Download Analytic and Elementary Number Theory: A Tribute to by Krishnaswami Alladi (auth.), K. Alladi, P. D. T. A. Elliott, PDF

By Krishnaswami Alladi (auth.), K. Alladi, P. D. T. A. Elliott, A. Granville, G. Tenebaum (eds.)

ISBN-10: 1441950583

ISBN-13: 9781441950581

ISBN-10: 1475745079

ISBN-13: 9781475745078

This quantity incorporates a choice of papers in Analytic and basic quantity concept in reminiscence of Professor Paul Erdös, one of many maximum mathematicians of this century. Written through many top researchers, the papers care for the newest advances in a large choice of subject matters, together with arithmetical features, leading numbers, the Riemann zeta functionality, probabilistic quantity idea, homes of integer sequences, modular kinds, walls, and q-series.
Audience: Researchers and scholars of quantity conception, research, combinatorics and modular kinds will locate this quantity to be stimulating.

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Extra resources for Analytic and Elementary Number Theory: A Tribute to Mathematical Legend Paul Erdös

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6) is also quite interesting. 6) is 00 00 j=l j=l no+ iq2j-l)(l _ iq2j-1) =no+ q4j-2). 13) =2(mod 4). On the other hand, 0, if e is odd, 1, if i 1 -1, if e = O(mod 4), =-2(mod 4). 14) This means we cannot have chains of odd length consisting of odd parts differing by exactly 2. Since ab = i( -i) = 1, all even parts will have weight 1. Thus we have Theorem 11. Let Q2,4(n) denote the number of partitions of n into distinct parts = 2 (mod 4). 14). Then Q2,4(n) = L wJrr). rreG,u(rr)=n Acknowledgments I would like to thank George Andrews and Basil Gordon for discussions on various aspects of this paper.

If we let N; be the number of quadruples corresponding to each i, then by (7) and the Cauchy-Schwarz inequality, N; 2: LL:Sm(n) -1 m n Also, N; :;:: 0 for each i. If bj1 < bj3 , then (9) implies bj2 < bj4 and hence 14 L N· < -1. 4 . (10) I- ' Define It = { i = 1 (mod 3) : S; P/ :;:: lz = { i = 1 (mod 3) : S; P/ < ~1 4 }, ~1 4 }. A straightforward calculation shows that S·I + p,. 3 4 3 > -1 2 1 I- (i E It). 4 j2P; fori E ]z, hence by (10), 1 1 I:-::::-. P; 2 (12) . I lE 2 LetM, = III I, M2 inequality, = 1/zl and H = M 1 +M2 • By(8), (11), (12) and theCauchy-Schwarz 12BI + IB 2 1::::: ~I 413 M, + L P; iEh > ~M I [413 + lM22 - l = ~z 4 1\H- Mz) + 2Mi.

2) Lemma. Suppose that m ¢;. b; (mod q)forall i. Then there is a sequence (u;, v;, w;) of triplets of integers, 1 ::S i ::5 I, such that each u = u;, v = v;, w = w; is a solution of congruence (2) ,for i =f. j the sets {u;, v;, w;} and {u i, vi, w i} are disjoint and I ~ p /8. Proof: Since the numbers b; - bi, i =f. j are all incongruent modulo q and their number is q- 1, they represent each nonzero residue exactly once. Thus for each u there is exactly one pair ( v, w) such that (2) holds. This shows that there are altogether p solutions.

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