cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A069215 Numbers n such that phi(n) = reversal(n).

Original entry on oeis.org

1, 21, 63, 270, 291, 2991, 6102, 46676013, 69460293, 2346534651, 6313047393, 23400000651, 80050617822, 234065340651, 234659934651, 2340000000651, 2530227348360, 2934000006591
Offset: 1

Views

Author

Joseph L. Pe, Apr 11 2002

Keywords

Comments

If 10^n-3 is prime (n is in the sequence A089765) and m=3*(10^n-3) then m is in this sequence, for example 299999999999999991 is a term of this sequence because 299999999999999991=3*(10^17-3) and 17 is in the sequence A089675. So 3*(10^A089675-3) is a subsequence of this sequence, A101700 is this subsequence. - Farideh Firoozbakht, Dec 26 2004
A072395 is a subsequence of this sequence. If m is in the sequence and 10 doesn't divide m then reversal(m) is in the sequence A085331, so see Comments on A085331. - Farideh Firoozbakht, Jan 09 2005
If p=(79*10^(4n+1)-83)/101 is prime then 3p is in the sequence. The proof is easy. 21, 2346534651 & 3*(79*10^2697-83)/101 are the first three such terms. - Farideh Firoozbakht, Apr 22 2008, Aug 16 2008
a(19) > 10^13. - Giovanni Resta, Aug 07 2019

Examples

			phi(291) = 192.
phi(6102) = 2016 = reversal(6102), so 6102 belongs to the sequence.
		

Crossrefs

Programs

  • Mathematica
    Do[If[EulerPhi[n] == FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 1, 10^5}]
  • PARI
    for( n=1,1e9, A004086(n)==eulerphi(n) & print1(n","))

Extensions

More terms from Farideh Firoozbakht, Aug 31 2004
One more term from Farideh Firoozbakht, Jan 09 2005
a(11)-a(13) from Donovan Johnson, Feb 03 2012
a(14)-a(15) from Giovanni Resta, Oct 28 2012
a(16)-a(18) from Giovanni Resta, Aug 07 2019

A086947 Numbers k such that R(k+9) = 3.

Original entry on oeis.org

21, 291, 2991, 29991, 299991, 2999991, 29999991, 299999991, 2999999991, 29999999991, 299999999991, 2999999999991, 29999999999991, 299999999999991, 2999999999999991, 29999999999999991, 299999999999999991, 2999999999999999991, 29999999999999999991, 299999999999999999991
Offset: 1

Views

Author

Ray Chandler, Jul 24 2003

Keywords

Comments

If k is in this sequence then Reverse(k) = (2/3)*k - 2. Also A101703 is the sequence of all numbers k such that Reverse(k) = (2/3)*k - 2. So this sequence is a subsequence of A101703. - Farideh Firoozbakht, Dec 30 2004

Crossrefs

Programs

  • Magma
    [3*(10^n-3): n in [1..25] ]; // Vincenzo Librandi, Aug 22 2011
  • Mathematica
    Table[3*(10^n-3), {n, 17}]
    Table[FromDigits[PadRight[{3},n,0]],{n,2,20}]-9 (* Harvey P. Dale, Nov 27 2012 *)

Formula

a(n) = 3*(10^n - 3).
R(a(n)) = A086948(n).
From Chai Wah Wu, Aug 01 2020: (Start)
a(n) = 11*a(n-1) - 10*a(n-2) for n > 2.
G.f.: x*(60*x + 21)/((x - 1)*(10*x - 1)). (End)
From Elmo R. Oliveira, May 01 2025: (Start)
E.g.f.: 3*(2 - 3*exp(x) + exp(10*x)).
a(n) = 3*A173833(n). (End)

A072395 Numbers n such that reverse(phi(n)) = n.

