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%I #20 Nov 24 2022 17:03:57
%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 A190664(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
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