A182432 -id:A182432 - cp's OEIS frontend

A084159 Pell oblongs.

1, 3, 21, 119, 697, 4059, 23661, 137903, 803761, 4684659, 27304197, 159140519, 927538921, 5406093003, 31509019101, 183648021599, 1070379110497, 6238626641379, 36361380737781, 211929657785303, 1235216565974041, 7199369738058939, 41961001862379597, 244566641436218639

Offset: 0

Paul Barry, May 18 2003

Essentially the same as A046727.

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 60 at p. 123.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. E. Trier, "Almost Isosceles" Right-Angled Triangles, Eureka, No. 4, May 1940, pp. 9 - 11.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A046727 (same sequence except for first term).

Cf. A001333, A001654, A078057, A084158, A084175, A090390, A182432, A211955.

[Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)+2*(-1)^n)/4): n in [0..30]]; // Vincenzo Librandi, Aug 13 2011

b[n_]:= Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]];
Join[{1}, Table[b[n+1], {n,50}]*Table[b[n], {n,50}]] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2011 *)
LinearRecurrence[{5,5,-1},{1,3,21},30] (* Harvey P. Dale, Aug 04 2019 *)

[(lucas_number2(2*n+1, 2, -1) + 2*(-1)^n)/4 for n in range(31)] # G. C. Greubel, Oct 11 2022

a(n) = ((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) + 2*(-1)^n)/4.
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3). - Paul Curtz, May 17 2008
G.f.: (1-x)^2/((1+x)*(1-6*x+x^2)). - R. J. Mathar, Sep 17 2008
a(n) = A078057(n)*A001333(n). - R. J. Mathar, Jul 08 2009
a(n) = A001333(n)*A001333(n+1).
From Peter Bala, May 01 2012: (Start)
a(n) = (-1)^n*R(n,-4), where R(n,x) is the n-th row polynomial of A211955.
a(n) = (-1)^n*1/u*T(n,u)*T(n+1,u) with u = sqrt(-1) and T(n,x) the Chebyshev polynomial of the first kind.
a(n) = (-1)^n + 4*Sum_{k = 1..n} (-1)^(n-k)*8^(k-1)*binomial(n+k,2*k).
Recurrence equations: a(n) = 6*a(n-1) - a(n-2) + 4*(-1)^n, with a(0) = 1 and a(1) = 3; a(n)*a(n-2) = a(n-1)*(a(n-1)+4*(-1)^n).
Sum_{k >= 0} (-1)^k/a(k) = 1/sqrt(2).
1 - 2*(Sum_{k = 0..n} (-1)^k/a(k))^2 = (-1)^(n+1)/A090390(n+1). (End)
a(n) = (A001333(2*n+1) + (-1)^n)/2. - G. C. Greubel, Oct 11 2022
E.g.f.: exp(-x)*(1 + exp(4*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2. - Stefano Spezia, Aug 03 2024

A302329 a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).

Original entry on oeis.org

1, 61, 3781, 234361, 14526601, 900414901, 55811197261, 3459393815281, 214426605350161, 13290990137894701, 823826961944121301, 51063980650397625961, 3165142973362708688281, 196187800367837541047461, 12160478479832564836254301, 753753477949251182306719201

Offset: 0

Bruno Berselli, Apr 05 2018

Centered hexagonal numbers (A003215) with index in A145607. Example: 35 is a member of A145607, therefore A003215(35) = 3781 is a term of this sequence.
Also, centered 10-gonal numbers (A062786) with index in A182432. Example: 28 is a member of A182432 and A062786(28) = 3781.
a(n) is a solution to the Pell equation (4*a(n))^2 - 15*b(n)^2 = 1. The corresponding b(n) are A258684(n). - Klaus Purath, Jul 19 2025

Colin Barker, Table of n, a(n) for n = 0..500
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (62,-1).

