A224210 -id:A224210 - cp's OEIS frontend

A224197 Least b > p_n^2 such that [p_1^2,p_2^2,...,p_n^2] in base b is prime, where p_j denotes the j-th prime.

11, 26, 51, 124, 177, 312, 394, 668, 843, 978, 1398, 1730, 1911, 2242, 2859, 3496, 3724, 4532, 5073, 5358, 6269, 6906, 7927, 9422, 10205, 10766, 11522, 12060, 12923, 16142, 17220, 18788, 19409, 22806, 22965, 25562, 26570, 28038, 30636

Offset: 2

Zhi-Wei Sun, Apr 01 2013

nonn

Conjecture: (i) For any positive integer k and distinct positive integers a_1< a_2 < ... < a_n with a_n prime, there are infinitely many integers b > a_n^k such that [a_1^k,a_2^k,...,a_n^k] in base b is prime.
(ii) For positive integers k, m and n>m, let s_k(m,n) denote the smallest integer b > p_n^k such that [p_m^k,p_{m+1}^k,...,p_n^k] in base b is prime. Then we have the inequality s_k(m,n) <= (n+1)^k*(m+n+1)^k.
This is the k-th power version of the author's conjecture related to A217788. Note that s(m,n) defined there is identical with s_1(m,n). It seems that s_2(m,n) < p_{n+1}*p_{m+n+1}.
For example, [2^2,6^2,9^2,20^2,29^2] in base 900 and [37^2,38^2,60^2,90^2,101^2] in base 10268 are both prime. Also, s_3(1,15) = 103960 and s_5(3,5) = 161098.
Note that for any integer b>13^2 the number [2^2,5,6,156,13^2] in base b is composite since
4x^4+5x^3+6x^2+156x+169 = (4x+13)*(x^3-2x^2+8x+13).
Although 1, 2, 3, 113, 115 are pairwise relatively prime, [1,2,3,113,115] in any base b>115 is composite since x^4+2x^3+3x^2+113x+115 = (x+5)*(x^3-3x^2+18x+23).

			a(35) = s_2(1,35) = 22806 since [p_1^2,p_2^2,...,p_{35}^2] in base 22806 is prime. Note that p_{36}^2 = 22801 < 22806 < p_{35}*p_{37} = 23393 < p_{36}*p_{37} = 23707.
a(287) = s_2(1,287) = 3519434 since [p_1^2,p_2^2,...,p_{287}^2] in base 3519434 is prime. Note that p_{287}*p_{289} = 3519367 < 3519434 < p_{288}^2 = 3523129 < p_{288}*p_{289} = 3526883.

Zhi-Wei Sun, Table of n, a(n) for n = 2..300
Zhi-Wei Sun, A general conjecture involving k-th powers, a message to Number Theory List, April 1, 2013.

Cf. A000040, A217788, A218465, A224210.

A[n_,x_]:=A[n,x]=Sum[Prime[k]^2*x^(n-k),{k,1,n}]
Do[Do[Do[If[PrimeQ[A[n,b]]==True,Print[n," ",b];Goto[aa]],{b,Prime[n]^2+1,Prime[n+1]Prime[n+2]-1}];
Print[n," ",counterexample];Label[aa];Continue,{n,2,100}]]

A224416 Least prime p such that the polynomial sum_{k=0}^n C_k*x^{n-k} is irreducible modulo p, where C_k denotes the Catalan number binomial(2k,k)/(k+1).

Original entry on oeis.org

2, 3, 2, 3, 17, 7, 47, 3, 53, 5, 137, 109, 79, 11, 37, 7, 59, 13, 53, 251, 251, 101, 467, 149, 79, 3, 83, 61, 239, 31, 79, 73, 73, 373, 199, 5, 337, 167, 17, 683, 523, 269, 37, 163, 431, 163, 163, 7, 487, 7, 167, 163, 197, 1549, 137, 503, 139, 263, 151, 283

Offset: 1

Zhi-Wei Sun, Apr 06 2013

nonn

Conjecture: (i) a(n) does not exceed n^2+n+5 for each n>0, and the Galois group of sum_{k=0}^n C_k*x^{n-k} over the rationals is isomorphic to the symmetric group S_n.
(ii) For any positive integer n, the polynomial sum_{k=0}^n binomial(2k,k)*x^{n-k} is irreducible modulo some prime if and only if n is not of the form 2k(k+1), where k is a positive integer.
(iii) For any positive integer n, the polynomial sum_{k=0}^n T_k*x^{n-k} is irreducible modulo some prime not exceeding n^2+n+5, where T_k referes to the central trinomial coefficient A002426(k) which is the coefficient of x^k in the expansion of (x^2+x+1)^k.

			a(10) = 5 since sum_{k=0}^{10} C_k*x^{n-k} irreducible modulo 5 but reducible modulo any of 2 and 3.
Note also that a(11) = 137 coincides with 11^2+11+5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..350

Cf. A000040, A000108, A224480, A224417, A224418, A220072, A223934, A224210, A217785, A217788, A224197, A002426.

