Wilf)).
as(w) − 2n/3
√
≤t
lim Probw∈Sn
n→∞
n
1
=√
π
(Gaussian distribution)
Z
√
t 45/4
e
−s2
ds
−∞
Alternating Permutations – p. 26
Limiting distribution
Theorem (Pemantle, Widom, (Wilf)).
as(w) − 2n/3
√
≤t
lim Probw∈Sn
n→∞
n
1
=√
π
(Gaussian distribution)
Z
√
t 45/4
e
−s2
ds
−∞
Alternating Permutations – p. 26
Umbral enumeration
Umbral formula: involves E k , where E is an
indeterminate (the umbra). Replace E k with the
Euler number Ek . (Technique from 19th century,
modernized by Rota et al.)
Alternating Permutations – p. 27
Umbral enumeration
Umbral formula: involves E k , where E is an
indeterminate (the umbra). Replace E k with the
Euler number Ek . (Technique from 19th century,
modernized by Rota et al.)
Example.
(1 + E 2 )3 =
=
=
=
1 + 3E 2 + 3E 4 + E 6
1 + 3E2 + 3E4 + E6
1 + 3 · 1 + 3 · 5 + 61
80
Alternating Permutations – p. 27
Another example
(1 + t)E =
=
=
=
=
E 2
E 3
t +
t + ···
1 + Et +
2
3
1
1 + Et + E(E − 1)t2 + · · ·
2
1
1 + E1 t + (E2 − E1 ))t2 + · · ·
2
1
1 + t + (1 − 1)t2 + · · ·
2
1 + t + O(t3 ).
Alternating Permutations – p. 28
Alt. fixed-point free involutions
fixed point free involution w ∈ S2n : all cycles of
length two
Alternating Permutations – p. 29
Alt. fixed-point free involutions
fixed point free involution w ∈ S2n : all cycles of
length two
Let f (n) be the number of alternating
fixed-point free involutions in S2n .
Alternating Permutations – p. 29
Alt. fixed-point free involutions
fixed point free involution w ∈ S2n : all cycles of
length two
Let f (n) be the number of alternating
fixed-point free involutions in S2n .
n = 3 : 214365 = (1, 2)(3, 4)(5, 6)
645231 = (1, 6)(2, 4)(3, 5)
f (3) = 2
Alternating Permutations – p. 29
An umbral theorem
Theorem.
F (x) =
X
f (n)xn
n≥0
Alternating Permutations – p. 30
An umbral theorem
Theorem.
F (x) =
X
f (n)xn
n≥0
=
1+x
1−x
(E 2 +1)/4
Alternating Permutations – p. 30
An umbral theorem
Theorem.
F (x) =
X
f (n)xn
n≥0
=
1+x
1−x
(E 2 +1)/4
Proof. Uses representation theory of the
symmetric group Sn . In particular, there is a
“nice” representation of dimension En .
Alternating Permutations – p. 30
Ramanujan’s Lost Notebook
Theorem (Ramanujan, Berndt, implicitly) As
x → 0+,
X 1 − x n(n+1) X
∼
f (k)xk = F (x),
2
1
+
x
n≥0
k≥0
an analytic (non-formal) identity.
Alternating Permutations – p. 31
A formal identity
Corollary (via Ramanujan, Andrews).
Qn
2j−1
X
(1
−
q
)
j=1
n
F (x) = 2
,
q Q2n+1
j)
(1
+
q
j=1
n≥0
where q =
1−x 2/3
1+x
, a formal identity.
Alternating Permutations – p. 32
Simple result, hard proof
Recall: number of n-cycles in Sn is (n − 1)!.
Alternating Permutations – p. 33
Simple result, hard proof
Recall: number of n-cycles in Sn is (n − 1)!.
Theorem. Let b(n) be the number of
alternating n-cycles in Sn . Then if n is odd,
1X
b(n) =
µ(d)(−1)(d−1)/2 En/d .
n
d|n
Alternating Permutations – p. 33
Special case
Corollary. Let p be an odd prime. Then
1
Ep − (−1)(p−1)/2 .
b(p) =
p
Alternating Permutations – p. 34
Special case
Corollary. Let p be an odd prime. Then
1
Ep − (−1)(p−1)/2 .
b(p) =
p
Combinatorial proof?
