A105324 Numbers n such that 2*reversal(n)=sigma(n).

6, 73, 483, 4074, 4473, 4623, 7993, 42813, 69855, 253782, 799993, 7999993, 46000023, 426000213, 4600000023, 6718967838, 42600000213, 46000000023, 79999999993, 426000000213

Offset: 1

Author

Farideh Firoozbakht, Apr 16 2005

Keywords

• base
• more
• nonn

Comments

I. If p=8*10^n-7 is a prime then p is in the sequence because reversal(p)=4*10^n-3 & sigma(p)=8*10^n-6 so 2*reversal(p) =sigma(p). 73,7993,799993 & 7999993 are such terms.

II. If q=(2*10^n+1)/3 is a prime then (a): 69*q is in the sequence because 69*q=46*10^n+23; reversal (69*q)=32*10^n+64 & sigma(69*q)=96*q+96=64*10^n+128 so 2*reversal (69*q)=sigma(69*q). 483,4623 & 46000023 are such terms. (b): 639*q is in the sequence because 639*q=426*10^n+213; reversal (639*q)=312*10^n+624 & sigma(639*q)=936*q+936=624*10^n+1248 so 2*reversal(639*q)=sigma(639*q). 42813 & 426000213 are such terms.

a(21) > 10^12. - Giovanni Resta, Oct 28 2012

Examples

			253782 is in the sequence because reversal(253782)=287352; sigma(253782)=574704 & 2*287352=574704.
		

Crossrefs

Cf. A093170, A096507, A099190, A105322, A105323, A105325, A105326.

Programs

Mathematica

reversal[n_]:= FromDigits[Reverse[IntegerDigits[n]]]; Do[If[2* reversal[n]== DivisorSigma[1, n], Print[n]], {n, 1000000000}]
Select[Range[8*10^6],2*IntegerReverse[#]==DivisorSigma[1,#]&] (* The program generates the first 12 terms of the sequence. *) (* Harvey P. Dale, Oct 29 2022 *)

Extensions

a(15)-a(19) from Donovan Johnson, Dec 21 2008

a(20) from Giovanni Resta, Oct 28 2012