A160496 -id:A160496 - cp's OEIS frontend

A160596 Denominator of resilience R(n) = phi(n)/(n-1).

1, 1, 3, 1, 5, 1, 7, 4, 9, 1, 11, 1, 13, 7, 15, 1, 17, 1, 19, 5, 21, 1, 23, 6, 25, 13, 9, 1, 29, 1, 31, 8, 33, 17, 35, 1, 37, 19, 39, 1, 41, 1, 43, 11, 45, 1, 47, 8, 49, 25, 17, 1, 53, 27, 55, 14, 57, 1, 59, 1, 61, 31, 63, 4, 13, 1, 67, 17, 23, 1, 71, 1, 73, 37, 25, 19, 77, 1, 79, 40, 81, 1

Offset: 2

M. F. Hasler, May 23 2009

The resilience of a denominator, R(d), is the ratio of proper fractions n/d, 0 < n < d, that are resilient, i.e., such that gcd(n,d)=1. Obviously this is the case for phi(d) proper fractions among the d-1 possible ones.

a(n) = 1 if and only if n is prime. - Robert Israel, Dec 26 2016

			a(9)=4 since for the denominator d=9, among the 8 proper fractions n/9 (n=1,...,8), six cannot be canceled down by a common factor (namely 1/9, 2/9, 4/9, 5/9, 7/9, 8/9), thus R(9) = 6/8 = 3/4.

Robert Israel, Table of n, a(n) for n = 2..10000
Project Euler, Problem 245: resilient fractions, May 2009

[Denominator(EulerPhi(n)/(n-1)): n in [2..80]]; // Vincenzo Librandi, Jan 02 2017

seq(denom(numtheory:-phi(n)/(n-1)),n=2..100); # Robert Israel, Dec 26 2016

Denominator[Table[EulerPhi[n]/(n-1),{n,2,90}]] (* Harvey P. Dale, Apr 18 2012 *)

A160496(n)=denominator(eulerphi(n)/(n-1))

A160495 Irregular triangle of residue classes (mod pq) of primes r such that the cyclotomic polynomial Phi(pqr,x) is flat.

Original entry on oeis.org

1, 14, 1, 2, 10, 11, 19, 20, 1, 7, 8, 25, 26, 32, 1, 34, 1, 2, 8, 17, 19, 20, 22, 31, 37, 38, 1, 13, 20, 22, 23, 28, 29, 31, 38, 50, 1, 2, 53, 54, 1, 2, 7, 13, 16, 23, 28, 29, 34, 41, 44, 50, 55, 56, 1, 64, 1, 7, 8, 10, 17, 19, 28, 41, 50, 52, 59, 61, 62, 68

Offset: 1

T. D. Noe, May 15 2009

A polynomial is flat if its coefficients are 1, 0, or -1. The values of pq are in sequence A046388. Each row begins with 1 and ends with pq-1. For each number k in a row, the number pq-k is also in the row. Row n has 2*A160496(n) terms. For the pq in sequence A160497, the row has only two terms. By Kaplan's theorems 2 and 3, only the first prime r in each residue class 1..(p-1)(q-1)/2 needs to be checked to determine whether the residue class produces flat cyclotomic polynomials. Is there a simplier method of finding these residue classes?

			The second row (1,2,10,11,19,20) is for pq=21. If r is a prime with r mod pq equal to one of these 6 values, then Phi(21*r,x) is flat.

T. D. Noe, Rows n=1..100 of triangle, flattened
Nathan Kaplan, Flat cyclotomic polynomials of order three, J. Number Theory 127 (2007), 118-126.

Cf. A046388, A117223.

A160497 Values of pq such that there are only two residue classes (mod pq) of primes r that produce flat cyclotomic polynomial Phi(pqr,x).

Original entry on oeis.org

15, 35, 65, 77, 85, 91, 95, 115, 119, 133, 143, 161, 185, 187, 209, 215, 217, 221, 235, 247, 259, 265, 287, 299, 319, 323, 329, 335, 341, 365, 371, 377, 391, 403, 407, 413, 415, 427, 437, 451, 469, 473, 481, 485, 493, 511, 515, 517, 527

Offset: 1

T. D. Noe, May 15 2009

nonn

Except for 91, 95, 287, and 473, it appears that for prime p>3, pq is here only if q-1 and q+1 are not multiples of p.

A160496

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A160596 Denominator of resilience R(n) = phi(n)/(n-1).

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A160495 Irregular triangle of residue classes (mod pq) of primes r such that the cyclotomic polynomial Phi(pqr,x) is flat.

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A160497 Values of pq such that there are only two residue classes (mod pq) of primes r that produce flat cyclotomic polynomial Phi(pqr,x).

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