Jan Fricke - 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)))

A152925 a(n) = smallest number m such that in 1,2,..,m written in base n, no two of the n digits occurs the same number of times.

Original entry on oeis.org

1, 5, 13, 47, 105, 536, 1341, 9231, 24697, 212594, 592269, 6100559, 17464969, 209215572, 610839805, 8338210079, 24709115769, 378460880126

Offset: 2

Jan Fricke, Dec 15 2008

			In base 5 in 1,2,..,142_5 = 47, digit 1 occurs 43 times, 2 occurs 20, 3 occurs 19 times, 4 occurs 17 times, and 0 occurs 14.

def different(d):
    for i in range(len(d)):
        for j in range(i):
            if d[i] == d[j]:
                return False
    return True
def a(base):
    d = [0] * base
    n = 0
    while True:
        n += 1
        m = n
        while m > 0:
            d[m % base] += 1
            m //= base
        if different(d):
            break
    return n

Edited by Franklin T. Adams-Watters, Sep 11 2011

a(16)-a(19) from Michael S. Branicky, Mar 26 2022

A127699 Length of period of the sequence (1^1^1^..., 2^2^2^..., 3^3^3^..., 4^4^4^..., ...) modulo n.

Original entry on oeis.org

1, 2, 6, 4, 20, 6, 42, 8, 18, 20, 220, 12, 156, 42, 60, 16, 272, 18, 342, 20, 42, 220, 5060, 24, 100, 156, 54, 84, 2436, 60, 1860, 32, 660, 272, 420, 36, 1332, 342, 156, 40, 1640, 42, 1806, 220, 180, 5060, 237820, 48, 294, 100, 816, 156, 8268, 54, 220, 168

Offset: 1

Jan Fricke, Apr 11 2007

For any positive integers a and m the sequence a, a^a, a^a^a, a^a^a^a,... becomes eventually constant modulo m. So the remainder of a^a^a^... modulo n is well-defined.

Shapiro and Shapiro treat this problem. - T. D. Noe, Jan 30 2009

			a(10)=20 because the last digit of 1^1^1^.. is 1; the sequence 2,2^2,2^2^2,.. ends with 2,4,6,6,...; the sequence 3,3^3,3^3^3,... with 3,7,7,...; 4,4^4,4^4^4,... with 4,6,6,...; and so on. We get as last digits 1,6,7,6,5,6,3,6,9,0, 1,6,3,6,5,6,7,6,9,0 and then the pattern repeats.

Alois P. Heinz, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)
Daniel B. Shapiro and S. David Shapiro, Iterated Exponents in Number Theory, Integers 7 (2007), #A23.

Cf. A002322.

a:= proc(n) option remember; `if`(n=1, 1,
      ilcm(n, a(numtheory[lambda](n))))
    end:
seq(a(n), n=1..56);  # Alois P. Heinz, Jan 03 2023

nn=100; a=Table[0,{nn}]; a[[1]]=1; Do[a[[n]]=LCM[n,a[[CarmichaelLambda[n]]]], {n,2,nn}]; a (* T. D. Noe, Jan 30 2009 *)

from functools import lru_cache
from math import lcm
from sympy import reduced_totient
@lru_cache(maxsize=None)
def A127699(n): return 1 if n == 1 else lcm(n, A127699(reduced_totient(n))) # Chai Wah Wu, Jan 03 2023

a(n) = lcm(n, a(lambda(n))), where lambda is Carmichael's reduced totient function. - T. D. Noe, Jan 30 2009

Extension and correction from T. D. Noe, Jan 30 2009

Incorrect formula removed by T. D. Noe, Feb 02 2009

A119486 Numbers of children for which there is a subset which cannot be generated by a counting-out game.

Original entry on oeis.org

9, 12, 15, 18, 20, 21, 24, 25, 27, 28, 30, 33

Offset: 1

Jan Fricke, May 23 2006

The numbers were generated by an exhaustive search via a C-program.

			For 9 children 1,2,3,4,5,6,7,8,9, there is no possibility to select 3,4,6,7 (in any order) by a counting-out game, e.g. for selecting 3,4,6,7 the count-to number has to be 3 mod 9, 1 mod 8, 2 mod 7 and 1 mod 6, which is impossible.

Complement of A119485.

Conjecture (by J. Fricke and G. Woeginger): The sequence contains all numbers n with an odd prime divisor p satisfying n/p>2.

A119485 Number of children for which any subset can be generated by a counting-out game.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 16, 17, 19, 22, 23, 26, 29, 31, 32

Offset: 1

Jan Fricke, May 23 2006

The numbers were generated by an exhaustive search via a C-program.

