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Derived Sequences
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47
6
Journal of Integer Sequences, Vol. 6 (2003),
Article 03.1.1
23 11
Derived Sequences
G. L. Cohen
Department of Mathematical Sciences, Faculty of Science
University of Technology, Sydney
PO Box 123, Broadway, NSW 2007
Australia
and
D. E. Iannucci
Division of Science and Mathematics
University of the Virgin Islands
St. Thomas, VI 00802
USA
Abstract
We define a multiplicative arithmetic function D by assigning D(pa ) = apa−1 , when p is a
prime and a is a positive integer, and, for n ≥ 1, we set D 0 (n) = n and D k (n) = D(D k−1 (n))
when k ≥ 1. We term {D k (n)}∞
k=0 the derived sequence of n. We show that all derived
sequences of n < 1.5 · 1010 are bounded, and that the density of those n ∈ N with bounded
derived sequences exceeds 0.996, but we conjecture nonetheless the existence of unbounded
sequences. Known bounded derived sequences end (effectively) in cycles of lengths only 1
to 6, and 8, yet the existence of cycles of arbitrary length is conjectured. We prove the
existence of derived sequences of arbitrarily many terms without a cycle.
1
Introduction
Define a multiplicative arithmetic function D by assigning
D(pa ) = apa−1 ,
(1)
when p is a prime and a is a positive integer. The multiplicativity implies D(1) = 1 and,
for example, D(pa q b ) = abpa−1 q b−1 , where q b is another prime power, q 6= p. It is only
1
the shape of the definition (1) that encourages us to use freely terms from calculus. (Our
derivatives have no relationship to an earlier use of the term, in Apostol [1], for example.)
Writing D 0 (n) = n and D k (n) = D(D k−1 (n)) for k ≥ 1 and any positive integer n, we call
2
3
{Dk (n)}∞
k=0 , or {n, D(n), D (n), D (n), . . . }, the derived sequence of n, and denote this by
D(n). We refer to D k (n) for k ≥ 1 as the kth derivative of n and we refer to D j (n) for
0 ≤ j < kQ
as integrals of D k (n).
ai
If n = w
i=1 pi is the prime decomposition of n then the definition of D(n) may be given
differently as
¶
µ
n
n
D(n) =
,
τ
C(n)
C(n)
Q
Qw
where τ (n) = w
i=1 (ai + 1) is the number of divisors of n and C(n) =
i=1 pi is the core of
n.
Our intention is to initiate a study of the ultimate behaviour of derived sequences. Different forms of ultimate behaviour are indicated in the following examples:
D(52 17 · 37) = {52 17 · 37, 2 · 5, 1, . . . },
D(225 ) = {225 , 224 52 , 227 3 · 5, 226 33 , 226 33 13, . . . },
D(1316) = {1316 , 24 1315 , 25 3 · 5 · 1314 , 25 5 · 7 · 1313 , 24 5 · 1313 , 25 1313 , . . . },
D(232 ) = {232 , 236 , 237 32 , 237 3 · 37, 236 37, . . . },
D(38 ) = {38 , 23 37 , 22 37 · 7, 22 36 · 7, 23 36 , . . . }.
(2)
(3)
(4)
(5)
(6)
The underlined terms in each case form cycles. Precisely: if
D(n) = {. . . , D j (n), Dj+1 (n), . . . , D j+k−1 (n), Dj+k (n), . . . }
and Dj+k (n) = D j (n), where j ≥ 0 and k ≥ 1 is the smallest integer with this property,
then Dj (n), Dj+1 (n), . . . , D j+k−1 (n) is a derived k-cycle, which, if we need to, we describe as
being arrived at in j + k iterations of D. For example, in (3), D(226 33 13) = 226 33 13, and, in
(6), D(23 36 ) = 23 37 . We have 1-cycles in (2) and (3), and 2-, 3- and 4-cycles in (4), (5) and
(6), respectively. The 3-cycle in (5) is arrived at in four iterations of D. We will refer to the
element of a 1-cycle as a fixed point of D.
It is not known whether the ultimate behaviour of D(n) is a cycle for all n, or whether, for
some n, Dk (n) increases without bound as k increases. We will show, however, that cycles
result for more than 99.5% of values of n. Many iterations of D may be required before a
cycle is reached, if that is to be the case: for example, D(563 ) arrives in 531 iterations at
the fixed point 21403 3329 5106 715 23 · 47 · 53 · 61. Our most impressive example is D(1735 1939 ),
which, in 443507 iterations of D, arrives at the fixed point
24318267 31370053 5525835 7159649 1133429 1320597 171037 191349 23299 3131
· 43 · 61 · 712 479 · 1013 · 22807 · 105167 · 1370053 · 4318267.
