A245226 -id:A245226 - cp's OEIS frontend

A382209 Numbers k such that 10+k and 10*k are perfect squares.

90, 136890, 197402490, 284654260890, 410471246808090, 591899253243012090, 853518312705176632890, 1230772815021611461622490, 1774773545742851022483004890, 2559222222188376152809031436090, 3690396669622092669499600847844090, 5321549438372835441042271613559748890

Offset: 1

Emilio Martín, Mar 18 2025

The limit of a(n+1)/a(n) is 1441.99930651839... = 721+228*sqrt(10) = (19+6*sqrt(10))^2.

If 10*A158490(n) is a perfect square, then A158490(n) is a term.

			90 is a term because 10+90=100 is a square and 10*90=900 is a square.
(3,1) is a solution to x^2 - 10*y^2 = -1, from which a(n) = 100*y^2-10 = 10*x^2 = 90.

Emilio Martín, Table of n, a(n) for n = 1..100
Wikipedia, Negative Pell equation (in German)
Wikipedia, Pell's equation
Index entries for linear recurrences with constant coefficients, signature (1443,-1443,1).

Subsequence of A158490.
Cf. A097315, A005667, A173127, A081071.
Cf. A383734 = 2*A008843 (2+k and 2*k are squares).
Cf. 5*A075796^2 (5+k and 5*k are squares).
Cf. 5*A081071 (20+k and 20*k are squares).
Cf. A245226 (m such that k+m and k*m are squares).

CoefficientList[Series[ 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)),{x,0,11}],x] (* or *) LinearRecurrence[{1443,-1443,1},{90,136890,197402490},12] (* James C. McMahon, May 08 2025 *)

from itertools import islice
def A382209_gen(): # generator of terms
    x, y = 30, 10
    while True:
        yield x**2//10
        x, y = x*19+y*60, x*6+y*19
A382209_list = list(islice(A382209_gen(),30)) # Chai Wah Wu, Apr 24 2025

a(n) = 10 * ((1/2) * (3+sqrt(10))^(2*n-1) + (1/2) * (3-sqrt(10))^(2*n-1))^2.
a(n) = 10 * (sinh((2n-1) * arcsinh(3)))^2.
a(n) = 10 * A173127(n)^2 = 100 * A097315(n)^2 - 10 (negative Pell's equation solutions).
a(n+2) = 1442 * a(n+1) - a(n) + 7200.
G.f.: 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)). - Stefano Spezia, Apr 24 2025

A248194 Positive integers n such that the equation x^2 - ny^2 = n(n+1)/2 has integer solutions.

Original entry on oeis.org

1, 3, 8, 9, 17, 19, 24, 25, 33, 49, 51, 57, 67, 72, 73, 81, 88, 89, 96, 97, 99, 121, 129, 136, 147, 152, 163, 169, 177, 179, 193, 201, 211, 225, 233, 241, 243, 249, 264, 288, 289, 297, 313, 337, 339, 352, 361, 369, 387, 393, 408, 409, 441, 449, 451, 456, 457

