A006452 -id:A006452 - cp's OEIS frontend

A055979 Solutions (value of r) of the Diophantine equation 2x^2 + 3x + 2 = r^2.

4, 11, 134, 373, 4552, 12671, 154634, 430441, 5253004, 14622323, 178447502, 496728541, 6061962064, 16874148071, 205928262674, 573224305873, 6995498968852, 19472752251611, 237641036678294, 661500352248901, 8072799748093144, 22471539224211023

Offset: 0

Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 24 2000

A necessary condition on any solution of the equation is x = [r/sqrt(2)] where [] denotes the floor function. The sequence lists the radii of circles for which a "best" digital approximation, as drawn by Bresenham-like algorithms, contains a point sequence [(x-1,x), (x,x), (x,x-1)] that is multiply connected by king moves. - clarified by M. Douglas McIlroy, May 18 2015
Corresponding values of x for above equation are given by A056161(n). The numbers a(n) are also solutions (value of r) to the Diophantine equation:  2x^2 - x + 1 = r^2, (excluding r = 1 at x = 0). - Richard R. Forberg, Nov 24 2013
This sequence lists the degrees n of those Chebyshev polynomials T(n,x) of the first kind which have the following exceptional property: There are exactly two coefficients in the power form of T(n,x) whose absolute values are identical and coincide with the height of T(n,x). This property is exceptional because for all remaining degrees n there is only one coefficient in the power form of T(n,x) whose absolute value coincides with the height of T(n,x). Recall that the height of a polynomial in power form is the maximum of the absolute value of its coefficients. Example: T(4,x) = 1 - 8x^2 + 8x^4; T(11,x) = - 11x + 220x^3 - 1232x^5 + 2816x^7 - 2816x^9 + 1024x^11. - Heinz-Joachim Rack, Nov 14 2015

H.-J. Rack, On the length and height of Chebyshev polynomials in one and two variables, East Journal on Approximations, 16 (2010), 35 - 91. See Theorem 5.2.1, Remark (k), and Table 5.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Seon-Hong Kim and Kenneth B. Stolarsky, Translations and extensions of the Nicomachus identity, arXiv:2306.17402 [math.NT], 2023. See also J. Int. Seq. (2024), Vol. 27, Issue 6, Art. No. 24.6.3, p. 12.
Zenon Kulpa, On the properties of discrete circles, rings, and disks, Computer Graphics and Image Processing, 10(1979), 348-365.
M. D. McIlroy, Best approximate circles on integer grids, ACM Transactions on Graphics 2(1983), 237-263.
Heinz-Joachim Rack, A comment on the Integer Sequence A055979
Ville Salo, Subshifts with sparse traces, University of Turki, Finland (2019).
Index entries for linear recurrences with constant coefficients, signature (0,34,0,-1).

Cf. A006452.

I:=[4,11,134,373]; [n le 4 select I[n] else 34*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, May 19 2015

a:= n-> (Matrix([11,4,1,2]). Matrix([[0,1,0,0], [34,0,1,0], [0,0,0,1], [ -1,0,0,0]])^n)[1,2]: seq(a(n), n=0..25); # Alois P. Heinz, Jun 03 2009

LinearRecurrence[{0,34,0,-1},{4,11,134,373},20] (* Harvey P. Dale, Feb 21 2012 *)

Vec((4+11*x-2*x^2-x^3)/(1+x^4-34*x^2) + O(x^50)) \\ Altug Alkan, Nov 15 2015

a(n) = A006452(2n+3) if n=0, 2, 4, ... a(n) = A006452(2n+2) if n=1, 3, 5, ...
G.f.: (4+11*x-2*x^2-x^3) / (1+x^4-34*x^2). - Alois P. Heinz, Jun 03 2009
a(n) = 34*a(n-2) - a(n-4); a(0)=4, a(1)=11, a(2)=134, a(3)=373. - Harvey P. Dale, Feb 21 2012

