

A198472


a(n)=q(n) if 4  q(n)2, and a(n)=q(n)/2 if 4  q(n), where q(n) is the least practical number q>n with 2(n+1)q practical.


1



2, 2, 2, 6, 6, 4, 4, 6, 6, 8, 6, 18, 8, 18, 8, 18, 18, 10, 10, 12, 12, 14, 12, 30, 14, 30, 14, 30, 30, 16, 16, 18, 18, 20, 18, 42, 20, 42, 20, 42, 42, 54, 24, 24, 28, 54, 24, 28, 30, 54, 28, 32, 54, 28, 28, 30, 30, 32, 30, 66, 32, 66, 32, 66, 66, 78, 36, 36, 40, 78, 36, 40, 42, 78, 40, 44, 78, 40, 40, 42, 42, 44, 42, 90, 44, 90, 44, 90, 90, 52, 48, 48, 50, 50, 48, 52, 50, 54, 50, 56
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OFFSET

1,1


COMMENTS

Conjecture: If b(1)>=4 is an integer and b(k+1)=a(b(k)) for k=1,2,3,..., then b(n)=4 for some n>0.
This conjecture has the same flavor as the Collatz conjecture.


LINKS



EXAMPLE

a(20)=12 since 2(20+1)=24+18 with 24 and 18 both practical.


MATHEMATICA

f[n_]:=f[n]=FactorInteger[n]
Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]1}]
pr[n_]:=pr[n]=n>0&&(n<3Mod[n, 2]+Con[n]==0)
Do[Do[If[pr[2k]==True&&pr[2n+22k]==True, Print[n, " ", 2k/(1+Mod[k1, 2])]; Goto[aa]], {k, Ceiling[(n+1)/2], n}];
Label[aa]; Continue, {n, 1, 100}]


PROG



CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



