By Emil Grosswald (auth.), Marvin I. Knopp (eds.)
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Extra resources for Analytic Number Theory: Proceedings of a Conference Held at Temple University, Philadelphia, May 12–15, 1980
Example text
The next result is contained in (16) of [28]. E~TRY 19(i). For each positive integer n, 2(22n-l)B2n is an integer. PROOF. We give a somewhat easier proof than that in [28]. By the von Staudt-Clausen theorem, the denominator of B2n is the product of all those primes p such that (p-l)I2n. theorem, pl(2P-l-l). Let p be such an odd prime. Then by Fermat's But since (p-l)I2n, we have pI(22n-l), which completes the proof. ENTRY I g { i i ) . The numerator of B2n is divisible by the largest factor of 2n which is relatively prime to the denominator of B2n.
Andrews, Ramanujan and his "lost" notebook, Vinculum, 16 (1979), 91-94. 6. E. Andrews, Partitions: Yesterdayand Today, New Zealand Math. , Wellington, 1979. 7. E. , {to appear). 8. E. , (to appear). 47 9. E. , (to appear). I0. N. Bailey, On the basic bilateral hypergeometric series 2~2, Quart. J. , l (1950), 194-198. II. R. Bellman, A Brief Introduction to Theta Functions, Holt, Rinehart and Winston, New York, 1961. 12. T. Copson, An Introduction to the Theory of Functions of a Complex Variable, Oxford University Press, London, 1935.
In the f i r s t example below, Ramanujan has incorrectly written -I/2 instead of I/2 on the right side. EXAMPLE I. I f eP+eQ+eR = 2+eP+Q+R, then 1 +~+~+ l ~ (P+Q+R) = 17" PROOF. In terms of p,q, and r, we are given that 3+p+q+r = 2+(l+p)(l+q)(l+r), which may be written as -l = l + l ~ 1 :! l) and ignored all terms with powers of P,Q, and R greater than the f i r s t . The desired approximation now follows. 63 EXAMPLE 2. If eP+Q+R+S = eP+eQ+eR+eS-2 e-P+e-Q+e-R+e-S_2 , then 1 PROOF. The given equality is equivalent (1+p)(1+q)(l+r)(l+s) to : 2+p+q+r+s 2-p-q-r-s" Now cross multiply and ignore all products involving p2, q2, r 2, and s2.
