1
2
3
47
6
Journal of Integer Sequences, Vol. 17 (2014),
Article 14.7.2
23 11
Asymptotic Series for Hofstadter’s
Figure-Figure Sequences
Benoı̂t Jubin1
Mathematics Research Unit
University of Luxembourg
6 rue Coudenhove-Kalergi
L-1359 Luxembourg City
Grand-Duchy of Luxembourg
benoit.jubin@uni.lu
Abstract
We compute asymptotic series for Hofstadter’s figure-figure sequences.
1
Introduction
We consider disjoint partitions of the set of strictly positive integers into two subsets such
that one set, B, consists of the differences of consecutive elements of the other set, A,
and a given difference appears at most once. There are many such partitions. We call a
the (strictly increasing) sequence enumerating A, and b the (injective) sequence of its first
differences, both with offset 1. Hofstadter’s figure-figure sequences are the sequences a and b
corresponding to the partition with the set A lexicographically minimal. This is equivalent
to b being increasing. The sequences read
an = 1, 3, 7, 12, 18, 26, 35, 45, 56, 69, . . .
bn = 2, 4, 5, 6, 8, 9, 10, 11, 13, 14, . . .
(OEIS A005228),
(OEIS A030124).
These sequences were introduced by Hofstadter in [2, p. 73]. They appear as an example
of complementary sequences in [3]. Their asymptotic behavior does not seem to be given
1
Supported by the Luxembourgish FNR via the AFR Postdoc Grant Agreement PDR 2012-1.
1
anywhere in the literature except for the asymptotic equivalents mentioned by Hasler and
Wilson in the related OEIS entries [1]. In this article, we compute asymptotic series for these
sequences.
Pn−1
We have by definition bn = an+1 −an , so an = 1+ k=1
bk . Since the sequence a is strictly
increasing, given any n ≥ 1, there is a unique k ≥ 1 such that ak − k < n ≤ ak+1 − (k + 1).
This defines a sequence u by letting un be this k. Therefore,
a(un ) − un < n ≤ a(un + 1) − (un + 1).
(1)
The sequence u is non-decreasing (actually, un+1 − un ∈ {0, 1}) and u1 = 1. It reads
un = 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, . . .
(OEIS A225687).
The partition condition implies
b n = n + un .
As a consequence,
n−1
(n − 1)n X
+
uk .
an = 1 +
2
k=1
2
(2)
Bounds and asymptotic equivalents
Since un ≥ 1, we have an ≥ 21 n(n + 1). Therefore, the left inequality of (1) implies 12 un (un +
q
1) − un ≤ n − 1, or u2n − un − 2(n − 1) ≤ 0, so un ≤ 21 + 14 + 2(n − 1), and finally
This implies n + 1 ≤ bn < n +
√
1 ≤ un <
√
1
2n + .
2
2n + 12 , so
bn ∼ n.
Pn−1 √
( 2k + 21 ). Since the
The upper bound on u implies in turn an < 1 + 21 (n − 1)n + k=1
Rn√
Pn−1 √
√
function x is strictly increasing, we have k=1 k < 1 x dx = 32 (n3/2 − 1). Therefore
n2 n
n2 23/2 3/2 1
+ ≤ an <
+
n −
2
2
2
3
3
and in particular
n2
.
2
3/2 1
2
2
3
− 3 and the right inequality of (1) imply n < (un +1)
The relation an < n2 + 23 n2
+
2
3/2
23/2
4
3/2
2
(un + 1) − un − 3 , which implies un → +∞. Therefore 2n ≤ un + O(un ), but we saw
3
√
√
√
that un = O( n), so O(un 3/2 ) ⊆ O(n3/4 ) ⊆ o(n), so u2n ≥ 2n + o(n), so un ≥ 2n + o( n).
Combining this with the above upper bound, we obtain
√
un ∼ 2n
√
and in particular O(un ) = O( n).
an ∼
2
3
Asymptotic series
2
Since an ∼ n2 , we have an+1 − an = O(n). Now (1) gives a(un ) = n + O(un ). On the other
Pn−1
P n −1
2
2
hand, (2)√gives an = n2 + k=1
uk + O(n), therefore u2n + uk=1
uk = n + O(un ). Since
un = O( n),
√ we can increment the upper limit of the summation index by 1, and since
O(un ) = O( n), we obtain the main relation
u
n
√
u2n X
+
uk = n + O( n).
2
k=1
We are now ready to prove by induction that for all K ≥ 1, we have the asymptotic
expansion
K
1+(k−1)k/2 1/2k
X
n
k+1 2
1/2K
un =
(−1) Qk−1
.
