A157456 -id:A157456 - cp's OEIS frontend

A159673 Expansion of 56x^2/(1 - 783x + 783*x^2 - x^3).

0, 56, 43848, 34289136, 26814060560, 20968561068840, 16397387941772376, 12822736401904929248, 10027363468901712899616, 7841385409944737582570520, 6131953363213315887857247080, 4795179688647403079566784646096, 3749824384568905994905337736000048

Offset: 1

Paul Weisenhorn, Apr 19 2009

Previous name was: The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 13*n(j) + 1 = a(j)*a(j) and 15*n(j) + 1 = b(j)*b(j) with positive integer numbers.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (783,-783,1).

Cf. A157456, A159668, A159669.

b:= func< n | Evaluate(ChebyshevSecond(n),391) >;
[(14/195)*(-1 +b(n+1) -781*b(n)): n in [1..30]]; // G. C. Greubel, Sep 25 2022

for a from 1 by 2 to 100000 do b:=sqrt((15*a*a-2)/13): if (trunc(b)=b) then
n:=(a*a-1)/13: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:
# Second program
seq((14/195)*(simplify(ChebyshevU(n, 391) -781*ChebyshevU(n-1, 391)) -1), n=1..30); # G. C. Greubel, Sep 25 2022

CoefficientList[Series[56 x/(- x^3 + 783 x^2 - 783 x + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 26 2014 *)
LinearRecurrence[{783,-783,1},{0,56,43848},20] (* Harvey P. Dale, Jan 06 2019 *)

Vec(56*x^2/(-x^3+783*x^2-783*x+1) + O(x^100)) \\ Colin Barker, Feb 24 2014

a(n) = round(-((391+28*sqrt(195))^(-n)*(-1+(391+28*sqrt(195))^n)*(14+sqrt(195)+(-14+sqrt(195))*(391+28*sqrt(195))^n))/390) \\ Colin Barker, Jul 25 2016

def A159673(n): return (14/195)*(-1 + chebyshev_U(n, 391) - 781*chebyshev_U(n-1, 391))
[A159673(n) for n in range(1,30)] # G. C. Greubel, Sep 25 2022

The a(j) recurrence is a(1)=1, a(2)=27, a(t+2) = 28*a(t+1) - a(t) resulting in terms 1, 27, 755, 21113, ... (A159668).
The b(j) recurrence is b(1)=1, b(2)=29, b(t+2) = 28*b(t+1) - b(t) resulting in terms 1, 29, 811, 22679, ... (A159669).
The n(j) recurrence is n(0) = n(1) = 0, n(2) = 56, n(t+3) = 783*(n(t+2) -n(t+1)) + n(t) resulting in terms 0, 0, 56, 43848, 34289136, ... (this sequence).
G.f.: 56*x^2/((1-x)*(1 - 782*x + x^2)). - Vincenzo Librandi, Feb 26 2014
a(n) = -((391+28*sqrt(195))^(-n)*(-1+(391+28*sqrt(195))^n)*(14+sqrt(195)+(-14+sqrt(195))*(391+28*sqrt(195))^n))/390. - Colin Barker, Jul 25 2016
a(n) = (14/195)*(-1 + ChebyshevU(n, 391) - 781*ChebyshevU(n-1, 391)). - G. C. Greubel, Sep 25 2022

More terms and new name from Colin Barker, Feb 24 2014

A159675 Expansion of x(1 + x)/(1 - 32x + x^2).

Original entry on oeis.org

1, 33, 1055, 33727, 1078209, 34468961, 1101928543, 35227244415, 1126169892737, 36002209323169, 1150944528448671, 36794222701034303, 1176264181904649025, 37603659598247734497, 1202140842962022854879, 38430903315186483621631, 1228586765243005453037313

Offset: 1

Paul Weisenhorn, Apr 19 2009

Previous name was: The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 15*n(j) + 1 = a(j)*a(j) and 17*n(j) + 1 = b(j)*b(j) with positive integer numbers.

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (32,-1).

Cf. A157456, A159674, A159677.

