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Least numbers m such that GCD of two consecutive values of cototients, i.e., gcd(cototient(m+1), cototient(m)) equals 2n - 1.
0

%I #20 Aug 02 2017 12:16:12

%S 2,9,38,392,135,120,362,116,745,1183,294,528,1395,428,1378,2602,1185,

%T 203,2313,3042,1966,3549,1431,551,7838,4076,473,2635,903,2044,13178,

%U 942,6819,12418,1188,2264,3282,1775,1517,2127,24380,2884,2035,11481

%N Least numbers m such that GCD of two consecutive values of cototients, i.e., gcd(cototient(m+1), cototient(m)) equals 2n - 1.

%F a(n) = min{x; A049586(x) = 2n - 1}.

%e For n=104: 2n - 1 = 207, a(104) = 235148 because A049586(235148) = 207 and it is the smallest such number. Remark that Count[t=Table[f[w],{w,1,100000}],1]=83132. This suggests that majority of values in A049586 equals one.

%t With[{s = Array[# - EulerPhi@ # &, 10^5]}, Function[t, MapAt[# + 1 &, TakeWhile[#, # > 0 &], 1] &@ Table[First[FirstPosition[t, n] /. k_ /; MissingQ@ k -> {0}], {n, 1, Max@ t, 2}]]@ Map[GCD @@ # &@ # &, Partition[s, 2, 1]]] (* _Michael De Vlieger_, Jul 30 2017 *)

%Y Cf. A051953.

%K nonn

%O 1,1

%A _Labos Elemer_ and _Benoit Cloitre_, Apr 12 2002

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