A114931 -id:A114931 - cp's OEIS frontend

A114930 Numbers n such that phi(n)=2*reversal(n).

6180, 27630, 2914830, 4471740, 27000630, 637062480, 27000000630, 679410757980, 4412687534631, 4421625783741

Offset: 1

Farideh Firoozbakht, Jan 29 2006

If m>1 and p=3*10^m+7 is prime then 90*p is in the sequence because phi(90*p)=phi(90)*phi(p)=24*(3*10^m+6)=2*(36*10^m+72) =2*reversal(27*10^m+63)=2*reversal(9*p)=2*reversal(90*p). Note that 30 divides all known terms of this sequence. Next term is greater than 11*10^7.

a(11) > 10^13. - Giovanni Resta, Aug 12 2019

			637062480 is a term because phi(637062480) = 2*84260736 = 2*reversal(637062480).

Cf. A000010, A004086, A069215, A114930, A136539, A114931, A136538.

Do[If[EulerPhi[n]==2*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 110000000}]

a(6)-a(8) from Giovanni Resta, Oct 28 2012

a(9)-a(10) from Giovanni Resta, Aug 12 2019

A136538 Numbers n such that reversal(n)=2*phi(n).

Original entry on oeis.org

2, 4, 8, 42, 84, 2763, 4032, 8064, 67314, 86558, 291483, 2700063, 2700000063, 4039603962, 46420566582, 6739054689866

Offset: 1

Farideh Firoozbakht, Jan 04 2008

If m>1 and p=3*10^m+7 is prime then n=9*p is in the sequence (the proof is easy). If n is an even term of the sequence and the largest digit of n is less than 5(3) then 2n is (both numbers 2n & 4n are) in the sequence (the proof is easy).

a(17) > 10^13. - Giovanni Resta, Aug 12 2019

			Reversal(42)=24=2*12=2*phi(42), so 42 is in the sequence. [Example corrected Jan 25 2008]

Cf. A000010, A004086, A069215, A114930, A114931, A136539.

Do[If[FromDigits@Reverse@IntegerDigits@n==2*EulerPhi[n], Print[n]],{n,100000000}]

a(13)-a(15) from Giovanni Resta, Oct 28 2012

a(16) from Giovanni Resta, Aug 12 2019

cp's OEIS Frontend

A114930 Numbers n such that phi(n)=2*reversal(n).

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