A240670 Numbers n for which all exponents in the prime power factorization of (2*n)! are odious (A000069).

1, 3, 4, 5, 15, 16

Offset: 1

Author

Vladimir Shevelev, Apr 10 2014

Keywords

• nonn
• fini
• full

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)=93 in order for the above arguments to be true. So 16 is the last term of the sequence. - Vladimir Shevelev, Apr 11 2014

Examples

			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.
		

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

Crossrefs

Cf. A000069, A240537, A240606, A240619, A240620, A240668, A240669.

Programs

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 *)

PARI

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