Alternating Permutations – p. 34
Another such result
Recall: the expected number of fixed points of
w ∈ Sn is 1.
Alternating Permutations – p. 35
Another such result
Recall: the expected number of fixed points of
w ∈ Sn is 1.
J (n) = expected number of fixed points of
reverse alternating w ∈ Sn
1
Theorem. J(2n) =
(E2n − (−1)n ).
E2n
Similar (but more complicated) for n odd or for
alternating permutations.
Alternating Permutations – p. 35
Another such result
Recall: the expected number of fixed points of
w ∈ Sn is 1.
J (n) = expected number of fixed points of
reverse alternating w ∈ Sn
1
Theorem. J(2n) =
(E2n − (−1)n ).
E2n
Similar (but more complicated) for n odd or for
alternating permutations.
Combinatorial proof?
Alternating Permutations – p. 35
Descent sets
Let w = a1 a2 · · · an ∈ Sn . Descent set of w:
D(w) = {i : ai > ai+1 } ⊆ {1, . . . , n − 1}
Alternating Permutations – p. 36
Descent sets
Let w = a1 a2 · · · an ∈ Sn . Descent set of w:
D(w) = {i : ai > ai+1 } ⊆ {1, . . . , n − 1}
D(4157623) = {1, 4, 5}
D(4152736) = {1, 3, 5} (alternating)
D(4736152) = {2, 4, 6} (reverse alternating)
Alternating Permutations – p. 36
βn(S)
βn (S) = #{w ∈ Sn : D(w) = S}
Alternating Permutations – p. 37
βn(S)
βn (S) = #{w ∈ Sn : D(w) = S}
w D(w)
123
∅
213 {1}
312 {1}
132 {2}
231 {2}
321 {1, 2}
β3 (∅) = 1, β3 (1) = 2
β3 (2) = 2, β3 (1, 2) = 1
Alternating Permutations – p. 37
uS
Fix n. Let S ⊆ {1, · · · , n − 1}. Let uS = t1 · · · tn−1 ,
where
(
a, i 6∈ S
ti =
b, i ∈ S.
Alternating Permutations – p. 38
uS
Fix n. Let S ⊆ {1, · · · , n − 1}. Let uS = t1 · · · tn−1 ,
where
(
a, i 6∈ S
ti =
b, i ∈ S.
Example. n = 8, S = {2, 5, 6} ⊆ {1, . . . , 7}
uS = abaabba
Alternating Permutations – p. 38
A noncommutative gen. function
Ψn (a, b) =
X
βn (S)uS .
S⊆{1,...,n−1}
Alternating Permutations – p. 39
A noncommutative gen. function
Ψn (a, b) =
X
βn (S)uS .
S⊆{1,...,n−1}
Example. Recall
β3 (∅) = 1, β3 (1) = 2, β3 (2) = 2, β3 (1, 2) = 1
Thus
Ψ3 (a, b) = aa + 2ab + 2ba + bb
Alternating Permutations – p. 39
A noncommutative gen. function
Ψn (a, b) =
X
βn (S)uS .
S⊆{1,...,n−1}
Example. Recall
β3 (∅) = 1, β3 (1) = 2, β3 (2) = 2, β3 (1, 2) = 1
Thus
Ψ3 (a, b) = aa + 2ab + 2ba + bb
= (a + b)2 + (ab + ba)
Alternating Permutations – p. 39
The cd-index
Theorem. There exists a noncommutative
polynomial Φn (c, d), called the cd-index of Sn ,
with nonnegative integer coefficients, such that
Ψn (a, b) = Φn (a + b, ab + ba).
Alternating Permutations – p. 40
The cd-index
Theorem. There exists a noncommutative
polynomial Φn (c, d), called the cd-index of Sn ,
with nonnegative integer coefficients, such that
Ψn (a, b) = Φn (a + b, ab + ba).