Other instances of sequences of iterated arithmetic functions are given by Guy [3]. Iteration of the function σ(n) − n, for example, where σ is the sum-of-divisors function, has
been studied extensively. The situation is similar: there is an eventual iterate equal to 1, or
2
there is eventually a cycle, or the ultimate behaviour is unknown. Iteration of the function
σ itself was studied in Cohen and te Riele [2].
In the following, p, q, r and t, with and without subscripts, denote prime numbers, and
s, with and without subscripts, denotes a squarefree number. We include 1 as a squarefree
number. Other letters denote positive integers, unless specified otherwise.
2
General results
We are able to give a number of results of a general nature.
Qw a i
First,
it
is
easy
to
see
that
n
is
a
fixed
point
of
D
if
and
only
if
either
n
=
1
or
n
=
i=1 pi ,
Qw
Qw
where i=1 pi = i=1 ai . More
we have
1, below. ForQ
simplicity of
Q generally,
Q Proposition
ai0
a0
a1
0
notation,Qwe will write n =
p0 and D(n) =
p1 as shorthand for n = w
i=1 pi0 and
ai1
1
D(n) = w
i=1 pi1 , and so on.
Q ak−1
Q a0
Q a1
D k−1 (n) = pk−1
.
= Q
p1 , . . . ,Q
Proposition 1 Suppose n > 1 andQ
writeQn = Q
p0 , D(n) Q
k
We have D (n) = n if and only if p0 p1 · · · pk−1 = a0 a1 · · · ak−1 .
Proof. Note that
n=
D(n) =
p0
Y
pa00 ,
pa11 ,
so
Y
pa11 =
Y
D2 (n) =
..
.
Y
pa22 ,
so
Y
pa22 =
Dk−1 (n) =
Y
k−1
pk−1
, so
Y
k−1
pk−1
=
If, further, D k (n) = n then
Y
Y
Y
p1 · · ·
The converse is also clear.
Q
a
pa00 =
Y
pk−1
¤
Q
a
−1
a
Y
a0 p0a0 −1 ,
a1 p1a1 −1 ,
Y
a
−1
k−2
ak−2 pk−2
.
k−1
ak−1 pk−1
, and we have
Q ak−1
Q a1 Q a2
Q a0
pk−1
p2
p1
p0
= Q a0 −1 Q a1 −1 · · · Q ak−2 −1 Q ak−1
−1
p0
p1
pk−2
pk−1
Y
Y Y
a1 · · ·
ak−1 .
=
a0
It does not seem to be easy to use this result to determine cycles, but we can at least
identify those numbers which have a derivative equal to the fixed point 1 of D:
Proposition 2 The integer n > 1 has
• first derivative 1 if n = s,
• second derivative 1 if n = p2 s where p > 2 and p - s,
• third derivative 1 if n = p3 s where p > 3 and p - s, or n = p3 q 2 s where p > 3, q > 3,
p 6= q and (s, pq) = 1.
3
There are no other situations in which n has a derivative equal to 1.
The proof is a matter of recognising those situations in which 22 or 33 might arise as
exact factors of terms in the derived sequence, and avoiding them since these factors will
persist in subsequent differentiations.
It is also a matter of checking that 2- and 3-cycles are obtained in the following situations.
Proposition 3 For any p,
• if p2 + 1 is squarefree, then D(pp
2 +1
) = {pp
2 +1
2
, (p2 + 1)pp , . . . },
• if p3 + 2 and p3 + 1 are squarefree, then
D(pp
3 +2
) = {pp
3 +2
, (p3 + 2)pp
3 +1
3
, (p3 + 1)pp , . . . }.
The underlined terms are cycles. Notice that 13, 37, 61, . . . , are primes p such that
p + 2 and p3 + 1 are squarefree. Certainly, 2- and 3-cycles may arise in other ways, as in
the examples (4) and (5).
We can also give a general instance that leads to a 4-cycle:
3
Proposition 4 Let s be such that s ≡ 2 (mod 3), 4s−1 is squarefree and 2s−1 is squarefree.
Then D(34s ) results in a 4-cycle.
