A298271 -id:A298271 - cp's OEIS frontend

A157459 Expansion of 72x^2 / (1 - 323x + 323*x^2 - x^3).

0, 72, 23256, 7488432, 2411251920, 776415629880, 250003421569512, 80500325329753056, 25920854752758914592, 8346434730063040745640, 2687526062225546361181560, 865375045601895865259716752, 278648077157748243067267612656, 89723815469749332371794911558552

Offset: 1

Paul Weisenhorn, Mar 01 2009

This sequence is part of a solution of a more general problem involving two equations, three sequences a(n), b(n), c(n) and a constant A:
    A    * c(n) + 1 = a(n)^2,
   (A+1) * c(n) + 1 = b(n)^2; for details see comment in A157014.
A157459 is the c(n) sequence for A=4.

Colin Barker, Table of n, a(n) for n = 1..250
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A007805, A049629, A157014, A157459, A201003, A298271.

LinearRecurrence[{323,-323,1},{0,72,23256},20] (* Harvey P. Dale, Feb 28 2021 *)

concat(0, Vec(72*x^2/(1-323*x+323*x^2-x^3)+O(x^20))) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = -round((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)*(9+4*sqrt(5)+(-9+4*sqrt(5))*(161+72*sqrt(5))^n))/80 \\ Colin Barker, Jul 25 2016

4*a(n) + 1 = A007805(n-1)^2.
5*a(n) + 1 = A049629(n-1)^2.
G.f.: 72*x^2/(1 - 323*x + 323*x^2 - x^3).
c(1) = 0, c(2) = 72, c(3) = 323*c(2), c(n) = 323*(c(n-1) - c(n-2)) + c(n-3) for n>3.
a(n) = -((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)*(9+4*sqrt(5)+(-9+4*sqrt(5))*(161+72*sqrt(5))^n))/80. - Colin Barker, Jul 25 2016
a(n) = 72*A298271(n-1). - Greg Dresden, Dec 02 2021
a(n) = 2*A201003(n-1). - Amiram Eldar, Dec 01 2024

Edited by Alois P. Heinz, Sep 09 2011

A298101 Expansion of x(1 + x)/((1 - x)(1 - 322*x + x^2)).

Original entry on oeis.org

0, 1, 324, 104329, 33593616, 10817040025, 3483053294436, 1121532343768369, 361129931640120384, 116282716455774995281, 37442673568827908360100, 12056424606446130716956921, 3882131280602085262951768464, 1250034215929265008539752488489

Offset: 0

Bruno Berselli, Jan 12 2018

16*k*a(n) provides infinitely many x-values solutions (x,y) of x*(5*x + k) = y^2.
This follows from the fact that 5*16*a(n) + 1 is a perfect square: more precisely, 80*a(n) + 1 = A023039(n)^2.
This is a divisibility sequence, that is a(n) divides a(m) if n divides m. It is the case P1 = 324, P2 = 644, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Jan 19 2018

Colin Barker, Table of n, a(n) for n = 0..300
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A023039, A049660, A225786, A253368, A298271.

P:=proc(n) trunc(evalf(((2+sqrt(5))^(4*n)+(2-sqrt(5))^(4*n)-2)/320,1000));
end: seq(P(i),i=0..13); # Paolo P. Lava, Jan 18 2018

CoefficientList[x (1 + x)/((1 - x) (1 - 322 x + x^2)) + O[x]^20, x]

makelist(coeff(taylor(x*(1+x)/((1-x)*(1-322*x+x^2)), x, 0, n), x, n), n, 0, 20);

first(n) = Vec(x*(1 + x)/((1 - x)*(1 - 322*x + x^2)) + O(x^n), -n) \\ Iain Fox, Jan 12 2018

gf = x*(1+x)/((1-x)*(1-322*x+x^2))
print(taylor(gf, x, 0, 20).list())

G.f.: x*(1 + x)/((1 - x)*(1 - 322*x + x^2)).
a(n) = a(-n) = ((2 + sqrt(5))^(4*n) + (2 - sqrt(5))^(4*n) - 2)/320.
a(n) = A225786(n)/48. This is the case k=3 of the first comment. Example: for n = 2, 16*3*a(2) = A225786(2) = 15552 and 15552*(5*15552+3) = 34776^2.
a(n) = A049660(n)^2.
a(n)*(80*a(n) + 1) = 81*A253368(n)^2 for n>0.
a(n)*a(n-2) = (a(n-1) - 1)^2.
a(n) = 322*a(n-1) - a(n-2) + 2.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3). - Iain Fox, Jan 12 2018
a(n) = A298271(n)+A298271(n-1). - R. J. Mathar, Nov 20 2020

A201003 Triangular numbers, T(m), that are four-fifths of another triangular number: T(m) such that 5T(m) = 4T(k) for some k.

