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A115926
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Numbers n such that phi(n)=reversal(n)-n.
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0
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37, 397, 1853, 15503, 48776, 198683, 200882, 1061361, 3542805, 3564217, 3868867, 3962197, 4438616, 19844683, 198444683, 202195682, 309520655, 431092646, 439419646, 500729929, 535973599
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| All primes of the form 4*10^n-3 are in the sequence because if 4*10^n-3 is prime then phi(4*10^n-3)=(4*10^n-4) =(8*10^n-7)-(4*10^n-3)=reversal(4*10^n-3)-(4*10^n-3).
Also if n>1 and p=(94*10^n+113)/9 is prime then 19*p is in the sequence (the proof is easy). Next term is greater than 125*10^6.
Next term is greater than 6*10^8 - Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
If p=(1/303)*(232*10^(4n)+71) is prime then 7*p is in the sequence (the proof is easy). The first four such terms happen for n=2, 101, 104 & 444 and numbers of digits of these terms of the sequence are 9, 405, 417 & 1777 respectively. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Jan 02 2008
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EXAMPLE
| If n=37, phi(37)= 36 = 73-37
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MATHEMATICA
| Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]-n, Print[n]], {n, 125000000}]
Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]-n, Print[n]], {n, 600000000}] - Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
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CROSSREFS
| Cf. A072393.
Sequence in context: A124337 A027944 A201789 * A083818 A090023 A201609
Adjacent sequences: A115923 A115924 A115925 * A115927 A115928 A115929
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KEYWORD
| more,nonn,base
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AUTHOR
| Farideh Firoozbakht (mymontain(AT)yahoo.com), Jan 31 2006
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EXTENSIONS
| More terms from Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
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