A048655 -id:A048655 - cp's OEIS frontend

A111566 a(n) = ((1+sqrt(8))(2+sqrt(2))^n + (1-sqrt(8))(2-sqrt(2))^n)/2.

1, 6, 22, 76, 260, 888, 3032, 10352, 35344, 120672, 412000, 1406656, 4802624, 16397184, 55983488, 191139584, 652591360, 2228086272, 7607162368, 25972476928, 88675582976, 302757378048, 1033678346240, 3529198628864, 12049437822976, 41139354034176, 140458540490752

Offset: 0

Creighton Dement, Aug 06 2005

Binomial transform of A048655: generalized Pellian with second term equal to 5.

Floretion Algebra Multiplication Program, FAMP Code: 1vesseq[K*J] with K = + .5'i + .5'j + .5k' + .5'kk' and J = + .5i' + .5j' + 2'kk' + .5'ki' + .5'kj'.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2).

Cf. A007052, A006012, A111567, A007070.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((1+2*r)*(2+r)^n+(1-2*r)*(2-r)^n)/2: n in [0..23] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 27 2009

LinearRecurrence[{4,-2},{1,6},30] (* Harvey P. Dale, Jan 31 2015 *)

x='x+O('x^30); Vec((1+2*x)/(1-4*x+2*x^2)) \\ G. C. Greubel, Jan 27 2018

a(n) = 4*a(n-1) - 2*a(n-2), a(0) = 1, a(1) = 6.
Program "FAMP" returns: a(n) = A007052(n) - A006012(n) + A111567(n).
From R. J. Mathar, Apr 02 2008: (Start)
O.g.f.: (1+2*x)/(1-4*x+2*x^2).
a(n) = A007070(n) + 2*A007070(n-1). (End)
a(n) = Sum_{k=0..n} A207543(n,k)*2^k. - Philippe Deléham, Feb 25 2012
a(n) = 4*A007070(n) - A007052(n+1). - Yuriy Sibirmovsky, Sep 13 2016
E.g.f.: exp(2*x)*(cosh(sqrt(2)*x) + 2*sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, May 26 2024

Edited by N. J. A. Sloane, Jul 27 2009 using new definition from Al Hakanson (hawkuu(AT)gmail.com)

A048693 Generalized Pellian with 2nd term equal to 6.

Original entry on oeis.org

1, 6, 13, 32, 77, 186, 449, 1084, 2617, 6318, 15253, 36824, 88901, 214626, 518153, 1250932, 3020017, 7290966, 17601949, 42494864, 102591677, 247678218, 597948113, 1443574444, 3485097001, 8413768446

Offset: 0

Barry E. Williams

Pisano period lengths: 1, 2, 8, 4, 12, 8, 6, 8, 24, 12, 24, 8, 28, 6, 24, 16, 16, 24, 40, 12, ... (is this A175181?). - R. J. Mathar, Aug 10 2012

			a(n)=[ (5+sqrt(2))(1+sqrt(2))^n-(5-sqrt(2))(1-sqrt(2))^n ]/2*sqrt(2)

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1)

Cf. A001333, A000129, A048654, A048655.

with(combinat): a:=n->4*fibonacci(n-1,2)+fibonacci(n,2): seq(a(n), n=1..26); # Zerinvary Lajos, Apr 04 2008

a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{5},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2,1},{1,6},30] (* Harvey P. Dale, Mar 29 2013 *)

a[0]:1$
a[1]:6$
a[n]:=2*a[n-1]+a[n-2]$
A048693(n):=a[n]$
makelist(A048693(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=6.
G.f.: (1+4*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008
a(n) = 4*A000129(n) + A000129(n+1). - R. J. Mathar, Aug 10 2012

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

A048697 Generalized Pellian with second term equal to 10.

Original entry on oeis.org

1, 10, 21, 52, 125, 302, 729, 1760, 4249, 10258, 24765, 59788, 144341, 348470, 841281, 2031032, 4903345, 11837722, 28578789, 68995300, 166569389, 402134078, 970837545, 2343809168, 5658455881

Offset: 0

Barry E. Williams

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1).

