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Numbers k such that A215339(6*k+2) = 0.
0

%I #25 Dec 27 2021 04:57:39

%S 0,2755452,4570452,123725995972

%N Numbers k such that A215339(6*k+2) = 0.

%C Numbers k such that b(k) is an integer, where b(n) = Sum_{k=0..n} (2*n+k)!/((2*n-2*k)!*(3*k+1)!) (= A001608(6*n+2)/(6*n+2)).

%H Don Zagier, <a href="http://www-groups.mcs.st-andrews.ac.uk/~john/Zagier/Problems.html">Problems posed at the St Andrews Colloquium, 1996</a>, See "Fifth Day: Sequences and recursions, problem 2".

%e n | b(n)

%e --+--------------------------

%e 0 | 1 (= 2/2)

%e 1 | 5/4 (= 10/8)

%e 2 | 51/14

%e 3 | 277/20

%e 4 | 1497/26

%e 5 | 4045/16 (= 8090/32)

%e 6 | 43721/38

%e 7 | 118141/22 (= 236282/44)

%e 8 | 638471/25 (= 1276942/50)

%e 9 | 6900995/56

%Y Cf. A001608, A013998, A215339.

%K nonn,more

%O 1,2

%A _Seiichi Manyama_, Mar 02 2019