54 Jean-Marie De Koninck
193
the smallest prime number p such that Ω(p+1) = 2, Ω(p+2) = 3 and Ω(p+3) =
4 (see the number 61);
the smallest number n which requires seven iterations of the σI (n) function
in order to reach 1; if nk stands for the smallest number n which requires k
iterations of σI (n) in order to reach 1, then n3 = 5, n4 = 9, n5 = 17, n6 = 67,
n7 = 193, n8 = 1 069, n9 = 2 137, n10 = 4 273, n11 = 34 183, n12 = 205 097 and
n13 = 990 361.
194
the smallest number n 1 which divides σ24(n);
the sixth solution of φ(n) = φ(n + 1) (see the number 15).
195
the smallest number n such that the Moebius function µ takes successively,
starting with n, the values −1, 0, −1, 0, −1, 0; the sequence of numbers satis-
fying this property begins as follows: 195, 1491, 1547, 2139, 2715, 2749, 2751,
2847, 2967, . . .
196
the
only66
solution n
1012
of σ(n) = 2n + 7;
the smallest number n which does not produce a palindrome by the following
algorithm, often called the 196-algorithm: given a number n, reverse its digits
and add the number thus obtained with n, giving, say, the number n1; then
reverse the digits of n1 and add it to n1 to get, say n2, and so on until a
palindrome emerges; for example, by this process, starting with n = 19, then
n1 = 110 and n2 = 121, a palindrome, and the process stops; the sequence of
numbers n which do not produce a palindrome by this algorithm, that is which
are such that the indicated process is endless, begins as follows: 196, 887, 1675,
7436, 13783, 52514, 94039, 187088, 1067869, 1075547, . . .
66Since here σ(n) is odd, it is clear that n is necessarily a perfect square or twice a perfect
square. On the other hand, it is interesting to observe that the only odd numbers k 100 such that
σ(n) = 2n + k has a solution n 1012 are 3, 7, 17, 19, 31, 39, 41, 51, 59, 65, 71 and 89, which makes
us tend to believe that for each odd number k, equation σ(n) = 2n + k has only a finite number of
solutions.
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