[n le 2 select (33)^(n-1) else 32*Self(n-1) -Self(n-2): n in [1..30]]; // G. C. Greubel, Sep 25 2022

for a from 1 by 2 to 100000 do b:=sqrt((17*a*a-2)/15): if (trunc(b)=b) then
n:=(a*a-1)/15: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:

LinearRecurrence[{32,-1},{1,33},20] (* or *)
CoefficientList[Series[(1+x)/(1-32 x+x^2),{x,0,20}], x] (* Harvey P. Dale, Apr 22 2011 *)

Vec(x*(1+x)/(1-32*x+x^2) + O(x^20)) \\ Colin Barker, Feb 24 2014

a(n) = round((16+sqrt(255))^(-n)*(-15-sqrt(255)+(-15+sqrt(255))*(16+sqrt(255))^(2*n))/30) \\ Colin Barker, Jul 25 2016

def A159675(n): return chebyshev_U(n-1, 16) + chebyshev_U(n-2, 16)
[A159675(n) for n in range(1,30)] # G. C. Greubel, Sep 25 2022

The a(j) recurrence is a(1)=1; a(2)=31; a(t+2)=32*a(t+1)-a(t) resulting in terms 1, 31, 991, 31681... (A159674).
The b(j) recurrence is b(1)=1; b(2)=33; b(t+2)=32*b(t+1)-b(t) resulting in terms 1, 33, 1055, 33727... (this sequence).
The n(j) recurrence is n(0)=n(1)=0; n(2)=64; n(t+3)=1023*(n(t+2)-n(t+1))+n(t) resulting in terms 0, 0, 64, 65472, 66912384... (A159677).
G.f.: x*(1 + x)/(1 - 32*x + x^2). - Harvey P. Dale, Apr 22 2011
a(n) = (16+sqrt(255))^(-n)*(-15 - sqrt(255) + (-15 + sqrt(255))*(16 + sqrt(255))^(2*n))/30. - Colin Barker, Jul 25 2016
a(n) = ChebyshevU(n-1, 16) + ChebyshevU(n-2, 16). - G. C. Greubel, Sep 25 2022

More terms from Harvey P. Dale, Apr 22 2011

New name from Colin Barker, Feb 24 2014

A159677 Expansion of 64x^2/(1 - 1023x + 1023*x^2 - x^3).

Original entry on oeis.org

0, 0, 64, 65472, 66912384, 68384391040, 69888780730560, 71426265522241344, 72997573474949923072, 74603448665133299138304, 76244651538192756769423680, 77921959268584332285051862720, 79636166127841649402566234276224, 81388083860694897105090406378438272

Offset: 0

Paul Weisenhorn, Apr 19 2009

Previous name was: The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 15*n(j) + 1 = a(j)*a(j) and 17*n(j) + 1 = b(j)*b(j) with positive integer numbers.

Colin Barker, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (1023,-1023,1).

Cf. A157456, A159674, A159675.

I:=[0,0,64]; [n le 3 select I[n] else 1023*Self(n-1) - 1023*Self(n-2) +Self(n-3): n in [1..30]]; // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((17*a*a-2)/15): if (trunc(b)=b) then
n:=(a*a-1)/15: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:

RecurrenceTable[{a[0]==a[1]==0,a[2]==64,a[n]==1023(a[n-1]-a[n-2])+ a[n-3]}, a,{n,20}] (* Harvey P. Dale, Jan 01 2014 *)
LinearRecurrence[{1023,-1023,1},{0,0,64},20] (* Harvey P. Dale, Jan 01 2014 *)

concat([0, 0], Vec(64/(-x^3+1023*x^2-1023*x+1) + O(x^20))) \\ Colin Barker, Mar 04 2014

a(n) = round(-((511+32*sqrt(255))^(-n)*(-1+(511+32*sqrt(255))^n)*(16+sqrt(255)+(-16+sqrt(255))*(511+32*sqrt(255))^n))/510) \\ Colin Barker, Jul 25 2016

def A159677(n): return (16/255)*(-1 +chebyshev_U(n, 511) -1021*chebyshev_U(n-1, 511))
[A159677(n) for n in range(31)] # G. C. Greubel, Sep 25 2022

