A007015 -id:A007015 - cp's OEIS frontend

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

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).

A050473 Smallest k such that phi(k+n) = 2*phi(k).

Original entry on oeis.org

2, 1, 1, 2, 1, 4, 3, 4, 3, 5, 5, 8, 26, 7, 5, 8, 9, 12, 5, 10, 7, 8, 46, 16, 5, 13, 9, 14, 7, 25, 21, 13, 9, 17, 7, 24, 62, 19, 11, 20, 76, 28, 13, 16, 15, 23, 17, 32, 21, 25, 17, 26, 52, 36, 11, 28, 13, 26, 13, 45, 74, 28, 17, 26, 13, 39, 33, 31, 21, 32, 13, 48, 39, 37, 25, 38

Offset: 1

Jud McCranie, Dec 24 1999

Makowski proved that the sequence is well-defined.

It appears that k <= 2n, with equality for the n in A110196 only. Computations for n < 10^6 appear to show that k < n for all but a finite number of n. - T. D. Noe, Jul 15 2005

			phi(13+26) = 24 = 2*phi(13), so a(13) = 26.

Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B36, p. 138.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).
Andrzej Makowski, On the equation phi(n+k)=2*phi(n), Elem. Math., Vol. 29, No. 1 (1974), p. 13.

Cf. A000010, A007015, A050472, A110196.

Cf. A110179 (least k such that phi(n+k)=2*phi(n)).

Table[k=1; While[EulerPhi[n+k] != 2*EulerPhi[k], k++ ]; k, {n, 100}] (Noe)

f(n) = apply(x -> x - n, select(x -> x > n, invphi(2*eulerphi(n)))); \\ using Max Alekseyev's invphi.gp
lista(len) = {my(v = vector(len), c = 0, k = 1, s); while(c < len, s = f(k); for(i = 1, #s, if(s[i] <= len && v[s[i]] == 0, c++; v[s[i]] = k)); k++); v;} \\ Amiram Eldar, Nov 05 2024

A001259 A sequence of sorted odd primes 3 = p_1 < p_2 < ... < p_m such that p_i-2 divides the product p_1p_2...*p_(i-1) of the earlier primes and each prime factor of p_i-1 is a prime factor of twice the product.

Original entry on oeis.org

3, 5, 7, 17, 19, 37, 97, 113, 257, 401, 487, 631, 971, 1297, 1801, 19457, 22051, 28817, 65537, 157303, 160001

Offset: 1

N. J. A. Sloane

Old name was: A special sequence of primes.
Holt shows this sequence is complete. - T. D. Noe, Jul 28 2005
This sequence was used by Schinzel (1958) and Schinzel and Wakulicz (1959) to prove that there are at least two solutions k to phi(n+k) = phi(k) for all n <= 8*10^47 and 2*10^58, respectively. - Amiram Eldar, Mar 19 2021

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).

Jeffery J. Holt, The minimal number of solutions to phi(n)=phi(n+k), Math. Comp., 72 (2003), 2059-2061.
A. Schinzel and Andrzej Wakulicz, Sur l'équation phi(x+k)=phi(x), I., Acta Arith. 4 (1958), 181-184. - _Jonathan Sondow_, Oct 07 2012
A. Schinzel and Andrzej Wakulicz; Sur l'équation phi(x+k)=phi(x). II. Acta Arith. 5 1959 425-426.

Cf. A007015, A342701.

New name, giving a definition, by Jonathan Sondow, Oct 06 2012

A217199 Odd primes p such that 2p-1 is prime and no p is equal to 2q-1 with q in the sequence.

