A199773
y-values in the solution to 17*x^2 - 16 = y^2.
Original entry on oeis.org
1, 16, 103, 169, 1072, 6799, 11153, 70736, 448631, 735929, 4667504, 29602847, 48560161, 307984528, 1953339271, 3204234697, 20322311344, 128890789039, 211430929841, 1340964564176, 8504838737303, 13951237134809, 88483338924272, 561190465872959, 920570219967553
Offset: 1
-
I:=[1,16,103,169,1072,6799]; [n le 6 select I[n] else 66*Self(n-3)-Self(n-6): n in [1..30]]; // Vincenzo Librandi, Jan 06 2016
-
LinearRecurrence[{0,0,66,0,0,-1}, {1,16,103,169,1072,6799}, 50]
CoefficientList[Series[(x + 1) (x^4 + 15 x^3 + 88 x^2 + 15 x + 1) / (x^6 - 66 x^3 + 1), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 06 2016 *)
-
Vec(x*(x+1)*(x^4+15*x^3+88*x^2+15*x+1)/(x^6-66*x^3+1) + O(x^100)) \\ Colin Barker, Sep 01 2013
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
-
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
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
-
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
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
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.
-
# 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]);
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
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.
-
# 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]);
Showing 1-5 of 5 results.
Comments