A124240 -id:A124240 - cp's OEIS frontend

A124241 Terms of A068563 that are not terms of A124240.

136, 408, 620, 680, 820, 1224, 1240, 1314, 2040, 2312, 2460, 2480, 2628, 2856, 3100, 3400, 3672, 3924, 3942, 4100, 4112, 4656, 4960, 5304, 5334, 5784, 6120, 6200, 6820, 6936, 7380, 7480, 7848, 7884, 8224, 8568, 9020, 9060, 9198, 9492, 9920, 10200, 10668, 11016, 11560, 11568, 11826, 12300, 12336, 12400, 13140, 13640

Offset: 1

Alexander Adamchuk, Oct 22 2006, Oct 27 2006

nonn

A068563 contains A124240 as a subsequence. This sequence gives their set difference.
Note that a(2) = 3*a(1) and a(4) = 5*a(1). a(6) = 1224 = 9*a(1), a(7) = 1240 = 2*a(3), a(8) = 1314, a(9) = 2040 = 15*a(1), a(10) = 2312 = 17*a(1), a(11) = 2460 = 3*a(5), a(12)= 2480 = 4*a(3), a(13) = 2856 = 21*a(1). Numbers k such that there exists a(n) = k*a(1) are k = {1, 3, 5, 9, 15, 17, 21, ...}.
Many but not all terms belong to A124276.

Cf. A124240, A124239, A068563, A124276, A064896.

for(n=1,10^5, m=n\2^valuation(n,2); if( Mod(n,znorder(Mod(2,m))), next); p=factor(n)[,1]; g=1; for(i=1,#p, if( Mod(n,p[i]-1), g=0; break) ); if(g,next); print1(n,", ") ) /* Alekseyev */

a(13) corrected and terms a(14) onward provided by Max Alekseyev, Aug 25 2013

A289257 Terms k of A006521 such that 2*k is a term of A124240.

Original entry on oeis.org

1, 3, 9, 27, 81, 171, 243, 513, 729, 1539, 2187, 3249, 4617, 6561, 9747, 13203, 13851, 19683, 29241, 39609, 41553, 59049, 61731, 87723, 118827, 124659, 177147, 185193, 250857, 263169, 356481, 373977, 531441, 555579, 752571, 789507, 1063611, 1069443, 1121931, 1172889, 1594323, 1666737

Offset: 1

Michel Marcus, Jun 29 2017

nonn

Novák numbers n that are 2n Novák-Carmichael. See Kalmynin link.

Alexander Kalmynin, On Novák numbers, arXiv:1611.00417 [math.NT], 2016. See Theorem 7 p. 11.

Cf. A006521, A124240, A289258.

Reap[Do[If[PowerMod[2, n, n]+1 == n && Divisible[2n, CarmichaelLambda[2n]], Print[n]; Sow[n]], {n, 2 10^6}]][[2, 1]] (* Jean-François Alcover, Sep 25 2018 *)

isnov(n) = Mod(2, n)^n==-1; \\ A006521
isnovcar(n) = n%lcm(znstar(n)[2])==0; \\ A124240
isok(n) = isnov(n) && isnovcar(2*n);

from itertools import count, islice
from sympy.ntheory.factor_ import reduced_totient
def A289257gen(): return filter(lambda n:2*n % reduced_totient(2*n) == 0 and pow(2,n,n)==n-1, count(1))
A289257_list = list(islice(A289257gen(),20)) # Chai Wah Wu, Dec 11 2021

A014117 Numbers n such that m^(n+1) == m (mod n) holds for all m.

