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The second of four solutions to a Monthly problem asking if there exist finite sequences 1 < a(1) < a(2) < ... < a(n) such that Sum_i 1/a(i) = 1 and gcd(a(i), a(i+1)) = 1 for 1 <= i < n.
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%I #12 Aug 06 2021 18:03:26

%S 3,5,7,9,11,21,22,45,154

%N The second of four solutions to a Monthly problem asking if there exist finite sequences 1 < a(1) < a(2) < ... < a(n) such that Sum_i 1/a(i) = 1 and gcd(a(i), a(i+1)) = 1 for 1 <= i < n.

%H Daniel Ullman, Proposer, <a href="https://www.jstor.org/stable/2323959">Problem E3359</a>, Amer. Math. Monthly, 98:2 (1991), 168.

%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>

%Y Cf. A346603, A346605, A346606.

%K nonn,fini,full

%O 1,1

%A _N. J. A. Sloane_, Aug 06 2021