Those Fascinating Numbers 69
288
the second powerful number n such that n+1 is also powerful: here 288 = 25 ·32
and 289 = 172; the only numbers n 1015 satisfying this property75 are 8, 288,
675, 9 800, 12 167, 235 224, 332 928, 465 124, 1 825 200, 11 309 768, 384 199 200,
592 192 224, 4 931 691 075, 5 425 069 447, 13 051 463 048, 221 322 261 600,
443 365 544 448, 865 363 202 000, 8 192 480 787 000, 11 968 683 934 831,
13 325 427 460 800, 15 061 377 048 200, 28 821 995 554 247, 48 689 748 233 307 and
511 643 454 094 368;76
the seventh unitary hyperperfect number: a number n is said to be unitary
hyperperfect if there exists a positive integer k such that n =
1+k(σ∗(n)−n−1),
where
σ∗(n)
stands for the sum of the unitary divisors of n (that is the divisors
d such that (d, n/d) = 1): it is clear that a square-free number is hyperperfect
if and only if it is unitary hyperperfect: the sequence of numbers satisfying this
property begins as follows: 6, 21, 40, 52, 60, 90, 288, 301, 657, 697, 1333, 1909,
2041, 2176, 3856, 3901, 5536, 6517, 15025, 24601, . . . ;
the third solution w + s of the aligned houses problem (see the number 35).
289
the fourth number n such that the binomial coefficient
(
n
2
)
is a perfect square:
here
(
289
2
)
=
2042;
the sequence of numbers satisfying this property begins as
follows: 2, 9, 50, 289, 1 682, 9 801, 57 122, 332 929, 1 940 450, . . .
77
293
the third self contained number: a number n is said to be self contained if it is
odd and if n|f k(n) for a certain positive integer k, where f k(n) = f(f k−1(n)),
f 1(n) = f(n), and where the function f is defined by
f(n) =
3n + 1 if n is odd,
n/2 if n is even;
the only self contained numbers
107
are 31, 83, 293, 347, 671, 19 151, 925 957
and 2 025 797.
75It
is easy to prove that there exist infinitely many numbers n such that n and n + 1 are both
powerful. Indeed, since the Fermat-Pell equation
x2

2y2
= 1 has infinitely many solutions (x, y),
then the numbers
2y2
and
x2
are consecutive and moreover they are powerful; to see that
2y2
is
powerful, first observe that x is odd (since otherwise we would have that 2|1), and that y is even,
since
x2
1 = (x 1)(x + 1) =
2y2,
which implies that
4|2y2.
This observation was first made in
1970 by S.W. Golomb [92].
76It
is interesting to observe that in this list of 25 pairs of consecutive powerful numbers, the
second member of each pair is always a perfect square, except in the case of the eighth one where
we have 465 125 = 53 · 612, of the 20th where we have 11 968 683 934 832 = 24 · 233 · 78412, of
the
23rd
where we have 28 821 995 554 248 =
23
·
34
·
132
·
162232
and of the
25th
where we have
511 643 454 094 368 = 22 · 73 · 132 · 432 · 3372.
77This
sequence is infinite because there exist infinitely many triangular numbers which are perfect
squares (see the number 36).
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