Original entry on oeis.org

1, 2, 6, 42, 1806

Offset: 1

David Broadhurst

"Somebody incorrectly remembered Fermat's little theorem as saying that the congruence a^{n+1} = a (mod n) holds for all a if n is prime" (Zagier). The sequence gives the set of integers n for which this property is in fact true.
If i == j (mod n), then m^i == m^j (mod n) for all m. The latter congruence generally holds for any (m, n)=1 with i == j (mod k), k being the order of m modulo n, i.e., the least power k for which m^k == 1 (mod n). - Lekraj Beedassy, Jul 04 2002
Also, numbers n such that n divides denominator of the n-th Bernoulli number B(n) (cf. A106741). Also, numbers n such that 1^n + 2^n + 3^n + ... + n^n == 1 (mod n). Equivalently, numbers n such that B(n)*n == 1 (mod n). Equivalently, Sum_{prime p, (p-1) divides n} n/p == -1 (mod n). It is easy to see that for n > 1, n must be an even squarefree number. Moreover, the set P of prime divisors of all such n satisfies the property: if p is in P, then p-1 is the product of distinct elements of P. This set is P = {2, 3, 7, 43}, implying that the sequence is finite and complete. - Max Alekseyev, Aug 25 2013
In 2005, B. C. Kellner proved E. W. Weisstein's conjecture that denom(B_n) = n only if n = 1806. - Jonathan Sondow, Oct 14 2013
Squarefree numbers n such that b^n == 1 (mod n^2) for every b coprime to n. Squarefree terms of A341858. - Thomas Ordowski, Aug 05 2024
Conjecture: Numbers n such that gcd(d+1, n) > 1 for every proper divisor d of n. Verified up to 10^696. - David Radcliffe, May 29 2025

M. A. Alekseyev, J. M. Grau, and A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT], 2016-2018.
John H. Castillo and Jhony Fernando Caranguay Mainguez, The set of k-units modulo n, arXiv:1708.06812 [math.NT], 2017.
Yongyi Chen and Tae Kyu Kim, On Generalized Carmichael Numbers, arXiv:2103.04883 [math.NT], 2021.
J. Dyer-Bennet, A Theorem in Partitions of the Set of Positive Integers, Amer. Math. Monthly, 47(1940) pp. 152-4.
L. Halbeisen and N. Hungerbühler, On generalised Carmichael numbers, Hardy-Ramanujan Society, 1999, 22 (2), pp.8-22. (hal-01109575)
J. M. Grau, A. M. Oller-Marcen, and Jonathan Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv 1309.7941 [math.NT], 2013; Monatsh. Math., 177 (2015), 421-436.
B. C. Kellner, The equation denom(B_n) = n has only one solution, preprint 2005.
Don Reble, A014117 and related OEIS sequences (shows there are no other terms)
Jonathan Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + … + k^n = (k+1)^n modulo k and k^2, arXiv:1011.2154 [math.NT], 2010; see Prop. 2; Integers, 11 (2011), article A34.
Eric Weisstein's World of Mathematics, Bernoulli Number
D. Zagier, Problems posed at the St Andrews Colloquium, 1996

Squarefree terms of A124240. - Robert Israel and Thomas Ordowski, Jun 23 2017

Cf. A007018, A031971, A054377, A100016, A242927, A341858.

r[n_] := Reduce[ Mod[m^(n+1) - m, n] == 0, m, Integers]; ok[n_] := Range[n]-1 === Simplify[ Mod[ Flatten[ m /. {ToRules[ r[n][[2]] ]}], n], Element[C[1], Integers]]; ok[1] = True; A014117 = {}; Do[ If[ok[n], Print[n]; AppendTo[ A014117, n] ], {n, 1, 2000}] (* Jean-François Alcover, Dec 21 2011 *)
Select[Range@ 2000, Function[n, Times @@ Boole@ Map[Function[m, PowerMod[m, n + 1, n] == Mod[m, n]], Range@ n] > 0]] (* Michael De Vlieger, Dec 30 2016 *)

[n for n in range(1, 2000) if all(pow(m, n+1, n) == m for m in range(n))] # David Radcliffe, May 29 2025

For n <= 5, a(n) = a(n-1)^2 + a(n-1) with a(0) = 1. - Raphie Frank, Nov 12 2012

a(n+1) = A007018(n) = A054377(n) = A100016(n) for n = 1, 2, 3, 4. - Jonathan Sondow, Oct 01 2013

A055744 Numbers k such that k and phi(k) have the same prime factors.