More terms from Alois P. Heinz, Jun 03 2009

A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

Original entry on oeis.org

2, 2, 1, 3, 4, 8, 9, 19, 22, 46, 53, 111, 128, 268, 309, 647, 746, 1562, 1801, 3771, 4348, 9104, 10497, 21979, 25342, 53062, 61181, 128103, 147704, 309268, 356589, 746639, 860882, 1802546, 2078353, 4351731, 5017588, 10506008, 12113529, 25363747, 29244646, 61233502

Offset: 0

Raphie Frank, Dec 30 2015

This sequence gives all x in N | 2*x^2 - 7(-1)^x = y^2. The companion sequence to this sequence, giving y values, is A266505.
A266505(n)/a(n) converges to sqrt(2).
Alternatively, 1/4*(3*A002203(floor[n/2]) - A002203(n-(-1)^n)), where A002203 gives the Companion Pell numbers, or, in Lucas sequence notation, V_n(2, -1).
Alternatively, bisection of A266506.
Alternatively, A048654(n -1) and A078343(n + 1) interlaced.
Alternatively, A100525(n-1), A266507(n), A038761(n) and A253811(n) interlaced.
Let b(n) = (a(n) - a(n)(mod 2))/2, that is b(n) = {1, 1, 0, 1, 2, 4, 4, 9, 11, 23, 26, 55, 64, ...}. Then:
A006452(n) = {b(4n+0) U b(4n+1)} gives n in N such that n^2 - 1 is triangular;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that n^2 + n + 1 is triangular (indices of Sophie Germain triangular numbers);
A216162(n) = {b(4n+0) U b(4n+2) U b(4n+1) U b(4n+3)}, sequences A006452 and A216134 interlaced.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[2,2,1,3]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

LinearRecurrence[{0, 2, 0, 1}, {2, 2, 1, 3}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(1 - x) (2 + 4 x + x^2)/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 41}] (* Michael De Vlieger, Dec 31 2015 *)

Vec((1-x)*(2+4*x+x^2)/(1-2*x^2-x^4) + O(x^50)) \\ Colin Barker, Dec 31 2015

a(n) = 1/sqrt(8)*(+sqrt(2)*(1+sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n - 3*(1-sqrt(2))^(floor(n/2)-(-1)^n) + sqrt(2)*(1-sqrt(2))^(floor(n/2)-(-1)^n)*(-1)^n + 3*(1+sqrt(2))^(floor(n/2)-(-1)^n)).
a(n) = 1/4*((3*((1+sqrt(2))^floor(n/2)+(1-sqrt(2))^floor(n/2))) - (-1)^n*((1+sqrt(2))^(floor(n/2)-(-1)^n)+(1-sqrt(2))^(floor(n/2)-(-1)^n))).
a(2n) = (+sqrt(2)*(1+sqrt(2))^(n-1) - 3 *(1-sqrt(2))^(n-1) + sqrt(2)*(1-sqrt(2))^(n-1) + 3*(1 + sqrt(2))^(n-1))/sqrt(8) = A048654(n -1).
a(2n) = 1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) - ((1+sqrt(2))^(n-1)+(1-sqrt(2))^(n-1))) = A048654(n -1).
a(2n + 1) = (-sqrt(2)*(1+sqrt(2))^(n+1) - 3 *(1-sqrt(2))^(n+1) - sqrt(2)*(1-sqrt(2))^(n+1) + 3*(1+sqrt(2))^(n+1))/sqrt(8) = A078343(n + 1).
a(2n + 1) =1/4*((3*((1+sqrt(2))^n+(1-sqrt(2))^n)) + ((1+sqrt(2))^(n+1)+(1-sqrt(2))^(n+1))) =  A078343(n + 1).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A100525(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A266507(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038761(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A253811(n).
sqrt(2*a(n)^2 - 7(-1)^a(n))*sgn(2*n - 1) = A266505(n).
(a(2n + 1) + a(2n))/2 = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 1) - a(2n))/2 = A000129(n), where A000129 gives the Pell numbers.
(a(2n+2) + a(2n+1))*2 = A002203(n+2)
(a(2n+2) - a(2n+1))*2 = A002203(n-1).
G.f.: (1-x)*(2+4*x+x^2) / (1-2*x^2-x^4). - Colin Barker, Dec 31 2015

A217278 Sequences A124174 and A006454 interlaced.

