A179186 -id:A179186 - cp's OEIS frontend

A001274 Numbers k such that phi(k) = phi(k+1).

1, 3, 15, 104, 164, 194, 255, 495, 584, 975, 2204, 2625, 2834, 3255, 3705, 5186, 5187, 10604, 11715, 13365, 18315, 22935, 25545, 32864, 38804, 39524, 46215, 48704, 49215, 49335, 56864, 57584, 57645, 64004, 65535, 73124, 105524, 107864, 123824, 131144, 164175, 184635

Offset: 1

N. J. A. Sloane

Unlike totients, cototient(x + 1) = cototient(x) never holds (except 2 - phi(2) = 3 - phi(3) = 1) because cototient(x) is congruent to x modulo 2. - Labos Elemer, Aug 08 2001
Lal-Gillard and Firoozbakht ask whether there is another pair of consecutive integers in this sequence, apart from a(16) + 1 = a(17) = 5187, see link. - M. F. Hasler, Jan 05 2011
There are 5236 terms less than 10^12. - Jud McCranie, Feb 13 2012
Up to 10^13 there are 10755 terms, and no further consecutive pairs like (5186, 5187). - Giovanni Resta, Feb 27 2014
A051179(k) for k from 0 to 5 are in the sequence. No other members of A051179 are in the sequence, because phi(2^(2^k)-1) = Product_{j=1..k-1} phi(2^(2^j)+1) and phi(2^(2^5)+1) < 2^(2^5) so if k > 5, phi(2^(2^k)-1) < Product_{j=1..k-1} 2^(2^j) = 2^(2^k-1) = phi(2^(2^k)). - Robert Israel, Mar 31 2015
Number of terms < 10^k, k=1,2,3,...: 2, 3, 10, 17, 36, 68, 142, 306, 651, 1267, 2567, 5236, 10755, ..., . - Robert G. Wilson v, Apr 10 2019
Conjecture: Except for the first two terms, all terms are composite and congruent to either 2 or 3 (mod 6). - Robert G. Wilson v, Apr 10 2019
Paul Kinlaw has noticed that up to 10^13 the only counterexample to the above conjecture is a(7424) = 3044760173455. - Giovanni Resta, May 23 2019

			phi(3) = phi(4) = 2, so 3 is in the sequence.
phi(15) = phi(16) = 8, so 15 is in the sequence.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, pp 5, Ellipses, Paris 2008.
R. K. Guy, Unsolved Problems Number Theory, Sect. B36.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Giovanni Resta, Table of n, a(n) for n = 1..10755 (terms < 10^13, a(1)-a(2567) from T. D. Noe, a(2568)-a(5236) from J. McCranie)
R. Baillie, Table of phi(n) = phi(n+1), Math. Comp., 30 (1976), 189-190.
Jonathan Bayless and Paul Kinlaw, Consecutive coincidences of Euler's function, International Journal of Number Theory, Vol. 12, No. 4 (2016), pp. 1011-1026.
Farideh Firoozbakht, Puzzle 466. phi(n-1)=phi(n)=phi(n+1), in C. Rivera's Primepuzzles.
Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.
Dov Jarden, Recurring Sequences, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy] See p. 66.
Paul Kinlaw, Mitsuo Kobayashi and Carl Pomerance, On the equation phi(n) = phi(n+1), Acta Arithmetica, Vol. 196 (2020), pp. 69-92, alternative link.
V. L. Klee, Jr., Some remarks on Euler's totient function, Amer. Math. Monthly, 54 (1947), 332.
M. Lal and P. Gillard, On the equation phi(n) = phi(n+k), Math. Comp., 26 (1972), 579-583.
K. Miller, Solutions of phi(n) = phi(n+1) for 1 <= n <= 500000. Unpublished, 1972. [ Cf. (Untitled), Math. Comp., Vol. 27, p. 447, 1973 ].
Leo Moser, Some equations involving Euler's totient function, Amer. Math. Monthly, 56 (1949), 22-23.
J. Shallit, Letter to N. J. A. Sloane, Jul 17 1975.

