A033949 -id:A033949 - cp's OEIS frontend

A180634 Numbers n such that the discriminant of the n-th cyclotomic polynomial is a square.

1, 2, 8, 12, 15, 16, 20, 21, 24, 28, 30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 51, 52, 55, 56, 57, 60, 63, 64, 65, 66, 68, 69, 70, 72, 75, 76, 77, 78, 80, 84, 85, 87, 88, 90, 91, 92, 93, 95, 96, 99, 100

Offset: 1

Jan Fricke, Sep 13 2010

A number n is in this sequence if the Galois group of the n-th cyclotomic polynomial over the rationals contains only even permutations.
Essentially the same as A033949. - R. J. Mathar, Oct 15 2011
Also, numbers n such that the product of the elements in the group Z_n of invertible elements mod n (i.e., the product mod n of x such that 1 <= x < n and x is coprime to n) is 1. An equivalent characterization of the latter (apart from n=2): n such that the number of square roots of 1 mod n is divisible by 4. (See comments at A033949). - Robert Israel, Dec 08 2014
To see this, use Gauss's generalization of Wilson's theorem namely, the product of the units of Z_n is -1 if n is 4 or p^i or 2p^i for odd primes p, i >0,  and is equal to 1 otherwise. - W. Edwin Clark, Dec 09 2014

			n=5: The 5th cyclotomic polynomial is x^4+x^3+x^2+x+1 with discriminant 125, which is not a square. The Galois group is generated by (1243), that is an odd permutation. Hence 5 is not in the sequence. n=8: The 8th cyclotomic polynomial is x^4+1 with discriminant 256, which is a square. The Galois group is {id,(13)(57),(15)(37),(17)(35)}, that are all even permutations. Hence 8 is in the sequence.

Robert G. Wilson v, Table of n, a(n) for n = 1..1000
Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.

Cf. A004124, A033949.

m := proc(n) local k, r; r := 1;
for k from 1 to n do if igcd(n,k) = 1 then r := modp(r*k,n) fi od; r end:
[1, op(select(n -> m(n) = 1, [$1..100]))]; # Peter Luschny, May 25 2017

fQ[n_] := IntegerQ@ Sqrt@ Discriminant[ Cyclotomic[ n, x], x]; Select[ Range@ 100, fQ] (* Robert G. Wilson v, Dec 10 2014 *)

for(n=1,100,if(issquare(poldisc(polcyclo(n))),print(n)))

A254141 The average of a(n) consecutive Fibonacci numbers is never an integer.

Original entry on oeis.org

8, 16, 21, 28, 32, 40, 52, 55, 56, 64, 65, 68, 69, 80, 84, 85, 87, 88, 92, 93, 99, 104, 105, 112, 117, 119, 128, 132, 133, 136, 140, 141, 145, 148, 152, 153, 155, 156, 160, 161, 164, 165, 171, 172, 176, 184, 187, 188, 196, 200, 203, 204, 205, 207, 208, 209, 212