Original entry on oeis.org

1, 4, 8, 16, 18, 32, 36, 50, 54, 64, 72, 100, 108, 128, 144, 162, 200, 216, 250, 256, 288, 294, 324, 400, 432, 450, 486, 500, 512, 576, 578, 588, 648, 800, 864, 882, 900, 972, 1000, 1014, 1024, 1152, 1156, 1176, 1210, 1250, 1296, 1350, 1458, 1600, 1728, 1764

Offset: 1

Labos Elemer, Jul 11 2000

nonn

Contains products of suitable powers of 2 and Fermat primes. For x = 2^u*3^w, phi(x) = 2^u*3^(w-1) with suitable exponents. Analogous constructions are possible with {2,3,7} prime divisors, etc.
From Ivan Neretin, Mar 19 2015: (Start)
Also, numbers k that meet the following criteria for every prime p dividing k:
  1. All prime divisors of p-1 must also divide k;
  2. If k has no prime divisors of the form m*p+1, and k is divisible by p, then k must be divisible by p^2.
Also, numbers k for which {k, phi(k), phi(phi(k))} is a geometric progression.
(End)
All terms > 1 are even. An even number k is in the sequence iff 2*k is in the sequence. - Robert Israel, Mar 19 2015
For n > 1, the largest prime factor of a(n) has multiplicity >= 2. For all prime factors more than half of the largest prime factor of a(n), the multiplicity differs from 1.
If k = p1^a1 * p2^a2 * ... * pm^am is in the sequence, then so is p1^b1 * p2^b2 * ... * pm^bm for 1 <= i <= m and prime pi and bi >= ai.
If m * p^2 is not in the sequence, for a prime p and some m > 0, then neither is m * p^3. - David A. Corneth, Mar 22 2015
A027748(a(n),j) = A027748(A000010(a(n)),j) for j=1..A001221(a(n)); also numbers k such that k and phi(k) have the same squarefree kernel: A007947(a(n)) = A007947(A000010(a(n))). - Reinhard Zumkeller, Jun 01 2015
Pollack and Pomerance call these numbers "phi-perfect numbers". - Amiram Eldar, Jun 02 2020

			k = 578 = 2*17*17, phi(578) = 272 = 2*2*2*2*17 with 2 and 17 prime factors, so 578 is a term.
k = 588 = 2*2*3*7*7, phi(588) = 168 = 2*2*2*3*7, so 588 is a term.
k = 264196 = 2*2*257*257, phi(264196) = 512*257 = 131584, so 264196 is a term.

David A. Corneth, Table of n, a(n) for n = 1..117561 (terms <= 10^11; first 10000 terms from T. D. Noe)
Paul Pollack and Carl Pomerance, Prime-Perfect Numbers, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 12a, Paper A14, 2012.

Intersection of A073539 and A151999. - Amiram Eldar, Jun 02 2020
Cf. A001221, A000010, A110751, A110819, A027598, A081377.
Cf. A007947, A027748, A055742, A173557, A256248, subsequence of A124240.

a055744 n = a055744_list !! (n-1)
a055744_list = 1 : filter f [2..] where
   f x = all ((== 0) . mod x) (concatMap (a027748_row . subtract 1) ps) &&
         all ((== 0) . mod (a173557 x))
             (map fst $ filter ((== 1) . snd) $ zip ps $ a124010_row x)
         where ps = a027748_row x
-- Reinhard Zumkeller, Jun 01 2015

select(numtheory:-factorset = numtheory:-factorset @ numtheory:-phi,
[1, 2*i $ i=1..2000]); # Robert Israel, Mar 19 2015
isA055744 := proc(n)
    nfs := numtheory[factorset](n) ;
    phinfs := numtheory[factorset](numtheory[phi](n)) ;
    if nfs = phinfs then
        true;
    else
        false;
    end if;
end proc:
A055744 := proc(n)
    if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if isA055744(a) then
                return a;
            end if;
        end do:
    end if;
end proc: # R. J. Mathar, Sep 23 2016

Select[Range@ 1800,
First /@ FactorInteger@ # == First /@ FactorInteger@ EulerPhi@ # &] (* Michael De Vlieger, Mar 21 2015 *)

is(n)=factor(n)[,1]==factor(eulerphi(n))[,1] \\ Charles R Greathouse IV, Oct 31 2011

is(n)=my(f=factor(n)); f[,1]==factor(eulerphi(f))[,1] \\ Charles R Greathouse IV, May 26 2015

Corrected and extended by James Sellers, Jul 11 2000

A174824 a(n) = period of the sequence {m^m, m >= 1} modulo n.