Cf. A001333, A000129, A048654, A048655.

a048697 n = a048697_list !! n
a048697_list = 1 : 10 : zipWith (+)
                        a048697_list (map (* 2) $ tail a048697_list)
-- Reinhard Zumkeller, Sep 05 2014

with(combinat): a:=n->8*fibonacci(n-1,2)+fibonacci(n,2): seq(a(n), n=1..25); # Zerinvary Lajos, Apr 04 2008

a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{9},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2,1},{1,10},35] (* Harvey P. Dale, Jul 26 2011 *)

a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=10.
a(n) = ((9+sqrt(2))(1+sqrt(2))^n - (9-sqrt(2))(1-sqrt(2))^n) / 2*sqrt(2).
G.f.: (1+8*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008

A117584 Generalized Pellian triangle.

Original entry on oeis.org

1, 1, 2, 1, 3, 5, 1, 4, 7, 12, 1, 5, 9, 17, 29, 1, 6, 11, 22, 41, 70, 1, 7, 13, 27, 53, 99, 169, 1, 8, 15, 32, 65, 128, 239, 408, 1, 9, 17, 37, 77, 157, 309, 577, 985, 1, 10, 19, 42, 89, 186, 379, 746, 1393, 2378

Offset: 1

Gary W. Adamson, Mar 29 2006

			First few rows of the triangle are:
  1;
  1, 2;
  1, 3,  5;
  1, 4,  7, 12;
  1, 5,  9, 17, 29;
  1, 6, 11, 22, 41, 70;
  1, 7, 13, 27, 53, 99, 169;
  ...
The triangle rows are antidiagonals of the generalized Pellian array:
  1, 2,  5, 12, 29, ...
  1, 3,  7, 17, 41, ...
  1, 4,  9, 22, 53, ...
  1, 5, 11, 27, 65, ...
  ...
For example, in the row (1, 5, 11, 27, 65, ...), 65 = 2*27 + 11.

G. C. Greubel, Rows n = 1..100, flattened

Diagonals include A000129, A001333, A048654, A048655, A048693.

Cf. A117185.

P:= func< n | Round( ((1+Sqrt(2))^n - (1-Sqrt(2))^n)/(2*Sqrt(2)) ) >;
T:= func< n,k | P(k) + (n-1)*P(k-1) >;
[T(n-k+1, k): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 05 2021

T[n_, k_]:= Fibonacci[k, 2] + (n-1)*Fibonacci[k-1, 2];
Table[T[n-k+1, k], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jul 05 2021 *)

def T(n,k): return lucas_number1(k,2,-1) + (n-1)*lucas_number1(k-1,2,-1)
flatten([[T(n-k+1, k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Jul 05 2021

Antidiagonals of the generalized Pellian array. First row of the array = A000129: (1, 2, 5, 12, ...). n-th row of the array starts (1, n+1, ...); as a Pellian sequence.
From G. C. Greubel, Jul 05 2021: (Start)
T(n, k) = P(k) + (n-1)*P(k-1), where P(n) = A000129(n) (square array).
Sum_{k=1..n} T(n-k+1, k) = A117185(n). (End)

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).

A048695 Generalized Pellian with second term equal to 8.

Original entry on oeis.org

1, 8, 17, 42, 101, 244, 589, 1422, 3433, 8288, 20009, 48306, 116621, 281548, 679717, 1640982, 3961681, 9564344, 23090369, 55745082, 134580533, 324906148, 784392829, 1893691806, 4571776441, 11037244688

Offset: 0

Barry E. Williams

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1)

Cf. A001333, A000129, A048654, A048655.

with(combinat): a:=n->6*fibonacci(n-1,2)+fibonacci(n,2): seq(a(n), n=1..26); # Zerinvary Lajos, Apr 04 2008

a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{7},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2,1},{1,8},30] (* Harvey P. Dale, May 01 2013 *)

a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=8.
a(n) = ((7+sqrt(2))(1+sqrt(2))^n - (7-sqrt(2))(1-sqrt(2))^n)/2*sqrt(2).
G.f.: (1+6*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008

A111955 a(n) = A078343(n) + (-1)^n.