Offset: 1

Paolo P. Lava, Jan 26 2015

nonn

Subset of A033949 and A175594 (essentially the same sequence).
Numbers of the form 2^k, with k>=3, appear to be part of the sequence.
The file "List of indexes and steps (k, x, y)" (see Links) for k = 1, 2, 3, 4, ... consecutive Fibonacci numbers gives the minimum index to start to calculate the average ( x ) and the step to add to get all the other averages ( y ).
E.g.: for k = 7 we have 7, 6, 8. This means that we must start from the 6th Fibonacci number to add 7 consecutive Fibonacci numbers and get an average that is an integer. Fibonacci(6) + Fibonacci(7) + ... + Fibonacci(12) = 8 + 13 + 21 + 34 + 55 + 89 + 144 = 364 and 364 / 7 = 52.
Then 6 + 1*8 = 14, 6 + 2*8 = 22, 6 + 3*8 = 30, etc. are the other indexes:
Fibonacci(14) + Fibonacci (15) + ... + Fibonacci(20) = 377 + 610 + 987 + 1597 + 2584 + 4181 + 6765 = 17101 and 17101 / 7 = 2443;
Fibonacci(22) + Fibonacci(23) + ... + Fibonacci(28) = 17711 + 28657 + 46368 + 75025 + 121393 + 196418 + 317811 = 803383 and 803383 / 7 = 114769;
Fibonacci(30) + Fibonacci(31) + ... + Fibonacci(36) = 832040 + 1346269 + 2178309 + 3524578 + 5702887 + 9227465 + 14930352 = 37741900 and 37741900 / 7 = 5391700; etc.
In particular we note that:
x = 0 is A219612; x = 1 is A124456; x = 0 and y = k - 1 is A106535;
y = 1 is A141767; x = k - 1 and y = k + 1 is A000057;
x = y - 1 or y|k is A023172; y = k is A000351;
x = y - k + 1 appears to give only prime numbers: 3,11,19,31,59,71,79,131,179,191,239,251,271,311,359,379,419,431,439,479,491,499,571,599,631,659,719,739,751,839,971, etc.

Paolo P. Lava, Table of n, a(n) for n = 1..200
Paolo P. Lava, List of indexes and steps (k, x, y)

Cf. A000045, A000057, A000071, A023172, A033949, A106535, A124456, A141767, A175594, A219612.

with(numtheory); with(combinat):P:=proc(q) local a,b,k,j,n,ok;
for j from 1 to q do b:=0; ok:=1;
for n from 0 to q do a:=add(fibonacci(n+k),k=0..j-1)/j;
if type(a,integer) then ok:=0; break; fi; od;
if ok=1 then print(j); fi; od; end: P(20000);

A279400 Row lengths of the irregular triangle A279399.

Original entry on oeis.org

3, 3, 4, 5, 6, 6, 7, 7, 7, 10, 9, 9, 9, 10, 10, 10, 12, 12, 13, 13, 13, 14, 14, 14, 14, 16, 17, 16, 15, 17, 17, 16, 18, 19, 19, 19, 18, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23

Offset: 1

Wolfdieter Lang, Jan 25 2017

nonn

a(n) is the number of primes of the smallest positive restricted residue system modulo A033949(n).

Cf. A033949, A279399.

A380594 a(n) is the number of positive integers having 2*n primitive roots.

Original entry on oeis.org

6, 4, 4, 6, 2, 8, 0, 4, 2, 2, 2, 8, 0, 2, 0, 4, 0, 4, 0, 12, 0, 2, 0, 12, 0, 2, 4, 0, 0, 2, 0, 6, 0, 0, 0, 10, 0, 0, 0, 2, 2, 6, 0, 4, 0, 2, 0, 12, 0, 2, 0, 0, 0, 4, 0, 6, 0, 0, 0, 10, 0, 0, 0, 6, 2, 2, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0, 0, 2, 0, 8, 4, 2, 0, 6, 0

Offset: 1

David James Sycamore, Michael De Vlieger, and Amiram Eldar, Jan 27 2025

nonn

Let [n] be the set {k; A046144(k) = 2*n}; n >= 1, then a(n) = |[n]|.

If 2*n is a term in A378508, [n] is nonempty and a(n) > 0. Otherwise, if 2*n is not in A378508 then there is no number having 2*n primitive roots, so a(n) = 0; see Example, and A380604.

			For n = 1, 2*n = 2 and there are 6 distinct numbers having 2 primitive roots; [2] = {5,7,9,10,14,18}; so a(10) = 6.
For n = 5, 2*n = 10 and there are just 2 distinct numbers having 10 primitive roots; [5] = {23,46}; so a(5) = 2.
For n = 7, 2*n = 14 and there are no numbers having 14 primitive roots, so a(7) = 0.
The sets [n] listed in rows start as follows; length of row n = a(n):
  n          [n]                   a(n)
  1    {5,7,9,10,14,18}             6;
  2    {11,13,22,26}                4;
  3    {29,27,30,54}                4;
  4    {17,25,31,34,50,62}          6;
  5    {23,46}                      2;
  6    {29,37,43,49,58,74,86,98}    8;
  7    { }                          0;
  8    {41,61,82,122}               4;
  9    {81,162}                     2;
  10   {67,134}                     2;
  ...

Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).
David M. Bressoud, A Course in Computational Number Theory (web page), CNT.m, Computational Number Theory Mathematica package.

Cf. A046144, A007617, A010554, A033949, A231772, A378506, A378508, A380604.

a[n_] := Count[Join @@ PhiInverse[PhiInverse[2*n]], ?(IntegerQ[PrimitiveRoot[#]] &)]; Array[a, 100] (* _Amiram Eldar, Jan 27 2025, using David M. Bressoud's CNT.m *)

isA033948(n) = {my(f = factor(n)); lcm(znstar(f)[2]) == eulerphi(f);}
a(n) = {my(v = invphi(2*n), w, c = 0); for(i = 1, #v, c += vecsum(apply(x -> isA033948(x), invphi(v[i])))); c;} \\ Amiram Eldar, Jan 27 2025, using Max Alekseyev's invphi.gp

a(n) <= A378506(2*n), with equality iff n is in A007617.

A277030 Smallest m such that b^phi(n) == b^m (mod n) for every integer b, where phi(n) = A000010(n).

Original entry on oeis.org

0, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 2, 12, 6, 4, 4, 16, 6, 18, 4, 6, 10, 22, 4, 20, 12, 18, 6, 28, 4, 30, 8, 10, 16, 12, 6, 36, 18, 12, 4, 40, 6, 42, 10, 12, 22, 46, 4, 42, 20, 16, 12, 52, 18, 20, 6, 18, 28, 58, 4, 60, 30, 6, 16, 12, 10, 66, 16, 22, 12, 70, 6, 72, 36, 20, 18, 30

Offset: 1

Thomas Ordowski and Altug Alkan, Sep 25 2016

nonn

It suffices to check all bases 1 <= b <= n.
For n > 1; if A002322(n) = phi(n), then a(n) = phi(n). So a(p) = p-1 for all primes p.
Numbers n > 1 such that a(n) < phi(n) are A033949 > 8.
Conjecture: a(n) > A002322(n) only for n = 8 and 24.

Antti Karttunen (terms 1..4000) & Altug Alkan, Table of n, a(n) for n = 1..10000

Cf. A000010, A002322, A033949.

A277030(n) = { my(b,m=0); if(1==n,0,while(1, m=m+1; b=1; while(((b^eulerphi(n))%n) == ((b^m)%n), b=b+1; if(b>n, return(m))))); }; \\ (Following the description). - Antti Karttunen, Jul 28 2017

from sympy import totient
def a(n):
    m=0
    if n==1: return 0
    else:
        while True:
            m+=1
            b=1
            while (b**totient(n))%n==(b**m)%n:
                b+=1
                if b>n: return m
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 29 2017, after PARI code

Conjectured: a(n) = A002322(n), except for a(1) = 0 and a(8) = a(24) = 4.

A277254 Numbers k such that p = k - phi(k) < q = k - lambda(k), and p and q are both primes, where phi(k) = A000010(k) and lambda(k) = A002322(k).