Original entry on oeis.org

0, 0, 1, 3, 10, 15, 45, 120, 351, 528, 1540, 4095, 11935, 17955, 52326, 139128, 405450, 609960, 1777555, 4726275, 13773376, 20720703, 60384555, 160554240, 467889345, 703893960, 2051297326, 5454117903, 15894464365, 23911673955, 69683724540, 185279454480

Offset: 0

Raphie Frank, Sep 29 2012

a(2n) and 2*a(2n) + 1 are triangular.

a(2n + 1) is triangular and a(2n + 1)/2 is the harmonic mean of consecutive triangular numbers (therefore, a(2n + 1) + 1 is square).

			a(18) = 35*(52326 - 1540) + 45 = 1777555,
a(19) = 35*(139128 - 4095) + 120 = 4726275.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,34,0,-34,0,-1,0,1).

Cf. (sqrt(8a(2n) + 1) - 1)/2 = A216134(n) = A216162(2n + 1).
Cf. sqrt(a(2n+1) + 1) = A006452(n + 1) = A216162(2n + 2).
Cf. (sqrt(8a(2n+1) + 1) - 1)/2 = A006451(n).

LinearRecurrence[{0,1,0,34,0,-34,0,-1,0,1},{0,0,1,3,10,15,45,120,351,528},40] (* Harvey P. Dale, Aug 04 2019 *)

concat([0,0], Vec(-x^2*(3*x^5+x^4+12*x^3+9*x^2+3*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)*(x^4+6*x^2+1)) + O(x^100))) \\ Colin Barker, Jun 23 2015

a(n) = 35*(a(n-4) - a(n-8)) + a(n-12).
lim n --> infinity a(2n)/a(2n - 1) = (3 + sqrt(8))/2.
From Raphie Frank, Dec 21 2015: (Start)
a(2*n + 1) = 1/64*(((4 + sqrt(2)) * (1 - (-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2)) + (4 - sqrt(2)) * (1+(-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2))))^2 - 1.
a(2*n + 2) = 1/2*(3*(a(2*n + 1)) + sqrt((a(2*n + 1)) + 1) * sqrt(8*(a(2*n + 1)) + 1) + 1).
(End)

A226070 Numbers k such that k^2-1 is a triangular number and k^2+1 is a prime.

Original entry on oeis.org

1, 2, 4, 134, 2174, 154634

Offset: 1

Alex Ratushnyak, May 25 2013

Subsequence of A006452 that consists of the terms k such that k^2+1 is prime.

a(7) is too large to include here (see b-file). - Max Alekseyev, Jan 30 2014

Max Alekseyev, Table of n, a(n) for n = 1..12

Corresponding primes: A226069.
Indices of corresponding triangular numbers: A226071.
Cf. A226072-A226074.

C
```
// see A226069.
```

Select[Range[155000],PrimeQ[#^2+1]&&OddQ[Sqrt[8(#^2-1)+1]]&] (* Harvey P. Dale, Jul 19 2019 *)

Terms a(7)-a(12) from Max Alekseyev, Jan 30 2014

A258149 Triangle of the absolute difference of the two legs (catheti) of primitive Pythagorean triangles.

