A069215 -id:A069215 - cp's OEIS frontend

A089675 Numbers k such that 10^k - 3 is prime.

1, 2, 3, 17, 140, 990, 1887, 3530, 5996, 13820, 21873, 26045, 87720, 232599, 480684, 538640

Offset: 1

Michael Gottlieb (mzrg(AT)verizon.net), Jan 05 2004

Numbers k such that 9*R_k - 2 is a prime number, where R_k = 11...1 is the repunit (A002275) of length k.
If k is in the sequence (10^k-3 is prime) and m=3*(10^k-3) then phi(m)=reversal(m), i.e., m is in the sequence A069215. - Farideh Firoozbakht, Dec 25 2004
No further terms for k <= 407197, see Kamada link.

			10^2 - 3 = 97 is a prime number (in fact all terms are the largest less than 10^k).

Makoto Kamada, Prime numbers of the form 99...997.
Index entries for primes involving repunits.

Cf. A002275, A056662, A069215, A093172, A101700.

m = 1000; For[n = 1, n < m, If[PrimeQ[10^n - 3], Print[n]]; n++]

a(n) = A056662(n) + 1.

a(8) from Robert G. Wilson v, Jan 14 2004
a(9) and a(10) from Gabriel Cunningham (gcasey(AT)mit.edu), Mar 06 2004
a(11) from Gabriel Cunningham (gcasey(AT)mit.edu), Mar 13 2004
a(12) from Henri Lifchitz.
Edited by Patrick De Geest, Dec 28 2004
Edited by Ray Chandler, Dec 23 2010
a(15) from Paul Bourdelais, Jan 06 2021
a(16) from Paul Bourdelais, Jan 28 2021

A115921 Numbers k such that the decimal digits of phi(k) are a permutation of those of k.

Original entry on oeis.org

1, 21, 63, 291, 502, 2518, 2817, 2991, 4435, 5229, 5367, 5637, 6102, 6174, 6543, 6822, 7236, 7422, 8022, 8541, 8982, 17631, 18231, 18261, 20301, 20518, 20617, 21058, 22471, 22851, 25196, 25918, 27615, 29817, 34816, 35683, 43218, 44305

Offset: 1

Giovanni Resta, Feb 06 2006

Contains A069215 and A113781; is itself a subsequence of A082060. - M. F. Hasler, Nov 28 2007

There is some m > 1 such that a(n) > m*n for all n > 1. This follows from the positive density of numbers n such that n/phi(n) > 10. - Charles R Greathouse IV, Sep 07 2012

			phi(20301) = 13200, phi(6543) = 4356.

V. Raman, Table of n, a(n) for n = 1..10000

Cf. A069215, A113781, A082060, A115920, A114065, A175795.

Select[Range[45000],Sort[IntegerDigits[EulerPhi[#]]]==Sort[IntegerDigits[#]]&] (* Harvey P. Dale, Jul 25 2018 *)

for(n=1,10^5,if(vecsort(Vecsmall(Str(n)))==vecsort(Vecsmall(Str(eulerphi(n)))),print1(n", "))) \\ M. F. Hasler, Nov 28 2007

from sympy import totient
A115921_list = [n for n in range(1,10**4) if sorted(str(totient(n))) == sorted(str(n))] # Chai Wah Wu, Dec 13 2015

Edited by M. F. Hasler, Nov 28 2007

A085331 Numbers n such that phi(rev(n))=n.

Original entry on oeis.org

1, 12, 36, 192, 1992, 2016, 31067664, 39206496, 1564356432, 3937403136, 15600000432, 22871605008, 156043560432, 156439956432, 1560000000432, 1956000004392

Offset: 1

Labos Elemer, Jul 04 2003

rev(2*(10^k-4)) = 3*(10^k-3). If 10^k-3 is prime, then phi(3*(10^k-3)) = 2*(10^k-4), so 2*(10^k-4) is a term. 10^1-3=7 is prime, so 2*(10^1-4)=12 is a term, a(2). 10^2-3=97 is prime, so 2*(10^2-4)=192 is a term, a(4). 10^3-3=997 is prime, so 2*(10^3-4)=1992 is a term, a(5). 10^17-3 is prime, so 2*(10^17-4)=199999999999999992 is a term. 10^140-3 is prime, so 2*(10^140-4) is a term. 10^990-3 is prime, so 2*(10^990-4) is a term. Conjecture: sequence is infinite. - Ray Chandler, Jul 20 2003
Let f(m,n,r,t)=((9).(m).78.(0)(n).21.(9)(m))(r).(9)(t).7 where m, n, r & t are nonnegative integers; dot between numbers means concatenation and "(m)(n)" means number of m's is n. If r*t=0 & p=f(m,n,r,t) is prime then reversal(3*p) = 1.((9)(m).56.(0)(n).43.(9)(m))(r).(9)(t).2 is in the sequence. For example p1=f(0,0,0,0)=7 so reversal(3*p1) = 12 is in the sequence, p2=f(0,0,2,0)=(7821)(2).7=782178217 so reversal(3*p2) = 1.(5643)(2).2 = 1564356432 is in the sequence & p3=f(0,0,674,0) so reversal(3*p3) = 1.(5643)(674).2 is in the sequence. Primes of the form f(m,n,r,t) are a generalized form of primes of the form 10^j-3 that were already related to this sequence by Ray Chandler. For all n, A085331(n) = reversal(A072395(n)). - Farideh Firoozbakht, Jan 08 2005
The list is complete through 2050000000. - Farideh Firoozbakht, Jan 15 2005
a(13) > 10^11. - Donovan Johnson, Feb 03 2012
a(17) > 10^13. - Giovanni Resta, Aug 06 2019

