A104907 Numbers n such that d(n)*reversal(n)=sigma(n), where d(n) is number of positive divisors of n.

1, 73, 861, 7993, 8241, 799993, 7999993, 44908500, 82000041, 293884500, 6279090751, 8200000041, 62698513951, 79999999993, 82000000041, 374665576800, 597921764310, 7999999999993, 8200000000041

Offset: 1

Author

Farideh Firoozbakht, Apr 16 2005

Keywords

• base
• more
• nonn

Comments

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.

a(20) > 10^13. - Giovanni Resta, Jul 13 2015

Examples

			Let p=8*10^n-7 be a prime so d(p)=2; reversal(p)=4*10^n-3 and sigma(p)
=8*10^n-6 hence d(p)*reversal(p)=sigma(p) and this shows that p
is in the sequence. 73,7993,799993 and 7999993 are such terms.
Also let q=(2*10^n+1)/3 be a prime, so 123*q=82*10^n+41; reversal
(123*q)=14*10^n+28; d(123*q)=8 and sigma(123*q)=168*q+168=112*10^n
+224 hence d(123*q)*reversal(123*q)=sigma(123*q) and this shows
that 123*q is in the sequence. 861,8241 and 82000041 are such terms.
		

Crossrefs

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

Programs

Mathematica

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

Extensions

a(11)-a(15) from Donovan Johnson, Feb 06 2010

a(16) from Giovanni Resta, Feb 06 2014

a(17)-a(19) from Giovanni Resta, Jul 13 2015