Original entry on oeis.org

0, 1, 4, 7, 20, 45, 112, 267, 648, 1561, 3772, 9103, 21980, 53061, 128104, 309267, 746640, 1802545, 4351732, 10506007, 25363748, 61233501, 147830752, 356895003, 861620760, 2080136521, 5021893804, 12123924127, 29269742060, 70663408245

Offset: 0

Creighton Dement, Aug 25 2005

This sequence is a companion sequence to A111954 (compare formula / program code). Three other companion sequences (i.e., they are generated by the same floretion given in the program code) are A105635, A097076 and A100828.

Floretion Algebra Multiplication Program, FAMP Code: 4kbasejseq[J*D] with J = - .25'i + .25'j + .5'k - .25i' + .25j' + .5k' - .5'kk' - .25'ik' - .25'jk' - .25'ki' - .25'kj' - .5e and D = + .5'i - .25'j + .25'k + .5i' - .25j' + .25k' - .5'ii' - .25'ij' - .25'ik' - .25'ji' - .25'ki' - .5e. (an initial term 0 was added to the sequence)

Index entries for linear recurrences with constant coefficients, signature (1,3,1).

Cf. A078343, A000129, A001333, A111954, A111956, A007070, A077995, A100828, A097076, A105635, A048655.

LinearRecurrence[{1,3,1},{0,1,4},40] (* Harvey P. Dale, Mar 12 2015 *)

a(n) + a(n+1) = A048655(n).
a(n) = a(n-1) + 3*a(n-2) + a(n-3), n >= 3; a(n) = (-1/4*sqrt(2)+1)*(1-sqrt(2))^n + (1/4*sqrt(2)+1)*(1+sqrt(2))^n - (-1)^n;
G.f.: -x*(1+3*x) / ( (1+x)*(x^2+2*x-1) ). - R. J. Mathar, Oct 02 2012
E.g.f.: cosh(x) - exp(x)*cosh(sqrt(2)*x) - sinh(x) + 3*exp(x)*sinh(sqrt(2)*x)/sqrt(2). - Stefano Spezia, May 26 2024

A117895 Triangle T(n, k) = (k-n)A000129(k+1) + (3n-3k+1)A000129(k) with T(n,0) = 1, for 0 <= k <= n-1, read by rows.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 4, 8, 1, 5, 5, 11, 19, 1, 6, 6, 14, 26, 46, 1, 7, 7, 17, 33, 63, 111, 1, 8, 8, 20, 40, 80, 152, 268, 1, 9, 9, 23, 47, 97, 193, 367, 647, 1, 10, 10, 26, 54, 114, 234, 466, 886, 1562, 1, 11, 11, 29, 61, 131, 275, 565, 1125, 2139, 3771, 1, 12, 12, 32, 68, 148, 316, 664, 1364, 2716, 5164, 9104

Offset: 0

Gary W. Adamson, Mar 30 2006

Successive deletions of the right borders of triangle A117894 produces triangles whose row sums = generalized Pell sequences starting (1, 2...), (1, 3...), (1, 4...); etc. Row sums of A117894 = A000129: (1, 2, 5...). Row sums of A117895 = A001333: (1, 3, 7...). Deletion of the border of A117895 would produce a triangle with row sums of the Pell sequence A048654 (1, 4, 9...); and so on.

			First few rows of the triangle are:
  1;
  1, 2;
  1, 3, 3;
  1, 4, 4,  8;
  1, 5, 5, 11, 19;
  1, 6, 6, 14, 26, 46;
  1, 7, 7, 17, 33, 63, 111;
  1, 8, 8, 20, 40, 80, 152, 268;
...
Row 4, (1, 4, 4, 8) is produced by adding (0, 1, 1, 3) to row 4 of A117894: (1, 3, 3, 5).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000129, A001333, A048654, A048655, A048693, A117894.

