A070997 -id:A070997 - cp's OEIS frontend

A131751 Numbers that are both centered triangular and centered pentagonal.

1, 31, 1891, 117181, 7263301, 450207451, 27905598631, 1729696907641, 107213302675081, 6645495068947351, 411913480972060651, 25531990325198812981, 1582571486681354344141, 98093900183918770523731, 6080239239916282418127151, 376876738974625591153359601

Offset: 1

Richard Choulet, Sep 20 2007

We solve 0.5*(3*p^2+3*p+2)=0.5*(5*r^2+5*r+2), i.e., 3*(2*p+1)^2=5*(2*r+1)^2-2.

The Diophantine equation 3*X^2=5*Y^2-2is such that : X is given by A057080 which satisfies the new formula a(n+1)=4*a(n)+(15*a(n)^2+10)^0.5, Y is given by A070997 which satisfies the new formula a(n+1)=4*a(n)+(15*a(n)^2-6)^0.5 while r is given by the sequence 0,3,27,216,1704,... which satisfies a(n+2)=8*a(n+1)-a(n)+3 and a(n+1)=4*a(n)+1.5+0.5*(60*a(n)^2+60*a(n)+9)^0.5, p is given by the sequence 0,4,35,279,2200,... which satisfies a(n+2)=8*a(n+1)-a(n)+3 and a(n+1)=4*a(n)+1.5+0.5*sqrt(60*a(n)^2+60*a(n)+25).

Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).

Cf. A050807 A070997.

A131751 := proc(n) coeftayl(x*(1-32*x+x^2)/(1-x)/(1-62*x+x^2),x=0,n) ; end: seq(A131751(n),n=1..20) ; # R. J. Mathar, Oct 24 2007

LinearRecurrence[{63,-63,1},{1,31,1891},20] (* Harvey P. Dale, Oct 01 2017 *)

a(n+2) = 62*a(n+1) - a(n) - 30, a(n+1) = 31*a(n) - 15 + sqrt(960*a(n)^2 - 960*a(n)+225).
G.f.: f(z) = a(1)*z+a(2)*z^2+... = z*(1-32*z+z^2)/((1-z)*(1-62*z+z^2)).
A005891 INTERSECT A005448. - R. J. Mathar, Oct 24 2007

Corrected and extended by R. J. Mathar, Oct 24 2007

A145607 Numbers k such that (3(2k + 1)^2 + 2)/5 is a square.

Original entry on oeis.org

0, 4, 35, 279, 2200, 17324, 136395, 1073839, 8454320, 66560724, 524031475, 4125691079, 32481497160, 255726286204, 2013328792475, 15850904053599, 124793903636320, 982500325036964, 7735208696659395, 60899169248238199

Offset: 1

Richard Choulet, Oct 14 2008

Square roots of (3*(2*k+1)^2+2)/5 are listed in A070997, therefore (3*(2*a(n) + 1)^2 + 2)/5 = A070997(n-1)^2.

Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A070997, A131751.

Cf. A001091 (first differences).

a(n+2) = 8*a(n+1) - a(n) + 3.
From R. J. Mathar, Oct 24 2008: (Start)
G.f.: x^2*(4 - x)/((1 - x)*(1 - 8*x + x^2)).
a(n) = (A057080(n-1)-1)/2. (End)

a(4) corrected, extended, definition corrected by R. J. Mathar, Oct 24 2008

Offset changed by Bruno Berselli, Apr 06 2018

A273053 Numbers n such that 15*n^2 + 16 is a square.

Original entry on oeis.org

0, 4, 32, 252, 1984, 15620, 122976, 968188, 7622528, 60012036, 472473760, 3719778044, 29285750592, 230566226692, 1815244062944, 14291386276860, 112515846151936, 885835382938628, 6974167217357088, 54907502355918076, 432285851629987520, 3403379310683982084

Offset: 1

Vincenzo Librandi, May 14 2016

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A070997, similar sequences listed in A273052.