The a(j) recurrence is a(0)=1, a(1)=31, a(t+2) = 32*a(t+1) - a(t) resulting in terms 1, 31, 991, 31681, ... (A159674).
The b(j) recurrence is b(0)=1, b(1)=33, b(t+2) = 32*b(t+1) - b(t) resulting in terms 1, 33, 1055, 33727, ... (A159675).
The n(j) recurrence is n(-1) = n(0) = 0, n(1) = 64, n(t+3) = 1023*(n(t+2) -n(t+1)) + n(t) resulting in terms 0, 0, 64, 65472, 66912384, ... (this sequence).
a(n) = -((511+32*sqrt(255))^(-n)*(-1+(511+32*sqrt(255))^n)*(16+sqrt(255)+(-16+sqrt(255))*(511+32*sqrt(255))^n))/510. - Colin Barker, Jul 25 2016
a(n) = (16/255)*(-1 + ChebyshevU(n, 511) - 1021*ChebyshevU(n-1, 511)). - G. C. Greubel, Sep 25 2022

More terms from Harvey P. Dale, Jan 01 2014
New name from Colin Barker, Feb 24 2014
Offset changed to 0 by Colin Barker, Mar 04 2014

A159679 a(n) are solutions of the 2 equations: 7a(n) +1 = c(n)^2 and 9a(n) +1 = b(n)^2.

Original entry on oeis.org

0, 32, 8160, 2072640, 526442432, 133714305120, 33962907058080, 8626444678447232, 2191082985418538880, 556526451851630428320, 141355527687328710254432, 35903747506129640774197440, 9119410511029241427935895360, 2316294366053921193054943224032

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..400
Index entries for linear recurrences with constant coefficients, signature (255,-255,1).

Cf. A077412, A157456, A159678.

R:=PowerSeriesRing(Integers(), 50); [0] cat Coefficients(R!(32*x^2/((1-x)*(1-254*x+x^2)))); // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((9*a*a-2)/7): if (trunc(b)=b) then
n:=(a*a-1)/7: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
# Second program
seq((8/63)*(simplify(ChebyshevU(n,127) -253*ChebyshevU(n-1,127)) -1), n=1..30); # G. C. Greubel, Sep 27 2022

LinearRecurrence[{255,-255,1}, {0, 32, 8160}, 50] (* or *) CoefficientList[Series[32*x^2/((1-x)*(x^2-254*x+1)), {x,0,50}], x] (* G. C. Greubel, Jun 03 2018 *)

concat(0, Vec(32*x^2/(-x^3+255*x^2-255*x+1) + O(x^100))) \\ Colin Barker, Mar 18 2014

a(n) = round((-16+(8+3*sqrt(7))*(127+48*sqrt(7))^(-n)+(8-3*sqrt(7))*(127+48*sqrt(7))^n)/126) \\ Colin Barker, Jul 25 2016

[(8/63)*(-1 + chebyshev_U(n, 127) - 253*chebyshev_U(n-1, 127)) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

G.f.: 32*x^2 / ((1-x)*(1-254*x+x^2)).
c(n) = A157456(n).
b(n) = A159678(n).
a(n+3) = 255*(a(n+2) -a(n+1)) + a(n).
a(n) = 2*A077412(n-2)*A077412(n-1). - Johannes Boot, Jan 17 2011
a(n) = (-16+(8+3*sqrt(7))*(127+48*sqrt(7))^(-n)+(8-3*sqrt(7))*(127+48*sqrt(7))^n)/126. - Colin Barker, Jul 25 2016
a(n) = (8/63)*(-1 + ChebyshevU(n, 127) - 253*ChebyshevU(n-1, 127)). - G. C. Greubel, Sep 27 2022

More terms from Colin Barker, Mar 18 2014

A159683 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 3n(j) + 1 = a(j)a(j) and 5n(j) + 1 = b(j)b(j) with positive integer numbers.

Original entry on oeis.org

0, 16, 1008, 62496, 3873760, 240110640, 14882985936, 922505017408, 57180428093376, 3544264036771920, 219687189851765680, 13617061506772700256, 844038126230055650208, 52316746764756677612656, 3242794261288683956334480, 201000927453133648615125120

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..559
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).

