A100525 -id:A100525 - cp's OEIS frontend

A129346 a(2n) = A100525(n), a(2n+1) = A001653(n+1); a Pellian-related sequence.

4, 5, 22, 29, 128, 169, 746, 985, 4348, 5741, 25342, 33461, 147704, 195025, 860882, 1136689, 5017588, 6625109, 29244646, 38613965, 170450288, 225058681, 993457082, 1311738121, 5790292204, 7645370045, 33748296142, 44560482149, 196699484648, 259717522849

Offset: 0

Creighton Dement, Apr 10 2007

Summation of -a(n) and A129345 returns twice Pell numbers A000129 (apart from signs; starting from 2nd term of A000129).

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

Cf. A129345, A000129, A001542, A038761.

LinearRecurrence[{0,6,0,-1},{4,5,22,29},30] (* Harvey P. Dale, Apr 08 2018 *)

Vec((4+5*x-2*x^2-x^3)/((x^2-2*x-1)*(x^2+2*x-1)) + O(x^40)) \\ Colin Barker, May 26 2016

O.g.f.: (4 + 5*x - 2*x^2 - x^3) / ((x^2 - 2*x - 1)*(x^2 + 2*x - 1)).
From Colin Barker, May 26 2016: (Start)
a(n) = (-(-1-sqrt(2))^(1+n)+(-1+sqrt(2))^(1+n)+(1-sqrt(2))^n*(-4+3*sqrt(2))+(1+sqrt(2))^n*(4+3*sqrt(2)))/(2*sqrt(2)).
a(n) = 6*a(n-2)-a(n-4) for n>3. (End)
E.g.f.: 2*cosh(sqrt(2)*x)*(sinh(x) + 2*cosh(x)) + (sinh(sqrt(2)*x)*(5*sinh(x) + 3*cosh(x)))/sqrt(2). - Ilya Gutkovskiy, May 26 2016

A048654 a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 9, 22, 53, 128, 309, 746, 1801, 4348, 10497, 25342, 61181, 147704, 356589, 860882, 2078353, 5017588, 12113529, 29244646, 70602821, 170450288, 411503397, 993457082, 2398417561, 5790292204

Offset: 0

Barry E. Williams

Generalized Pellian with second term equal to 4.

The generalized Pellian with second term equal to s has the terms a(n) = A000129(n)*s + A000129(n-1). The generating function is -(1+s*x-2*x)/(-1+2*x+x^2). - R. J. Mathar, Nov 22 2007

T. D. Noe, Table of n, a(n) for n = 0..300
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
A. F. Horadam, Basic Properties of a Certain Generalized Sequence of Numbers, Fibonacci Quarterly, Vol. 3, No. 3, 1965, pp. 161-176.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1).

Cf. A000129, A001333, A048655, A038761, A084214, A100525.

a048654 n = a048654_list !! n
a048654_list =
   1 : 4 : zipWith (+) a048654_list (map (* 2) $ tail a048654_list)
-- Reinhard Zumkeller, Aug 01 2011

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!((1+2*x)/(1-2*x-x^2))); // G. C. Greubel, Jul 26 2018

LinearRecurrence[{2,1},{1,4},30] (* Harvey P. Dale, Jul 27 2011 *)

a[0]:1$
a[1]:4$
a[n]:=2*a[n-1]+a[n-2]$
A048654(n):=a[n]$
makelist(A048654(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n)=(([0, 1; 1,2]^n)*[1,4]~)[1] \\ Charles R Greathouse IV, May 18 2015

[lucas_number1(n+1,2,-1) +2*lucas_number1(n,2,-1) for n in (0..40)] # G. C. Greubel, Aug 09 2022

a(n) = ((3+sqrt(2))*(1+sqrt(2))^n - (3-sqrt(2))*(1-sqrt(2))^n)/2*sqrt(2).
a(n) = 2*A000129(n+2) - 3*A000129(n+1). - Creighton Dement, Oct 27 2004
G.f.: (1+2*x)/(1-2*x-x^2). - Philippe Deléham, Nov 03 2008
a(n) = binomial transform of 1, 3, 2, 6, 4, 12, ... . - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
E.g.f.: exp(x)*cosh(sqrt(2)*x) + 3*exp(x)*sinh(sqrt(2)*x)/sqrt(2). - Vaclav Kotesovec, Feb 16 2015
a(n) is the denominator of the continued fraction [4, 2, ..., 2, 4] with n-1 2's in the middle. For the numerators, see A221174. - Greg Dresden and Tongjia Rao, Sep 02 2021
a(n) = A001333(n) + A000129(n). - G. C. Greubel, Aug 09 2022

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

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).

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).

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

A129346 a(2n) = A100525(n), a(2n+1) = A001653(n+1); a Pellian-related sequence.

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A048654 a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=4.

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

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