Proof. Write 4s − 1 = s1 and 2s − 1 = 3s2 (where 3 - s2 ). If s is odd, then
D(34s ) = {34s , 22 34s−1 s, 22 34s−2 s1 , 23 34s−2 s2 , 23 34s−1 s2 , 22 34s−1 s1 , . . . },
while if s is even, then
D(34s ) = {34s , 23 34s−1 (s/2), 22 34s−1 s1 , 22 34s−2 s1 , 23 34s−2 s2 , 23 34s−1 s2 , . . . }.
The underlined terms in each case are 4-cycles (in fact, algebraically, the same 4-cycle).
¤
The smallest permissible value of s in this proposition is s = 2, as in the example (6);
thereafter, s may take the values 11, 17, 26, 29, 35, . . . .
Other examples of 4-cycles are not difficult to find, and we need not always start at a
prime power. For example:
D(210 310 ) = {210 310 , 211 39 52 , 211 310 5 · 11, 211 39 5 · 11, 210 310 · 11, . . . }.
It was initially more difficult to find examples of derived k-cycles for k > 4, but, having
found a few, patterns were detected suggesting infinite families of these. Some are described
in the following propositions. (We have other, more general, examples.)
Proposition 5 Let s be such that (s, 2 · 3 · 5 · 47) = 1, and suppose (s1 , 5 · 23 · 47) = 1, where
s1 = (3s − 1)/2. Put n = 23s 345 55 23s1 . Then n, D(n), . . . , D 4 (n) is a 5-cycle.
In this, s may take the values 1, 13, 23, 29, 41, 53, 61, . . . .
4
Proposition 6 (a) Let s be such that (s, 2 · 3 · 5 · 7 · 53) = 1, and suppose (s 1 , 5 · 53 · 107) = 1,
where s1 = 6s + 1. Put n = 26s 3105 56 7s1 . Then n, D(n), . . . , D 5 (n) is a 6-cycle.
(b) Let s be such that (s, 2 · 3 · 5 · 13 · 43) = 1, and suppose (s1 , 7 · 13 · 43 · 131) = 1, where
s1 = 30s + 1. Put n = 230s 3129 56 7 · 43s1 . Then n, D(n), . . . , D 5 (n) is a 6-cycle.
Proof. As usual, the proofs are a matter of straightforward verification. We will demonstrate
this in the case of Proposition 6(a). We have
n = 26s 3105 56 7s1 ,
D(n) = 26s+1 3107 56 7s,
D2 (n) = 26s+1 3107 55 107s1 ,
D3 (n) = 26s 3106 55 107s1 ,
D4 (n) = 26s+1 3106 55 53s,
D5 (n) = 26s+1 3105 55 53s1 ,
D6 (n) = 26s 3105 56 7s1 ,
since
since
since
since
since
since
(s1 , 2 · 3 · 5 · 7) = 1,
(s, 2 · 3 · 5 · 7) = 1,
(s1 , 2 · 3 · 5 · 107) = 1,
(s1 , 2 · 3 · 5 · 107) = 1,
(s, 2 · 3 · 5 · 53) = 1,
(s1 , 2 · 3 · 5 · 53) = 1.
But D6 (n) = n. We have also used the fact that s and s1 are squarefree.
¤
In Proposition 6(a), we may have s = 11, 13, 17, 23, 31, 37, 41, . . . . In Proposition 6(b),
s may take the values 1, 7, 11, 19, 23, 37, 41, . . . .
Proposition 7 Let s be such that s ≡ 7 (mod 10) and (s, 3 · 23 · 31 · 47 · 103 · 311) = 1, and
suppose (s1 , 3 · 5 · 31 · 47 · 103 · 311) = 1, where s1 = (2s + 1)/5. Put n = 22s 3311 546 103s1 .
Then n, D(n), . . . , D 7 (n) is an 8-cycle.
In this, s may take the values 17, 107, 167, 197, 227, . . . . Proposition 7 was found by
observing that D(513 2954 ) arrives in 428 iterations of D at the 8-cycle beginning with 2929 n,
with s = 557. Two other examples of 8-cycles turned up in our searches:
2159 316725 55 79 · 8363,
2159 316725 56 7 · 79 · 8363,
2158 316727 56 7 · 43 · 53 · 389,
2158 316727 55 43 · 53 · 389,
2158 316726 57 53 · 223,
2159 316727 57 53 · 223,
2159 316727 55 43 · 79 · 389,
2158 316726 55 43 · 79 · 389,
287 3149325 55 43 · 197 · 379,
287 3149325 56 7 · 43 · 197 · 379,
286 3149327 56 7 · 29 · 31 · 4817,
286 3149327 55 29 · 31 · 4817,
286 3149326 57 11 · 29 · 181,
287 3149327 57 11 · 29 · 181,
287 3149327 55 31 · 43 · 4817,
286 3149326 55 31 · 43 · 4817.
and
These occur in D(316695 ) and D(3149319 ), respectively. It is not difficult to determine a twoparameter family, containing general exponents on 2 and 3, that includes both of these
8-cycles.