Cf. A057080, A070997, A157456, A245031.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(16*x^2/((1-x)*(1-62*x+x^2)))); // G. C. Greubel, Jun 02 2018

for a from 1 by 2 to 100000 do b:=sqrt((5*a*a-2)/3): if (trunc(b)=b) then
n:=(a*a-1)/3: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
# Second program
seq((4/15)*(simplify(ChebyshevU(n, 31) - 61*ChebyshevU(n-1, 31)) -1), n=1..30); # G. C. Greubel, Sep 27 2022

CoefficientList[Series[16*x/((1-x)*(1-62*x+x^2)), {x, 0, 30}], x] (* G. C. Greubel, Jun 02 2018 *)
LinearRecurrence[{63,-63,1},{0,16,1008},30] (* Harvey P. Dale, May 07 2022 *)

concat(0, Vec(16*x^2/((1-x)*(1-62*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 25 2015

[(4/15)*(-1 + chebyshev_U(n, 31) - 61*chebyshev_U(n-1, 31)) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

The a(j) recurrence is a(1)=1, a(2)=7, a(t+2) = 8*a(t+1) - a(t) resulting in terms 1, 7, 55. 433, 3409, ... (A070997).
The b(j) recurrence is b(1)=1, b(2)=9, b(t+2) = 8*b(t+1) - b(t) resulting in terms 1, 9, 71, 559, 4401, ... (A057080).
The n(j) recurrence is n(0) = n(1) = 0, n(2)=16, n(t+3) = 63*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 16, 1008, 62496, ... (this sequence).
From Colin Barker, Sep 25 2015: (Start)
a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3) for n>3.
G.f.: 16*x^2 / ((1-x)*(1-62*x+x^2)). (End)
a(n) = (-8+(4+sqrt(15))*(31+8*sqrt(15))^(-n) -(-4+sqrt(15))*(31+8*sqrt(15))^n)/30. - Colin Barker, Mar 03 2016
a(n) = (4/15)*(-1 + ChebyshevU(n, 31) - 61*ChebyshevU(n-1, 31)). - G. C. Greubel, Sep 27 2022

A266698 x-values of solutions to the Diophantine equation x^2 - 7*y^2 = 2.

Original entry on oeis.org

3, 45, 717, 11427, 182115, 2902413, 46256493, 737201475, 11748967107, 187246272237, 2984191388685, 47559815946723, 757972863758883, 12080006004195405, 192522123203367597, 3068273965249686147, 48899861320791610755, 779329507167416085933, 12420372253357865764173, 197946626546558436140835

Offset: 1

Sture Sjöstedt, Jan 03 2016

A159678 gives the y-values of solutions to the Diophantine equation x^2 - 7*y^2 = 2.

G. C. Greubel, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Cf. A041008, A077412, A157456, A159678.

[n: n in [1..2*10^7] | IsSquare((n^2-2)/7)]; // Vincenzo Librandi, Jan 06 2016

Mathematica
```
LinearRecurrence[{16,-1}, {3, 45}, 20 ]
```

lista(nn) = {print1(x = 3, ", "); print1(y = 45, ", "); for (n=2, nn, z = 16*y - x; print1(z, ", "); x = y; y = z;);} \\ Michel Marcus, Jan 05 2016

[3*(chebyshev_U(n-1, 8) - chebyshev_U(n-2, 8)) for n in (1..30)] # G. C. Greubel, Jun 25 2022

a(1)=3, a(2)=45, a(n) = 16*a(n-1) - a(n-2).
a(n) = A041008(4n-2). - Robert Israel, Jan 05 2016
From R. J. Mathar, Jan 12 2016: (Start)
G.f.: 3*x*(1-x) / ( 1-16*x+x^2 ).
a(n) = 3*A157456(n). (End)
From G. C. Greubel, Jun 25 2022: (Start)
a(n) = 3*(ChebyshevU(n-1, 8) - ChebyshevU(n-2, 8)).
E.g.f.: exp(8*x)*(3*cosh(3*sqrt(7)*x) - sqrt(7)*sinh(3*sqrt(7)*x)) - 3. (End)

A159665 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 11n(j) + 1 = a(j)a(j) and 13n(j) + 1 = b(j)b(j); with positive integer numbers.

Original entry on oeis.org

0, 48, 27600, 15842400, 9093510048, 5219658925200, 2996075129554800, 1719741904705530048, 987128857225844692800, 566610244305730148137200, 325233293102631879186060048, 186683343630666392922650330400, 107155914010709406905722103589600

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..363
Index entries for linear recurrences with constant coefficients, signature (575,-575,1).

Cf. A157456, A159661, A159664.