Offset: 1

Colin Barker, Oct 03 2014

nonn

All odd squares are in this sequence. Proof: Set n = (2k + 1)^2, then we have x^2 - (2k + 1)^2 * y^2 = (2k + 1)^2 * (2k^2 + 2k + 1). Rearranging gives x^2 = (2k + 1)^2 * (y^2 + 2k^2 + 2k + 1). As 2k^2 + 2k + 1 is odd, a careful selection of y makes the RHS square. So [(2k+1) * (k(k + 1) + 1), k(k + 1)]. E.g., if k=2, then (5*7)^2 - 25*6^2 = 1225 - 900 = 325 = 25*26/2. - Jon Perry, Nov 07 2014
No even squares are in the sequence. Proof: Rearrange the equation to read x^2 = n(n + 1 + 2y^2)/2, with n = 4k^2. n + 1 + 2y^2 is always odd and so the RHS contains an odd exponent of 2, and therefore cannot be square. - Jon Perry, Nov 15 2014
From Jon Perry, Nov 15 2014: (Start)
Odd squares + 8 are always in this sequence. Proof: Let m = 4k^2 + 4k + 9 and let n = (m+1)/2 = 2k^2 + 2k + 5.
Rearranging the equation x^2 - m*y^2 = m(m + 1)/2, we get x^2  = m(m + 1 + 2y^2)/2, and so x^2 = m(n + y^2) = (4k^2 + 4k + 9)(2k^2 + 2k + 5 + y^2).
We aim to find a y such that the last bracket on the RHS is z^2 * (4k^2 + 4k + 9), so that x equals z*m. We claim that if we let Y = ((n-3)/2)^2*m - n, then Y is a square, and letting Y = y^2, we have y^2 + n = Y + n = z^2 * m as required, with z = (n-3)/2 = k^2 + k + 1.
To prove that Y is a square, Y = [(n^2 - 6n + 9)*(2n - 1) - 4n]/4 = [2n^3 - 13n^2 + 20n - 9]/4 = [(n-1)^2*(2n-9)]/4, and with n as it is, 2n - 9 = 4k^2 + 4k + 1 = (2k + 1)^2, and so we arrive at Y = [(n-1)^2*(2k+1)^2]/4 = [(n-1)(2k+1)/2]^2 = [(k^2 + k + 2)(2k + 1)]^2, a square as required, with y = (k^2 + k + 2)(2k + 1). Also GCD(n-3,2n-1)=1 as required.
This gives a solution as [(k^2 + k + 1)*(4k^2 + 4k + 9), (k^2 + k + 2)*(2k + 1)]. E.g., if k=4, n=45 and a solution is [21*89, 22*9] = [1869, 198]. To validate, 1869^2 - 89*198^2 = 3493161 - 3489156 = 4005 = 89*45.
(End)

			3 is in the sequence because x^2 - 3*y^2 = 6 has integer solutions, including (x, y) = (3, 1) and (9, 5).

Lars Blomberg, Table of n, a(n) for n = 1..10000

Cf. A134419, A245226.

More terms from Lars Blomberg, Nov 02 2014

"Positive" added to definition by N. J. A. Sloane, Nov 02 2014

A383734 Numbers k such that 2+k and 2*k are squares.

Original entry on oeis.org

2, 98, 3362, 114242, 3880898, 131836322, 4478554082, 152139002498, 5168247530882, 175568277047522, 5964153172084898, 202605639573839042, 6882627592338442562, 233806732499933208098, 7942546277405390632802, 269812766699283348307202, 9165691521498228451812098

Offset: 1

Emilio Martín, May 07 2025

The limit of a(n+1)/a(n) is 33.97056... = 17+12*sqrt(2) = (3+2*sqrt(2))^2 (see A156164).

			98 is a term becouse 98+2=100 is a square and 98*2=196 is a square.

Eric Weisstein's World of Mathematics, NSW numbers.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A002315, A088165, A008843, A008844, A075870, A031396, A156164.
Cf. A382209 (10+k and 10*k are squares).
Cf. A245226 (m such that k+m and k*m are squares).

LinearRecurrence[{35, -35, 1}, {2, 98, 3362}, 20] (* Amiram Eldar, May 07 2025 *)

from itertools import islice
def A383734_gen(): # generator of terms
    x, y = 1, 7
    while True:
        yield 2*x**2
        x, y = y, 6*y - x
A383734_list = list(islice(A383734_gen(), 100))

a(n) = (1/2) * ((3+2*sqrt(2))^(2*n-1) + (3-2*sqrt(2))^(1-2*n)) - 1.
a(n) = -2*sqrt(2)*sinh(n*log(17+12*sqrt(2))) + 3*cosh(n*log(17+12*sqrt(2))) - 1.
a(n) = 2*A002315(n-1)^2.
a(n) = A075870(n)^2 - 2.
a(n) = 34*a(n-1) - a(n-2) + 32.
G.f.: 2 * (1 + 14*x + x^2) / ((1 - x)*(1 - 34*x + x^2)). - Stefano Spezia, May 08 2025

A383898 a(n) is the smallest nonnegative integer k such that n + k and n*k are squares, or -1 if there is no such number.