Original entry on oeis.org

3, 7, 19, 31, 79, 97, 139, 199, 211, 229, 271, 307, 331, 337, 367, 379, 439, 499, 547, 577, 601, 607, 619, 691, 727, 811, 829, 937, 967, 1009, 1069, 1171, 1279, 1297, 1399, 1429, 1459, 1531, 1609, 1627, 1759, 1867, 2011, 2029, 2089, 2131, 2179, 2221, 2281

Offset: 1

Michel Marcus, Sep 27 2012

nonn

At each step, the smallest possible p is chosen.
These are the primes described in lemma 2 of the paper by Holt. - T. D. Noe, Sep 28 2012
This sequence was used by Holt (2003) to prove that there are at least two solutions k to phi(n+k) = phi(k) for all even n <= 1.38*10^26595411. - Amiram Eldar, Mar 19 2021

Michel Marcus, Table of n, a(n) for n = 1..1000
Jeffery J. Holt, The minimal number of solutions to phi(n)=phi(n+k), Math. Comp., 72 (2003), 2059-2061.
A. Schinzel and Andrzej Wakulicz, Sur l'équation phi(x+k)=phi(x), I., Acta Arith. 4 (1958), 181-184.

Cf. A007015, A110581, A217198, A342701.

t = {}; p = 2; Do[p = NextPrime[p]; If[PrimeQ[2*p - 1] && ! MemberQ[2*t - 1, p], AppendTo[t, p]], {PrimePi[2281]}]; t

intab(val, tab) = {for (ii=1, length(tab),if (tab[ii] == val, return (1);););return(0);}
lista(nn) = {tab = []; for (i=1, nn, len = length(tab); if (len == 0, p = 3, p = nextprime(tab[len]+1)); while (! isprime(2*p-1) || intab((p+1)/2, tab) , p = nextprime(p+1);); tab = concat(tab, p); print1(p, ", "););}

A342701 a(n) is the second smallest k such that phi(n+k) = phi(k), or 0 if no such solution exists.

Original entry on oeis.org

3, 7, 5, 14, 9, 34, 7, 16, 15, 26, 11, 68, 39, 28, 15, 32, 33, 72, 25, 40, 35, 56, 17, 101, 45, 37, 45, 56, 29, 152, 31, 61, 39, 56, 35, 144, 37, 61, 39, 74, 41, 128, 35, 88, 45, 161, 47, 192, 49, 82, 51, 74, 95, 216, 43, 97, 75, 203, 59, 304, 91, 88, 63, 122

Offset: 1

Amiram Eldar, Mar 18 2021

nonn

Sierpiński (1956) proved that there is at least one solution for all n>=1.
Schinzel (1958) proved that there are at least two solutions k to phi(n+k) = phi(k) for all n <= 8*10^47. Schinzel and Wakulicz (1959) increased this bound to 2*10^58.
Schinzel (1958) observed that under the prime k-tuple conjecture there is a second solution for all even n.
Holt (2003) proved that there is a second solution for all even n <= 1.38 * 10^26595411.

			a(1) = 3 since the solutions to the equation phi(1+k) = phi(k) are k = 1, 3, 15, 104, 164, ... (A001274), and 3 is the second solution.

Richard K. Guy, Unsolved Problems in Number Theory, 3rd edition, Springer, 2004, section B36, page 138-142.
József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 3, p. 217-219.
Wacław Sierpiński, Sur une propriété de la fonction phi(n), Publ. Math. Debrecen, Vol. 4 (1956), pp. 184-185.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Jeffery J. Holt, The minimal number of solutions to phi(n)=phi(n+k), Math. Comp., Vol. 72, No. 244 (2003), pp. 2059-2061.
Andrzej Schinzel, Sur l'équation phi(x + k) = phi(x), Acta Arith., Vol. 4, No. 3 (1958), pp. 181-184.
Andrzej Schinzel and Andrzej Wakulicz, Sur l'équation phi(x+k)=phi(x). II, Acta Arith., Vol. 5, No. 4 (1959), pp. 425-426.
Eric Weisstein's World of Mathematics, k-Tuple Conjecture.

Cf. A000010, A001259, A001274, A007015, A217199.

f[n_, 0] = 0; f[n_, k0_] := Module[{k = f[n, k0 - 1] + 1}, While[EulerPhi[n + k] != EulerPhi[k], k++]; k]; Array[f[#, 2] &, 100]

a(n) = my(k=1, nb=0); while ((nb += (eulerphi(n+k)==eulerphi(k))) != 2, k++); k; \\ Michel Marcus, Mar 19 2021

A217140 a(n) = m/n where m is the least number divisible by n such that phi(m) = phi(m+6n).

