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%I #34 May 05 2021 09:19:25
%S 3,8,24,80,288,1088,4224,4374,16640,66048,263168,1050624,4198400
%N Numbers n such that A007947(n) = A007947(m+1) and A007947(m) = A007947(n+1), where n > m.
%C For every k >= 0, the sequence includes 4^k + 2^(k+1), with m = 2^k + 1. - _David Wasserman_, Jan 29 2004
%C So a(13) <= 4198400. - _Michel Marcus_, Aug 10 2014
%C Are there other terms like 4374 that are not of this form? - _Michel Marcus_, Aug 10 2014
%F G.f.: Conjecture: Q(0)/x - 1/x where Q(k)= 1 + 2^k*x/(1 - 2*x/(2*x + 2^k*x/Q(k+1) )); (continued fraction ). - _Sergei N. Gladkovskii_, Apr 10 2013
%e With n=3 and m=2, rad(3) = rad(3) and rad(2) = rad(4), so 3 is in the sequence.
%p rad:= n -> convert(numtheory:-factorset(n),`*`):
%p count:= 0: lastr:= rad(1):
%p for n from 2 to 10^7 do
%p newr:= rad(n);
%p P[lastr,newr]:= n-1;
%p if assigned(P[newr,lastr]) then
%p count:= count+1; A[count]:= n-1; M[count]:= P[newr,lastr];
%p fi;
%p lastr:= newr;
%p od:
%p seq(A[n],n=1..count); # _Robert Israel_, Aug 10 2014
%t (* Recomputation up to a(13), assuming m of the form 2^k+1 *)
%t rad[n_] := rad[n] = Select[Divisors[n], SquareFreeQ][[-1]];
%t okQ[n_] := Module[{r = rad[n], r1 = rad[n+1], k, m}, For[k = 0, k < Log[2, n-1], k++, m = 2^k+1; If[r == rad[m+1] && rad[m] == r1, Return[True]]]; False];
%t Reap[For[n = 1, n <= 5*10^6, n++, If[okQ[n], Print[n]; Sow[n]]]][[2, 1]] (* _Jean-François Alcover_, Apr 11 2019 *)
%o (PARI) lista(nn) = {v = vector(nn, i, rad(i)); for (n=1, nn-1, ok = 0; if (n % 2, ma = 2, ma = 1); forstep (m = ma, n-1, 2, if ((v[n] == v[m+1]) && (v[m] == v[n+1]), ok = 1; break);); if (ok, print1(n, ", ")););} \\ _Michel Marcus_, Aug 10 2014
%Y Cf. A007947 (rad(n)), A087914 (similar sequence), A091697 (the values of m).
%K nonn,more
%O 1,1
%A _Naohiro Nomoto_, Oct 29 2003
%E More terms from _David Wasserman_, Jan 29 2004
%E a(13) confirmed by _Robert Israel_, Aug 10 2014