A267509 -id:A267509 - cp's OEIS frontend

A267510 Integers in A267509 such that (B0 + B1 + ... + Bm) is congruent to 0 mod m.

20, 22, 24, 26, 28, 30, 33, 36, 39, 40, 42, 44, 46, 48, 50, 55, 60, 62, 63, 64, 66, 68, 69, 70, 77, 80, 82, 84, 86, 88, 90, 93, 96, 99, 110, 121, 130, 132, 143, 150, 154, 156, 165, 169, 170, 176, 187, 190, 198, 200, 202, 204, 206, 208, 220, 222, 224, 226, 228, 231, 240

Offset: 1

Abdul Gaffar Khan, Jan 16 2016

If Bi is congruent to 0 mod m for all i=1,2,...,m then the integer n = (Bm,...,B1,B0) is a member of this sequence if and only if n is a member of A267509.

			22 is a term, as f(x)=B0+B1x=2+2x=2*(1+x)=g(x)*h(x) with g(x)=2, h(x)=1+x, and neither g(x) nor h(x) is a unit in the ring of integers implies that f(x) is reducible over the ring of integers and 2+2=4=0 mod 1.
121 is a term, as f(x)=B0+B1x+B2x^2=1+2x+1x^2=1+2x+x^2=(1+x)*(1+x)=g(x)*h(x) with g(x)=1+x=h(x) and neither g(x) nor h(x) is a unit in the ring of integers implies that f(x) is reducible over the ring of integers and 1+2+1=4=0 mod 2.

Cf. A267509.

okQ[n_] := MatchQ[Factor[(id = IntegerDigits[n]).x^Range[lg = Length[id] - 1, 0, -1]][[0]], Times | Power] && Divisible[Total[id], lg]; Select[ Range[240], okQ] (* Jean-François Alcover, Feb 01 2016 *)

A121719 Strings of digits which are composite regardless of the base in which they are interpreted. Exclude bases in which numbers are not interpretable.

Original entry on oeis.org

4, 6, 8, 9, 20, 22, 24, 26, 28, 30, 33, 36, 39, 40, 42, 44, 46, 48, 50, 55, 60, 62, 63, 64, 66, 68, 69, 70, 77, 80, 82, 84, 86, 88, 90, 93, 96, 99, 100, 110, 112, 114, 116, 118, 120, 121, 130, 132, 134, 136, 138, 140, 143, 144

Offset: 1

Tanya Khovanova, Sep 08 2006

Comments from Franklin T. Adams-Watters:
"Think of these as polynomials. E.g. 121 is the polynomial n^2+2n+1. There are three cases:
"(1) If the coefficients (digits) all have a common factor, the result will be divisible by that factor.
"(2) If the polynomial can be factored, the numbers will be composite. n^2+2n+1 = (n+1)^2, so it is always composite.
"(3) Otherwise, look at the polynomial modulo primes up to its degree. For example, 112 (n^2+n+2, degree 2) modulo 2 is always 0, so it is always divisible by 2.
"Note that condition (1) is really a special case of condition (2), where one of the factors is a constant.
"If none of the above conditions apply, the polynomial will (probably) have prime values."
From Iain Fox, Sep 02 2020: (Start)
lim_{k->infinity} (1/k)*Sum_{i=1..k} a_c(i) > .3 if it exists, where a_c(n) is the characteristic function of a(n) (1 if n is in a(n), otherwise 0).
If the Bunyakovsky conjecture is true, the list of reasons a number is in this sequence detailed by Franklin T. Adams-Watters above is a complete list.
If the Bunyakovsky conjecture and the Extended Riemann Hypothesis are true, the above limit equals 4340435807/13235512500 = 0.3279386... (proof by Ravi Fernando in link by Iain Fox).
All members of A008592 except 1 and 10 are in this sequence.
(End)

			String 55 in every base in which it is interpretable is divisible by 5. String 1001 in base a is divisible by a+1. Hence 55 and 1001 both belong to this sequence.

Iain Fox, Table of n, a(n) for n = 1..10000
Iain Fox, What percentage of positive integers, written in base 10, are composite regardless of what base they are interpreted in?, Math StackExchange, September 2020.
Ed Pegg, Jr., Bouniakowsky Conjecture.
Eric W. Weisstein, Extended Riemann Hypothesis.
Wikipedia, Bunyakovsky conjecture.
Wikipedia, Generalized Riemann Hypothesis.

Supersequence: A002808.

Cf. A008592, A267509.

is(n)=if(n<10, return(!isprime(n)&&n>1)); if(content(n=digits(n))>1, return(1)); if(vecsum(factor(n*=vectorv(#n, i, x^(#n-i)))[,2])>1, return(1)); forprime(p=2, #n-1, for(x=1, p, if(eval(n)%p, next(2))); return(1)); for(x=vecmax(Vec(n))+1, +oo, if(isprime(eval(n)), return(0))) \\ Iain Fox, Aug 31 2020

More terms from Franklin T. Adams-Watters, Sep 12 2006

A348139 Three-digit numbers abc such that the quadratic equation ax^2 + bx + c = 0 has a rational root.