Original entry on oeis.org

1, 2, 6, 4, 20, 6, 42, 8, 18, 20, 110, 12, 156, 42, 60, 16, 272, 18, 342, 20, 42, 110, 506, 24, 100, 156, 54, 84, 812, 60, 930, 32, 330, 272, 420, 36, 1332, 342, 156, 40, 1640, 42, 1806, 220, 180, 506, 2162, 48, 294, 100, 816, 156, 2756, 54, 220, 168, 342

Offset: 1

Franklin T. Adams-Watters, Mar 30 2010

nonn

This is a divisibility sequence: if n divides m, a(n) divides a(m).

We have the equality n = a(n) for numbers n in A124240, which is related to Carmichael's function (A002322). The largest values of a(n) occur when n is prime, in which case a(n) = n*(n-1). - T. D. Noe, Feb 21 2014

			For n=3, 1^1 == 1 (mod 3), 2^2 == 1 (mod 3), 3^3 == 0 (mod 3), etc. The sequence of residues 1, 1, 0, 1, 2, 0, 1, 1, 0, ... has period 6, so a(3) = 6. - _Michael B. Porter_, Mar 13 2018

T. D. Noe, Table of n, a(n) for n = 1..1000
José María Grau and Antonio M. Oller-Marcén, On the last digit and the last non-zero digit of n^n in base b, arXiv:1203.4066 [math.NT], 2012. (See page 3)
Index to divisibility sequences

Cf. A000312, A009262, A127699.

Cf. A002322, A124240, A268336.

Table[LCM[n, CarmichaelLambda[n]], {n, 100}] (* T. D. Noe, Feb 20 2014 *)

a(n)=local(ps);ps=factor(n)[,1]~;for(k=1,#ps,n=lcm(n,ps[k]-1));n

a(n) = lcm(n, lcm(znstar(n)[2])); \\ Michel Marcus, Mar 18 2016; corrected by Michel Marcus, Nov 13 2019

apply( {A174824(n)=lcm(lcm([p-1|p<-factor(n)[,1]]),n)}, [1..99]) \\ [...] = znstar(n)[2], but 3x faster. - M. F. Hasler, Nov 13 2019

a(n) = lcm(n, A173614(n)) = lcm(n, A002322(n)) = lcm(n, A011773(n)).
If n and m are relatively prime, a(n*m) = lcm(a(n), a(m)); a(p^k) = (p-1)*p^k for p prime and k > 0.
a(n) = n*A268336(n). - M. F. Hasler, Nov 13 2019

A226960 Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 2 (mod n).

Original entry on oeis.org

1, 4, 12, 84, 3612

Offset: 1

José María Grau Ribas, Jun 24 2013

Also, numbers n such that B(n)*n == 2 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -2 (mod n). - Max Alekseyev, Aug 25 2013

M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT]

Cf. A031971, A014117.
Subsequence of A124240.
Solutions to 1^n+2^n+...+n^n == m (mod n): A005408 (m=0), A014117 (m=1), this sequence (m=2), A226961 (m=3), A226962 (m=4), A226963 (m=5), A226964 (m=6), A226965 (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 2 &]

is(n)=if(n%2,return(n==1)); Mod(sumdiv(n/2,d, if(isprime(2*d+1), n/(2*d+1)))+n/2,n)==-2 \\ Charles R Greathouse IV, Nov 13 2013

a(1)=1 prepended by Max Alekseyev, Aug 25 2013

A068563 Numbers k such that 2^k == 4^k (mod k).

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 40, 42, 48, 54, 60, 64, 72, 80, 84, 96, 100, 108, 120, 126, 128, 136, 144, 156, 160, 162, 168, 180, 192, 200, 216, 220, 240, 252, 256, 272, 288, 294, 300, 312, 320, 324, 336, 342, 360, 378, 384, 400, 408, 420, 432, 440

Offset: 1

Benoit Cloitre, Mar 25 2002

If k is in the sequence then 2k is also in the sequence, but the converse is not true.
Contains A124240 as a subsequence. Their difference is given by A124241. - T. D. Noe, May 30 2003
Also, integers k such that A007733(k) divides k. Also, integers k such that for every odd prime divisor p of k, A007733(p) = A002326((p-1)/2) divides k. Also, integers k such that A000265(k) divides 2^k-1. - Max Alekseyev, Aug 25 2013

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
Eric Weisstein's World of Mathematics, Carmichael Function.

Cf. A000265, A002322, A002326, A007733, A124240, A124241.