Original entry on oeis.org

15, 33, 35, 65, 77, 87, 91, 95, 119, 123, 143, 185, 215, 221, 247, 255, 259, 287, 329, 341, 377, 395, 407, 427, 437, 455, 473, 485, 511, 515, 537, 573, 595, 635, 705, 713, 717, 721, 749, 767, 779, 793, 795, 803, 805, 815, 817, 843, 869, 871, 885, 899, 923, 965, 1001

Offset: 1

Thomas Ordowski, Oct 07 2016

nonn

Numbers k such that p = A051953(k) < q = A277127(k), and p and q are both primes.
If k is such number, then b^p == b^q (mod k) for every integer b.
Problem: are there infinitely many such numbers?
Suppose p^2 divides k. Then p divides k - phi(k), and so the only way k - phi(k) can be prime is if k = p^2. But then k - phi(k) = k - A002322(k). Hence all terms in this sequence are squarefree. - Charles R Greathouse IV, Oct 08 2016
All terms are odd composites. - Robert Israel, Oct 09 2016
It seems that gpf(k) < p = k - phi(k). - Thomas Ordowski, Oct 09 2016

			For n=15, A051953(15) = 7, A277127(15) = 11, 7 < 11 and both are primes, thus 15 is included in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Subsequence of A033949 and of A024556.

Cf. A000010, A002322, A050530, A051953, A277127.

filter:= proc(n) uses numtheory;
  local p,q;
  p:= n-phi(n);
  q:= n-lambda(n);
  pRobert Israel, Oct 09 2016

Select[Range[10^3], And[#1 < #2, Times @@ Boole@ PrimeQ@ {#1, #2} == 1] & @@ {# - EulerPhi@ #, # - CarmichaelLambda@ #} &] (* Michael De Vlieger, Oct 08 2016 *)

is(n)=my(f=factor(n),p=n-eulerphi(f),q=n-lcm(znstar(f)[2])); p < q && isprime(p) && isprime(q) \\ Charles R Greathouse IV, Oct 08 2016

More terms from Altug Alkan, Oct 07 2016

A284406 Odd numbers k such that lambda(k) < phi(k) and gcd(lambda(k), k-1) = gcd(phi(k), k-1).

Original entry on oeis.org

15, 35, 39, 45, 51, 55, 63, 75, 85, 87, 95, 99, 111, 115, 117, 119, 123, 135, 143, 147, 153, 155, 159, 165, 171, 175, 183, 187, 195, 203, 205, 207, 215, 219, 221, 231, 235, 245, 247, 255, 259, 261, 267, 275, 279, 285, 287, 291, 295, 299, 303, 315, 319, 323, 325, 327, 333, 335, 339, 351

Offset: 1

Thomas Ordowski and Altug Alkan, Mar 26 2017

nonn

If odd n is in A033949 and n-1 is squarefree, then n is in the sequence.
The set of such numbers has a positive natural density.
The density is 1/2. Almost all odd numbers have this property.
The number of terms below 10^k for k = 1, 2, ... are 0, 12, 204, 2507, 27801, 296583, 3102205, 32054920, 328714616, 3353406273, .... Apparently the asymptotic density of this sequence is less than 1/2. - Amiram Eldar, Jul 14 2020

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..439 from Indranil Ghosh)

Subsequence of A257591.
A264012 is a subsequence.
Cf. A000010, A002322, A280262.

Select[Range[1, 351, 2], Function[k, And[#1 < #2, GCD[#1, k - 1] == GCD[#2, k - 1]] & @@ {CarmichaelLambda@ k, EulerPhi@ k}]] (* Michael De Vlieger, Mar 26 2017 *)

cp's OEIS Frontend

A180634 Numbers n such that the discriminant of the n-th cyclotomic polynomial is a square.

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A254141 The average of a(n) consecutive Fibonacci numbers is never an integer.

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A279400 Row lengths of the irregular triangle A279399.

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A380594 a(n) is the number of positive integers having 2*n primitive roots.

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A277030 Smallest m such that b^phi(n) == b^m (mod n) for every integer b, where phi(n) = A000010(n).

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A277254 Numbers k such that p = k - phi(k) < q = k - lambda(k), and p and q are both primes, where phi(k) = A000010(k) and lambda(k) = A002322(k).

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A284406 Odd numbers k such that lambda(k) < phi(k) and gcd(lambda(k), k-1) = gcd(phi(k), k-1).

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