			Having 6 children 1,2,3,4,5,6, then the children 2,4,6 can be counted-out by counting to 42: first selected child is 6, then 2 and finally 4.

Complement of A119486.

Conjecture (by J. Fricke and G. Woeginger): The sequence contains exactly: powers of 2, primes and doubled primes.

A074051 For each n there are uniquely determined numbers a(n) and b(n) and a polynomial p_n(x) such that for all integers m we have Sum_{i=1..m}i^n(i+1)! = a(n)Sum_{i=1..m} (i+1)! + p_n(m)(m+2)! + b(n). The sequence b(n) is A074052.

Original entry on oeis.org

1, -1, 0, 3, -7, 0, 59, -217, 146, 2593, -15551, 32802, 160709, -1856621, 7971872, 1299951, -287113779, 2262481448, -7275903849, -36989148757, 698330745002, -4867040141851, 10231044332629, 184216198044034, -2679722886596295, 17971204188130391, -17976259717948832

Offset: 0

Jan Fricke, Aug 14 2002

If a(n)=0 then Sum_{i>=1}i^n(i+1)! = b(n) in the p-adic numbers. The only known numbers n with a(n)=0 are 2 and 5.

a(n)*(-1)^n gives the alternating row sums of the Sheffer triangle A143494 (2-restricted Stirling2). - Wolfdieter Lang, Oct 06 2011

			a(2)=0 because Sum_{i=1..m}i^2(i+1)! = (m-1)(m+2)!+2.
a(3)=3 because Sum_{i=1..m}i^3(i+1)! = 3*Sum_{i=1..m}(i+1)!+(m^2-m-1)(m+2)!+2.

Robert Israel, Table of n, a(n) for n = 0..545

Cf. A074052, A143494.

alias(S2 = combinat[stirling2]);
A074051 := proc(n) local k;
1 + add((-1)^(n+k) * (S2(n+1, k+1) - S2(n+2, k+1)), k = 0..n) end:
seq(A074051(i), i = 0..26); # Peter Luschny, Apr 17 2011

A[a_] := Module[{p, k}, p[n_] = 0; For[k = a - 1, k >= 0, k--, p[n_] = Expand[p[n] + n^k Coefficient[n^a - (n + 2)p[n] + p[n - 1], n^(k + 1)]] ]; Expand[n^a - (n + 2)p[n] + p[n - 1]] ]
(* Second program: *)
a[n_] := (-1)^n (BellB[n+2, -1] - BellB[n+1, -1]);
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 21 2018, after Peter Luschny *)

from itertools import accumulate
def A074051_list(size):
    if size < 1: return []
    L, accu = [], [1]
    for n in range(size-1):
        accu = list(accumulate([-accu[-1]] + accu))
        L.append(-(-1)**n*accu[-2])
    return L
print(A074051_list(28)) # Peter Luschny, Apr 25 2016

From Vladeta Jovovic, Jan 27 2005: (Start)
Second inverse binomial transform of A000587.
E.g.f.: exp(1 - 2*x - exp(-x)).
G.f.: Sum_{k >= 0}((x/(1+2*x))^k/Product_{l=0..k}(1 + l*x/(1+2*x)))/(1+2*x).
a(n) = Sum_{k=0..n} (-1)^(n-k)*(k^2-3*k+1)*Stirling2(n, k). (End)
a(n) = (-1)^n*(A000587(n+2)-A000587(n+1)). - Peter Luschny, Apr 17 2011
From Sergei N. Gladkovskii, Sep 28 2012 to Apr 22 2013: (Start)
Continued fractions:
G.f.: 1/U(0) where U(k)= x*k + 1 + x + x^2*(k+1)/U(k+1).
G.f.: -1/U(0) where U(k)= -x*k - 1 - x + x^2*(k+1)/U(k+1).
G.f.: 1/(U(0) - x) where U(k)= 1 + x + x*(k+1)/(1 - x/U(k+1)).
G.f.: 1/(U(0) + x) where U(k)= 1 + x*(2*k+1) - x*(k+1)/(1 + x/U(k+1)).
G.f.: 1/G(0) where G(k)= 1 + 2*x/(1 + 1/(1 + 2*x*(k+1)/G(k+1))).
G.f.: 1 - 2*x/(G(0) + 2*x) where G(k)= 1 + 1/(1 + 2*x*(k+1)/(1 + 2*x/G(k+1))).
G.f.: (G(0) - 1)/(x-1) where G(k) =  1 - 1/(1+k*x+2*x)/(1-x/(x-1/G(k+1))).
G.f.: (G(0)-2-2*x)/x^2 where G(k) = 1 + 1/(1+k*x)/(1-x/(x+1/G(k+1) )).
G.f.: (S-2-2*x)/x^2 where S = sum(k>=0, (2 + x*k)*x^k/prod(i=0..k, (1+x*i))).
G.f.: (G(0)-2)/x where G(k) = 1 + 1/(1+k*x+x)/(1-x/(x+1/G(k+1))).
G.f.: (1+x)/x/Q(0) - 1/x, where Q(k)=  1 + x - x/(1 + x*(k+1)/Q(k+1)). (End)
a(n) = exp(1) * (-1)^n * Sum_{k>=0} (-1)^k * (k+2)^n / k!. - Ilya Gutkovskiy, Sep 02 2021