We have no examples of derived k-cycles with k = 7 or k > 8.
5
3
Bounded derived sequences
We have two results on the number of bounded derived sequences, the first resulting largely
from a direct search, the second of a much more theoretical nature.
Proposition 8 For all n < 1.5 · 1010 , the derived sequence D(n) is bounded.
Q w ai
Proof. The proof involved
a
direct
incremental
investigation
of
all
numbers
n
=
i=1 pi <
P
1.5 · 1010 for which w
(a
−
1)
≥
8.
(An
initial
factorisation
of
each
n
determined
whether
i=1 i
this condition was satisfied.)
also that the
P In all cases, D(n) resulted in a cycle. We showed
3 2
same is true of all n with w
(a
−
1)
≤
7.
For
example,
suppose
n
=
p
q
s,
where p 6= q
i=1 i
2
2
and (s, pq) = 1. If p and q are both greater than 3, then D(n) = 2 · 3p q, D (n) = 2p and
D3 (n) = 1 (or use Proposition 2); then the numbers p3 22 s (p > 3), p3 32 s (p > 3), 23 q 2 s
(q > 3), 33 q 2 s (q > 3), 23 32 s and 33 22 s must be separately and similarly considered. ¤
It would seem probable that D(n) is unbounded for some n, and in that case for all
numbers ns, where (n, s) = 1, as well. Then the set of such numbers would have positive
density in N. We show now that this density is less than 0.004. We have computed lower
bounds for the densities of 45 classes of integers, including the known result for the set of
squarefree numbers, and have shown that integers in these 45 classes have bounded derived
sequences. In each case, the density was computed to within 10−6 and then truncated to
five decimal digits. Those densities (33 of the 45) which gave a positive lower bound (to
that number of digits) are given in Table 1. We refer, for example, to the type p 3 q 2 S
as the set of integers of the form p3 q 2 s, where p 6= q and (s, pq) = 1. How the given
densities were obtained will be illustrated shortly by the determination of such for the type
p3 q 2 S. The 45 classes of integers were all possible types of the form pa q b · · · S such that
(a − 1) + (b − 1) + · · · ≤ 7 (precisely as considered in the proof of Proposition 8), and Table 1
shows that their cumulative density is at least 0.996.
We show now how we obtain that the density of the class p3 q 2 S is 0.01447, truncated to
five decimal digits.
Let x > 0 be given. In general, the number of positive integers n ≤ x, not divisible by
the prime squares p21 , p22 , . . . , p2l and not divisible by the primes q1 , q2 , . . . , qm (all these
being different primes) is
x·
l µ
Y
i=1
1
1− 2
pi
¶ Y
¶
m µ
1
·
+ O(1).
1−
q
i
i=1
√
Note that the positive integer n ≤ x is squarefree if p2 - n for all primes p ≤ x. Fix
distinct primes p and q. As above, the number of squarefree positive integers n ≤ x/p 3 q 2
which are divisible by neither p nor q is
6
x
p3 q 2
µ
1
1−
p
x
= 3 2
pq
=x·
µ
¶µ
¶
1
1−
q
1
1−
p
¶µ
r≤
1
1−
q
Y
√
x/p3 q 2
r6=p, q
¶µ
1
1
·
2
p (p + 1) q(q + 1)
r≤
¶
µ
1
1 − 2 + O(1)
r
1
1− 2
p
¶−1 µ
Y
µ
¶
1
1 − 2 + O(1),
r
√
x/p3 q 2
1
1− 2
q
¶−1
r≤
Y
√
x/p3 q 2
µ
1
1− 2
r
¶
+ O(1)
the products being taken over primes
Q r.
In general, for y > 0 we have r≤y (1 − 1/r2 ) = 6/π 2 + O(1/y). Applying this above, the
number of squarefree positive integers n ≤ x/p3 q 2 which are divisible by neither p nor q is
then
Ãr
Ã
!!
