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Smallest k>n such that tau(n) divides tau(k)=A000005(k).
3

%I #13 Sep 15 2018 05:30:56

%S 2,3,5,9,6,8,8,10,12,14,12,18,14,15,21,48,18,20,20,28,22,24,24,30,28,

%T 27,30,32,30,40,32,44,34,35,38,100,38,39,40,42,42,54,44,45,50,51,48,

%U 80,50,52,54,60,54,56,56,66,58,60,60,72,62,65,68,192,66,70,68,72,70,78,72

%N Smallest k>n such that tau(n) divides tau(k)=A000005(k).

%H Ivan Neretin, <a href="/A069805/b069805.txt">Table of n, a(n) for n = 1..10000</a>

%o (PARI) for(s=1,100,n=s+1; while(frac(numdiv(n)/numdiv(s))>0,n++); print1(n,","); )

%o (PARI) a(n) = my(k=n+1); while(numdiv(k)%numdiv(n), k++); k; \\ _Michel Marcus_, Sep 15 2018

%Y Cf. A000005.

%Y Cf. A140635 (without k>n constraint).

%K easy,nonn

%O 1,1

%A _Benoit Cloitre_, May 01 2002