Original entry on oeis.org

1, 0, 7, 7, 0, 17, 0, 1, 0, 31, 23, 0, 0, 0, 49, 0, 17, 0, 23, 0, 71, 47, 0, 7, 0, 41, 0, 97, 0, 41, 0, 7, 0, 0, 0, 127, 79, 0, 31, 0, 0, 0, 89, 0, 161, 0, 73, 0, 17, 0, 47, 0, 119, 0, 199, 119, 0, 0, 0, 1, 0, 73, 0, 0, 0, 241

Offset: 2

Wolfdieter Lang, Jun 10 2015

For primitive Pythagorean triangles characterized by certain (n,m) pairs and references see A225949.
Here a(n,m) = 0 for non-primitive Pythagorean triangles, and for primitive Pythagorean triangles a(n,m) = abs(n^2 - m^2 - 2*n*m) = abs((n-m)^2 - 2*m^2).
The number of non-vanishing entries in row n is A055034(n).
D(n,m):= n^2 - m^2 - 2*n*m >= 0 if 1 <= m <= floor(n/(sqrt(2)+1)), and D(n,m) < 0 if n/(sqrt(2)+1)+1 <= m <= n-1, for n >= 2.
The Pell equation (n-m)^2 - 2*m^2 = +/- N is important here to find the representations of +N or -N in the triangle D(n,m). For instance, odd primes N have to be of the +1 (mod 8) (A007519) or -1 (mod 8) (A007522) form, that is, from A001132. See the Nagell reference, Theorem 110, p. 208 with Theorem 111, pp. 210-211. E.g., N = +7 appears for m = 1, 3, 9, 19, 53, ... (A077442) for n = 4, 8, 22, 46, 128, ... (2*A006452).
  N = -7 appears for n = 3, 9, 19, 53, 111, ... (A077442) and m = 2, 4, 8, 22, 46, ... (2*A006452).
For the  signed version 2*n*m - (n^2 - m^2) see A278717. - Wolfdieter Lang, Nov 30 2016

			The triangle a(n,m) begins:
n\m   1  2  3  4  5  6  7   8   9  10  11 ...
2:    1
3:    0  7
4:    7  0 17
5:    0  1  0 31
6:   23  0  0  0 49
7:    0 17  0 23  0 71
8:   47  0  7  0 41  0 97
9:    0 41  0  7  0  0  0 127
10:  79  0 31  0  0  0 89   0 161
11:   0 73  0 17  0 47  0 119   0 199
12: 119  0  0  0  1  0 73   0   0   0 241
...
a(2,1) = |1^2 - 2*1^2| = 1 for the primitive Pythagorean triangle (pPt) [3,4,5] with |3-4| = 1.
a(3,2) = |1^2 - 2*2^2| = 7 for the pPt [5,12,13] with |5 - 12| = 7.
a(4,1) = |3^2 - 2*1^2| = 7 for the pPt [15, 8, 17] with |15 - 8| = 7.

See also A225949.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 208, 210-211.

Cf. A249866, A222946, A225949, A222951, A258150, A278717 (signed).

a[n_, m_] /; n > m >= 1 && CoprimeQ[n, m] && (-1)^(n+m) == -1 := Abs[n^2 - m^2 - 2*n*m]; a[, ] = 0; Table[a[n, m], {n, 2, 12}, {m, 1, n-1}] // Flatten (* Jean-François Alcover, Jun 16 2015, after given formula *)

a(n,m) = abs(n^2 - m^2 -2*n*m) = abs((n-m)^2 - 2*m^2) if n > m >= 1, gcd(n,m) = 1, and n and m are integers of opposite parity (i.e., (-1)^(n+m) = -1); otherwise a(n,m) = 0.