Cf. A003215, A145607; A062786, A182432.
Fourth row of the array A188646.
First bisection of A041449, A042859.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k: A000012 (k=1), A001570 (k=2), A077420 (k=3), this sequence (k=4), A302330 (k=5), A302331 (k=6), A302332 (k=7), A253880 (k=8).

Mathematica
```
LinearRecurrence[{62, -1}, {1, 61}, 20]
```

x='x+O('x^99); Vec((1-x)/(1-62*x+x^2)) \\ Altug Alkan, Apr 06 2018

G.f.: (1 - x)/(1 - 62*x + x^2).
a(n) = a(-1-n).
a(n) = cosh((2*n + 1)*arccosh(4))/4.
a(n) = ((4 + sqrt(15))^(2*n + 1) + 1/(4 + sqrt(15))^(2*n + 1))/8.
a(n) = (1/4)*T(2*n+1, 4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022
E.g.f.: exp(31*x)*(4*cosh(8*sqrt(15)*x) + sqrt(15)*sinh(8*sqrt(15)*x))/4. - Stefano Spezia, Jul 25 2025

A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 10, 4, 1, 10, 30, 28, 8, 1, 15, 70, 112, 72, 16, 1, 21, 140, 336, 360, 176, 32, 1, 28, 252, 840, 1320, 1056, 416, 64, 1, 36, 420, 1848, 3960, 4576, 2912, 960, 128, 1, 45, 660, 3696, 10296, 16016, 14560, 7680, 2176, 256

Offset: 0

Peter Bala, Apr 30 2012

Let b(n,x) = Sum_{k = 0..n} binomial(n+k,2*k)*x^k denote the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients (in ascending powers of x) of the related polynomial sequence R(n,x) := (1/2)*b(n,2*x) + 1/2. Several sequences already in the database are of the form (R(n,x))n>=0 for a fixed value of x. These include A101265 (x = 1), A011900 (x = 2), A182432 (x = 3), A054318 (x = 4) as well as signed versions of A133872 (x = -1), A109613(x = -2), A146983 (x = -3) and A084159 (x = -4).

The polynomials R(n,x) factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x). The coefficients of P(n,x) occur in several tables in the database, although without the connection to the Morgan-Voyce polynomials being noted - see A211956 for more details. In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u * T(2*n+1,u), where u = sqrt((x+2)/2). Hence R(n,x) = 1/u * T(n,u) * T(n+1,u).

			Triangle begins
.n\k.|..0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1....1
..2..|..1....3....2
..3..|..1....6...10....4
..4..|..1...10...30...28....8
..5..|..1...15...70..112...72...16
..6..|..1...21..140..336..360..176...32

Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials

Cf. A011900, A084159, A085478, A101265 (row sums), A109613, A112373, A123519, A133872 (alt row sums), A146983, A182432, A204021, A208513, A211956, A211957.

T(n,0) = 1; T(n,k) = 2^(k-1)*binomial(n+k,2*k) for k > 0.
O.g.f. for column k (except column 0): 2^(k-1)*x^k/(1-x)^(2*k+1).
O.g.f.: (1-t*(x+2)+t^2)/((1-t)*(1-2*t(x+1)+t^2)) = 1 + (1+x)*t + (1+3*x+2*x^2)*t^2 + ....
Removing the first column from the triangle produces the Riordan array (x/(1-x)^3, 2*x/(1-x)^2).
The row polynomials R(n,x) := 1/2*b(n,2*x) + 1/2 = 1 + x*Sum_{k = 1..n} binomial(n+k,2*k)*(2*x)^(k-1).
Recurrence equation: R(n,x) = 2*(1+x)*R(n-1,x) - R(n-2,x) - x with initial conditions R(0,x) = 1, R(1,x) = 1+x.
Another recurrence is R(n,x)*R(n-2,x) = R(n-1,x)*(R(n-1,x) + x).
With P(n,x) as defined in the Comments section we have (x+2)/x - {Sum_{k = 0..2n} 1/R(k,x)}^2 = 2/(x*P(2*n+1,x)^2); (x+2)/x - {Sum_{k = 0..2n+1} 1/R(k,x)}^2 = (x+2)/(x*P(2*n+2,x)^2); consequently Sum_{k >= 0} 1/R(k,x) = sqrt((x+2)/x) for either x > 0 or x <= -2.
Row sums R(n,1) = A101265(n+1); Alt. row sums R(n,-1) = A133872(n+1);
R(n,2) = A011900(n); R(n,-2) = (-1)^n * A109613(n); R(n,3) = A182432;
R(n,-3) = (-1)^n * A146983(n); R(n,4) = A054318(n+1); R(n,-4) = (-1)^n * A084159(n).

