A304119 Numerators of record low values of the ratio n*3^A006667(n)/2^A006666(n).

1, 27, 1701, 6561, 1760826122505, 115093142840908791, 460166680231540515, 1840049047529878113

Offset: 1

Author

Michel Lagneau, May 03 2018

Keywords

• nonn
• frac
• more

Comments

This has been verified for n up to 10^7.

Conjecture: Consider A006666 and A006667, the sequences giving the number of halving and tripling steps to reach 1 in 3x+1 problem. There exists a rational constant c such that c <= n*3^A006667(n)/2^A006666(n) <= 1 where c = 1840049047529878113/2305843009213693952 is the last term in the sequence of the ratios.

Note that n*3^A006667(n)/2^A006666(n) = 1 if n is a power of 2 (A000079).

It seems that n*3^A006667(n)/2^A006666(n) = c for n = 993*2^k, k = 0, 1, 2, ... In this case, c = 993*2^k*3^32/2^(61+k), where 32 = A006667(993*2^k) and 61+k = A006666(993*2^k). For example, c = 993*3^32/2^61 = 1986*3^32/2^62 = 3972*3^32/2^63 = 7944*3^32/2^64 = ...

Examples

			For n=1 to 10 the ratios are: 1, 1, 27/32, 1, 15/16, 27/32, 1701/2048, 1, 6561/8192, 15/16, so the low records are 1, 27/32, 1701/2048, 6561/8192, ...
		

Crossrefs

Cf. A006666, A006667, A127789 (for the indices where these records occur).

Programs

Mathematica

q=1; Collatz[n_]:=NestWhileList[If[EvenQ[#], #/2, 3 #+1]&, n, #>1&]; nn=5000; t={}; n=0; While[Length[t]
				

PARI

ht(n) = my(h, t); while(n>1, if(n%2, n=3*n+1; t++, n>>=1; h++)); return([h, t]);
lista(nn) = {m = 2; for (n=1, nn, v = ht(n); newm = n*3^v[2]/2^v[1]; if (newm < m, print1(numerator(newm), ", "); m = newm));} \\ Michel Marcus, May 06 2018