Original entry on oeis.org

1, 21, 63, 291, 2991, 6102, 46676013, 69460293, 2346534651, 6313047393, 23400000651, 80050617822, 234065340651, 234659934651, 2340000000651, 2934000006591
Offset: 1

Views

Author

Joseph L. Pe, Jul 21 2002

Keywords

Comments

For all n, a(n) = reversal(A085331(n)), so see Comment on A085331. This sequence is a subsequence of A069215 and if m is a term of A069215 and 10 doesn't divide m then m is in this sequence. - Farideh Firoozbakht, Jan 09 2005
a(13) > 10^11. - Donovan Johnson, Feb 03 2012
a(17) > 10^13. - Giovanni Resta, Aug 06 2019

Examples

			reverse(phi(6102)) = reverse(2016) = 6102, so 6102 is a term of the sequence.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[10^5], FromDigits[Reverse[IntegerDigits[EulerPhi[#]]]] == # &] (* corrected by Harvey P. Dale, Oct 03 2011 *)

Extensions

More terms from Robert G. Wilson v, Jul 23 2002
More terms from Farideh Firoozbakht, Jan 09 2005
a(10)-a(12) from Donovan Johnson, Feb 03 2012
a(13)-a(16) from Giovanni Resta, Aug 06 2019

A086948 a(n) = k where R(k+8) = 2.

Original entry on oeis.org

12, 192, 1992, 19992, 199992, 1999992, 19999992, 199999992, 1999999992, 19999999992, 199999999992, 1999999999992, 19999999999992, 199999999999992, 1999999999999992, 19999999999999992, 199999999999999992, 1999999999999999992, 19999999999999999992, 199999999999999999992
Offset: 1

Views

Author

Ray Chandler, Jul 24 2003

Keywords

Crossrefs

Programs

  • Magma
    [2*(10^n-4): n in [1..25] ]; // Vincenzo Librandi, Aug 22 2011
  • Mathematica
    Table[10*FromDigits[PadRight[{1},n,9]]+2,{n,20}] (* Harvey P. Dale, Dec 15 2017 *)

Formula

a(n) = 2*(10^n - 4).
R(a(n)) = A086947(n).
From Chai Wah Wu, Aug 01 2020: (Start)
a(n) = 11*a(n-1) - 10*a(n-2) for n > 2.
G.f.: x*(60*x + 12)/((x - 1)*(10*x - 1)). (End)
From Elmo R. Oliveira, May 01 2025: (Start)
E.g.f.: 2*(3 - 4*exp(x) + exp(10*x)).
a(n) = 12*A246057(n-1) = 6*A323639(n). (End)

A102278 Numbers k such that 78*10^k + 217 is prime.

Original entry on oeis.org

1, 2, 8, 10, 13, 21, 22, 36, 57, 80, 149, 484, 505, 642, 806, 974, 1674, 34177
Offset: 1

Views

Author

Farideh Firoozbakht, Jan 04 2005

Keywords

Comments

If n is a term greater than 2 in this sequence and m = 3*(78*10^n + 217) then phi(m) = reversal(m) (m is in the sequence A069215) because phi(m) = 2*(78*10^n + 216) = 156*10^n + 432 = reversal(234*10^n + 651) = reversal(m).
For example since 8>2 & 8 is in this sequence, for m = 3* (78*10^8 + 217) = 23400000651 phi(m) = reversal(m), so 23400000651 is a term of A069215.
Let f(n,m,r,t) = ((9)(n).78.(0)(m).21.(9)(n))(r).(9)(t).7 where dot between numbers means concatenation and "(m)(n)" means number of m's is n.
In fact I proved that for nonnegative integers n, m, r & t such that r*t = 0 if p = f(n,m,r,t) is prime then phi(3*p) = reversal (3*p). (3*p is in the sequence A069215, some special cases:
Case I, p = f(0,0,0,n-1) = (9)(n-1).7 = 10^n - 3 (see A089675). Case II, p = f(0,n-3,0,0) = 78.(0)(n-3).217 = 78*10^n + 217. Case III, p = f(0,0,n,0) = (7821)(n).7. In this case I found only three such prime p1 = (78217)(0).7 = 7, p2 = (7821)(2).7 = 782178217 & p3 = (7821)(674).7, p3 is a prime with length 2697.
Next term is greater than 8280.
Next term is greater than 24000. - Michael S. Branicky, Mar 22 2023

Examples

			8 is in the sequence because 78.(8-3)(0).217 = 7800000217 is prime.
		

Crossrefs

Programs

  • Mathematica
    Do[If[PrimeQ[78*10^n + 217], Print[n]], {n, 8280}]
  • PARI
    is(n)=ispseudoprime(78*10^n+217) \\ Charles R Greathouse IV, May 22 2017

Extensions

a(18) from Michael S. Branicky, Oct 15 2024
Showing 1-5 of 5 results.