1
p3 q 2
1
6
x· 2
·
+O
+ O(1)
p (p + 1) q(q + 1) π 2
x
µr
¶
1
1
6x
x
·
+O
.
= 2· 2
π p (p + 1) q(q + 1)
p3 q 2
To find the number of positive integers n ≤ x of the form n = p3 q 2 s, where p,
q are
√
3
any p
two distinct primes and (s, pq) = 1, we sum our result above over primes p ≤ x and
q ≤ x/p3 . Therefore the proportion of these integers p3 q 2 s which are at most x is given by
¶¶
µr
µ
1
1
6x
x
1 X X
·
·
+O
x √
π 2 p2 (p + 1) q(q + 1)
p3 q 2
√
3
p≤
=
x q≤
X
p≤
√
3
x/p3
q6=p
X
x q≤
√
x/p3
q6=p
1
1
6
· 2
·
+O
2
π p (p + 1) q(q + 1)
µX
p≤
√
3
X
x q≤
√
x/p3
1
¶
p
.
xp3 q 2
P
P
In general, p≤y 1/p3/2 = O(1) and q≤y 1/q = O(log log y), and so
µX X
¶
µ
¶
¶
µ
log log x
1
1 X 1 X 1
p
p
√
O
=O
.
=O √
3
3q2
x √
x
√ q
√
√
xp
p
3
3
q≤ x
p≤ x
p≤ x
3
q≤
x/p
Thus the required density is
X X 6
X
1
1
6 X
1
1
lim
·
·
=
≥ 0.0144.
2
2
2
2
x→∞
π p (p + 1) q(q + 1)
π p p (p + 1) q6=p q(q + 1)
√
√
3
p≤ x
3
q≤
x/p
q6=p
The double sum was estimated on a computer.
We have therefore shown the following:
7
Type
S
Density
p2 S
P
3
p S
2 2
p q S
p4 S
3 2
p q S
2 2 2
p q r S
5
p S
p4 q 2 S
p3 q 3 S
3 2 2
p q r S
2 2 2 2
p q r t S
p6 S
5 2
p q S
4 3
p q S
4 2 2
p q r S
p3 q 3 r 2 S
3 2 2 2
p q r t S
7
p S
6 2
p q S
p5 q 3 S
p5 q 2 r 2 S
4 4
p q S
4 3 2
p q r S
p3 q 3 r 3 S
8
p S
7 2
p q S
6 3
p q S
p6 q 2 r 2 S
p5 q 4 S
5 3 2
p q r S
4 4 2
p q r S
p4 q 3 r 3 S
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
6
π2
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
Pp
p
Truncated density
0.60792
1
p(p+1)
1
p(p+1)
P
1
1
q>p q(q+1)
p(p+1)
1
p3 (p+1)
P
1
1
q6=p q(q+1)
p2 (p+1)
P
P
1
1
1
q>p q(q+1)
r>q r(r+1)
p(p+1)
1
p4 (p+1)
P
1
1
q6=p q(q+1)
p3 (p+1)
P
1
1
q>p q 2 (q+1)
p2 (p+1)
P
P
1
1
1
q6=p q(q+1)
r>q, r6=p r(r+1)
p2 (p+1)
P
P
P
1
1
1
1
q>p q(q+1)
r>q r(r+1)
t>r t(t+1)
p(p+1)
1
p5 (p+1)
P
1
1
q6=p q(q+1)
p4 (p+1)
P
1
1
q6=p q 2 (q+1)
p3 (p+1)
P
P
1
1
1
q6=p q(q+1)
p3 (p+1)
P
Pr>q, r6=p r(r+1)
1
1
1
2
p2 (p+1)
Pq>p q (q+1)
P r6=q,p r(r+1)
P
1
1
1
1
2
q6=p q(q+1)
r>q, r6=p r(r+1)
t>r, t6=p t(t+1)
p (p+1)
1
p6 (p+1)
P
1
1
p5 (p+1)
Pq6=p q(q+1)
1
1
q6=p q 2 (q+1)
p4 (p+1)
P
P
1
1
1
q6=p q(q+1)
r>q, r6=p r(r+1)
p4 (p+1)
P
1
1
q>p q 3 (q+1)
p3 (p+1)
P
P
1
1
1
q6=p q 2 (q+1)
p3 (p+1)
P
Pr6=q, r6=p1 r(r+1)
1
1
q>p q 2 (q+1)
r>q r 2 (r+1)
p2 (p+1)
1
7
p (p+1)
P
1
1
q6=p q(q+1)
p6 (p+1)
P
1
1
2
p5 (p+1)
Pq6=p q (q+1)
P
1
1
1
q6=p q(q+1)
r>q, r6=p r(r+1)
p5 (p+1)
P
1
1
q6=p q 3 (q+1)
p4 (p+1)
P
P
1
1
1
q6=p q 2 (q+1)
r6=q, r6=p r(r+1)
p4 (p+1)
P
P
1
1
1
3
p3 (p+1)
Pq>p q (q+1)
Pr6=q, r6=p r(r+1)
1
1
1
q6=p q 2 (q+1)
r>q, r6=p r 2 (r+1)
p3 (p+1)
Total :
Table 1: Densities giving a positive lower bound
8
0.20075
0.07417
0.02212
0.03206
0.01447
0.00107
0.01474
0.00586
0.00216
0.00091
0.00002
0.00699
0.00259
0.00163
0.00035
0.00023
0.00002
0.00338
0.00120
0.00068
0.00015
0.00029
0.00016
0.00001
0.00165
0.00057
0.00030
0.00006
0.00023
0.00006
0.00002
0.00001
0.99683
Proposition 9 The density in N of integers n for which D(n) is unbounded is less than
0.004.
