OFFSET
1,1
COMMENTS
Numbers k such that k, k+1 and k+2 are all in A359565.
There must be 3 or more divisors of k that have the same Euler totient value, and ditto for k+1 and k+2, but those values may differ as among k, k+1, and k+2. - Harvey P. Dale, Sep 01 2024
MATHEMATICA
q[n_] := Max[Tally[EulerPhi[Divisors[n]]][[;; , 2]]] > 2; seq[kmax_] := Module[{s = {}, q1 = 0, q2 = 0, q3}, Do[q3 = q[k]; If[q1 && q2 && q3, AppendTo[s, k-2]]; q1=q2; q2=q3, {k, 3, kmax}]; s]; seq[10^8]
SequencePosition[Table[If[Max[Tally[EulerPhi[Divisors[n]]][[;; , 2]]]>2, 1, 0], {n, 88*10^6}], {1, 1, 1}] [[;; , 1]] (* The program generates the first 3 terms of the sequence. *) (* Harvey P. Dale, Sep 01 2024 *)
PROG
(PARI) is(k) = vecmax(matreduce(apply(x->eulerphi(x), divisors(k)))[, 2]) > 2;
lista(kmax) = {my(q1 = 0, q2 = 0, q3); for(k = 3, kmax, q3 = is(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3); }
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Jun 08 2024
STATUS
approved