I:=[0,48,27600]; [n le 3 select I[n] else 575*Self(n-1) -575*Self(n-2) +Self(n-3): n in [1..31]]; // G. C. Greubel, Jun 26 2022

for a from 1 by 2 to 100000 do b:=sqrt((13*a*a-2)/11): if (trunc(b)=b) then
n:=(a*a-1)/11: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:

LinearRecurrence[{575,-575,1}, {0,48,27600}, 30] (* G. C. Greubel, Jun 26 2022 *)

concat(0, Vec(-48*x^2/((x-1)*(x^2-574*x+1)) + O(x^30))) \\ Colin Barker, Sep 25 2015

a(n) = round((-24+(12+sqrt(143))*(287+24*sqrt(143))^(-n)-(-12+sqrt(143))*(287+24*sqrt(143))^n)/286) \\ Colin Barker, Jul 26 2016

[(12/143)*(chebyshev_U(n,287) -573*chebyshev_U(n-1,287) -1) for n in (1..30)] # G. C. Greubel, Jun 26 2022

The a(j) recurrence is a(1)=1; a(2)=23; a(t+2) = 24*a(t+1) - a(t) resulting in terms 1, 23, 551, 13201, ... (A159664).
The b(j) recurrence is b(1)=1; b(2)=25; b(t+2) = 24*b(t+1) - b(t) resulting in terms 1, 25, 599, 14351, ... (A159661).
The n(j) recurrence is n(0)=n(1)=1; n(2)=48; n(t+3) = 575*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 48, 27600, 15842400 as listed above.
From Colin Barker, Sep 25 2015: (Start)
a(n) = 575*a(n-1) - 575*a(n-2) + a(n-3) for n > 3.
G.f.: 48*x^2 / ((1-x)*(1-574*x+x^2)). (End)
a(n) = (-24 + (12 + sqrt(143))*(287 + 24*sqrt(143))^(-n) - (-12 + sqrt(143))*(287 + 24*sqrt(143))^n)/286. - Colin Barker, Jul 26 2016
From G. C. Greubel, Jun 25 2022: (Start)
a(n) = (12/143)*(ChebyshevU(n, 287) - 573*ChebyshevU(n-1, 287) - 1).
E.g.f.: (12/143)*(exp(287*x)*( (sqrt(143)/12)*sinh(24*sqrt(143)*x) + cosh(24*sqrt(143)*x) ) - exp(x)). (End)

A159681 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 5n(j)+1=a(j)a(j) and 7n(j)+1=b(j)b(j) with positive integer numbers.

Original entry on oeis.org

0, 24, 3432, 487344, 69199440, 9825833160, 1395199109304, 198108447688032, 28130004372591264, 3994262512460271480, 567157146764985958920, 80532320578115545895184, 11435022364945642531157232, 1623692643501703123878431784, 230552920354876897948206156120

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..450
Index entries for linear recurrences with constant coefficients, signature (143,-143,1).

Cf. A077417, A077416, A157456.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients( R!(24*x^2/((1-x)*(1-142*x+x^2)))); // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((7*a*a-2)/5): if (trunc(b)=b) then
n:=(a*a-1)/5: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
# Second program
seq((6/35)*(simplify(ChebyshevU(n,71) -141*ChebyshevU(n-1,71)) -1), n=1..30); # G. C. Greubel, Sep 27 2022

LinearRecurrence[{143,-143,1}, {0, 24, 3432}, 30] (* or *) CoefficientList[Series[24*x^2/((1-x)*(1-142*x+x^2)), {x,0,30}], x] (* G. C. Greubel, Jun 03 2018 *)

concat(0, Vec(-24*x^2/((x-1)*(x^2-142*x+1)) + O(x^20))) \\ Colin Barker, Jul 26 2016

a(n) = round((-12+(6+sqrt(35))*(71+12*sqrt(35))^(-n)-(-6+sqrt(35))*(71+12*sqrt(35))^n)/70) \\ Colin Barker, Jul 26 2016

[(6/35)*(-1 + chebyshev_U(n, 71) - 141*chebyshev_U(n-1, 71)) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