Cf. A000010, A001494, A051953, A003276, A003275, A007015, A179186, A179187, A179188, A179189, A179202, A217139.

import Data.List (elemIndices)
a001274 n = a001274_list !! (n-1)
a001274_list = map (+ 1) $ elemIndices 0 $
                           zipWith (-) (tail a000010_list) a000010_list
-- Reinhard Zumkeller, May 20 2014, Mar 31 2011

[n: n in [1..3*10^5] | EulerPhi(n) eq EulerPhi(n+1)]; // Vincenzo Librandi, Apr 14 2015

select(n -> numtheory:-phi(n) = numtheory:-phi(n+1), [$1..10^5]); # Robert Israel, Mar 31 2015

Reap[For[n = 1; k = 2; f1 = 1, k <= 10^9, k++, f2 = EulerPhi[k]; If[f1 == f2, Print["a(", n, ") = ", k - 1]; Sow[k - 1]; n++]; f1 = f2]][[2, 1]] (* Jean-François Alcover, Mar 29 2011, revised Dec 26 2013 *)
Flatten[Position[Partition[EulerPhi[Range[200000]],2,1],{x_,x_}]] (* Harvey P. Dale, Dec 27 2015 *)
Select[Range[1000], EulerPhi[#] == EulerPhi[# + 1] &] (* Alonso del Arte, Oct 03 2014 *)
SequencePosition[EulerPhi[Range[200000]],{x_,x_}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 01 2018 *)
k = 8; lst = {1, 3}; While[k < 200000, If[ !PrimeQ[k +1], ep = EulerPhi[k +1]; If[ EulerPhi[k] == ep, AppendTo[lst, k]]; If[ep == EulerPhi[k +2], AppendTo[lst, k +1]]]; k += 6]; lst (* Robert G. Wilson v, Apr 10 2019 *)

is(n)=eulerphi(n)==eulerphi(n+1) \\ Charles R Greathouse IV, Feb 27 2014

list(lim)=my(v=List(),old=1); forfactored(n=2,lim\1+1, my(new=eulerphi(n)); if(old==new, listput(v,n[1]-1)); old=new); Vec(v) \\ Charles R Greathouse IV, Jul 17 2022

Conjecture: a(n) ~ C*n^3*log(n), where C = 9/Pi^2 = 0.91189... - Thomas Ordowski, Oct 21 2014

Sum_{n>=1} 1/a(n) is in the interval (1.4324884, 7.8358) (Kinlaw et al., 2020; an upper bound 441702 was given by Bayless and Kinlaw, 2016). - Amiram Eldar, Oct 15 2020

More terms from David W. Wilson

A001494 Numbers k such that phi(k) = phi(k+2).

Original entry on oeis.org

4, 7, 8, 10, 26, 32, 70, 74, 122, 146, 308, 314, 386, 512, 554, 572, 626, 635, 728, 794, 842, 910, 914, 1015, 1082, 1226, 1322, 1330, 1346, 1466, 1514, 1608, 1754, 1994, 2132, 2170, 2186, 2306, 2402, 2426, 2474, 2590, 2642, 2695, 2762, 2906, 3242, 3314

Offset: 1

N. J. A. Sloane

If p and 2p-1 are odd primes then 2*(2p-1) is a solution of the equation. Other terms (7,8,32,70,...) are not of this form.
There are 506764111 terms under 10^12. - Jud McCranie, Feb 13 2012
If 2^(2^m) + 1 is a Fermat prime in A019434, so, m = 0, 1, 2, 3, 4, then k = 2^(2^m + 1) is a term; this subsequence consists of {4, 8, 32, 512, 131072} and, in this case, phi(k) = phi(k+2) = 2^(2^m). - Bernard Schott, Apr 22 2022

D. M. Burton, Elementary Number Theory, section 7-2.
R. K. Guy, Unsolved Problems Number Theory, Sect. B36.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Jud McCranie, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.
M. F. Hasler, Table of n, a(n) for n = 1..17286. (Terms up to 10^7.)
V. L. Klee, Jr., Some remarks on Euler's totient function, Amer. Math. Monthly, 54 (1947), 332.
Leo Moser, Some equations involving Euler's totient function, Amer. Math. Monthly, 56 (1949), 22-23.