More terms from Vladeta Jovovic, Jan 27 2005

A081257 a(n) is the greatest prime factor of (n^3 - 1).

Original entry on oeis.org

7, 13, 7, 31, 43, 19, 73, 13, 37, 19, 157, 61, 211, 241, 13, 307, 17, 127, 421, 463, 13, 79, 601, 31, 37, 757, 271, 67, 29, 331, 151, 1123, 397, 97, 43, 67, 1483, 223, 547, 1723, 139, 631, 283, 109, 103, 61, 181, 43, 2551, 379, 919, 409, 2971, 79, 103, 3307, 163

Offset: 2

Jan Fricke, Mar 14 2003

The record values here (as well as those for A081256) appear to match the terms of A002383 for n > 1. - Bill McEachen, Jun 19 2023

			a(7)=19 because 7^3 - 1 = 342 = 2*3*3*19.

Jon E. Schoenfield, Table of n, a(n) for n = 2..10000 (first 999 terms from R. J. Mathar).

Cf. A081256, A081258, A002383.

Cf. A096175 (n^3-1 is an odd semiprime), A096176 ((n^3-1)/(n-1) is prime).

A081257 := proc(n)
        A006530( n^3-1) ;
end proc: # R. J. Mathar, Jul 18 2015

FactorInteger[#][[-1,1]]&/@(Range[2,60]^3-1) (* Harvey P. Dale, Oct 09 2017 *)

a(n)=my(f=factor(n^3-1)); f[#f~,1] \\ Charles R Greathouse IV, Mar 08 2017

from sympy import primefactors
def A081257(n): return max(primefactors(n-1)+primefactors(n*(n+1)+1)) # Chai Wah Wu, Oct 15 2022

a(n) = A006530(A068601(n)). - Michel Marcus, Jun 19 2023

More terms from Hugo Pfoertner, Jun 21 2004

A081258 Numbers k > 1 such that k^3 - 1 (or equivalently k^2 + k + 1) has no prime factor greater than k.

Original entry on oeis.org

16, 18, 22, 30, 49, 67, 68, 74, 79, 81, 87, 100, 102, 121, 135, 137, 146, 149, 154, 158, 159, 163, 165, 169, 172, 178, 181, 191, 211, 221, 229, 230, 235, 256, 262, 263, 269, 273, 277, 291, 292, 301, 305, 313, 315, 324, 326, 334, 352, 361, 372, 373, 380, 393

Offset: 1

Jan Fricke, Mar 14 2003

One might also include 1 as a term here. - R. J. Mathar, Oct 11 2011

			16 is a term: 16^3 - 1 = 4095 = 3*3*5*7*13.

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

Cf. A081256, A081257.

isA081258 := proc(n)
        numtheory[factorset](n^3-1) ;
        if max(op(%)) <= n then
                true;
        else
                false;
        end if;
end proc;
for n from 1 to 400 do
        if isA081258(n) then
                printf("%d,",n);
        end if;
end do: # R. J. Mathar, Oct 11 2011

Select[Range[2, 1000], FactorInteger[#^3 - 1][[-1, 1]] <= #&] (* Jean-François Alcover, Jun 15 2020 *)

Name changed by Robert Israel, Nov 11 2016

A081256 Greatest prime factor of n^3 + 1.

Original entry on oeis.org

2, 3, 7, 13, 7, 31, 43, 19, 73, 13, 37, 19, 157, 61, 211, 241, 13, 307, 7, 127, 421, 463, 13, 79, 601, 31, 37, 757, 271, 67, 19, 331, 151, 1123, 397, 97, 43, 67, 1483, 223, 547, 1723, 139, 631, 283, 109, 103, 61, 181, 43, 2551, 379, 919, 409, 2971, 79, 103, 3307, 163

Offset: 1

Jan Fricke, Mar 14 2003

Record values appear to match the terms of A002383 for n>1. - Bill McEachen, Oct 18 2023

R. J. Mathar and T. D. Noe, Table of n, a(n) for n = 1..10000 (first 5000 terms from R. J. Mathar)
J. Buchmann, K. Győry, M. Mignotte, and N. Tzanakis, Lower bounds for P(x^3+k), an elementary approach, Publ. Math. Debrecen, Vol. 38, No. 1-2 (1991), pp. 145-163.