The a(j) recurrence is a(1)=1, a(2)=11, a(t+2) = 12*a(t+1) - a(t) resulting in terms 1, 11, 131, 1561, ... (A077417).
The b(j) recurrence is b(1)=1, b(2)=13, b(t+2) = 12*b(t+1) - b(t) resulting in terms 1, 13, 155, 1847, ... (A077416).
The n(j) recurrence is n(0)=n(1)=0, n(2)=24, n(t+3) = 143*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 24, 3432, 487344, ... (this sequence).
G.f.: 24*x^2/((1-x)*(1-142*x+x^2)). - R. J. Mathar, Apr 20 2009
a(n) = (-12+(6+sqrt(35))*(71+12*sqrt(35))^(-n)-(-6+sqrt(35))*(71+12*sqrt(35))^n)/70. - Colin Barker, Jul 26 2016
a(n) = (6/35)*(ChebyshevU(n, 71) - 141*ChebyshevU(n-1, 71) - 1). - G. C. Greubel, Sep 27 2022

More terms from R. J. Mathar, Apr 20 2009

A243470 Numerators of the rational convergents to the periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))).

Original entry on oeis.org

1, 7, 15, 112, 239, 1785, 3809, 28448, 60705, 453383, 967471, 7225680, 15418831, 115157497, 245733825, 1835294272, 3916322369, 29249550855, 62415424079, 466157519408, 994730462895, 7429270759673, 15853271982241, 118402174635360, 252657621252961, 1887005523406087

Offset: 1

Peter Bala, Jun 06 2014

The sequence of convergents to the simple periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))) begins [0/1, 1/2, 7/15, 15/32, 112/239, 239/510, ...]. The present sequence is the sequence of numerators of the convergents. It is a strong divisibility sequence, that is gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. The sequence is closely related to A041111, the Lehmer numbers U_n(sqrt(R),Q) with parameters R = 14 and Q = -1.

See A243469 for the sequence of denominators to the convergents.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Peter Bala, Notes on 2-periodic continued fractions and Lehmer sequences
Leonhard Euler, Introductio in analysin infinitorum, Vol.1, Chapter 18, section 378. French and German translations.
Eric Weisstein's World of Mathematics, Lehmer Number
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,16,0,-1).

Cf. A041111, A077412, A243469.

I:=[1,7,15,112]; [n le 4 select I[n] else 16*Self(n-2) -Self(n-4): n in [1..31]]; // G. C. Greubel, May 21 2022

LinearRecurrence[{0,16,0,-1},{1,7,15,112},30] (* Harvey P. Dale, Nov 06 2017 *)

Vec(x*(1+7*x-x^2)/(1-16*x^2+x^4)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2015

def b(n): return chebyshev_U(n,8) # b=A077412
def A243470(n): return 7*((n-1)%2)*b(n//2 -1) +(n%2)*(b((n-1)//2) -b((n-1)//2 -1))
[A243470(n) for n in (1..30)] # G. C. Greubel, May 21 2022

Let alpha = ( sqrt(14) + sqrt(18) )/2 and beta = ( sqrt(14) - sqrt(18) )/2 be the roots of the equation x^2 - sqrt(14)*x - 1 = 0. Then a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, while a(n) = 7*(alpha^n - beta^n)/(alpha^2 - beta^2) for n even.
a(2*n + 1) = Product_{k = 1..n} (14 + 4*cos^2(k*Pi/(2*n+1)));
a(2*n) = 7*Product_{k = 1..n-1} (14 + 4*cos^2(k*Pi/(2*n))).
Recurrence equations: a(0) = 0, a(1) = 1 and for n >= 2, a(2*n) = 7*a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 2*a(2*n) + a(2*n - 1).
Fourth-order recurrence: a(n) = 16*a(n - 2) - a(n - 4) for n >= 5.
O.g.f.: x*(1 + 7*x - x^2)/(1 - 16*x^2 + x^4).
a(2n-1) = A157456(n), a(2n) = 7*A077412(n-1). - Ralf Stephan, Jun 13 2014
a(n) = (1/2)*( 7*(1+(-1)^n)*ChebyshevU((n-2)/2, 8) + (1-(-1)^n)*(ChebyshevU((n- 1)/2, 8) - ChebyshevU((n-3)/2, 8)) ). - G. C. Greubel, May 21 2022

A159680 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 9n(j) + 1 = a(j)a(j) and 11n(j) + 1 = b(j)b(j) with positive integer numbers.

