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
D. Berend, G. Kolesnik, Regularity of patterns in the factorization of n!, J. Number Theory, 124 (2007), no. 1, 181-192.
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]]] (* Jean-François 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
Vladimir Shevelev, Apr 10 2014
STATUS
approved