A104907 - OEIS

%I #16 Jan 31 2023 16:38:14

%S 1,73,861,7993,8241,799993,7999993,44908500,82000041,293884500,

%T 6279090751,8200000041,62698513951,79999999993,82000000041,

%U 374665576800,597921764310,7999999999993,8200000000041

%N Numbers n such that d(n)*reversal(n)=sigma(n), where d(n) is number of positive divisors of n.

%C All primes of the form 8*10^n-7 are in the sequence, so 8*10^A099190-3 is a subsequence of this sequence. A105322 is this subsequence. Also if p=(2*10^n+1)/3 is prime then 123*p is in the sequence, so 123*A093170 is a subsequence of this sequence. A105323 is this subsequence.

%C a(20) > 10^13. - _Giovanni Resta_, Jul 13 2015

%e Let p=8*10^n-7 be a prime so d(p)=2; reversal(p)=4*10^n-3 and sigma(p)

%e =8*10^n-6 hence d(p)*reversal(p)=sigma(p) and this shows that p

%e is in the sequence. 73,7993,799993 and 7999993 are such terms.

%e Also let q=(2*10^n+1)/3 be a prime, so 123*q=82*10^n+41; reversal

%e (123*q)=14*10^n+28; d(123*q)=8 and sigma(123*q)=168*q+168=112*10^n

%e +224 hence d(123*q)*reversal(123*q)=sigma(123*q) and this shows

%e that 123*q is in the sequence. 861,8241 and 82000041 are such terms.

%t reversal[n_]:= FromDigits[Reverse[IntegerDigits[n]]]; Do[If[DivisorSigma[0, n]*reversal[n] == DivisorSigma[1, n], Print[n]], {n, 1125000000}]

%t Select[Range[8*10^6],DivisorSigma[0,#]IntegerReverse[#]==DivisorSigma[1,#]&] (* The program generates the first 7 terms of the sequence. *) (* _Harvey P. Dale_, Jan 31 2023 *)

%Y Cf. A056657, A093170, A096507, A099190, A105322, A105323, A105324.

%K base,more,nonn

%O 1,2

%A _Farideh Firoozbakht_, Apr 16 2005

%E a(11)-a(15) from _Donovan Johnson_, Feb 06 2010

%E a(16) from _Giovanni Resta_, Feb 06 2014

%E a(17)-a(19) from _Giovanni Resta_, Jul 13 2015