Cf. A081257, A081258, A096173, A096174.

Cf. A006530, A001093, A002383, A014442, A096172.

A081256 := proc(n)
    A006530(n^3+1) ;
end proc:
seq(A081256(n),n=1..20) ; # R. J. Mathar, Feb 13 2014

Table[Max[Transpose[FactorInteger[n^3 + 1]][[1]]], {n, 25}]

a(n)=my(f=factor(n^3+1)); f[#f~,1] \\ Charles R Greathouse IV, Mar 08 2017

A081256(n)=vecmax(factor(n^3+1)[,1]) \\ It seems slightly slower to get the last element using ...[-1..-1][1]. - M. F. Hasler, Jun 15 2018

a(n) = A006530(A001093(n)). - M. F. Hasler, Jun 13 2018

a(n) >= 31 for n >= 70 (Buchmann et al., 1991). - Amiram Eldar, Oct 25 2024

More terms from Harvey P. Dale, Mar 22 2003

More terms from Hugo Pfoertner, Jun 20 2004

A074052 The lowest order term in an expansion of Sum_{i=1..m} i^n*(i+1)! in a special factorial basis.

Original entry on oeis.org

0, -2, 2, 2, -14, 26, 34, -398, 1210, 450, -23406, 118634, -166286, -1983342, 18159658, -68002894, -112926670, 3497644570, -24969255550, 64943618962, 607880756218, -9318511004702, 60525142971954, -80108659182870, -3000122066181358

Offset: 0

Jan Fricke, Aug 14 2002

For each n there unique numbers a(n) and b(n) and a polynomial p_n such that for all integers m: Sum_{i=1..m} i^n * (i+1)! = a(n) + b(n) * Sum_{i=1..m}(i+1)! + p_n(m)*(m+2)! The sequence b(n) is A074051(n), and this sequence here are the a(n).

			a(0) = 0 because Sum_{i=1..m} (i+1)! = 0 + 1*Sum_{i=1..m} (i+1)! + 0*(m+2)!.
a(1) = -2 because Sum_{i=1..m} i*(i+1)! = -2 -1*Sum_{i=1..m} (i+1)! + 1*(m+2)!.
a(2) = 2 because Sum_{i=1..m} i^2*(i+1)! = 2 +0*Sum_{i=1..m} (i+1)! + (m-1)*(m+2)!.
a(3) = 2 because Sum_{i=1..n} i^3*(i+1)! = 2 +3*Sum_{i=1..m} (i+1)! + (m^2-m-1)*(m+2)!.
a(4)=-14 because Sum_{i=1..n} i^4*(i+1)! = -14 -7*Sum_{i=1..n} (i+1)! + (m^3-m^2-2*m+7)*(m+2)!.

Cf. A074051, A197184.

A[a_] := Module[{p, k}, p[n_] = 0; For[k = a - 1, k >= 0, k--, p[n_] = Expand[p[n] + n^k Coefficient[n^a - (n + 2)p[n] + p[n - 1], n^(k + 1)]] ]; -2 p[0] ]

More terms from R. J. Mathar, Oct 11 2011

cp's OEIS Frontend

User: Jan Fricke

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

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A152925 a(n) = smallest number m such that in 1,2,..,m written in base n, no two of the n digits occurs the same number of times.

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A127699 Length of period of the sequence (1^1^1^..., 2^2^2^..., 3^3^3^..., 4^4^4^..., ...) modulo n.

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A119486 Numbers of children for which there is a subset which cannot be generated by a counting-out game.

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A119485 Number of children for which any subset can be generated by a counting-out game.

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A074051 For each n there are uniquely determined numbers a(n) and b(n) and a polynomial p_n(x) such that for all integers m we have Sum_{i=1..m}i^n(i+1)! = a(n)*Sum_{i=1..m} (i+1)! + p_n(m)*(m+2)! + b(n). The sequence b(n) is A074052.

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A081257 a(n) is the greatest prime factor of (n^3 - 1).

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A074051 For each n there are uniquely determined numbers a(n) and b(n) and a polynomial p_n(x) such that for all integers m we have Sum_{i=1..m}i^n(i+1)! = a(n)Sum_{i=1..m} (i+1)! + p_n(m)(m+2)! + b(n). The sequence b(n) is A074052.