Nikolay Osipov - cp's OEIS frontend

User: Nikolay Osipov

Nikolay Osipov has authored 3 sequences.

A376008 Primes p such that there exists a cyclic permutation of the nonzero residues modulo p such that v^2 - 4uw == 0 (mod p) for any three consecutive residues u,v,w.

3, 17, 251, 257, 433, 641, 1459, 3457, 3889, 21169, 39367, 54001, 65537, 110251, 114689, 139969, 210913, 246241, 274177, 319489, 629857, 746497, 974849, 995329, 1161217, 1299079, 1492993, 1769473, 2020001, 2424833, 2555521, 2654209, 5038849, 5304641, 5419387, 5746001, 6049243, 6561001

Offset: 1

Nikolay Osipov, Dmitry Petukhov, and Max Alekseyev, Sep 05 2024

nonn

In other words, for any three consecutive residues u,v,w, the quadratic polynomial u*x^2 + v*x + w has zero discriminant modulo p.

It is shown that all suitable permutations q for prime p = a(n) can be constructed by starting with q(1) = 1, q(2) = a primitive root modulo p, and then defining q(k) = q(k-1)^2/(4*q(k-2)) mod p for k >= 3. Hence, the number of suitable permutations (up to cyclic rotations) is given by A046144(a(n)).

			For a(2) = 17, a suitable cyclic permutation is (1, 3, 15, 6, 4, 12, 9, 7, 16, 14, 2, 11, 13, 5, 8, 10).

Max Alekseyev, Table of n, a(n) for n = 1..100
N. Osipov et al., Residues on a circle (in Russian), dxdy.ru, 2024.

Contains Fermat primes (A019434) as a subsequence.

Cf. A001918, A002326, A046144, A376010.

forprime(p=3,10^8, s=(p-1)/znorder(Mod(2,p)); if(factor(p-1)[,1]==factor(2*s)[,1] && !(p%4==1 && s%2==1),print1(p,", ")) );

An odd prime p is a term iff for s:=(p-1)/A002326((p-1)/2), radicals of p-1 and 2s coincide, excluding the case p==1 (mod 4) and s==1 (mod 2).

A371957 Positive integers k such that the parametric Pell-type equation x^2 - mxy + y^2 = - m^2 - k has no integer solutions (x,y) for all integers m >= 1, excluding the cases k==1 (mod 4), k==3 (mod 9), and k==6 (mod 9).

Original entry on oeis.org

4, 8, 14, 18, 19, 26, 38, 44, 47, 54, 63, 68, 74, 79, 98, 99, 103, 110, 118, 119, 124, 126, 134, 143, 144, 154, 158, 166, 179, 180, 194, 198, 199, 206, 207, 208, 214, 215, 224, 234, 238, 239, 250, 254, 263, 274, 278, 279, 287, 299, 306, 308, 314, 319, 324, 326, 334, 342, 351, 359, 362, 368, 374, 378, 383, 404, 406, 414, 418

Offset: 1

Orlov Nikita and Nikolay Osipov, Apr 14 2024

nonn

For k == 1 (mod 4), k == 3 (mod 9), or k == 6 (mod 9), the equation x^2 - m*x*y + y^2 = - m^2 - k (*) has no integer solutions modulo 4 or 9. For other positive integer k, it suffice to check that the equation (*) has no integer solutions (x,y) for all integers m with 3<=m<=2*k+8 (see references for the proof of similar assertions). This condition can be verified by an algorithm similar to brute-force search for the general Pell equation x^2 - Dy^2 = N (see, for example, sect. 4.4.5 in: Andreescu T., Andrica D. Quadratic Diophantine Equations. New York: Springer, 2015).

For any other positive integer k, the equation (*) has integer solutions (x,y) for infinitely many integers m >= 1.

N. Osipov, A Pell-Type Diophantine Equation, Amer. Math. Monthly, 128 (2021), p. 858-860.
N. Osipov, A Pell-type Equation in Disguise, Amer. Math. Monthly, 129 (2022), p. 389-390.

Orlov Nikita, Pascal program.

Cf. A370721 (for the equation x^2-m*x*y+y^2=m^2+k)

check:=proc(k) local flag,m,y,mm,yy; flag:=0;
for m from 3 to 2*k+8 while flag=0 do
for y from 1 to evalf(sqrt((m^2+k)/(m-2)))+1 while flag=0 do
if issqr((m^2-4)*y^2-4*(m^2+k))=true then flag:=1; mm:=m; yy:=y; fi; od; od;
if flag=0 then return 0 else return [mm,yy]; fi; end proc:
for k from 1 to 2000 do
if k mod 4<>1 and k mod 9<>3 and k mod 9<>6 and check(k)=0 then print(k); fi; od:

Pascal
```
// see link
```

Edited by Nikolay Osipov, Jun 11 2024

A370721 Positive integers k == 2 (mod 4) such that the parametric Pell-type equation x^2 - mxy + y^2 = m^2 + k has no integer solutions (x,y) for all integer m >= 1.