A[n_,x_]:=A[n,x]=Sum[Binomial[2k,k]/(k+1)*x^(n-k),{k,0,n}]
Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n^2+n+5]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

A224417 Least prime p such that sum_{k=0}^n B_k*x^{n-k} is irreducible modulo p, where B_k refers to the Bell number A000110(k).

Original entry on oeis.org

2, 3, 2, 11, 3, 2, 193, 113, 2, 29, 71, 167, 19, 3, 7, 13, 199, 5, 101, 59, 13, 41, 3, 359, 7, 11, 2, 31, 197, 139, 3, 59, 2, 139, 83, 37, 23, 193, 587, 199, 67, 47, 401, 41, 571, 73, 1063, 229, 1163, 47, 53, 239, 347, 223, 577, 499, 271, 269, 11, 179

Offset: 1

Zhi-Wei Sun, Apr 06 2013

nonn

Conjecture: a(n) < 4n^2-1 for all n>0.

			a(5) = 3 since the polynomial sum_{k=0}^5 B_5*x^{5-k} = x^5+x^4+2*x^3+5*x^2+15*x+52 is irreducible modulo 3 but reducible modulo 2.
Note also that a(7) = 193 < 4*7^2-1 = 195.

Zhi-Wei Sun, Table of n, a(n) for n = 1..300

Cf. A000040, A000110, A220072, A223934, A224210, A224416, A218465, A217785, A217788, A224197.

A[n_,x_]:=A[n,x]=Sum[BellB[k]*x^(n-k),{k,0,n}]
Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[4n^2-2]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

A224418 Least prime q such that sum_{k=0}^n p(k)*x^{n-k} is irreducible modulo q, where p(k) refers to the partition number A000041(k).

Original entry on oeis.org

2, 3, 2, 11, 2, 13, 19, 19, 13, 29, 73, 47, 19, 43, 7, 59, 13, 29, 3, 13, 179, 29, 173, 19, 3, 163, 23, 3, 101, 71, 131, 977, 5, 157, 43, 13, 73, 2, 89, 197, 151, 151, 313, 3, 13, 31, 23, 97, 173, 241, 181, 109, 487, 157, 17, 29, 89, 109, 257, 317

Offset: 1

Zhi-Wei Sun, Apr 06 2013

nonn

Conjecture: a(n) < n^2 for all n > 1.

			a(2) = 3 since sum_{k=0}^2 p(k)*x^{n-k} = x^2 + x + 2  is irreducible modulo 3 but reducible modulo 2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..400

Cf. A000040, A000041, A224417, A224416, A220072, A223934, A224210, A217788, A224197.

A[n_,x_]:=A[n,x]=Sum[PartitionsP[k]*x^(n-k),{k,0,n}]
Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[Max[1,n^2-1]]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

A224480 Least prime q such that x^n + sum_{k=1}^n p_k*x^{n-k} is irreducible modulo q, where p_k denotes the k-th prime.

Original entry on oeis.org

2, 11, 2, 2, 2, 2, 2, 53, 13, 3, 5, 2, 2, 2, 2, 421, 29, 19, 7, 2, 29, 37, 2, 743, 41, 23, 13, 47, 5, 2, 269, 139, 211, 31, 73, 307, 2, 2, 5, 89, 23, 839, 181, 379, 173, 89, 2, 353, 101, 307, 3, 29, 389, 2, 863, 71, 503, 619, 193, 2

Offset: 1

Zhi-Wei Sun, Apr 07 2013

nonn

Conjecture: a(n) <= (n+4)*(n+5)+1 for all n>0.

			a(10) = 3 since P(x) = x^{10} + 2*x^9 + 3*x^8 + 5*x^7 + 7*x^6
+ 11*x^5 + 13*x^4 + 17*x^3 + 19*x^2 + 23*x + 29 is irreducible modulo 3, but reducible modulo 2, for,
    P(x)==(x+1)^2*(x^3+x+1)*(x^5+x^3+1) (mod 2).
Note also that a(16) = 421 = (16+4)*(16+5)+1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..500

Cf. A000040, A220947, A220072, A223934, A224210, A224416, A224417, A224418, A217788, A224197, A218465, A217785.