Select[Range[500], PowerMod[2,#,# ] == PowerMod[4,#,# ] & ]

isok(k) = Mod(2, k)^k == Mod(4, k)^k; \\ Amiram Eldar, Apr 19 2025

Comment and Mathematica program corrected by T. D. Noe, Oct 17 2008

A268336 a(n) = A174824(n)/n, where A174824(n) = lcm(A002322(n), n) and A002322(n) is the Carmichael lambda function (also known as the reduced totient function or the universal exponent of n).

Original entry on oeis.org

1, 1, 2, 1, 4, 1, 6, 1, 2, 2, 10, 1, 12, 3, 4, 1, 16, 1, 18, 1, 2, 5, 22, 1, 4, 6, 2, 3, 28, 2, 30, 1, 10, 8, 12, 1, 36, 9, 4, 1, 40, 1, 42, 5, 4, 11, 46, 1, 6, 2, 16, 3, 52, 1, 4, 3, 6, 14, 58, 1, 60, 15, 2, 1, 12, 5, 66, 4, 22, 6, 70, 1, 72, 18, 4, 9, 30, 2, 78, 1, 2

Offset: 1

Juri-Stepan Gerasimov, Feb 01 2016

Cf. A002322, A068563, A124240, A174824.

[1] cat [Lcm(n, CarmichaelLambda(n))/n: n in [2..100]]: // Feb 03 2016

Table[LCM[n, CarmichaelLambda@ n]/n, {n, 100}] (* Michael De Vlieger, Feb 03 2016, after T. D. Noe at A174824 *)

a(n)=my(ps); ps=factor(n)[, 1]~; m = n; for(k=1, #ps, m=lcm(m, ps[k]-1)); m/n \\ Michel Marcus, Feb 21 2016

apply( {A268336(n)=lcm(lcm([p-1|p<-factor(n)[,1]]),n)/n}, [1..99]) \\ [...] = znstar(n)[2], but 3x faster. - M. F. Hasler, Nov 13 2019

a(n) = A174824(n)/n.

a(A124240(n)) = 1. - Michel Marcus, Feb 21 2016

More terms from Vincenzo Librandi, Feb 03 2016

A276976 Smallest m such that b^m == b^n (mod n) for every integer b.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 4, 3, 2, 1, 2, 1, 2, 3, 4, 1, 6, 1, 4, 3, 2, 1, 4, 5, 2, 9, 4, 1, 2, 1, 8, 3, 2, 11, 6, 1, 2, 3, 4, 1, 6, 1, 4, 9, 2, 1, 4, 7, 10, 3, 4, 1, 18, 15, 8, 3, 2, 1, 4, 1, 2, 3, 16, 5, 6, 1, 4, 3, 10, 1, 6, 1, 2, 15, 4, 17, 6, 1, 4, 27, 2, 1

Offset: 1

Thomas Ordowski, Sep 23 2016

It suffices to check all bases 0 < b < n for n > 2.
The congruence n == a(n) (mod A002322(n)) is always true.
a(n) = 1 iff n is a prime or a Carmichael number.
We have a(n) > 0 for n > 1, and a(n) <= n/2.
If n > 2 then a(n) is odd iff n is odd.
Conjecture: a(n) <= n/3 for every n >= 9.
Professor Andrzej Schinzel proved this conjecture (in a letter to the author). - Thomas Ordowski, Sep 30 2016
Note: a(n) = n/3 for n = A038754 >= 3.
Numbers n such that a(n) > A270096(n) are A290960.
Information from Carl Pomerance: a(n) > A002322(n) if and only if 8|n and n is in A050990; such n = 8, 24, 56, ... - Thomas Ordowski, Jun 21 2017
Number of integers k < n such that b^k == b^n (mod n) for every integer b is f(n) = (n - a(n))/lambda(n). For n > 1, f(n) = floor((n-1)/lambda(n)) if and only if a(n) <= lambda(n), where lambda(n) = A002322(n). - Thomas Ordowski, Jun 21 2017
a(n) >= A051903(n); numbers n such that a(n) = A051903(n) are 1, primes, Carmichael numbers, and A327295. - Thomas Ordowski, Dec 06 2019

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A002322, A002997, A038754, A050990, A051903, A124240, A219175, A270096, A290960, A327295.

