

A240670


Numbers n for which all exponents in the prime power factorization of (2*n)! are odious (A000069).


13




OFFSET

1,2


COMMENTS

The next term, if it exists, must be more than 45000.  Peter J. C. Moses, Apr 11 2014
The sequence is finite.
Proof. For sufficiently large n, we always have a prime in (n/4, n/3]. Such primes p divide n! and, at the same time, for them we have 3<=n/p<4. Thus floor(n/p)=3, and in case sqrt(n)<n/4, floor(n/p^2)=0. Therefore, they involve in n! with exponent 3. Since 3 is evil, we are done. Moreover, numerically, using estimate for (4/3)Ramanujan number (see Shevelev, Greathouse IV, and Moses link, Proposition 8), it is sufficient to consider n>=93 in order for the above arguments to be true. So 16 is the last term of the sequence.  Vladimir Shevelev, Apr 11 2014


LINKS

Table of n, a(n) for n=1..6.
D. Berend, G. Kolesnik, Regularity of patterns in the factorization of n!, J. Number Theory, 124 (2007), no. 1, 181192.
Vladimir Shevelev, Charles R. Greathouse IV, Peter J. C. Moses, On intervals (kn, (k+1)n) containing a prime for all n>1, Journal of Integer Sequences, Vol. 16 (2013), Article 13.7.3. arXiv:1212.2785


EXAMPLE

32! = 2^31*3^14*5^7*7^4*11^2*13^2*17*19*23*29*31, and all exponents: 31,14,7,4,2,2,1,1,1,1,1 are odious, so 16 is in the sequence.


MATHEMATICA

odiousQ[n_] := OddQ[DigitCount[n, 2][[1]]];
For[n = 1, True, n++, If[AllTrue[FactorInteger[(2 n)!][[All, 2]], odiousQ], Print[n]]] (* JeanFrançois Alcover, Sep 20 2018 *)


PROG

(PARI) isok(n) = {f = factor((2*n)!); sum(i=1, #f~, hammingweight(f[i, 2]) % 2) == #f; } \\ Michel Marcus, Apr 11 2014


CROSSREFS

Cf. A000069, A240537, A240606, A240619, A240620, A240668, A240669.
Sequence in context: A221173 A330138 A051530 * A048040 A051033 A242787
Adjacent sequences: A240667 A240668 A240669 * A240671 A240672 A240673


KEYWORD

nonn,fini,full


AUTHOR

Vladimir Shevelev, Apr 10 2014


STATUS

approved