Cf. A000010, A001274, A007015, A179186, A179187, A179188, A179189, A179202, A217139.

import Data.List (elemIndices)
a001494 n = a001494_list !! (n-1)
a001494_list = map (+ 1) $ elemIndices 0 $
               zipWith (-) (drop 2 a000010_list) a000010_list
-- Reinhard Zumkeller, Feb 08 2013

[n: n in [1..4000] | EulerPhi(n) eq EulerPhi(n+2)]; // Vincenzo Librandi, Sep 07 2016

Select[Range[3500], EulerPhi[#]==EulerPhi[#+2]&] (* Harvey P. Dale, Apr 24 2011 *)
Flatten[Position[Partition[EulerPhi[Range[3500]],3,1],?(#[[1]]==#[[3]]&),{1},Heads->False]] (* This program is more efficient than the first program above because it only has to calculate phi of each number once. *) (* _Harvey P. Dale, Aug 20 2014 *)
SequencePosition[EulerPhi[Range[4300]],{x_,,x}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Sep 04 2020 *)

op=[0,c=0]; for( n=1,1e7,if( op[bittest(n,0)+1]+0==op[bittest(n,0)+1]=eulerphi(n), write("b001494.txt",c++," "n-2))) \\ M. F. Hasler, Jan 05 2011

A000010(a(n)) = A000010(a(n) + 2). - Reinhard Zumkeller, Feb 08 2013

More terms from Jud McCranie, Dec 24 1999

A179188 Numbers n such that phi(n) = phi(n+6), with Euler's totient function phi=A000010.

Original entry on oeis.org

24, 34, 36, 39, 43, 44, 57, 72, 78, 82, 84, 93, 96, 108, 111, 146, 178, 201, 216, 222, 225, 226, 228, 306, 327, 364, 366, 381, 399, 417, 432, 438, 442, 466, 471, 482, 516, 527, 540, 543, 562, 576, 597, 610, 626, 633, 648, 706, 714, 732, 738, 802, 818, 866, 898, 912, 921, 924, 942, 948, 972, 1011

Offset: 1

M. F. Hasler, Jan 05 2011

nonn

There are 1385502728 terms under 10^12. - Jud McCranie, Feb 13 2012

Jud McCranie, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
F. Firoozbakht, Puzzle 466. phi(n-1)=phi(n)=phi(n+1), in C. Rivera's Primepuzzles.
Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.

Cf. A000010, A001274, A001494, A179186, A179187, A007015.

[n: n in [1..1000] | EulerPhi(n) eq EulerPhi(n+6)]; // Vincenzo Librandi, Sep 08 2016

Flatten[Position[Partition[EulerPhi[Range[1200]],7,1],?(#[[1]] == #[[7]]&),{1},Heads->False]] (* _Harvey P. Dale, Jan 30 2016 *)
Select[Range[1000], EulerPhi[#] == EulerPhi[# + 6] &] (* Vincenzo Librandi, Sep 08 2016 *)

{op=vector(N=6); for( n=1,1e4,if( op[n%N+1]+0==op[n%N+1]=eulerphi(n),print1(n-N,",")))}

A000010(a(n)) = A000010(a(n)+6).

A217139 Numbers n such that phi(n) = phi(n+12), with Euler's totient function phi = A000010.