A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

Original entry on oeis.org

-1, 1, 3, 5, 5, 11, 13, 27, 31, 65, 75, 157, 181, 379, 437, 915, 1055, 2209, 2547, 5333, 6149, 12875, 14845, 31083, 35839, 75041, 86523, 181165, 208885, 437371, 504293, 1055907, 1217471, 2549185, 2939235, 6154277, 7095941, 14857739, 17131117, 35869755, 41358175, 86597249, 99847467

Offset: 0

Raphie Frank, Dec 30 2015

a(n)/A266504(n) converges to sqrt(2).
Alternatively, bisection of A266506.
Alternatively, A135532(n) and A048655(n) interlaced.
Alternatively, A255236(n-1), A054490(n), A038762(n) and A101386(n) interlaced.
Let b(n) = (a(n) - (a(n) mod 2))/2, that is b(n) = {-1, 0, 1, 2, 2, 5, 6, 13, 15, 32, 37, 78, 90, ...}. Then:
A006451(n) = {b(4n+0) U b(4n+1)} gives n in N such that triangular(n) + 1 is square;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that triangular(n) follows form n^2 + n + 1 (twice a triangular number + 1).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[-1,1,3,5]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

a:=proc(n) option remember; if n=0 then -1 elif n=1 then 1 elif n=2 then 3 elif n=3 then 5 else 2*a(n-2)+a(n-4); fi; end:  seq(a(n), n=0..50); # Wesley Ivan Hurt, Jan 01 2016

LinearRecurrence[{0, 2, 0, 1}, {-1, 1, 3, 5}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(-1 + 3 x) (1 + x)^2/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 42}] (* Michael De Vlieger, Dec 31 2015 *)

my(x='x+O('x^40)); Vec((-1+3*x)*(1+x)^2/(1-2*x^2-x^4)) \\ G. C. Greubel, Jul 26 2018

G.f.: (-1 + 3*x)*(1 + x)^2/(1 - 2*x^2 - x^4).
a(n) = (-(1+sqrt(2))^floor(n/2)*(-1)^n - sqrt(8)*(1-sqrt(2))^floor(n/2) - (1-sqrt(2))^floor(n/2)*(-1)^n + sqrt(8)*(1+sqrt(2))^floor(n/2))/2.
a(n) = 3*(((1+sqrt(2))^floor(n/2)-(1-sqrt(2))^floor(n/2))/sqrt(8)) - (-1)^n*(((1+sqrt(2))^(floor(n/2)-(-1)^n)-(1-sqrt(2))^(floor(n/2)-(-1)^n))/sqrt(8)).
a(n) = (3*A000129(floor(n/2)) - A000129(n-(-1)^n)), where A000129 gives the Pell numbers.
a(n) = sqrt(2*A266504(n)^2 - 7*(-1)^A266504(n))*sgn(2*n-1), where A266504 gives all x in N such that 2*x^2 - 7*(-1)^x = y^2. This sequence gives associated y values.
a(2n) = (-(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n - (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n) = A135532(n).
a(2n) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) - (((1+sqrt(2))^(n-1)-(1-sqrt(2))^(n-1))/sqrt(8)) = A135532(n).
a(2n+1) = (+(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n + (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n + 1) = A048655(n).
a(2n+1) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) + (((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/sqrt(8)) = A048655(n).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A255236(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A054490(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038762(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A101386(n).
(sqrt(2*(a(2n + 1) )^2 + 14*(-1)^floor(n/2)))/2 = A266504(n).
(a(2n + 1) + a(2n))/8 = A000129(n), where A000129 gives the Pell numbers.
a(2n + 1) - a(2n) = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 2) + a(2n + 1))/2 = A000129(n+2).
(a(2n + 2) - a(2n + 1))/2 = A000129(n-1).

A077447 Numbers k such that (k^2 - 14)/2 is a square.

Original entry on oeis.org

4, 8, 16, 44, 92, 256, 536, 1492, 3124, 8696, 18208, 50684, 106124, 295408, 618536, 1721764, 3605092, 10035176, 21012016, 58489292, 122467004, 340900576, 713790008, 1986914164, 4160273044, 11580584408, 24247848256, 67496592284

Offset: 1

Gregory V. Richardson, Nov 09 2002

nonn

The equation "(n^2 - 14)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A156649 (R1).

