A114927 -id:A114927 - cp's OEIS frontend

A072393 Numbers n such that n - reverse(n) = phi(n).

91, 874, 3411, 9093, 40112, 44252, 54081, 67284, 80224, 90933, 91503, 4961782, 5400081, 5726691, 8750834, 9076921, 9155055, 54000081, 62023914, 90766921, 93079231, 430770922, 540000081, 636355044, 808618664, 907666921, 928709013, 4050394312, 4262971312

Offset: 1

Joseph L. Pe, Jul 21 2002

If m>1 and p=2*10^m+3 is prime then n=27*p is in the sequence because n-reversal(n)=27*(2*10^m+3)-reversal(27*(2*10^m+3))= (54*10^m+81)-(18*10^m+45)=36*10^m+36=18*(2*10^m+2)=phi(27)* phi(2*10^m+3)=phi(27*(2*10^m+3))=phi(n). Also if m>2 and p=(389*10^m+109)/3 is prime then 7*p is in the sequence (the proof is easy). Next term is greater than 2*10^8. - Farideh Firoozbakht, Jan 27 2006

a(51) > 10^12. - Giovanni Resta, Oct 28 2012

			91 - 19 = 72 = phi(91), so 91 is a term of the sequence.

Giovanni Resta, Table of n, a(n) for n = 1..50

Cf. A114926, A114927.

Select[Range[10^5], # - FromDigits[Reverse[IntegerDigits[n]]] == EulerPhi[ # ] &]

More terms from Farideh Firoozbakht, Jan 27 2006

a(22)-a(29) from Donovan Johnson, Dec 04 2011

A114928 Numbers n such that sigma(n)=4*reversal(n).

Original entry on oeis.org

42, 402, 492, 4000002, 57906504, 400000002, 4000000002, 6279090751, 62698513951, 400000000002

Offset: 1

Farideh Firoozbakht, Jan 28 2006

If p=(2*10^n+1)/3 is prime then m=6*p is in the sequence because sigma(m)=sigma(6*p)=12*(2*10^n+4)/3=4*(2*10^n+4)=4* reversal(4*10^n+2)=4*reversal(6*(2*10^n+1)/3)=4*reversal(6*p) =4*reversal(m). Next term is greater than 5*10^8.

a(11) > 10^12. - Giovanni Resta, Oct 28 2012

			492 is in the sequence because sigma(492)=sigma(4*3*41)=7*4*42
=4*294=4*reversal(492).

Cf. A069216, A105324, A114927.

Do[If[DivisorSigma[1, n]==4*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 500000000}]

a(7)-a(9) from Donovan Johnson, Dec 21 2008

a(10) from Giovanni Resta, Oct 28 2012

cp's OEIS Frontend

A072393 Numbers n such that n - reverse(n) = phi(n).

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