Original entry on oeis.org

24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 60, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36, 24, 24, 24, 24, 36

Offset: 1

Michel Marcus and Jonathan Sondow, Oct 01 2012

nonn

			A179188(1)=24 is divisible by 1 and the quotient is 24, so a(1)=24.
A217139(1)=48 is divisible by 2 and the quotient is 24, so a(2)=24.
The first solution to phi(n)=phi(n+18) to be divisible by 3 is 72 and the quotient is 24, so a(3)=24.

Michel Marcus, Table of n, a(n) for n = 1..1000

Cf. A007015, A179188, A217139, A217141, A217068.

A241928 a(n) = smallest k such that lambda(n+k) = lambda(k).

Original entry on oeis.org

1, 4, 3, 4, 3, 6, 7, 4, 3, 5, 5, 9, 13, 7, 5, 8, 17, 6, 9, 4, 3, 11, 23, 16, 5, 13, 9, 14, 7, 10, 31, 13, 9, 17, 5, 36, 37, 10, 13, 20, 41, 14, 5, 16, 15, 23, 9, 36, 7, 10, 17, 13, 52, 9, 5, 7, 13, 14, 45, 20, 61, 31, 9, 16, 7, 18, 45, 17, 23, 10, 71, 45, 39

Offset: 1

Michel Lagneau, May 02 2014

nonn

Lambda(n) is the Carmichael lambda function(A002322).
It is highly probable that a solution exists for each n>0.
The corresponding values of lambda(k) are 1, 2, 2, 2, 2, 2, 6, 2, 2, 4, 4, 6, 12, 6, 4, 2, 16, 2, 6, 2, 2, 10, 22, 4, 4, 12, 6, 6, 6, 4, 30, ...

			a(29) = 7 because lambda(29+7) = lambda(7) = 6.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A002322, A007015, A173695.

with(numtheory):for n from 1 to 70 do:ii:=0:for k from 1 to 10^8 while(ii=0) do:if lambda(k) = lambda(k+n) then ii:=1:printf(`%d, `,k):else fi:od:od:

klambda[n_]:=Module[{k=1}, While[CarmichaelLambda[n+k]!= CarmichaelLambda [k], k++]; k]; Array[klambda, 70]

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

A330429 Numbers k such that phi(k) = phi(k+9), where phi (A000010) is Euler's totient function.

Original entry on oeis.org

9, 15, 1005079920836, 13695542245376, 26160864154416, 27402841561095, 27599063056565, 110263115897935, 124632211478775, 127400054266476, 154090744843026, 205849483744896, 231019991767556, 339938754880725, 459718637643265, 632733228632505, 646552697065275, 683008674773416, 884965354448175

Offset: 1

Giovanni Resta, Mar 01 2020

nonn

a(20) > 10^15.

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.

Cf. A000010, A007015.
Cf. A001274, A001494, A330251, A179186, A179187, A179188, A179189, A179202.
Cf. A276503, A276504, A217139.

cp's OEIS Frontend

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

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

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A050473 Smallest k such that phi(k+n) = 2*phi(k).

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A001259 A sequence of sorted odd primes 3 = p_1 < p_2 < ... < p_m such that p_i-2 divides the product p_1*p_2*...*p_(i-1) of the earlier primes and each prime factor of p_i-1 is a prime factor of twice the product.

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A217199 Odd primes p such that 2p-1 is prime and no p is equal to 2q-1 with q in the sequence.

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A342701 a(n) is the second smallest k such that phi(n+k) = phi(k), or 0 if no such solution exists.

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A217140 a(n) = m/n where m is the least number divisible by n such that phi(m) = phi(m+6n).

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A241928 a(n) = smallest k such that lambda(n+k) = lambda(k).

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A001259 A sequence of sorted odd primes 3 = p_1 < p_2 < ... < p_m such that p_i-2 divides the product p_1p_2...*p_(i-1) of the earlier primes and each prime factor of p_i-1 is a prime factor of twice the product.