Original entry on oeis.org

100, 110, 120, 121, 130, 132, 140, 143, 144, 150, 154, 156, 160, 165, 168, 169, 170, 176, 180, 187, 190, 198, 200, 210, 220, 230, 231, 240, 242, 250, 252, 253, 260, 264, 270, 273, 275, 276, 280, 286, 288, 290, 294, 297, 299, 300, 310, 320, 330, 340, 341, 350, 352, 360, 363, 370, 372, 374, 380, 384, 385, 390, 396, 400, 410, 420, 430, 440, 441, 450, 451, 460, 462, 470

Offset: 1

Bernard Schott, Oct 02 2021

Inequalities: 1 <= a <= 9, 0 <= b, c <= 9.
If the quadratic equation ax^2 + bx + c = 0 has a rational root, then b^2-4ac is a square, the two roots are rational and nonpositive.
Proposition: these three-digit numbers abc are all composite.
The Olympiad problem proposed in Changhua, Taiwan, 2010 (see Reference) asked for a proof that the three-digit number abc is not a prime number.
If abc is a term with a, b, c >= 1 then cba is another term.
The total number of terms is 147.
The first 19 terms are also the first 19 terms of A033828, then A033828(20) = 182 while a(20) = 187.
Also, the first 23 terms are the first 23 3-digit terms of A267509, from A267509(39) to A267509(61), then A267509(62) = 202 while a(24) = 210.

			x^2 + 2x = x*(x+2), whose roots are {-2, 0}, so 120 is a term.
2x^2 = 0 has double root {0}, so 200 is a term.
4x^2 + 7x + 3 = 4*(x+1)*(x+3/4), whose roots are {-3/4, -1}, so 473 = 11*43 is a term.

Xiong Bin and Lee Peng Yee, Mathematical Olympiad in China (2009-2010), Problems and Solutions, Changhua, Taiwan, 2010, First Day, Problem 1, p. 147, East China Normal university Press - World Scientific, 2013.

Michel Marcus, Table of n, a(n) for n = 1..147
Index to sequences related to Olympiads.

Cf. A033828, A267509.

Select[Range[100, 999], (d = (#[[2]]^2 - 4*#[[1]]*#[[3]])&@ IntegerDigits[#]) >= 0 && IntegerQ @ Sqrt[d] &] (* Amiram Eldar, Oct 02 2021 *)

isok(m) = my(d=digits(m)); (#d==3) && issquare(d[2]^2 - 4*d[1]*d[3]); \\ Michel Marcus, Oct 03 2021

from math import isqrt
def ok(n):
    s = str(n)
    if len(s) != 3: return False
    a, b, c = list(map(int, s))
    D = b**2 - 4*a*c
    return D >= 0 and isqrt(D)**2 == D
def afull(): return [m for m in range(100, 1000) if ok(m)]
print(afull()) # Michael S. Branicky, Oct 02 2021

A267521 Integers whose base-10 representation (Bm,...,B1,B0) is such that the polynomial f(x) = B0 + B1x + ... + Bmx^m is irreducible over the ring of integers, 0 <= Bi <= 9.

Original entry on oeis.org

1, 2, 3, 5, 7, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 23, 25, 27, 29, 31, 32, 34, 35, 37, 38, 41, 43, 45, 47, 49, 51, 52, 53, 53, 56, 57, 58, 59, 61, 65, 67, 71, 72, 73, 74, 75, 76, 78, 79, 81, 83, 85, 87, 89, 91, 92, 94, 95, 97, 98, 101, 102, 103, 104, 105, 106, 107, 108, 109, 111, 112, 113, 114, 115, 116, 117, 118, 119

Offset: 1

Abdul Gaffar Khan, Jan 16 2016

			11 is a member as f(x) = B0 + B1*x = 1 + 1*x has no factorization other than the trivial one, i.e., 1*(1+x), hence f(x) is irreducible over the ring of integers.
114 is a member as f(x) = B0 + B1*x + B2*x^2 = 4 + 1*x + 1*x^2 = 4 + x + x^2 is irreducible over the ring of integers.

okQ[n_] := If[n<10, !CompositeQ[n], !MatchQ[Factor[(id = IntegerDigits[n]). x^Range[Length[id]-1, 0, -1]][[0]], Times|Power]]; Select[Range[120], okQ] (* Jean-François Alcover, Feb 01 2016 *)

Integers in A000027 but not in A267509.

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A267510 Integers in A267509 such that (B0 + B1 + ... + Bm) is congruent to 0 mod m.

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A121719 Strings of digits which are composite regardless of the base in which they are interpreted. Exclude bases in which numbers are not interpretable.

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A348139 Three-digit numbers abc such that the quadratic equation ax^2 + bx + c = 0 has a rational root.

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A267521 Integers whose base-10 representation (Bm,...,B1,B0) is such that the polynomial f(x) = B0 + B1*x + ... + Bm*x^m is irreducible over the ring of integers, 0 <= Bi <= 9.

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A267521 Integers whose base-10 representation (Bm,...,B1,B0) is such that the polynomial f(x) = B0 + B1x + ... + Bmx^m is irreducible over the ring of integers, 0 <= Bi <= 9.