A054318 a(n)-th star number (A003154) is a square.

Original entry on oeis.org

1, 5, 45, 441, 4361, 43165, 427285, 4229681, 41869521, 414465525, 4102785725, 40613391721, 402031131481, 3979697923085, 39394948099365, 389969783070561, 3860302882606241, 38213059042991845, 378270287547312205

Offset: 1

Ignacio Larrosa Cañestro, Feb 27 2000

A two-way infinite sequence which is palindromic.
Also indices of centered hexagonal numbers (A003215) which are also centered square numbers (A001844). - Colin Barker, Jan 02 2015
Also positive integers y in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0. - Colin Barker, Jan 02 2015

			a(2) = 5 because the 5th Star number (A003154) 121=11^2 is the 2nd that is a square.

Colin Barker, Table of n, a(n) for n = 1..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

A031138 is 3*a(n)-2. Cf. A003154, A006061, A182432, A211955.
Quintisection of column k=2 of A233427.
Cf. A001844, A003215, A253475.

a:=[1,5,45];; for n in [4..30] do a[n]:=11*a[n-1]-11*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) )); // G. C. Greubel, Jul 23 2019

CoefficientList[Series[x(1-6x+x^2)/((1-x)(1-10x+x^2)), {x,0,30}], x] (* Michael De Vlieger, Aug 11 2016 *)
LinearRecurrence[{11,-11,1},{1,5,45},30] (* Harvey P. Dale, Nov 05 2016 *)

a(n)=if(n<1,a(1-n),1/2+subst(poltchebi(n)+poltchebi(n-1),x,5)/12)

Vec(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ Colin Barker, Jan 02 2015

(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 1/2 + (3 - sqrt(6))/12*(5 + 2*sqrt(6))^n + (3 + sqrt(6))/12*(5 - 2*sqrt(6))^n.
From Michael Somos, Mar 18 2003: (Start)
G.f.: x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)).
12*a(n)*a(n-1) + 4 = (a(n) + a(n-1) + 2)^2.
a(n) = a(1-n) = 10*a(n-1) - a(n-2) - 4.
a(n) = 12*a(n-1)^2/(a(n-1) + a(n-2)) - a(n-1).
a(n) = (a(n-1) + 4)*a(n-1)/a(n-2). (End)
From Peter Bala, May 01 2012: (Start)
a(n+1) = 1 + (1/2)*Sum_{k = 1..n} 8^k*binomial(n+k,2*k).
a(n+1) = R(n,4), where R(n,x) is the n-th row polynomial of A211955.
a(n+1) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(3) and T(n,x) the Chebyshev polynomial of the first kind.
Sum {k>=0} 1/a(k) = sqrt(3/2). (End)
A003154(a(n)) = A006061(n). - Zak Seidov, Oct 22 2012
a(n) = (4*a(n-1) + a(n-1)^2) / a(n-2), n >= 3. - Seiichi Manyama, Aug 11 2016
2*a(n) = 1+A072256(n). - R. J. Mathar, Feb 07 2022

More terms from James Sellers, Mar 01 2000

A146983 a(n) = A002531(n)*A002531(n+1).

