A072394 Numbers n such that sigma(n)=reversal(n)-n.

1563, 1633, 18673, 32207, 1405313, 1567563, 1656833, 193613415, 325933027, 376491249, 2287850446, 2432416646, 13823276223, 14055445313, 19087920283, 23804849568, 36303512827, 148868530953

Offset

1,1

Comments

If (58*1000^n-169)/111 is prime then (58*1000^n-169)/37 is in the sequence (the proof is easy). Next term is greater than 12*10^8. - Farideh Firoozbakht, Jan 29 2006

From Farideh Firoozbakht, May 25 2010: (Start)

If p = 156/101*(10^(4n)-1)-1 is prime then 91*p is in the sequence (the proof is easy).

A178321 gives numbers n such that (58*1000^n-169)/111 = 58/111*(10^(3n)-1)-1 is prime and A178322 gives numbers n such that 156/101*(10^(4n)-1)-1 is prime. (End)

a(19) > 10^12. - Giovanni Resta, Oct 28 2012

Links

Table of n, a(n) for n=1..18.

Example

reverse(1563) - 1563 = 3651 - 1563 = 2088 = sigma(1563), so 1563 is a term of the sequence.
376491249 is in the sequence because sigma(376491249)=565703424 =942194673-376491249=reversal(376491249)-376491249.

Mathematica

Select[Range[10^6], FromDigits[Reverse[IntegerDigits[n]]] - # == DivisorSigma[1, # ] &]
Do[If[DivisorSigma[1, n]==FromDigits[Reverse[IntegerDigits[n]]]- n, Print[n]], {n, 1200000000}] (* Farideh Firoozbakht *)

Crossrefs

Cf. A072234.

Cf. A178321, A178322. [From Farideh Firoozbakht, May 25 2010]

Sequence in context: A035865 A031800 A280354 * A243528 A237673 A246734

Adjacent sequences: A072391 A072392 A072393 * A072395 A072396 A072397

Keyword

base,nonn

Author

Joseph L. Pe, Jul 21 2002

Extensions

More terms from Farideh Firoozbakht, Jan 29 2006

a(11)-a(17) from Donovan Johnson, Dec 21 2008

a(18) from Giovanni Resta, Oct 28 2012

Status

approved