A199773 -id:A199773 - cp's OEIS frontend

A199772 x-values in the solution to 17*x^2 - 16 = y^2.

1, 4, 25, 41, 260, 1649, 2705, 17156, 108809, 178489, 1132036, 7179745, 11777569, 74697220, 473754361, 777141065, 4928884484, 31260608081, 51279532721, 325231678724, 2062726378985, 3383672018521, 21460361911300, 136108680404929, 223271073689665

Offset: 1

Sture Sjöstedt, Nov 10 2011

When are both n+1 and 17*n+1 perfect squares? This problem gives the equation 17*x^2-16=y^2.

			a(7) = 66*41-1 = 2705.

Index entries for linear recurrences with constant coefficients, signature (0,0,66,0,0,-1).

Cf. A199773, A199774, A199798.

LinearRecurrence[{0,0,66,0,0,-1}, {1,4,25,41,260,1649}, 50]

Vec(-x*(x-1)*(x^4+5*x^3+30*x^2+5*x+1)/(x^6-66*x^3+1) + O(x^100)) \\ Colin Barker, Sep 01 2013

a(n) = 66*a(n-3) - a(n-6), a(1)=1, a(2)=4, a(3)=25, a(4)=41, a(5)=260, a(6)=1649.

G.f.: -x*(x-1)*(x^4+5*x^3+30*x^2+5*x+1) / (x^6-66*x^3+1). - Colin Barker, Sep 01 2013

More terms from T. D. Noe, Nov 10 2011

A199774 x-values in the solution to 17*x^2 + 16 = y^2.

Original entry on oeis.org

0, 3, 5, 32, 203, 333, 2112, 13395, 21973, 139360, 883867, 1449885, 9195648, 58321827, 95670437, 606773408, 3848356715, 6312798957, 40037849280, 253933221363, 416549060725, 2641891279072, 16755744253243, 27485925208893, 174324786569472, 1105625187492675

Offset: 1

Sture Sjöstedt, Nov 10 2011

When are both n-1 and 17*n-1 perfect squares? This problem gives the equation 17*x^2+16=y^2.

			a(7)=66*32-0=2112.

Index entries for linear recurrences with constant coefficients, signature (0,0,66,0,0,-1).

Cf. A199772, A199773, A199798.

LinearRecurrence[{0,0,66,0,0,-1}, {0,3,5,32,203,333}, 50]

Vec(x^2*(3*x^4+5*x^3+32*x^2+5*x+3)/(x^6-66*x^3+1) + O(x^100)) \\ Colin Barker, Sep 01 2013

a(n) = 66*a(n-3) - a(n-6), a(1)=0, a(2)=3, a(3)=5, a(4)=32, a(5)=203, a(6)=333.

G.f.: x^2*(3*x^4+5*x^3+32*x^2+5*x+3) / (x^6-66*x^3+1). - Colin Barker, Sep 01 2013

More terms from T. D. Noe, Nov 10 2011

A199798 y-values in the solution to 17*x^2 + 16 = y^2.

Original entry on oeis.org

4, 13, 21, 132, 837, 1373, 8708, 55229, 90597, 574596, 3644277, 5978029, 37914628, 240467053, 394459317, 2501790852, 15867181221, 26028336893, 165080281604, 1046993493533, 1717475775621, 10892796795012, 69085703391957, 113327372854093, 718759508189188

Offset: 1

Sture Sjöstedt, Nov 10 2011

When are both n-1 and 17*n-1 perfect squares? This problem gives the equation 17*x^2+16=y^2.

			a(7)=66*132-4=8708.

Index entries for linear recurrences with constant coefficients, signature (0,0,66,0,0,-1).

Cf. A199774, A199772, A199773.

LinearRecurrence[{0,0,66,0,0,-1}, {4,13,21,132,837,1373}, 50]

Vec(-x*(13*x^5+21*x^4+132*x^3-21*x^2-13*x-4)/(x^6-66*x^3+1) + O(x^100)) \\ Colin Barker, Sep 01 2013

a(n) = 66*a(n-3) - a(n-6), a(1)=4, a(2)=13, a(3)=21, a(4)=132, a(5)=837, a(6)=1373.

G.f.: -x*(13*x^5+21*x^4+132*x^3-21*x^2-13*x-4) / (x^6-66*x^3+1). - Colin Barker, Sep 01 2013

More terms from T. D. Noe, Nov 10 2011

A200409 The y-values in the solution to 19*x^2 - 18 = y^2.