Original entry on oeis.org

0, 40, 15960, 6352080, 2528111920, 1006182192120, 400457984351880, 159381271589856160, 63433345634778399840, 25246312181370213280200, 10047968814839710107119800, 3999066341994023252420400240, 1591618356144806414753212175760, 633460106679290959048526025552280

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..350
Index entries for linear recurrences with constant coefficients, signature (399,-399,1).

Cf. A075839, A083043, A157456.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(40*x^2/((1-x)*(1-398*x+x^2)))); // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((9*a*a-2)/7): if (trunc(b)=b) then
n:=(a*a-1)/7: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:

LinearRecurrence[{399,-399,1}, {0,40,15960}, 50] (* G. C. Greubel, Jun 03 2018 *)

a(n) = round((-20+(10+3*sqrt(11))*(199+60*sqrt(11))^(-n)+(10-3*sqrt(11))*(199+60*sqrt(11))^n)/198) \\ Colin Barker, Jul 26 2016

concat(0, Vec(-40*x^2/((x-1)*(x^2-398*x+1)) + O(x^20))) \\ Colin Barker, Jul 26 2016

[(10/99)*(chebyshev_U(n, 199) -397*chebyshev_U(n-1, 199) -1) for n in (1..30)] # G. C. Greubel, Jun 26 2022

The a(j) recurrence is a(1)=1; a(2)=19; a(t+2) = 20*a(t+1) - a(t) resulting in terms 1, 19, 379, 7561, ... (A075839).
The b(j) recurrence is b(1)=1; b(2)=21; b(t+2) = 20*b(t+1) - b(t) resulting in terms 1, 21, 419, 8359, ... (A083043).
The n(j) recurrence is n(0)=n(1)=0; n(2)=40; n(t+3) = 399*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 40, 15960, 6352080 as listed above
G.f.: 40*x^2/((1-x)*(1-398*x+x^2)). - R. J. Mathar, Apr 20 2009
a(n) = (-20 + (10 + 3*sqrt(11))*(199 + 60*sqrt(11))^(-n) + (10 - 3*sqrt(11))*(199 + 60*sqrt(11))^n)/198. - Colin Barker, Jul 26 2016
From G. C. Greubel, Jun 26 2022: (Start)
a(n) = (10/99)*( ChebyshevU(n, 199) - 397*ChebyshevU(n-1, 199) - 1 ).
E.g.f.: (10/99)*(exp(199*x)*( (3*sqrt(11)/10)*sinh(60*sqrt(11)*x) + cosh(60*sqrt(11)*x) ) - exp(x)). (End)

More terms from R. J. Mathar, Apr 20 2009

cp's OEIS Frontend

A159673 Expansion of 56*x^2/(1 - 783*x + 783*x^2 - x^3).

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A159675 Expansion of x*(1 + x)/(1 - 32*x + x^2).

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A159677 Expansion of 64*x^2/(1 - 1023*x + 1023*x^2 - x^3).

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A159679 a(n) are solutions of the 2 equations: 7*a(n) +1 = c(n)^2 and 9*a(n) +1 = b(n)^2.

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A159683 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 3*n(j) + 1 = a(j)*a(j) and 5*n(j) + 1 = b(j)*b(j) with positive integer numbers.

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A266698 x-values of solutions to the Diophantine equation x^2 - 7*y^2 = 2.

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A159673 Expansion of 56x^2/(1 - 783x + 783*x^2 - x^3).

A159675 Expansion of x(1 + x)/(1 - 32x + x^2).

A159677 Expansion of 64x^2/(1 - 1023x + 1023*x^2 - x^3).

A159679 a(n) are solutions of the 2 equations: 7a(n) +1 = c(n)^2 and 9a(n) +1 = b(n)^2.

A159683 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 3n(j) + 1 = a(j)a(j) and 5n(j) + 1 = b(j)b(j) with positive integer numbers.

A159665 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 11n(j) + 1 = a(j)a(j) and 13n(j) + 1 = b(j)b(j); with positive integer numbers.

A159681 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 5n(j)+1=a(j)a(j) and 7n(j)+1=b(j)b(j) with positive integer numbers.

A159680 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 9n(j) + 1 = a(j)a(j) and 11n(j) + 1 = b(j)b(j) with positive integer numbers.