

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



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



KEYWORD

nonn,fini,full


AUTHOR



STATUS

approved