Original entry on oeis.org

14, 94, 114, 118, 154, 158, 214, 238, 254, 294, 358, 414, 478, 574, 594, 598, 614, 654, 658, 694, 718, 758, 790, 814, 834, 862, 874, 878, 934, 958, 994, 1014, 1054, 1106, 1174, 1198, 1294, 1414, 1434, 1454, 1486, 1494, 1498, 1558, 1634, 1678, 1738, 1774, 1794, 1834, 1894, 1918, 1978

Offset: 1

Orlov Nikita and Nikolay Osipov, Mar 07 2024

nonn

For a positive integer k == 2 (mod 4), it suffice to check that the equation x^2-m*x*y+y^2 = m^2+k (*) has no integer solutions (x,y) for all integer m with 1 <= m <= k/2 (see references for the proof of some similar assertions). This condition can be verified by an algorithm similar to brute force search for the general Pell equation x^2-Dy^2 = N (see, for example, sect. 4.4.5 in: Andreescu T., Andrica D. Quadratic Diophantine Equations. New York: Springer, 2015).

Also, the equation (*) has no integer solutions (x,y) for all integer m >= 1 when k = 1 or k = 4. For any other positive integer k, the equation (*) has integer solutions (x,y) for infinitely many integers m >= 1.

N. Osipov, A Pell-Type Diophantine Equation, Amer. Math. Monthly, 128 (2021), p. 858-860.
N. Osipov, A Pell-type Equation in Disguise, Amer. Math. Monthly, 129 (2022), p. 389-390.

Orlov Nikita, Pascal program.

Cf. A371957 (for the equation x^2-m*x*y+y^2=-m^2-k).

check:=proc(k) local flag,y,m,yy,mm; flag:=0;
for y from 0 to evalf(2*sqrt((k+1)/3)+1) while flag=0 do
if issqr(-3*y^2+4*k+4)=true then flag:=1; mm:=1; yy:=y; fi; od;
for m from 3 to k/2 while flag=0 do
if m mod 4<>2 then for y from 0 to evalf(sqrt((m^2+k)/(m+2)))+1 while flag=0 do
if issqr((m^2-4)*y^2+4*(m^2+k))=true then flag:=1; mm:=m; yy:=y; fi; od; fi; od;
if flag=0 then return 0 else return [mm,yy]; fi; end proc:
for k from 1 to 2000 do if k mod 4=2 and check(k)=0 then print(k); fi; od:

Pascal
```
(* see link *)
```

Edited by Nikolay Osipov, Jun 11 2024

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User: Nikolay Osipov

A376008 Primes p such that there exists a cyclic permutation of the nonzero residues modulo p such that v^2 - 4uw == 0 (mod p) for any three consecutive residues u,v,w.

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A371957 Positive integers k such that the parametric Pell-type equation x^2 - mxy + y^2 = - m^2 - k has no integer solutions (x,y) for all integers m >= 1, excluding the cases k==1 (mod 4), k==3 (mod 9), and k==6 (mod 9).

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A370721 Positive integers k == 2 (mod 4) such that the parametric Pell-type equation x^2 - mxy + y^2 = m^2 + k has no integer solutions (x,y) for all integer m >= 1.

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cp's OEIS Frontend

User: Nikolay Osipov

A376008 Primes p such that there exists a cyclic permutation of the nonzero residues modulo p such that v^2 - 4*u*w == 0 (mod p) for any three consecutive residues u,v,w.

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Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A371957 Positive integers k such that the parametric Pell-type equation x^2 - m*x*y + y^2 = - m^2 - k has no integer solutions (x,y) for all integers m >= 1, excluding the cases k==1 (mod 4), k==3 (mod 9), and k==6 (mod 9).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Pascal

Extensions

A370721 Positive integers k == 2 (mod 4) such that the parametric Pell-type equation x^2 - m*x*y + y^2 = m^2 + k has no integer solutions (x,y) for all integer m >= 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Pascal

Extensions

A376008 Primes p such that there exists a cyclic permutation of the nonzero residues modulo p such that v^2 - 4uw == 0 (mod p) for any three consecutive residues u,v,w.

A371957 Positive integers k such that the parametric Pell-type equation x^2 - mxy + y^2 = - m^2 - k has no integer solutions (x,y) for all integers m >= 1, excluding the cases k==1 (mod 4), k==3 (mod 9), and k==6 (mod 9).

A370721 Positive integers k == 2 (mod 4) such that the parametric Pell-type equation x^2 - mxy + y^2 = m^2 + k has no integer solutions (x,y) for all integer m >= 1.