With[{nn = 83}, Table[SelectFirst[Range[nn/4 + 10], Function[m, AllTrue[Range[2, n - 1], PowerMod[#, m , n] == PowerMod[#, n , n] &]]] - Boole[n == 1], {n, nn}]] (* Michael De Vlieger, Aug 15 2017 *)
a[1] = 0; a[8] = a[24] = 4; a[n_] := If[(rem = Mod[n, (lam = CarmichaelLambda[n])]) >= Max @@ Last /@ FactorInteger[n], rem, rem + lam]; Array[a, 100] (* Amiram Eldar, Nov 30 2019 *)

a(n)=if(n<3, return(n-1)); forstep(m=1,n,n%2+1, for(b=0,n-1, if(Mod(b,n)^m-Mod(b,n)^n, next(2))); return(m)) \\ Charles R Greathouse IV, Sep 23 2016

def a(n): return next(m for m in range(0, n+1) if all(pow(b,m,n)==pow(b,n,n) for b in range(1, 2*n+1))) # Nicholas Stefan Georgescu, Jun 03 2022

a(p) = 1 for prime p.
a(2*p) = 2 for prime p.
a(3*p) = 3 for odd prime p.
a(p^k) = p^(k-1) for odd prime p and k >= 1.
a(2*p^k) = 2*p^(k-1) for odd prime p and k >= 1.
a(2^k) = 2^(k-2) for k >= 4.
From Thomas Ordowski, Jul 09 2017: (Start)
Full description of the function:
a(n) = lambda(n) if lambda(n)|n except n = 1, 8, and 24;
a(n) = lambda(n)+2 if lambda(n)|(n-2) and 8|n;
a(n) = n mod lambda(n) otherwise;
where lambda(n) = A002322(n). (End)
For n <> 8 and 24, a(n) = A(n) if A(n) >= A051903(n) or a(n) = A002322(n) + A(n) otherwise, where A(n) = A219175(n). - Thomas Ordowski, Nov 30 2019

More terms from Altug Alkan, Sep 23 2016

A140470 Numbers n such that p+1 divides n for every prime p that divides n.

Original entry on oeis.org

1, 12, 24, 36, 48, 60, 72, 96, 108, 120, 132, 144, 168, 180, 192, 216, 240, 264, 288, 300, 324, 336, 360, 384, 396, 432, 480, 504, 528, 540, 552, 576, 600, 612, 648, 660, 672, 720, 768, 792, 840, 864, 900, 960, 972, 1008, 1056, 1080, 1104, 1140, 1152, 1176

Offset: 1

Leroy Quet, Jun 28 2008

nonn

Every term is a multiple of 12.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A027748, A124240, A141766.

a140470 n = a140470_list !! (n-1)
a140470_list = filter
   (\x -> all (== 0) $ map ((mod x) . (+ 1)) $ a027748_row x) [1..]
-- Reinhard Zumkeller, Aug 27 2013

a = {}; For[n = 2, n < 1500, n++, b = Select[Range[n], PrimeQ[ # ] && Mod[n, # ] == 0 &]; c = 1; For[d = 1, d < Length[b] + 1, d++, If[Mod[n, b[[d]] + 1] > 0, c = 0; Break]]; If[c == 1, AppendTo[a, n]]]; a (* Stefan Steinerberger, Jul 01 2008 *)
eppQ[n_]:=Union[Differences/@Select[Partition[Divisors[n],2,1], PrimeQ[ #[[1]]]&]]=={{1}}; Join[{1},Select[Range[1200],eppQ]] (* Harvey P. Dale, May 27 2016 *)

More terms from Stefan Steinerberger, Jul 01 2008

a(1)=1 prepended by Max Alekseyev, Aug 27 2013

cp's OEIS Frontend

A124241 Terms of A068563 that are not terms of A124240.

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A289257 Terms k of A006521 such that 2*k is a term of A124240.

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A014117 Numbers n such that m^(n+1) == m (mod n) holds for all m.

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Mathematica

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A055744 Numbers k such that k and phi(k) have the same prime factors.

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Haskell

Maple

Mathematica

PARI

PARI

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A174824 a(n) = period of the sequence {m^m, m >= 1} modulo n.

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Mathematica

PARI

PARI

PARI

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A226960 Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 2 (mod n).

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Mathematica

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A068563 Numbers k such that 2^k == 4^k (mod k).

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