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

1,2

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

Links

Table of n, a(n) for n=1..19.

Example

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.

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 *)

Crossrefs

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

Sequence in context: A100412 A169830 A197341 * A296024 A254136 A123811

Adjacent sequences: A104904 A104905 A104906 * A104908 A104909 A104910

Keyword

base,more,nonn

Author

Farideh Firoozbakht, Apr 16 2005

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

Status

approved