Original entry on oeis.org

1, 39, 571, 911, 13299, 194141, 309739, 4521621, 66007369, 105310349, 1537337841, 22442311319, 35805208921, 522690344319, 7630319841091, 12173665722791, 177713179730619, 2594286303659621, 4139010540540019, 60421958418066141, 882049712924430049

Offset: 1

Sture Sjöstedt, Nov 17 2011

When are both n+1 and 19*n+1 perfect squares? This gives the equation 19*x^2 - 18 = y^2.

			a(7) = 340*911 - 1 = 309739.

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (0,0,340,0,0,-1).

Cf. A200407, A199773, A199774, A199798.

I:=[1, 39, 571, 911, 13299, 194141]; [n le 6 select I[n] else 340*Self(n-3)-Self(n-6): n in [1..30]]; // Vincenzo Librandi, Nov 18 2011

LinearRecurrence[{0, 0, 340, 0, 0, -1}, {1, 39, 571, 911, 13299,194141}, 50]

Vec(x*(x+1)*(x^4+38*x^3+533*x^2+38*x+1)/(x^6-340*x^3+1) + O(x^100)) \\ Colin Barker, Sep 01 2013

a(n) = 340*a(n-3) - a(n-6), a(1)=1, a(2)=39, a(3)=571, a(4)=911, a(5)=13299, a(6)=194141.

G.f.: x*(x+1)*(x^4 + 38*x^3 + 533*x^2 + 38*x + 1) / (x^6 - 340*x^3 + 1). - Colin Barker, Sep 01 2013

A350544 a(n) is the least prime p such that there exists a prime q with p^2 + n = (n+1)*q^2, or 0 if there is no such p.

Original entry on oeis.org

7, 5, 0, 11, 7, 13, 5, 0, 41, 23, 17, 10496997797584752004430879, 41, 11, 7

Offset: 1

J. M. Bergot and Robert Israel, Jan 04 2022

a(16) > 10^1000 if it is not 0.

If it is not 0, then a(16) = A199773(k) where k is the smallest index such that both p = A199773(k) and q = A199772(k) are prime. If such an index exists, a(16) > 10^10000. - Jon E. Schoenfield, Jan 11 2022

			a(3) = 0 as the only positive integer solution of p^2 + 3 = 4*q^2 is p=1, q=1, and 1 is not prime.
a(4) = 11 as 11^2 + 4 = 125 = (4+1)*5^2 with 11 and 5 prime.

Robert Israel, Table of n, a(n) for n = 1 .. 179 with conjectured 0 values as -1.

Cf. A199772, A199773, A350550.

# Returned values of -1 indicate that either a(n) = 0 or a(n) > 10^1000.
f:= proc(n) local m,x,y,S,cf,i,c,a,b,A,M,Sp;
  m:= n+1;
  if issqr(m) then
    S:= [isolve(x^2+n=m*y^2)];
    S:= map(t -> subs(t,[x,y]),S);
    S:= select(t -> andmap(isprime,t),S);
    if S = [] then return 0
    else return min(map(t -> t[1],S))
    fi;
  fi;
  cf:= NumberTheory:-ContinuedFraction(sqrt(m));
  for i from 1 do
    c:= Convergent(cf,i);
    if numer(c)^2 - m*denom(c)^2 = 1 then break fi
  od;
  a:= numer(c); b:= denom(c);
  A:= <|>;
  M:= floor(sqrt(n)*(1+sqrt(a+b*sqrt(m)))/(2*sqrt(m)));
  S:= select(t -> issqr(m*t^2-m+1), [$0..M]);
  S:= select(t -> igcd(t[1],t[2])=1,map(t -> , S));
  S:= map(t -> (t, <-t[1],t[2]>), S);
  if nops(S) = 0 then return 0 fi;
  for i from 0 do
    Sp:= select(t -> isprime(t[1]) and isprime(t[2]),S);
    if nops(Sp)>0 then return min(map(t -> t[1],Sp)) fi;
    S:= map(t -> A.t,S);
    if min(map(t -> t[1],S))>10^1000 then break fi;
  od;
  -1
end proc:
map(f, [$1..20]);

a(n)^2 + n = (n+1)*A350550(n)^2 if a(n) > 0.

A350550 a(n) is the least prime q such that there exists a prime p with p^2 + n = (n+1)*q^2, or 0 if there is no such q.

Original entry on oeis.org

5, 3, 0, 5, 3, 5, 2, 0, 13, 7, 5, 2911343369048029930623841, 11, 3, 2

Offset: 1

J. M. Bergot and Robert Israel, Jan 04 2022

a(16) > 10^1000 if it is not 0.