Original entry on oeis.org

48, 68, 72, 78, 86, 88, 114, 143, 144, 156, 157, 164, 168, 186, 192, 203, 216, 222, 247, 273, 292, 356, 402, 432, 444, 450, 452, 456, 612, 654, 728, 732, 762, 798, 834, 864, 876, 884, 932, 942, 964, 1032, 1054, 1080, 1086, 1124, 1147, 1152, 1194, 1209, 1220

Offset: 1

Michel Marcus, Sep 27 2012

Most of numbers n in this sequence are divisible by 2, and it appears that n/2 belongs to A179188. The other ones are listed in sequence A217141.

Proof of the comment: If n is even and not a multiple of 4 then phi(n)=phi(n/2). If n is a multiple of 4 then phi(n)=2 * phi(n/2). So when k is a multiple of 4 and phi(n)=phi(n+k), then phi(n/2)=phi(n/2+k/2). QED. This also applies to A179186, A179202. - Jud McCranie, Dec 30 2012

Jud McCranie, Table of n, a(n) for n = 1..10000
Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.

Cf. A000010, A179188, A217141, A007015.

[n: n in [1..3000] | EulerPhi(n) eq EulerPhi(n+12)]; // Vincenzo Librandi, Sep 08 2016

Select[Range[1, 5000], EulerPhi[#] == EulerPhi[# + 12] &] (* Vincenzo Librandi, Jun 24 2014 *)

{op=vector(N=12); for( n=1, 1e4, if( op[n%N+1]+0==op[n%N+1]=eulerphi(n), print1(n-N, ", ")))}

A179187 Numbers n such that phi(n)=phi(n+5), with Euler's totient function phi=A000010.

Original entry on oeis.org

5, 9, 15, 21, 15556, 21016, 25930, 25935, 27027, 36304, 46683, 129675, 266128, 307923, 329175, 430348, 503139, 636400, 684411, 812170, 1014778, 1252713, 1777545, 1871788, 1892452, 1911987, 2622160, 2629930, 2731360, 2947035, 3397480, 4200100, 5451537

Offset: 1

M. F. Hasler, Jan 05 2011

nonn

There are only 43 terms below 10^7, and 1843 terms below 10^12. [Jud McCranie, Feb 13 2012]

Jud McCranie, Table of n, a(n) for n = 1..1843 (terms < 10^12)
F. Firoozbakht, Puzzle 466. phi(n-1)=phi(n)=phi(n+1), in C. Rivera's Primepuzzles.
Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.

Cf. A000010, A001274, A001494, A179186, A007015.

[n: n in [1..5000000] | EulerPhi(n) eq EulerPhi(n+5)]; // Vincenzo Librandi, Sep 08 2016

Select[Range[5000000], EulerPhi[#] == EulerPhi[# + 5] &] (* Vincenzo Librandi, Sep 08 2016 *)
Position[Partition[EulerPhi[Range[55*10^5]],6,1],?(#[[1]] == #[[6]]&), {1}, Heads->False]//Flatten (* _Harvey P. Dale, Nov 10 2016 *)

{op=vector(N=5); for( n=1,1e7,if( op[n%N+1]+0==op[n%N+1]=eulerphi(n),print1(n-N,",")))}

A000010(a(n)) = A000010(a(n)+5).

A179189 Numbers n such that phi(n) = phi(n+7), with Euler's totient function phi = A000010.

Original entry on oeis.org

5, 7, 21, 45, 75, 105, 285, 488, 585, 765, 1148, 1275, 1358, 1785, 2528, 3465, 4088, 6825, 9405, 12375, 14348, 15345, 16208, 16988, 23648, 25905, 25935, 42698, 50018, 52845, 54615, 61448, 62865, 68445, 78195, 80025, 82005, 88328, 93555, 98475

Offset: 1

M. F. Hasler, Jan 05 2011

nonn

There are 40 terms below 10^5, 81 terms below 10^6 and 162 terms below 10^7.  There are 6606 terms below 10^12. [Jud McCranie, Feb 13 2012]
Farideh Firoozbakht asks whether there is some a(n+1) = a(n)+7, cf. link.
For n < 10^13, the only n such that phi(n-7) = phi(n) = phi(n+7) is 30057431145. - Giovanni Resta, Feb 27 2014

Jud McCranie, Table of n, a(n) for n = 1..6606 (terms < 10^12)
F. Firoozbakht, Puzzle 466. phi(n-1)=phi(n)=phi(n+1), in C. Rivera's Primepuzzles.
Kevin Ford, Solutions of phi(n) = phi(n+k) and sigma(n) = sigma(n + k), arXiv:2002.12155 [math.NT], 2020.

