By Jeffrey Stopple
This undergraduate-level creation describes these mathematical houses of best numbers that may be deduced with the instruments of calculus. Jeffrey Stopple will pay designated awareness to the wealthy historical past of the topic and historical questions about polygonal numbers, excellent numbers and amicable pairs, in addition to to the real open difficulties. The end result of the ebook is a quick presentation of the Riemann zeta functionality, which determines the distribution of major numbers, and of the importance of the Riemann speculation.
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Additional resources for A Primer of Analytic Number Theory: From Pythagoras to Riemann
Compute s(n) for as many n as possible. Determine whether n is deﬁcient, perfect, or abundant. Look for patterns when n is odd or even. Is 32 2. Products and Divisibility s(n) odd or even? ) You can compute s(n) easily in Mathematica by ﬁrst deﬁning it as a function s[n ] := DivisorSigma[1, n] - n The input s returns the answer 7. In Maple, after loading the numtheory package (see p. xiii), you can deﬁne the function using s:=n->sigma(n)-n; The input s(12); returns the answer 16. 15. The number 284 = 22 · 71 isn’t perfect; s(284) = 220 = 22 · 5 · 11.
13) misses one case. There is no factorial power whose difference is n −1 . In other words, n −1 is not a factorial power. ) So we make a deﬁnition instead, deﬁning the nth harmonic number to be Hn = 1≤k≤n 1 1 1 = 1 + + ··· + . 17) Notice that after changing the variable slightly, we can also write Hn = 0≤k If n factors as p1k1 p2k2 . . ptkt , then (n) = (k1 + 1)(k2 + 1) . . (kt + 1), (n) = p1k1 +1 − 1 p2k2 +1 − 1 ptkt +1 − 1 ... 3) Proof. This is just a combination of the previous lemmas. Because p1k1 , p2k2 , . . , ptkt are all relatively prime, (n) = p1k1 p2k2 . . ptkt = (k1 + 1)(k2 + 1) . . (kt + 1), and similarly for (n). Theorem (Euclid). If p is a prime number such that Mp = 2 p − 1 is also a prime, then n = 2 p−1· Mp is a perfect number. Proof. To show s(n) = n, we show that (n) = s(n) + n is just 2n.
If n factors as p1k1 p2k2 . . ptkt , then (n) = (k1 + 1)(k2 + 1) . . (kt + 1), (n) = p1k1 +1 − 1 p2k2 +1 − 1 ptkt +1 − 1 ... 3) Proof. This is just a combination of the previous lemmas. Because p1k1 , p2k2 , . . , ptkt are all relatively prime, (n) = p1k1 p2k2 . . ptkt = (k1 + 1)(k2 + 1) . . (kt + 1), and similarly for (n). Theorem (Euclid). If p is a prime number such that Mp = 2 p − 1 is also a prime, then n = 2 p−1· Mp is a perfect number. Proof. To show s(n) = n, we show that (n) = s(n) + n is just 2n.