Original entry on oeis.org

0, 2, -1, 0, 20, -1, -1, 8, 0, 90, -1, -1, 4212, -1, -1, 0, 272, 18, -1, 5, -1, -1, -1, -1, 0, 650, -1, -1, 142100, -1, -1, 32, -1, -1, -1, 0, 1332, -1, -1, 360, 41984, -1, -1, -1, 180, -1, -1, -1, 0, 50, -1, 117, 1755572, -1, -1, -1, -1, 568458, -1, -1, 53872730964

Offset: 1

Gonzalo Martínez, May 15 2025

sign

a(m^2) = 0, for all positive integers m.
If m is not in A245226 then a(m) = -1. Indeed, if (u, v) is the smallest solution of the equation x^2 - m*y^2 = m, then m + m*v^2 = u^2 and m*(m*v^2) = (m*v)^2. Therefore, a(m) = m*v^2.
The sequence A383734 contains the numbers k such that 2 + k and 2*k are squares, where A383734(2) = a(2). Similarly, A382209(1) = a(10).
a(181) <=  223502910856088814900 = 181*1111225770^2. - Michel Marcus, May 24 2025

			If n = 5, then 20 satisfies that 5 + 20 = 5^2 and 5*20 = 10^2, where 20 is the smallest integer greater than 5 with this property. So, a(5) = 20.

Michel Marcus, Table of n, a(n) for n = 1..180

Cf. A382209, A383734, A245226.

isok(m) = if (issquare(4*m), 1, #qfbsolve(Qfb(1, 0, -m), m, 2)); \\ A245226
a(n) = if (!isok(n), return(-1)); my(k=0); while (!(issquare(n+k) && issquare(n*k)), k++); k; \\ Michel Marcus, May 21 2025

isok(m) = if (issquare(4*m), 1, #qfbsolve(Qfb(1, 0, -m), m, 2)); \\ A245226
a(n) = if (!isok(n), return(-1)); my(x = sqrtint(n)); while (! issquare(n*(x^2 - n)), x++); x^2-n; \\ Michel Marcus, May 24 2025

A248220 Values of n such that the equation x^2 - ny^2 = n(n+1)*(n+2)/6 has integer solutions.

Original entry on oeis.org

1, 2, 4, 7, 14, 16, 25, 31, 39, 48, 49, 52, 57, 64, 73, 74, 79, 84, 86, 97, 100, 112, 121, 127, 129, 134, 169, 192, 194, 196, 199, 217, 241, 244, 254, 256, 266, 273, 289, 292, 302, 304, 326, 336, 337, 350, 361, 372, 399, 400, 409, 448, 457, 484, 487, 489, 511

Offset: 1

Colin Barker, Oct 04 2014

nonn

			7 is in the sequence because x^2-7*y^2=84 has integer solutions, including (x,y) = (14,4) and (28,10).

Cf. A000292, A134419, A245226, A248194.

Select[Range[300],Length[FullSimplify[Solve[x^2-#*y^2==#*(#+1)*(#+2)/6,{x,y},Integers]/.C[1]->1]]>0&] (* Vaclav Kotesovec, Oct 05 2014, Mathematica version >= 8 *)

More terms from Vaclav Kotesovec, Oct 05 2014

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A382209 Numbers k such that 10+k and 10*k are perfect squares.

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A248194 Positive integers n such that the equation x^2 - n*y^2 = n*(n+1)/2 has integer solutions.

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A383734 Numbers k such that 2+k and 2*k are squares.

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A383898 a(n) is the smallest nonnegative integer k such that n + k and n*k are squares, or -1 if there is no such number.

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A248220 Values of n such that the equation x^2 - n*y^2 = n*(n+1)*(n+2)/6 has integer solutions.

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A248194 Positive integers n such that the equation x^2 - ny^2 = n(n+1)/2 has integer solutions.

A248220 Values of n such that the equation x^2 - ny^2 = n(n+1)*(n+2)/6 has integer solutions.