LinearRecurrence[{0,6,0,-1},{4,8,16,44},40] (* Harvey P. Dale, Jul 22 2013 *)

Limit_{k -> oo} a(2*k+1)/a(2*k) = 2.09383632135605431360 = (9 + 4*sqrt(2))/7 = R1.
Limit_{k -> oo} a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2.
Limit_{k -> oo} a(n)/a(n-2) = 3 + 2*sqrt(2) = RG (Grand Ratio); RG = R1*R2.
For n = 2*k-1, a(n) = [ 2*[(3+2*sqrt(2))^n + (3-2*sqrt(2))^n] - [(3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1)] + [(3+2*sqrt(2))^(n-2) + (3-2*sqrt(2))^(n-2)] ] / 4.
For n = 2*k, a(n) = [ 5*[(3+2*sqrt(2))^n + (3-2*sqrt(2))^n] + [(3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1)] ] / 4.
a(n) = 6*a(n-2) - a(n-4) = 4*A006452(n).
G.f.: -4*x*(x-1)*(x^2+3*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Jul 03 2011

A078699 Primes p such that p^2-1 is a triangular number.

Original entry on oeis.org

2, 11, 23, 373, 12671, 901273, 19472752251611, 53072032161200090602953513048447623, 5027153581127740201460650182713355379768873, 11604855412241025458500993236724193227031777965785837784548351709747881343573

Offset: 1

Jason Earls, Dec 18 2002

nonn

Equivalently, primes in A006452.

The sequence of corresponding triangular numbers begins 3, 120, 528, 139128, 160554240, 812293020528, 379188080252621270252095320, ... [Shreevatsa R, Jul 12 2013]

Joerg Arndt, Table of n, a(n) for n = 1..14

Cf. A000217, A006452.

a[n_] := a[n]=If[n<4, {1, 2, 4, 11}[[n+1]], 6a[n-2]-a[n-4]]; Select[a/@Range[200], ProvablePrimeQ] (* First do <

default(primelimit,10^7) istri(n) = t=floor(sqrt(2*n)); if(2*n==t*(t+1),1,0) forprime(p=2,5*10^6,if(istri(p^2-1),print1(p" ")))

istriang(n)=issquare(8*n+1);
forprime(p=2,10^10,if(istriang(p^2-1),print1(p,", ")));
\\ Joerg Arndt, Jul 15 2013

/* much more efficient: */
N=1166; f=( 1+x-4*x^2-2*x^3 ) / ( (x^2+2*x-1)*(x^2-2*x-1) )+O(x^N);
for(n=0,N-1,my(c=polcoeff(f,n)); if(isprime(c), print1(c,", ")));
\\ Joerg Arndt, Jul 15 2013

Edited by Dean Hickerson, Dec 19 2002

A156066 Numbers n with property that n^2 is a square arising in A154138.

Original entry on oeis.org

2, 3, 9, 16, 52, 93, 303, 542, 1766, 3159, 10293, 18412, 59992, 107313, 349659, 625466, 2037962, 3645483, 11878113, 21247432, 69230716, 123839109, 403506183, 721787222, 2351806382, 4206884223, 13707332109, 24519518116, 79892186272

Offset: 1

Zak Seidov, Oct 21 2009

Except for the first term, positive values of x (or y) satisfying x^2 - 6xy + y^2 + 23 = 0. - Colin Barker, Feb 08 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A154138.

a:=[2,3,9,16];; for n in [5..30] do a[n]:=6*a[n-2]-a[n-4]; od; a; # Muniru A Asiru, Sep 28 2018

