A130282 -id:A130282 - cp's OEIS frontend

A130280 a(n) = smallest integer k>1 such that n(k^2-1)+1 is a perfect square, or 0 if no such number exists.

2, 5, 3, 0, 2, 3, 5, 2, 0, 3, 7, 5, 4, 11, 3, 2, 4, 13, 9, 7, 2, 5, 19, 4, 0, 5, 21, 3, 11, 9, 11, 14, 2, 29, 5, 3, 6, 31, 21, 2, 13, 11, 13, 169, 3, 7, 41, 6, 0, 7, 5, 11, 22, 419, 3, 2, 5, 23, 461, 27, 8, 55, 7, 4, 2, 3, 49, 29

Offset: 1

M. F. Hasler, May 20 2007, May 25 2007

A084702(n) = a(n)^2-1, resp. a(n) = sqrt(A084702(n)+1). See A130283 for values where A130280(n)=0.

			a( (2k)^2 ) <= k since (2k)^2(k^2-1)+1 = (2k^2-1)^2 (but k=1 is excluded since with k^2-1=0 this would be a trivial solution for any n).

M. F. Hasler, Table of n, a(n) for n = 1..1000

Cf. A084702, A094357, A130281, A130282, A130283, A130284.

See also A306767.

A130280:=proc(n) local x,y,z; if n=1 then return 2 fi; isolve(n*(x^2-1)+1=y^2,z); select(has,`union`(%),x); map(rhs,%); simplify(eval(%,z=1) union eval(%,z=0)) minus {-1,1}; if %={} then 0 else (min@op@map)(abs,%) fi end;

$MaxExtraPrecision = 100;
r[n_, c_] := Reduce[k > 1 && j > 1 && n*(k^2 - 1) + 1 == j^2, {j, k}, Integers] /. C[1] -> c // Simplify;
a[n_] := If[rn = r[n,0] || r[n,1] || r[n,2]; rn === False, 0, k /. {ToRules[rn]} // Min];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 800}] (* Jean-François Alcover, May 12 2017 *)

{A130280(n,L=10^15)=if(issquare(n),L=2+sqrtint(n>>2)); for( k=2, L, if( issquare(n*(k^2-1)+1),return(k)))}

If n=(2k)^2, then A130280(n) <= k, since (2k)^2(k^2-1)+1 = (2k^2-1)^2. See A130281 for the cases where equality does not hold. If n=k^2-1, then A130280(n) <= k-1 since (k^2-1)((k-1)^2-1)+1 = (k^2-k-1)^2. See A130282 for the cases where equality does not hold.

A144743 Recurrence sequence a(n)=a(n-1)^2-a(n-1)-1, a(0)=3.

Original entry on oeis.org

3, 5, 19, 341, 115939, 13441735781, 180680260792773944179, 32645356640144805339284259388335434039861, 1065719310162246533488642668727242229836148490441005113524301742665845135502859459

Offset: 0

Artur Jasinski, Sep 20 2008

nonn

a(0)=3 is the smallest integer generating an increasing sequence of the form a(n)=a(n-1)^2-a(n-1)-1.

Conjecture: A130282 and this sequence are disjoint. If this is true, for n >= 1, a(n+1) is the smallest m such that (m^2-1) / (a(n)^2-1) + 1 is a square. - Jianing Song, Mar 19 2022

Cf. A000058, A082732, A144744, A144745, A144746, A144747, A144748.

a = {3}; k = 3; Do[k = k^2 - k - 1; AppendTo[a, k], {n, 1, 10}]; a

a(n,s=3)=for(i=1,n,s=s^2-s-1);s \\ M. F. Hasler, Oct 06 2014

a(n) = a(n-1)^2-a(n-1)-1, a(0)=3.

a(n) ~ c^(2^n), where c = 2.07259396780115004655284076205241023281287049774423620992171834046728756... . - Vaclav Kotesovec, May 06 2015

Edited by M. F. Hasler, Oct 06 2014

cp's OEIS Frontend

A130280 a(n) = smallest integer k>1 such that n(k^2-1)+1 is a perfect square, or 0 if no such number exists.

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