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A304119
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Numerators of record low values of the ratio n*3^A006667(n)/2^A006666(n).
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1
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OFFSET
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1,2
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COMMENTS
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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.
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 = ...
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LINKS
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EXAMPLE
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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, ...
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MATHEMATICA
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q=1; Collatz[n_]:=NestWhileList[If[EvenQ[#], #/2, 3 #+1]&, n, #>1&]; nn=5000; t={}; n=0; While[Length[t]<nn, n++; c=Collatz[n]; ev=Length[Select[c, EvenQ]]; od=Length[c]-ev-1; If[n*3^od/2^ev<q, Print[n, " ", od, " ", ev, " ", n*3^od/2^ev]; q=n*3^od/2^ev]]
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,frac,more
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AUTHOR
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STATUS
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approved
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