Original entry on oeis.org

0, 36, 11628, 3744216, 1205625960, 388207814940, 125001710784756, 40250162664876528, 12960427376379457296, 4173217365031520372820, 1343763031112773180590780, 432687522800947932629858376, 139324038578874121533633806328, 44861907734874666185897455779276

Offset: 0

Charlie Marion, Dec 20 2011

Also, numbers m such that 8*m+1 and 10*m+1 are squares. Example: 8*1205625960+1 = 98209^2 and 12056259601 = 109801^2. - Bruno Berselli, Mar 03 2016

			5*0 = 4*0;
5*36 = 4*45;
5*11628 = 4*14535;
5*3744216 = 4*4680270.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008, A298271.

m:=20; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(36*x/((1-x)*(1-322*x+x^2)))); // G. C. Greubel, Jul 15 2018

triNums = Table[(n^2 + n)/2, {n, 0, 4999}]; Select[triNums, MemberQ[triNums, (5/4)#] &] (* Alonso del Arte, Dec 20 2011 *)
CoefficientList[Series[-36 x/((x - 1) (x^2 - 322 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 11 2014 *)
LinearRecurrence[{323,-323,1},{0,36,11628},20] (* Harvey P. Dale, Dec 21 2015 *)

concat(0, Vec(36*x/((1-x)*(1-322*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016

For n>1, a(n) = 322*a(n-1) - a(n-2) + 36. See A200993 for generalization.
G.f.: 36*x / ((1-x)*(x^2-322*x+1)). - R. J. Mathar, Aug 10 2014
From Colin Barker, Mar 02 2016: (Start)
a(n) = (-18+(9-4*sqrt(5))*(161+72*sqrt(5))^(-n)+(9+4*sqrt(5))*(161+72*sqrt(5))^n)/160.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n>2. (End)
a(n) = 36*A298271(n). - Amiram Eldar, Dec 01 2024

A201004 Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4T(m) = 5T(k) for some k.

Original entry on oeis.org

0, 45, 14535, 4680270, 1507032450, 485259768675, 156252138480945, 50312703331095660, 16200534220474321620, 5216521706289400466025, 1679703788890966475738475, 540859403501184915787322970, 174155048223592651917042257910, 56077384668593332732371819724095

Offset: 0

Charlie Marion, Feb 15 2012

			4*0 = 5*0.
4*45 = 5*36.
4*14535 = 5*11628.
4*4680270 = 5*3744216.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008, A298271.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(45*x/((1-x)*(1-322*x+x^2)))); // G. C. Greubel, Jul 15 2018

LinearRecurrence[{323, -323, 1}, {0, 45, 14535}, 20] (* T. D. Noe, Feb 15 2012 *)
CoefficientList[Series[-45 x/((x - 1) (x^2 - 322 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 11 2014 *)

concat(0, Vec(45*x/((1-x)*(1-322*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016

For n > 1, a(n) = 322*a(n-1) - a(n-2) + 45. See A200994 for generalization.
G.f.: 45*x / ((1-x)*(x^2-322*x+1)). - R. J. Mathar, Aug 10 2014
From Colin Barker, Mar 02 2016: (Start)
a(n) = (-18 + (9-4*sqrt(5))*(161+72*sqrt(5))^(-n) + (9+4*sqrt(5))*(161+72*sqrt(5))^n)/128.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2. (End)
a(n) = 45*A298271(n). - Amiram Eldar, Dec 01 2024

a(7) corrected by R. J. Mathar, Aug 10 2014

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A157459 Expansion of 72*x^2 / (1 - 323*x + 323*x^2 - x^3).

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A298101 Expansion of x*(1 + x)/((1 - x)*(1 - 322*x + x^2)).

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A201003 Triangular numbers, T(m), that are four-fifths of another triangular number: T(m) such that 5*T(m) = 4*T(k) for some k.

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A201004 Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4*T(m) = 5*T(k) for some k.

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A157459 Expansion of 72x^2 / (1 - 323x + 323*x^2 - x^3).

A298101 Expansion of x(1 + x)/((1 - x)(1 - 322*x + x^2)).

A201003 Triangular numbers, T(m), that are four-fifths of another triangular number: T(m) such that 5T(m) = 4T(k) for some k.

A201004 Triangular numbers, T(m), that are five-quarters of another triangular number; T(m) such that 4T(m) = 5T(k) for some k.