A[n_,x_]:=A[n,x]=Sum[x^n+Prime[k]*x^(n-k),{k,1,n}]
Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n^2+9n+21]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

A220947 Least prime p such that sum_{k=0}^n F(k+1)*x^{n-k} is irreducible modulo p, where F(j) denotes the Fibonacci number A000045(j).

Original entry on oeis.org

2, 3, 2, 11, 3, 2, 5, 3, 2, 11, 5, 41, 181, 31, 73, 89, 5, 7, 71, 11, 29, 5, 193, 41, 89, 61, 2, 43, 3, 31, 13, 191, 2, 61, 103, 97, 103, 47, 383, 367, 89, 17, 191, 1627, 193, 163, 5, 337, 349, 23, 149, 193, 199, 233, 173, 617, 593, 59, 113, 151

Offset: 1

Zhi-Wei Sun, Apr 07 2013

nonn

Conjecture: a(n) <= n^2+12 for all n>0.

Such a phenomenon happens quite often. In fact, for many interesting integer sequences a(k) (k=1,2,3,...), each of the polynomials x^n + sum_{k=0}^n a(k)*x^{n-k} (n>0) is irreducible modulo some prime not exceeding a*n^2+b*n+c, where a, b, c are suitable nonnegative constants.

			a(2) = 3 since x^2+x+2 is irreducible modulo 3 but reducible modulo 2.
Note also that a(13) = 181 = 13^2+12.

Zhi-Wei Sun, Table of n, a(n) for n = 1..450

Cf. A000040, A000045, A224416, A224417, A224418, A224480, A224210, A220072, A223934, A218465.

A[n_,x_]:=A[n,x]=Sum[Fibonacci[k+1]*x^(n-k),{k,0,n}]
Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n^2+12]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

A220949 Least prime p such that sum_{k=0}^n (2k+1)*x^(n-k) is irreducible modulo p.

Original entry on oeis.org

2, 2, 3, 2, 5, 3, 71, 23, 11, 2, 5, 2, 13, 23, 47, 47, 269, 2, 7, 19, 53, 101, 7, 53, 113, 11, 23, 2, 43, 347, 53, 283, 191, 17, 41, 2, 239, 677, 3, 281, 37, 641, 613, 41, 17, 269, 181, 137, 383, 41, 127, 2, 71, 739, 71, 353, 59, 2, 83, 2

Offset: 1

Zhi-Wei Sun, Apr 07 2013

nonn

Conjecture: a(n) <= n^2+22 for all n>0.

We have similar conjectures with 2k+1 in the definition replaced by (2k+1)^m (m=2,3,...).

			a(5) = 5 since f(x) = x^5+3*x^4+5*x^3+7*x^2+9*x+11 is irreducible modulo 5, but f(x)==(x+1)*(x^2+x+1)^2 (mod 2) and f(x)==(x+1)^4*(x-1) (mod 3).
Note also that a(7) = 71 = 7^2+22.

Zhi-Wei Sun, Table of n, a(n) for n = 1..500
Zhi-Wei Sun, A family of polynomials and a related conjecture on primes, a message to Number Theory List, March 30, 2013.

Cf. A000040, A218465, A224210, A224416, A224417, A224418, A224480, A220072, A223934.

A[n_,x_] := A[n,x] = Sum[(2k+1)*x^(n-k), {k,0,n}]; Do[Do[If[IrreduciblePolynomialQ[A[n,x], Modulus->Prime[k]] == True, Print[n," ",Prime[k]]; Goto[aa]], {k,1,PrimePi[n^2+22]}]; Print[n," ",counterexample]; Label[aa]; Continue,{n,1,100}]

cp's OEIS Frontend

A224197 Least b > p_n^2 such that [p_1^2,p_2^2,...,p_n^2] in base b is prime, where p_j denotes the j-th prime.

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A224416 Least prime p such that the polynomial sum_{k=0}^n C_k*x^{n-k} is irreducible modulo p, where C_k denotes the Catalan number binomial(2k,k)/(k+1).

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A224417 Least prime p such that sum_{k=0}^n B_k*x^{n-k} is irreducible modulo p, where B_k refers to the Bell number A000110(k).

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A224418 Least prime q such that sum_{k=0}^n p(k)*x^{n-k} is irreducible modulo q, where p(k) refers to the partition number A000041(k).

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A224480 Least prime q such that x^n + sum_{k=1}^n p_k*x^{n-k} is irreducible modulo q, where p_k denotes the k-th prime.

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A220947 Least prime p such that sum_{k=0}^n F(k+1)*x^{n-k} is irreducible modulo p, where F(j) denotes the Fibonacci number A000045(j).

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A220949 Least prime p such that sum_{k=0}^n (2k+1)*x^(n-k) is irreducible modulo p.

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