			phi[{1,21,63,291,2991,6102}] = {1,12,36,192,1992,2016}

Cf. A000010, A004086, A069215, A101700, A102278.

v = {1}; Do[ If[ n == EulerPhi[ FromDigits[ Reverse[ IntegerDigits [ n ] ] ] ], v = Append[ v, n ]; Print[ v ], If[ Mod[ n, 1000000 ] == 0, Print[ -n ] ] ], {n, 2, 2050000000, 2} ] (Firoozbakht)

The terms 31067664, 39206496, 1564356432 are from Farideh Firoozbakht, Jan 08 2005
a(10)-a(12) from Donovan Johnson, Feb 03 2012
a(13)-a(16) from Giovanni Resta, Aug 06 2019

A101700 Numbers of the form 3*(10^n-3), where 10^n-3 is prime.

Original entry on oeis.org

21, 291, 2991, 299999999999999991

Offset: 1

Farideh Firoozbakht, Dec 31 2004

nonn

a(5) = 3*(10^140-3) is 141 digits long and is too large to include.

If m is in this sequence then phi(m)=r(m), so this sequence is a subsequence of A069215. a(n)=3*(10^A089675(n)-3), so a(4)=3*(10^17-3), a(5)=3*(10^140-3), a(6)=3*(10^990-3), a(7)=3*(10^1887-3), a(8)=3*(10^3530-3), a(9)=3*(10^5996-3), a(10)=3*(10^13820-3), a(11)=3*(10^21873-3) & a(12)=3*(10 ^26045-3).

			299999999999999991 is in the sequence because (10^17-3) is prime and 3*(10^17-3)=299999999999999991.

Cf. A086947, A089675, A069215.

Do[If[PrimeQ[10^n-3], Print[3*(10^n-3)]], {n, 150}]
3#&/@Select[10^Range[20]-3,PrimeQ] (* Harvey P. Dale, Mar 23 2022 *)

a(n) = 3*(10^A089675(n) - 3).

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

Joseph L. Pe, Jul 21 2002

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

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

Cf. A085331, A069215, A102278.

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

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

A082060 Numbers n such that n and phi(n) have the same distinct decimal digits.

Original entry on oeis.org

1, 21, 63, 101, 233, 291, 502, 677, 1021, 1031, 1051, 1061, 1091, 1201, 1226, 1301, 1601, 1801, 1901, 2011, 2201, 2333, 2383, 2393, 2518, 2633, 2677, 2700, 2767, 2817, 2833, 2991, 3011, 3023, 3122, 3203, 3253, 3323, 3623, 3677, 3767, 3823, 3923, 3989

Offset: 1

Labos Elemer, Apr 04 2003

Contains A113781 as a subsequence. - M. F. Hasler, Nov 28 2007

Numbers n such that n and phi(n) have the same decimal digits = A115921. - Jaroslav Krizek, Nov 13 2014

			n=502 is a member since phi[502]=250

Harvey P. Dale, Table of n, a(n) for n = 1..2500

Cf. A000005, A000010, A082059, A115921, A113781, A113622, A069215, A072395, A113781.

Select[Range[4000],Union[IntegerDigits[#]]==Union[IntegerDigits[ EulerPhi[ #]]]&] (* Harvey P. Dale, Jan 31 2022 *)

for(n=1,10^4,if(Set(Vec(Str(n)))==Set(Vec(Str(eulerphi(n)))),print1(n", "))) \\ M. F. Hasler, Nov 28 2007

from sympy import totient
A082060_list = [n for n in range(1,10**4) if set(str(totient(n))) == set(str(n))] # Chai Wah Wu, Dec 13 2015

Definition and comment corrected by Jaroslav Krizek, Nov 13 2014

A101703 Numbers n such that reversal(n) = (2/3)*n - 2.