Cf. A000010, A001274, A001494, A179186, A179187, A179188, A007015.

[n: n in [1..100000] | EulerPhi(n) eq EulerPhi(n+7)]; // Vincenzo Librandi, Sep 08 2016

Select[Range[100000], EulerPhi[#] == EulerPhi[# + 7] &] (* Vincenzo Librandi, Sep 08 2016 *)

{op=vector(N=7); for( n=1, 1e5, if( op[n%N+1]+0==op[n%N+1]=eulerphi(n), print1(n-N, ", ")))}

A000010(a(n)) = A000010(a(n)+7).

A179202 Numbers n such that phi(n) = phi(n+8), with Euler's totient function phi=A000010.

Original entry on oeis.org

13, 16, 19, 25, 28, 32, 40, 70, 104, 128, 175, 182, 209, 280, 296, 488, 551, 584, 657, 715, 806, 910, 1232, 1256, 1544, 1602, 2022, 2048, 2216, 2288, 2504, 2540, 2590, 2717, 2912, 3176, 3368, 3640, 3656, 4060, 4328, 4904, 5246, 5288, 5320, 5384, 5864, 5969

Offset: 1

M. F. Hasler, Jan 05 2011

nonn

Among the 5596 terms below 10^7, a(6)=32 is the only term such that a(n+1) = a(n)+8.
There are 141741552 terms under 10^12. - Jud McCranie, Feb 13 2012
If a(n) is even then a(n)/2 is in A179186 - see comment at A217139. - Jud McCranie, Dec 31 2012

M. F. Hasler and Jud McCranie, Table of n, a(n) for n = 1..10000 (first 5596 terms from M. F. Hasler)
F. Firoozbakht, Puzzle 466. phi(n-1)=phi(n)=phi(n+1), in C. Rivera's Primepuzzles.
Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.

Cf. A000010, A001274, A001494, A179186, A179187, A179188, A179189, A007015.

[n: n in [1..10000] | EulerPhi(n) eq EulerPhi(n+8)]; // Vincenzo Librandi, Sep 08 2016

Select[Range[6000], EulerPhi[#] == EulerPhi[# + 8] &] (* Vincenzo Librandi, Sep 08 2016 *)

{op=vector(N=8); for( n=1, 1e4, if( op[n%N+1]+0==op[n%N+1]=eulerphi(n), print1(n-N, ", ")))}

A000010(a(n)) = A000010(a(n)+8).

A276503 Numbers n such that phi(n) = phi(n+10), with Euler's totient function phi = A000010.

Original entry on oeis.org

20, 26, 35, 100, 130, 160, 370, 400, 610, 730, 793, 1000, 1570, 1843, 1930, 2500, 2560, 2770, 2860, 3130, 3970, 4000, 4171, 4210, 4570, 5410, 5767, 6130, 6400, 6610, 6730, 7330, 7570, 8770, 9106, 9640, 9970, 9991, 10498, 10660, 10930, 11248

Offset: 1

Vincenzo Librandi, Sep 08 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..1200
Kevin Ford, Solutions of phi(n) = phi(n+k) and sigma(n) = sigma(n + k), arXiv:2002.12155 [math.NT], 2020.

Cf. A000010.

Cf. numbers n such that phi(n)=phi(n+k): A001274 (k=1), A001494 (k=2), A179186 (k=4), A179187 (k=5), A179188 (k=6), A179189 (k=7), A179202 (k=8), this sequence (k=10), A276504 (k=11), A217139 (k=12).