Original entry on oeis.org

1, 2, 10, 35, 133, 494, 1846, 6887, 25705, 95930, 358018, 1336139, 4986541, 18610022, 69453550, 259204175, 967363153, 3610248434, 13473630586, 50284273907, 187663465045, 700369586270, 2613814880038, 9754889933879, 36405744855481, 135868089488042

Offset: 0

Paul Barry, Nov 04 2008

a(n+1) is the Hankel transform of A051960 aerated.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,3,-1).

Cf. A001654, A182432, A211955.

a:=[1,2,10];; for n in [4..30] do a[n]:=3*a[n-1]+3*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 09 2020

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x+x^2)/((1+x)*(1-4*x+x^2)) )); // G. C. Greubel, Jan 09 2020

seq(coeff(series((1-x+x^2)/((1+x)*(1-4*x+x^2)), x, n+1), x, n), n = 0..30); # G. C. Greubel, Jan 09 2020

LinearRecurrence[{3,3,-1}, {1,2,10}, 30] (* G. C. Greubel, Jan 09 2020 *)

my(x='x+O('x^30)); Vec((1-x+x^2)/((1+x)*(1-4*x+x^2))) \\ G. C. Greubel, Jan 09 2020

def A146983_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-x+x^2)/((1+x)*(1-4*x+x^2)) ).list()
A146983_list(30) # G. C. Greubel, Jan 09 2020

From Peter Bala, May 01 2012: (Start)
a(n) = (-1)^n + 3*Sum_{k = 1..n} (-1)^(n-k)*6^(k-1)*binomial(n+k,2*k).
a(n) = (-1)^n*R(n,-3), where R(n,x) is the n-th row polynomial of A211955.
a(n) = (-1)^n*1/u*T(n,u)*T(n+1,u) with u = sqrt(-1/2) and T(n,x) denotes the Chebyshev polynomial of the first kind  Cf. A182432.
Recurrence: a(n) = 4*a(n-1) -a(n-2) +3*(-1)^n, with a(0) = 1 and a(1) = 2; a(n)*a(n-2) = a(n-1)*(a(n-1)+3*(-1)^n).
Sum_{k>=0} (-1)^k/a(k) = 1/sqrt(3). (End)
From Colin Barker, Jul 29 2013: (Start)
a(n) = 3*a(n-1) + 3*a(n-2) - a(n-3).
G.f.: (1-x+x^2)/((1+x)*(1-4*x+x^2)). (End)

More terms from Colin Barker, Jul 29 2013

A253470 Indices of centered triangular numbers (A005448) which are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 5, 36, 280, 2201, 17325, 136396, 1073840, 8454321, 66560725, 524031476, 4125691080, 32481497161, 255726286205, 2013328792476, 15850904053600, 124793903636321, 982500325036965, 7735208696659396, 60899169248238200, 479458145289246201, 3774765993065731405

Offset: 1

Colin Barker, Jan 01 2015

Also indices of pentagonal numbers (A000326) which are also centered pentagonal numbers (A005891).

Also positive integers x in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0, the corresponding values of y being A182432.

			5 is in the sequence because the 5th centered triangular number is 31, which is also the 4th centered pentagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A005448, A005891, A182432, A253654.

Vec(x*(4*x-1)/((x-1)*(x^2-8*x+1)) + O(x^100))

a(n) = 9*a(n-1)-9*a(n-2)+a(n-3).
G.f.: x*(4*x-1) / ((x-1)*(x^2-8*x+1)).
a(n) = (6-(4-sqrt(15))^n*(3+sqrt(15))+(-3+sqrt(15))*(4+sqrt(15))^n)/12. - Colin Barker, Mar 03 2016

cp's OEIS Frontend

A084159 Pell oblongs.

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A302329 a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).

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A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

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A054318 a(n)-th star number (A003154) is a square.

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A146983 a(n) = A002531(n)*A002531(n+1).

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A253470 Indices of centered triangular numbers (A005448) which are also centered pentagonal numbers (A005891).

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