(3)
+o n
j
2
j=1 (2 + 1)
k=1
√
Indeed, the case K = 1 reduces to un ∼ 2n, which we already proved. We also prove
the case
since it is slightly different from the general case. We write
√ K = 2 separately √
un = 2n + vn with vn = o( n). We have
√
u2n
v2
− n = 2n vn + n .
2
2
√
We do not know a priori that vn2 = O( n), and that is why we have to prove this case
separately. We also have
un
un
un
X
√
√ X
X
23/2 3/2
3/2
k+
vk =
un + o O(un )
+
vk .
uk = 2
3
k=1
k=1
k=1
k=1
un
X
P un
√ 3/2 3/4
3/2
n)
We have
v
=
o
O(
⊆
o(n
)
and
o
O(u
)
⊆ o(n3/4 ). We also have
k
n
k=1
√
un 3/2 ∼ ( 2n)3/2 = (2n)3/4 . Therefore,
u
n
√
v 2 29/4 3/4
u2n X
+
n + o(n3/4 ).
uk − n = 2n vn + n +
2
2
3
k=1
√
√
This has to be O( n) by the main relation. Dividing the right-hand side by 2n, we obtain
vn 2
27/4 1/4
vn + √ +
n = o(n1/4 ).
3
2 2n
√
Since vn = o( n), we have
v√n 2
2 2n
= o(vn ), so
vn +
27/4 1/4
n = o(n1/4 ) + o(vn ),
3
3
1/4
2
so vn ∼ − 23 n2
, as desired.
Now, suppose that the expansion holds for some K ≥ 2. We prove it for K + 1. It will
be convenient to denote the coefficients of the expansion by
αk = (−1)
K
so α1 = 2. We write vn = o n1/2
k+1
21+(k−1)k/2
,
Qk−1 j
j=1 (2 + 1)
for the remainder in (3). Then (3) gives
!2
!2
K
K
n 1/2k
n 1/2k
X
X
√
1
v
n
2n +
u2n =
αk
αk
= 2n 1 + √
+ vn
+√
2
2
2n k=2
2n
k=2
!
K
vn2
vn
2 X n 1/2k
1/4+1/4−1
1/4−1
αk
+O
+O n
vn .
+ 2√ + O n
= 2n 1 + √
2
n
2n k=2
2n
K
⊆ o n1/4 . Therefore
Since K ≥ 2, we have vn = o n1/2
On the other hand,
K
n 1/2+1/2k √
X
√
u2n
−n=2
αk
+ 2n vn + O( n).
2
2
k=2
un
X
K
X
u 1+1/2k
2k
n
1+1/2K
αk
uk =
2 k
+ o un
2 +1
2
k=1
k=1
Therefore
K
n 1/2+1/2k+1
X
2k+1
1/2+1/2K+1
α
+
o
n
.
=
k
k +1
2
2
k=1
un
K
K
n 1/2+1/2k X
X
u2n X
2k+1 n 1/2+1/2k+1
+
uk − n = 2
αk
+
αk k
2
2
2 +1 2
k=1
k=2
k=1
√
K+1
+ 2n vn + o n1/2+1/2
n 1/2
2K+1 n 1/2+1/2K+1
K+1
= αK K
+2
vn + o n1/2+1/2
2 +1 2
2
since the terms in the sums cancel out except for the last in the second sum. This expression
1/2K+1
√
K
has to be O( n) by the main relation, so vn ∼ − 2K2 +1 αK n2
, as desired.
From the expansion of un , we find that of bn = n + un , and that of an by term-by-term
integration. We obtain
bn = n +
K
X
k=1
(−1)
k+1
21+(k−1)k/2 n 1/2k
1/2K
+o n
Qk−1 j
2
j=1 (2 + 1)
4
and
K
an =
4
2k(k+1)/2 n 1+1/2k
n2 X
K
+
(−1)k+1 Qk
+ o n1+1/2 .
j
2
2
j=1 (2 + 1)
k=1
Acknowledgments
I would like to thank Neil J. A. Sloane for useful comments on a first version of this article,
and Clark Kimberling and Maximilian Hasler for their advice.
References
[1] The On-line Encyclopedia of Integer Sequences (OEIS), published electronically at
http://oeis.org, 2013.
[2] Douglas R. Hofstadter, Gödel, Escher, Bach: an Eternal Golden Braid, Basic Books,
1979.
[3] Clark Kimberling, Complementary equations, J. Integer Sequences, 10 (2007),
Article 07.1.4.
2010 Mathematics Subject Classification: Primary 41A60.
Keywords: Hofstadter sequence, asymptotic series.
(Concerned with sequences A005228, A030124, A225687.)
Received April 8 2014; revised version received May 23 2014. Published in Journal of Integer
Sequences, June 10 2014.
Return to Journal of Integer Sequences home page.
5