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 A042996 Numbers k such that binomial(k, floor(k/2)) is divisible by k. 4
 1, 2, 3, 5, 7, 9, 11, 12, 13, 15, 17, 19, 21, 23, 25, 27, 29, 30, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 56, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 84, 85, 87, 89, 90, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 LINKS Ivan Neretin, Table of n, a(n) for n = 1..10000 EXAMPLE For n=12, binomial(12,6) = 924 = 12*77 is divisible by 12, so 12 is in the sequence. For n=13, binomial(13,6) = 1716 = 13*132 is divisible by 13, so 13 is in the sequence. From David A. Corneth, Apr 22 2018: (Start) For n = 20, we wonder if 20 = 2^2 * 5 divides binomial(20, 10) = 20! / (10!)^2. The exponent of 2 in the prime factorization of 20! is 10 + 5 + 2 + 1 = 18. The exponent of 2 in the prime factorization of 10! is 5 + 2 + 1 = 8. Therefore, the exponent of 2 in binomial(20, 10) is 18 - 2*8 = 2. The exponent of 5 in the prime factorization of 20! is 4. The exponent of 5 in the prime factorization of 10! is 2. Therefore, exponent of 5 in binomial(20, 10) is 4 - 2*2 = 0. So binomial(20, 10) is not divisible by 20, and 20 is not in the sequence. (End) MATHEMATICA Select[Range, Divisible[Binomial[#, Floor[#/2]], #]&] (* Harvey P. Dale, Sep 15 2011 *) PROG (PARI) isok(n) = (binomial(n, n\2) % n) == 0; \\ Michel Marcus, Apr 22 2018 CROSSREFS Cf. A001405, A020475. Sequence in context: A092875 A155498 A069149 * A073040 A087268 A106765 Adjacent sequences:  A042993 A042994 A042995 * A042997 A042998 A042999 KEYWORD nonn AUTHOR STATUS approved

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Last modified August 11 20:13 EDT 2022. Contains 356067 sequences. (Running on oeis4.)