I:=[2,3,9,16]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Feb 11 2014

seq(coeff(series(-x*(x-1)*(x+2)*(2*x+1)/((x^2-2*x-1)*(x^2+2*x-1)),x,n+1), x, n), n = 1..30); # Muniru A Asiru, Sep 28 2018

a[1]=2;a[2]=3;a[3]=9;a[4]=16;a[n_]:=a[n]=6*a[n-2]-a[n-4];A1=Table[a[n],{n,25}]
CoefficientList[Series[-(x - 1) (x + 2) (2 x + 1)/((x^2 - 2 x - 1) (x^2 + 2 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 11 2014 *)

Vec(-x*(x-1)*(x+2)*(2*x+1)/((x^2-2*x-1)*(x^2+2*x-1)) + O(x^100)) \\ Colin Barker, Feb 08 2014

a(n) = sqrt((A154138(n)^2 + A154138(n) + 6)/2).
a(1..4) = (2,3,9,16); a(n>4) = 6*a(n-2) - a(n-4).
G.f.: -x*(x-1)*(x+2)*(2*x+1) / ((x^2-2*x-1)*(x^2+2*x-1)). - Colin Barker, Feb 08 2014
a(n) = A006452(n-1) - A006452(n) + A006452(n+1). - Carl Najafi, Sep 27 2018

A217975 Integers k such that 2*k^2 - 7 is a square.

Original entry on oeis.org

2, 4, 8, 22, 46, 128, 268, 746, 1562, 4348, 9104, 25342, 53062, 147704, 309268, 860882, 1802546, 5017588, 10506008, 29244646, 61233502, 170450288, 356895004, 993457082, 2080136522, 5790292204, 12123924128, 33748296142, 70663408246, 196699484648

Offset: 1

Sture Sjöstedt, Oct 16 2012

a(n) gives y-values solving the Diophantine equation x^2 + 7 = 2*y^2. A077446(n) gives the x-values. - Sture Sjöstedt, Oct 16 2012

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 28 = 0. - Colin Barker, Feb 08 2014

			Since 2(4^2) - 7 = 25 = 5^2, and 4 is the second number with this property, a(2) = 4.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A077442 (2*n^2 + 7 is a square).

I:=[2, 4, 8, 22]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..31]]; // Vincenzo Librandi, Oct 16 2012

LinearRecurrence[{0, 6, 0, -1}, {2, 4, 8, 22}, 50] (* Sture Sjöstedt, Oct 16 2012 *)

Vec(2*x*(1-x)*(x^2+3*x+1)/(x^2-2*x-1)/(x^2+2*x-1)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

a(n) = 6*a(n - 2) - a(n - 4) with a(1)=2, a(2)=4, a(3)=8, a(4)=22. - Sture Sjöstedt, Oct 16 2012
a(n)*a(n+3)-a(n+1)*a(n+2) = 10-2*(-1)^n. - Bruno Berselli, Oct 25 2012
a(n) = 2*A006452(n). - R. J. Mathar, Oct 17 2012
G.f.: -2*x*(x - 1)*(x^2 + 3*x + 1)/((x^2 - 2*x - 1)*(x^2 + 2*x - 1)). - Colin Barker, Oct 24 2012
a(n) = a(-n+1) = ((4+sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))+(4-sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2)))/4. - Bruno Berselli, Oct 25 2012
a(2n-1) = A078343(2n-1), a(2n) = A100525(n-1). - Bruno Berselli, Oct 25 2012

cp's OEIS Frontend

A055979 Solutions (value of r) of the Diophantine equation 2*x^2 + 3*x + 2 = r^2.

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A266504 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = a(1) = 2, a(2) = 1, a(3) = 3.

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A217278 Sequences A124174 and A006454 interlaced.

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A226070 Numbers k such that k^2-1 is a triangular number and k^2+1 is a prime.

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A258149 Triangle of the absolute difference of the two legs (catheti) of primitive Pythagorean triangles.

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A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

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A077447 Numbers k such that (k^2 - 14)/2 is a square.

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A055979 Solutions (value of r) of the Diophantine equation 2x^2 + 3x + 2 = r^2.