[n: n in [1..20000] | EulerPhi(n) eq EulerPhi(n+10)];

Select[Range[15000], EulerPhi[#] == EulerPhi[# + 10] &]
SequencePosition[EulerPhi[Range[12000]],{x_,,,_,,,_,,,_,x_}][[;;,1]] (* Harvey P. Dale, Apr 29 2025 *)

A192221 Numbers n such that phi(n-2)=phi(n+2) where phi=A000010.

Original entry on oeis.org

10, 16, 18, 22, 37, 54, 66, 93, 142, 150, 246, 294, 405, 457, 618, 630, 774, 803, 1013, 1026, 1110, 1146, 1254, 1272, 1297, 1458, 1590, 1686, 1822, 1830, 2032, 2166, 2454, 2625, 2646, 2662, 2694, 2934, 3030, 3218, 3323, 3510, 3990, 4266, 4342, 4374, 4614, 4806, 4854, 4950, 5182, 5286

Offset: 1

Juri-Stepan Gerasimov, Jun 26 2011

nonn

Cf. A000010, A066812.

Select[Range[3, 5000], EulerPhi[# - 2] == EulerPhi[# + 2] &] (* Alonso del Arte, Jun 28 2011 *)
Flatten[Position[Partition[EulerPhi[Range[5300]],5,1],?(#[[1]] == #[[5]]&), 1,Heads -> False]]+2 (* _Harvey P. Dale, Jul 01 2019 *)

for(n=3,1e4,if(eulerphi(n-2)==eulerphi(n+2),print1(n", "))) \\ Charles R Greathouse IV, Jun 27 2011

a(n) = A179186(n)+2. - R. J. Mathar, Jun 26 2011

A330251 Numbers k such that phi(k) = phi(k+3), where phi (A000010) is Euler's totient function.

Original entry on oeis.org

3, 5, 8720288051472, 9134280520365, 41544070492925, 42466684755492, 51363581614342, 68616494581632, 113312918293575, 210911076210835, 215517565688425, 294988451482725, 383617980270525, 432759876053505, 442863123838135, 532068058516992, 892813363927485, 923102743748185, 929531173876305

Offset: 1

Michel Marcus and Giovanni Resta, Feb 29 2020

nonn

10^15 < a(20) <= 1089641067389872.
Also terms: 1248817919303952, 1332436545865422, 1394926716616125, 1868522795664525, 1950445682260072.
a(4) and a(9) appear in Kevin Ford's paper.

Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.
S. W. Graham, J. J. Holt, and C. Pomerance, On the solutions to phi(n) = phi(n+k), Number Theory in Progress, K. Gyory, H. Iwaniec, and J. Urbanowicz, eds., vol. 2, de Gruyter, Berlin and New York, 1999, pp. 867-882.
Mathematics StackExchange, Conjecture on the gap between integers having the same number of co-primes, Sep 25 2019.

Cf. A000010, A007015.
Cf. A001274, A001494, A179186, A179187, A179188, A179189, A179202, A330429.
Cf. A276503, A276504, A217139.

Select[Range[100000], EulerPhi[#] == EulerPhi[# + 3] &] (* Alonso del Arte, Mar 01 2020 *)

isok(k) = eulerphi(k) == eulerphi(k+3); \\ Michel Marcus, Feb 29 2020

cp's OEIS Frontend

A001274 Numbers k such that phi(k) = phi(k+1).

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A001494 Numbers k such that phi(k) = phi(k+2).

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A179188 Numbers n such that phi(n) = phi(n+6), with Euler's totient function phi=A000010.

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A217139 Numbers n such that phi(n) = phi(n+12), with Euler's totient function phi = A000010.

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A179187 Numbers n such that phi(n)=phi(n+5), with Euler's totient function phi=A000010.

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A179189 Numbers n such that phi(n) = phi(n+7), with Euler's totient function phi = A000010.

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