If it is not 0, then a(16) = A199772(k) where k is the smallest index such that both q = A199772(k) and p = A199773(k) are prime. If such an index exists, a(16) > 10^10000. - Jon E. Schoenfield, Jan 11 2022

			a(3) = 0 as the only positive integer solution of p^2 + 3 = 4*q^2 is p=1, q=1, and 1 is not prime.
a(4) = 5 as 11^2 + 4 = 125 = (4+1)*5^2 with 11 and 5 prime.

Robert Israel, Table of n, a(n) for n = 1..179 with conjectured 0 values as -1.

Cf. A199772, A199773, A350544.

# Returned values of -1 indicate that either a(n) = 0 or a(n) > 10^1000.
f:= proc(n) local m, x, y, S, cf, i, c, a, b, A, M, Sp;
  m:= n+1;
  if issqr(m) then
    S:= [isolve(x^2+n=m*y^2)];
    S:= map(t -> subs(t, [x, y]), S);
    S:= select(t -> andmap(isprime, t), S);
    if S = [] then return 0
    else return min(map(t -> t[2], S))
    fi;
  fi;
  cf:= NumberTheory:-ContinuedFraction(sqrt(m));
  for i from 1 do
    c:= Convergent(cf, i);
    if numer(c)^2 - m*denom(c)^2 = 1 then break fi
  od;
  a:= numer(c); b:= denom(c);
  A:= <|>;
  M:= floor(sqrt(n)*(1+sqrt(a+b*sqrt(m)))/(2*sqrt(m)));
  S:= select(t -> issqr(m*t^2-m+1), [$0..M]);
  S:= select(t -> igcd(t[1], t[2])=1, map(t -> , S));
  S:= map(t -> (t, <-t[1], t[2]>), S);
  if nops(S) = 0 then return 0 fi;
  for i from 0 do
    Sp:= select(t -> isprime(t[1]) and isprime(t[2]), S);
    if nops(Sp)>0 then return min(map(t -> t[2], Sp)) fi;
    S:= map(t -> A.t, S);
    if min(map(t -> t[2], S))>10^1000 then break fi;
  od;
  -1
end proc:
map(f, [$1..20]);

A350544(n)^2 + n = (n+1)*a(n)^2 if a(n) > 0.

A200407 The x-values in the solution to 19*x^2 - 18 = y^2.

Original entry on oeis.org

1, 9, 131, 209, 3051, 44539, 71059, 1037331, 15143129, 24159851, 352689489, 5148619321, 8214278281, 119913388929, 1750515426011, 2792830455689, 40770199546371, 595170096224419, 949554140655979, 13861747932377211, 202356082200876449, 322845614992577171

Offset: 1

Sture Sjöstedt, Nov 17 2011

When are both n+1 and 19*n+1 perfect squares? This gives the equation 19*x^2-18=y^2.

			a(7)=340*209-1=71059.

Index entries for linear recurrences with constant coefficients, signature (0,0,340,0,0,-1).

Cf. A199773, A199774, A199798, A200409.

LinearRecurrence[{0, 0, 340, 0, 0, -1}, {1, 9, 131, 209, 3051, 44539}, 50]

Vec(-x*(x-1)*(x^4+10*x^3+141*x^2+10*x+1)/(x^6-340*x^3+1) + O(x^100)) \\ Colin Barker, Sep 01 2013

a(n) = 340*a(n-3)+a(n-6), a(1)=1, a(2)=9, a(3)=131, a(4)=209, a(5)=3051, a(6)=44539.

G.f.: -x*(x-1)*(x^4+10*x^3+141*x^2+10*x+1) / (x^6-340*x^3+1). - Colin Barker, Sep 01 2013

cp's OEIS Frontend

A199772 x-values in the solution to 17*x^2 - 16 = y^2.

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A199774 x-values in the solution to 17*x^2 + 16 = y^2.

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A199798 y-values in the solution to 17*x^2 + 16 = y^2.

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A200409 The y-values in the solution to 19*x^2 - 18 = y^2.

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A350544 a(n) is the least prime p such that there exists a prime q with p^2 + n = (n+1)*q^2, or 0 if there is no such p.

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A350550 a(n) is the least prime q such that there exists a prime p with p^2 + n = (n+1)*q^2, or 0 if there is no such q.

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A200407 The x-values in the solution to 19*x^2 - 18 = y^2.

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