Original entry on oeis.org

21, 291, 885, 2991, 29991, 234651, 299991, 2340651, 2999991, 8221845, 23400651, 29346591, 29999991, 234000651, 293406591, 299999991, 2340000651, 2346534651, 2934006591, 2993465991, 2999999991, 23400000651, 23465934651, 29340006591, 29934065991, 29999999991, 82277815845, 234000000651

Offset: 1

Farideh Firoozbakht, Dec 31 2004

Numbers of the form 3*(10^n-3) are in the sequence, so A086947 is an infinite subsequence of this sequence. Also A101700 is a subsequence of this sequence.
Let f(r,s,t,z) = 2.(9)(r+s).(34.(0)(t).65)(z).(9)(s).1 where the dot between numbers means concatenation and "(m)(n)" means number of m's is n, for example f(0,2,1,3)= 299340653406534065991, it is interesting that all numbers of the form f(r,s,t,z) where r, s, t & z are nonnegative integers and r*z=0 are in this sequence.
Except for 885 & 8221845 all known terms of this sequence are of the form f(r,s,t,z).
For all r, s & t we have f(r,s,t,0)=f(r,s,0,0)=f(r+2s,0,0,0)=A086947(r+2s+1)= 3*(10^(r+2s+1)-3).
a(1) = 21 = f(0,0,0,0), a(2) = 291 = f(1,0,0,0), a(4) = 2991 = f(2,0,0,0) = f(0,1,0,0), a(5) = 29991 = f(3,0,0,0) = f(1,1,0,0), a(6) = 234651 = f(0,0,0,1), a(7) = 299991 = f(4,0,0,0) = f(0,2,0,0), a(8) = 2340651 = f(0,0,1,1), etc. Next term is greater than 11*10^8.
From David Wasserman, Mar 27 2008: (Start)
234653406534651 is a term that doesn't fit the f(r,s,t,z) format.
We may redefine f so that t is a vector of length z, which must be symmetrical to produce a member. For example f(0,0,[0,1,0],3) = 234653406534651 is a member, but f(0,0,[1,0,0],3) = 234065346534651 is not a member.
23465934651 is another member that doesn't fit the pattern. In general there may be any number of 9's between a 5 and a 3, provided that the 9's are symmetrical. So 2346593465934651 is a member, but 23465993465934651 is not. (End)

			f(0,1,2,3) = 2934006534006534006591 is in the sequence because reversal(2934006534006534006591) = 1956004356004356004392 = (2/3)*2934006534006534006591-2.

Giovanni Resta, Table of n, a(n) for n = 1..56 (terms < 10^14)

Cf. A086947, A069215, A101700, A101704.

Do[If[FromDigits[Reverse[IntegerDigits[n]]] == 2/3*n - 2, Print[n]], {n, 1100000000}]

More terms from David Wasserman, Mar 27 2008

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

Farideh Firoozbakht, Jan 04 2005

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

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

Cf. A089675, A069215, A085331, A101700.

Do[If[PrimeQ[78*10^n + 217], Print[n]], {n, 8280}]

is(n)=ispseudoprime(78*10^n+217) \\ Charles R Greathouse IV, May 22 2017

a(18) from Michael S. Branicky, Oct 15 2024

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

Original entry on oeis.org

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

A114931 Numbers n such that phi(n)=4*reversal(n).

Original entry on oeis.org

10, 20, 40, 50, 80, 210, 420, 630, 711, 831, 840, 2910, 29910, 40320, 80640, 98361, 673140, 865580, 8656341, 466760130, 694602930, 821412711, 23465346510, 40396039620, 63130473930, 234000006510, 464205665820, 2340653406510, 2346599346510

Offset: 1

Farideh Firoozbakht, Jan 29 2006

If p=10^m-3 is prime then 30*p is in the sequence because phi(30*p)=phi(30)*phi(p)=8*(10^m-4)=4*(2*10^m-8)=4*reversal (3*10^m-9)=4*reversal(3*p)=4*reversal(30*p). Next term is greater than 55*10^7.
Let f(m,n)=(78*10^(m+3)+210)*(10^(n*(m+4))-1)/(10^(m+4)-1)+7, if p=f(m,n) is prime then 30*p is a term of the sequence. - Jahangeer Kholdi, Nov 13 2013
Also if p=(1/101)*(680*10000^n+27) is prime then 60*p is in the sequence. - Jahangeer Kholdi, Nov 13 2013
a(30) > 10^13. - Giovanni Resta, Aug 12 2019

			20 is in the sequence because phi(20)=4*2=4*reversal(20).

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

Do[If[EulerPhi[n]==4*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 550000000}]

a(21)-a(27) from Giovanni Resta, Oct 28 2012

a(28)-a(29) from Giovanni Resta, Aug 12 2019

cp's OEIS Frontend

A089675 Numbers k such that 10^k - 3 is prime.

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A115921 Numbers k such that the decimal digits of phi(k) are a permutation of those of k.

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A085331 Numbers n such that phi(rev(n))=n.

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A101700 Numbers of the form 3*(10^n-3), where 10^n-3 is prime.

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

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A082060 Numbers n such that n and phi(n) have the same distinct decimal digits.

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A101703 Numbers n such that reversal(n) = (2/3)*n - 2.

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