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A problem posed by the Slovak mathematician Stefan Znám in 1972 asking whether, for all integers , there exist integers all greater than 1 such that is a proper divisor of for each . The answer is negative for (Jának and Skula 1978) and affirmative for (Sun Qi 1983). Sun Qi also gave a lower bound for the number of solutions.All solutions for have now been computed, summarized in the table below. The numbers of solutions for , 3, ... terms are 0, 0, 0, 2, 5, 15, 93, ... (OEIS A075441), and the solutions themselves are given by OEIS A075461.known solutions references20--Jának and Skula (1978)30--Jának and Skula (1978)40--Jának and Skula (1978)522, 3, 7, 47, 3952, 3, 11, 23, 31652, 3, 7, 43, 1823, 1936672, 3, 7, 47, 403, 194032, 3, 7, 47, 415, 81112, 3, 7, 47, 583, 12232, 3, 7, 55, 179, 243237152, 3, 7, 43, 1807, 3263447, 2130014000915Jának and Skula (1978)2, 3, 7, 43, 1807, 3263591, 71480133827Cao, Liu, and Zhang..

A multiple of a number is any quantity with an integer. If and are integers, then is called a factor of .

Vorobiev's theorem states that if , then , where is a Fibonacci number and means divides . The theorem was discovered by Vorobiev in 1942, but not published until 1967. It was used by Y. Matiyasevich in his negative solution to the Hilbert's tenth problem.Note that the converse does not hold. For example, , but . The plot above shows values of for which and (black) and for which but (red).

A homework problem proposed in Steffi's math class in January 2003 asked students to prove that no ratio of two unequal numbers obtained by permuting all the digits 1, 2, ..., 7 results in an integer. If such a ratio existed, then some permutation of 1234567 would have to be divisible by . can immediately be restricted to , since a ratio of two permutations of the first seven digits must be less than , and the permutations were stated to be unequal, so . The case can be eliminated by the divisibility test for 3, which says that a number is divisible by 3 iff the sum of its digits is divisible by 3. Since the sum of the digits 1 to 7 is 28, which is not divisible by 3, there is no permutation of these digits that is divisible by 3. This also eliminates as a possibility, since a number must be divisible by 3 to be divisible by 6.This leaves only the cases , 4, and 5 to consider. The case can be eliminated by noting that in order to be divisible by 5, the last digits of the numerator..

A number satisfies the Carmichael condition iff for all prime divisors of . This is equivalent to the condition for all prime divisors of .

Porter's constant is the constant appearing in formulasfor the efficiency of the Euclidean algorithm,(1)(2)(3)(OEIS A086237), where is the Euler-Mascheroni constant, is the Riemann zeta function, and is the Glaisher-Kinkelin constant (Knuth 1998, p. 357). The notation is generally used for this constant (Knuth 1998, p. 357, Finch 2003, pp. 156-157), though other authors use (Ustinov 2010) or (Dimitrov et al. 2000).The related constant originally considered by Porter (1975) and Knuth (1976) was denoted and , respectively, and defined by(4)(5)Knuth (1976) suggested be called the Lochs-Porter constant due to the work of Lochs (1961).

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