Although we cannot exhibit an unbounded derived sequence, we do have the following
result.
Proposition 10 For any positive integer M , there exists an integer n such that D(n) requires more than M iterations of D before a cycle is possible.
Proof. Let p be a large prime, with “large” to be qualified shortly. Take n = p 4p . Then
D(n) = {p4p , 22 p4p , 24 p4p , 27 p4p , 28 7p4p , 212 p4p , 215 3p4p , . . . }.
The exact factor p4p will persist in all terms until the exponent on 2 or some other prime
(not p) equals p. Until this happens, if it will ever happen, there can be no derived k-cycle
in D(n),
Q i for any k. For suppose there were such a cycle and let the ith term of the cyclek
, 1 ≤Qi ≤ k, Q
where the notation is as in Proposition
be pai Q
Q Q 1. ItQfollows that at most 2
2k
divides p1 p2 · · · pk , while at least 2 divides a1 a2 · · · ak . By Proposition 1,
this is a contradiction. The proof is non-constructive, in that we can say only that, whatever
the value of M , a prime p exists such that p exceeds all exponents on primes other than p
resulting from M iterations of D. ¤
4
Further work
(1) Find examples of integers n for which D(n) results in a k-cycle for k = 7 or k > 8. Do
derived k-cycles exist for all positive integers k?
(2) Can it be shown that D(n) is unbounded for some n, as we have conjectured above?
q
We suspect this to be the case for “most” sequences D(pq p ) (p 6= q), for reasons suggested in the proof of Proposition 10 (where we considered q = 2). It is not known
whether D(74046 ), D(111674 ) and D(13504 ) are bounded, but this is the case for all smaller
exponents on the respective primes. (It took a few weeks for referee’s comments to be
returned. We are grateful for those. In that time we left a program running, checking
D(31124 ) for boundedness. After k = 48218701 iterations, we found no cycle and D k (31124 ) =
21516268557 3780548532 5348780008 · · · 12710414 131131 139139 149141851 · · · 62763353 · 348779999, with all
primes up to 131 present. The program was then permanently halted.)
(3) As in the calculus, differentiation is a craft, but integration is an art. Can a technique be
developed for finding integrals of a given positive integer or of showing that certain integers,
the primes being examples, have no integrals? In Proposition 2, we have given all integrals
of 1, but even to identify all integrals of 4 seems to be very difficult.
References
[1] T. M. Apostol. Introduction to Analytic Number Theory, Springer–Verlag, Berlin, 1980.
9
[2] G. L. Cohen and H. J. J. te Riele. Iterating the sum-of-divisors function, Experiment.
Math., 5 (1996), 91–100. Errata in Experiment. Math., 6 (1997), 177.
[3] R. K. Guy. Problems in Number Theory, second edition, Springer–Verlag, New York,
1994.
2000 Mathematics Subject Classification: Primary 11Y55; Secondary 11A25, 11B83.
Keywords: Arithmetic functions, density, unbounded sequences, cycles.
Received October 25, 2002; revised version received December 1, 2002. Published in Journal
of Integer Sequences December 23, 2002. Revised, February 10 2004.
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