A353702 - OEIS

%I #18 May 05 2023 09:13:55

%S 12,15,21,26,27,33,38,57,62,69,74,85,88,93,106,108,129,133,134,145,

%T 166,177,178,205,213,217,218,226,237,248,249,253,254,262,265,278,309,

%U 314,328,362,375,376,393,398,417,422,424,445,459,466,469,488,489,493,502

%N Composite k such that tau(k') = (tau(k))', where tau(k) is the number of divisors of k (A000005) and k' is the arithmetic derivative of k (A003415).

%C Since for any prime number p, p' = 1 and (tau(p))' = 2' = 1 = tau(1) = tau (p'), the sequence requires only composite numbers that satisfy the given relation.

%C For p in A092109 the number m = 3*p is a term. Indeed, (tau(m))' = (tau(3*p))' = 4' = 4 and tau(m') = tau((3*p)') = tau(p + 3) = 4, so m is a term.

%C If p is in A045536 then p, p + 2 and 2*p + 1 are prime numbers and m = 3*(2*p + 1) is a term. Indeed, tau(m') = tau((3*(2*p + 1))') = tau(2*p + 4) = tau(2*(p+2)) = 4 and (tau(m))' = (tau((3*(2*p + 1)))' = 4' = 4, so m is a term.

%C If k is in A174100 then the numbers 2*k + 1 and 6*k + 1 are prime numbers and the numbers m = 2*(6*k + 1) is a term. Indeed, (tau(m))' = (tau(2*(6*k + 1)) )' = 4' = 4 and tau(m') = tau(2*(6*k + 1))') = tau(6*k + 3) = tau(2*(2*k + 1)) = 4, so m is a term.

%e 12' = 16, (tau(12)) = 6' = 5 and tau(12') = tau(16) = tau(2^4) = 5, so 12 is a term.

%e 15' = 8, (tau(15))’ = 4' = 4 and tau(15') = tau(8) = tau(2^3) = 4, so 15 is a term.

%p isA353702 := proc(n)

%p     if not isprime(n) and numtheory[tau](A003415(n)) = A003415( numtheory[tau](n) ) then

%p         true ;

%p     else

%p         false;

%p     end if;

%p end proc:

%p for n from 2 to 1000 do

%p     if isA353702(n) then

%p         printf("%d,",n) ;

%p     end if;

%p end do: # _R. J. Mathar_, May 05 2023

%t d[0] = d[1] = 0; d[n_] := n*Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); Select[Range[500], CompositeQ[#] && DivisorSigma[0, d[#]] == d[DivisorSigma[0, #]] &] (* _Amiram Eldar_, May 07 2022 *)

%o (Magma) f:=func<h |h eq 1 select 0 else  h*(&+[Factorisation(h)[i][2] / Factorisation(h)[i][1]: i in [1..#Factorisation(h)]])>; [p:p in [3..550]|not IsPrime(p) and  #Divisors(Floor(f(p))) eq Floor(f(#Divisors(p)))];

%o (PARI) ad(n) = vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]); \\ A003415

%o isok(k) = (k>1) && !isprime(k) && numdiv(ad(k)) == ad(numdiv(k)); \\ _Michel Marcus_, May 08 2022

%Y Cf. A000005, A003415, A045536, A092109, A174100, A260961.

%K nonn

%O 1,1

%A _Marius A. Burtea_, May 07 2022