Pell:= func< n | Round(((1+Sqrt(2))^n - (1-Sqrt(2))^n)/(2*Sqrt(2))) >;
[k eq 0 select 1 else (k-n)*Pell(k+1) + (3*n-3*k+1)*Pell(k): k in [0..n-1], n in [0..12]]; // G. C. Greubel, Sep 27 2021

T[n_, k_]:= T[n, k]= If[k==0, 1, (k-n)*Fibonacci[k+1, 2] + (3*n-3*k +1)*Fibonacci[k, 2]]; Table[T[n, k], {n,0,12}, {k,0,n-1}]//Flatten (* G. C. Greubel, Sep 27 2021 *)

def P(n): return lucas_number1(n, 2, -1)
def A117895(n,k): return 1 if (k==0) else (k-n)*P(k+1) + (3*n-3*k+1)*P(k)
flatten([[A117895(n,k) for k in (0..n-1)] for n in (0..12)]) # G. C. Greubel, Sep 27 2021

Delete right border of triangle A117894. Alternatively, let row 1 = 1 and using the heading 0, 1, 1, 3, 7, 17, 41, 99, 239...(i.e. A001333 starting with 0, 1, 1, 3...); add the first n terms of the heading to n-th row of triangle A117894.
From G. C. Greubel, Sep 27 2021: (Start)
T(n, k) = (k-n)*A000129(k+1) + (3*n-3*k+1)*A000129(k) with T(n,0) = 1.
T(n, 1) = n+1 for n >= 1.
T(n, 2) = n+1 for n >= 2.
T(n, n) = 2*[n=0] + A078343(n). (End)

New name and more terms added by G. C. Greubel, Sep 27 2021

A135246 Shifted Pell recurrence: a(n) = 2*a(n-2) + a(n-4).

Original entry on oeis.org

1, 3, 5, 7, 11, 17, 27, 41, 65, 99, 157, 239, 379, 577, 915, 1393, 2209, 3363, 5333, 8119, 12875, 19601, 31083, 47321, 75041, 114243, 181165, 275807, 437371, 665857, 1055907, 1607521, 2549185, 3880899, 6154277, 9369319, 14857739, 22619537, 35869755, 54608393, 86597249, 131836323, 209064253, 318281039, 504725755, 768398401, 1218515763, 1855077841, 2941757281, 4478554083

Offset: 0

Paul Curtz, Feb 15 2008

Mix A048655(n) and A001333(n+2).

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. A001333, A048655, A109543.

LinearRecurrence[{0, 2, 0, 1}, {1, 3, 5, 7}, 25] (* G. C. Greubel, Oct 04 2016 *)

Vec((1 + 3*x + 3*x^2 + x^3)/(1 - 2*x^2 - x^4) + O(x^50)) \\ Michel Marcus, Oct 05 2016

G.f.: (1 + 3*x + 3*x^2 + x^3)/(1 - 2*x^2 - x^4). - G. C. Greubel, Oct 04 2016

a(n) = 2*a(n-2) + a(n-4). - Wesley Ivan Hurt, Dec 30 2023

cp's OEIS Frontend

A111566 a(n) = ((1+sqrt(8))*(2+sqrt(2))^n + (1-sqrt(8))*(2-sqrt(2))^n)/2.

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A048693 Generalized Pellian with 2nd term equal to 6.

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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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A048697 Generalized Pellian with second term equal to 10.

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Haskell

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A117584 Generalized Pellian triangle.

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Sage

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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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A048695 Generalized Pellian with second term equal to 8.

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A111566 a(n) = ((1+sqrt(8))(2+sqrt(2))^n + (1-sqrt(8))(2-sqrt(2))^n)/2.

A117895 Triangle T(n, k) = (k-n)A000129(k+1) + (3n-3k+1)A000129(k) with T(n,0) = 1, for 0 <= k <= n-1, read by rows.