Example. Recall
Ψ3 (a, b) = aa + 2ab + 2ba + b2 = (a + b)2 + (ab + ba).
Therefore
Φ3 (c, d) = c2 + d.
Alternating Permutations – p. 40
Small values of Φn(c, d)
Φ1
Φ2
Φ3
Φ4
Φ5
Φ6
=
=
=
=
=
=
1
c
c2 + d
c3 + 2cd + 2dc
c4 + 3c2 d + 5cdc + 3dc2 + 4d2
c5 + 4c3 d + 9c2 dc + 9cdc2 + 4dc3
+12cd2 + 10dcd + 12d2 c.
Alternating Permutations – p. 41
The set Sµ
Recall: w = a1 · · · an ∈ Sn is simsun if for all
1 ≤ k ≤ n, the restriction b1 · · · bk of w to
1, 2, . . . , k has no double descents
bi−1 > bi > bi+1 .
Alternating Permutations – p. 42
The set Sµ
Recall: w = a1 · · · an ∈ Sn is simsun if for all
1 ≤ k ≤ n, the restriction b1 · · · bk of w to
1, 2, . . . , k has no double descents
bi−1 > bi > bi+1 .
Given a cd-monomial µ, let c → 0, d → 10, and
remove final 0.
Example.
µ = cd2 c2 d → 01010001,
the characteristic vector of Sµ = {2, 4, 8}.
Alternating Permutations – p. 42
The coefficients of Φn(c, d)
Theorem (Simion-Sundaram) Let µ be a
cd-monomial of degree n − 1. The coefficient of µ
in Φn (c, d) is the number of simsun permutations
w ∈ Sn−1 with descent set Sµ .
Alternating Permutations – p. 43
The case n = 4
w D(w)
µ
123
∅
c3 → 00
213
1
dc → 10
312
1
dc → 10
132
2
cd → 01
231
2
cd → 01
⇒ Φ4 (c, d) = c3 + 2dc + 2cd
Alternating Permutations – p. 44
Expansion of cd-monomials
Let m = m(c, d) be a cd-monomial, e.g., c2 d2 cd.
Expand m(a + b, ab + ba):
(a + b)2 (ab + ba)2 (a + b)(ab + ba) = a9 + ba8 + · · · .
Alternating Permutations – p. 45
Expansion of cd-monomials
Let m = m(c, d) be a cd-monomial, e.g., c2 d2 cd.
Expand m(a + b, ab + ba):
(a + b)2 (ab + ba)2 (a + b)(ab + ba) = a9 + ba8 + · · · .
Key observation: m(a + b, ab + ba) is a sum of
distinct monomials, always including ababab · · ·
and bababa · · · .
Alternating Permutations – p. 45
Alternating permutations
The coefficient of ababab · · · and of bababa · · · in
Ψn (a, b) is
βn (2, 4, 6, . . . ) = βn (1, 3, 5, . . . ) = En .
Alternating Permutations – p. 46
Alternating permutations
The coefficient of ababab · · · and of bababa · · · in
Ψn (a, b) is
βn (2, 4, 6, . . . ) = βn (1, 3, 5, . . . ) = En .
Hence:
Theorem. (a) Φn (1, 1) = En (can interpret
coefficients combinatorially).
(b) For all S ⊆ {1, . . . , n − 1},
βn (S) ≤ βn (1, 3, 5, . . . ) = En .
Alternating Permutations – p. 46
An example
Φ5 = 1c4 + 3c2 d + 5cdc + 3dc2 + 4d2
1 + 3 + 5 + 3 + 4 = 16 = E5
Alternating Permutations – p. 47
Comments
N OTE . (b) is due originally to Niven and later de
Bruijn (different proofs).
Alternating Permutations – p. 48
Comments
N OTE . (b) is due originally to Niven and later de
Bruijn (different proofs).
N OTE . Not hard to show that
βn (S) < En
unless S = {1, 3, 5, . . . } or S = {2, 4, 6, . . . }.
Alternating Permutations – p. 48
Alternating Permutations – p. 49