A130281 -id:A130281

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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.

+10
8

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

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

LINKS

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

FORMULA

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.

EXAMPLE

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).

MAPLE

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;

MATHEMATICA

$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 *)

PROG

(PARI) {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)))}

CROSSREFS

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

See also A306767.

KEYWORD

easy,nonn

AUTHOR

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

STATUS

approved

A130284

Integers j > 0 such that (2j+1)^2(m^2-1) + 1 is a square for some integer m > 1.

+10
5

7, 17, 31, 49, 71, 97, 104, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 594, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1455, 1457, 1567, 1681, 1799, 1921, 1952, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

All terms > 4 in A130283 are odd squares, but not all odd squares are in that sequence: This sequence here gives the exceptions as (2a(n)+1)^2. The sequence consists mainly of the subsequences: (1) A056220(k) = 2k^2-1 with k>1: {7,17,31,49,...}, for which m=k gives (1+2*A056220(k))^2(k^2-1)+1 = k^2(4k^2-3)^2; (2) 2*A079414(k) = 2k^2(4k^2-3) with k>1: {104,594,1952,4850,...}, for which m=k gives (1+4*A079414(k))^2(k^2-1)+1 = k^2(16k^4-20k^2+5)^2. A third subsequence starts {1455,20195,...}; up to 20195, all terms are in one of these subsequences.

LINKS

Table of n, a(n) for n=1..48.

FORMULA

A130284 = { P[k](m) ; k=1,2,3,..., m=2,3,4,... } where P[k] = (sqrt((X^2 Q[k]^2 - 1)/(X^2 - 1))-1)/2 and Q[0] = Q[-1] = 1, Q[k+1] = (4X^2 -2)*Q[k] - Q[k-1]. Furthermore, (2P[k](m)+1)^2 (m^2 - 1)+1 = m^2 Q[k](m)^2, thus A130280(P[k](m)) <= m. So far, no case is known where we have strict inequality.

EXAMPLE

Up to k=17, a(k)=P[1](k+1) with P[1] = 2x^2 - 1, A130280(a(k)) = k+1.

a(18) = P[2](2) < P[1](19) with P[2] = 2x^2*(4x^2 - 3), A130280(a(18)) = 2.

a(106) = P[1](100) < a(107) = P[3](3) < a(108) = P[4](2) < a(109) = P[1](101).

MATHEMATICA

r[n_] := Reduce[m>1 && k>1 && (2n+1)^2*(m^2-1)+1 == k^2, {m, k}, Integers];

Reap[For[n=1, n <= 5000, n++, If[r[n] =!= False, Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, May 12 2017 *)

PROG

(PARI) A130284( LIM=9999, START=1 )={ local(N); for( n=START, LIM, N=(2*n+1)^2; for( m=2, sqrtint(n>>1+1), if(!issquare( N*(m^2-1)+1 ), next); print1(n", "); next(2))) }

(PARI) {Q(k, x=x)=if(m>0, (4*x^2-2)*Q(k-1, x)-Q(k-2, x), 1)} {P(k, x=x)=if(type(x=(x^2*Q(k, x)^2-1)/(x^2-1))!="t_POL", sqrtint(x)\2, ((-1)^k*Pol(sqrt(x))-1)/2)}

CROSSREFS

Cf. A084702, A130280, A130284, A130288.

Cf. A130280, A130283, A130281.

KEYWORD

nonn

AUTHOR

M. F. Hasler, May 24 2007, May 29 2007

STATUS

approved

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