A190644 - OEIS

%I #28 Jul 25 2024 08:17:07

%S 6,34,39,7,11,19,29,41,79,71,179,199,181,239,883,419,701,839,881,1429,

%T 2351,1259,1871,2161,4049,3079,3361,2521,6481,4159,6299,5279,11551,

%U 5039,20789,7561,25919,10079,16561,13441,38611,13859,23761,21839,100673,20161

%N Least number k>1 such that (tau(k-1)+tau(k+1))/tau(k) = n where tau = A000005.

%p with(numtheory):

%p a:= proc(n) local k;

%p       for k from 2 while (tau(k-1)+tau(k+1)) /tau(k)<>n do od; k

%p     end:

%p seq(a(n), n=1..50);  # _Alois P. Heinz_, May 19 2011

%t tau = DivisorSigma[0, #]&;

%t a[n_] := For[k=2, True, k++, If[(tau[k-1]+tau[k+1])/tau[k]==n, Return[k]]];

%t Array[a, 50] (* _Jean-François Alcover_, Mar 27 2017 *)

%t Module[{nn=300000,tau},tau=(#[[1]]+#[[3]])/#[[2]]&/@Partition[DivisorSigma[ 0,Range[nn]],3,1];Flatten[Table[Position[tau,n,1,1],{n,50}]]+1] (* _Harvey P. Dale_, Nov 24 2022 *)

%o (Sage)

%o def A190644(n):

%o     tau = number_of_divisors

%o     return next((k for k in IntegerRange(2,infinity) if tau(k-1)+tau(k+1) == n*tau(k))) # _D. S. McNeil_, May 19 2011

%Y Cf. A000005 (number of divisors of n), A190612.

%K nonn

%O 1,1

%A _Juri-Stepan Gerasimov_, May 15 2011