[n: n in [0..2*10^7] |IsSquare(15*n^2+16)];

a:=proc(n) option remember; if n=1 then 0 elif n=2 then 4 else 8*a(n-1) - a(n-2); fi; end: seq(a(n), n=1..30); # Wesley Ivan Hurt, May 14 2016

Mathematica
```
LinearRecurrence[{8, -1}, {0, 4}, 30]
```

concat(0, Vec(4*x^2/(1-8*x+x^2) + O(x^50))) \\ Colin Barker, May 14 2016

O.g.f.: 4*x^2/(1 - 8*x + x^2).
E.g.f.: 4*(1 + (4*sqrt(15)*sinh(sqrt(15)*x) - 15*cosh(sqrt(15)*x))*exp(4*x)/15). - Ilya Gutkovskiy, May 14 2016
a(n) = 8*a(n-1) - a(n-2) for n>2.
a(n) = -(2*((4-sqrt(15))^n*(4+sqrt(15))+(-4+sqrt(15))*(4+sqrt(15))^n))/sqrt(15). - Colin Barker, May 14 2016
a(n+2) - a(n+1) = 4*070997(n) for n>0. - Wesley Ivan Hurt, May 14 2016

A145542 Numerators in continued fraction expansion of sqrt(3/5).

Original entry on oeis.org

1, 3, 7, 24, 55, 189, 433, 1488, 3409, 11715, 26839, 92232, 211303, 726141, 1663585, 5716896, 13097377, 45009027, 103115431, 354355320, 811826071, 2789833533, 6391493137, 21964312944, 50320119025, 172924670019, 396169459063, 1361433047208, 3119035553479

Offset: 1

Gary W. Adamson, Oct 12 2008

a(n)/A145543(n) tends to sqrt(3/5).

A strong divisibility sequence, that is gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. Related to the Lehmer sequence U_n(sqrt(R),Q) with parameters R = 6 and Q = -1. See A041023. - Peter Bala, Jun 06 2014

			[a(7), a(8)] = [433, 1488] X^4 * [1, 0] = [433, 1488].
a(5) = 55 = 2*a(4) + a(3) = 2*24 + 7.
G.f. = x + 3*x^2 + 7*x^3 + 24*x^4 + 55*x^5 + 189*x^6 + 433*x^7 + 1488*x^8 + ...

Peter Bala, Notes on 2-periodic continued fractions and Lehmer sequences
Eric Weisstein's World of Mathematics, Lehmer Number

Cf. A001090, A041203, A070997, A145543.

Numerator[Convergents[Sqrt[3/5], 30]] (* gives terms with 0 prepended *) (* Wesley Ivan Hurt, Jun 15 2014 *)

{a(n) = if( n<1, 0, polcoeff( x * (1 + 3*x - x^2) / (1 - 8*x^2 + x^4) + x * O(x^n), n))}; /* Michael Somos, Nov 14 2015 */

Numerators in continued fraction expansion of sqrt(3/5); i.e., of [1, 3, 2, 3, 2, 3, 2, 3, 2, ...].
[a(2*n - 1), a(2*n)] = X^n * [1,0], where X is the 2 X 2 matrix [1,2; 3,7].
Empirical G.f.: x*(1+3*x-x^2)/(1-8*x^2+x^4). - Colin Barker, Jan 04 2012
From Peter Bala, Jun 06 2014: (Start)
a(2*n + 1) = Product_{k=1..n} (6 + 4*cos^2(k*Pi/(2*n+1))).
a(2*n) = 3*Product_{k=1..n-1} (6 + 4*cos^2(k*Pi/(2*n))).
a(2*n + 1) = A070997(n); a(2*n) = 3*A001090(n). (End)

More terms from Wesley Ivan Hurt, Jun 15 2014

A158197 Expansion of (1-x^2c(x)^4)/(1-4x*c(x)^2), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 4, 23, 140, 866, 5388, 33603, 209796, 1310510, 8188328, 51169094, 319779544, 1998527188, 12490460620, 78064190235, 487896926580, 3049340393430, 19058321475960, 119114304522450, 744463650984360, 4652895041524380

Offset: 0

Paul Barry, Mar 13 2009

Apply the inverse of the Riordan array (1/(1-x^2),x/(1+x)^2) to 4^n. Hankel transform is A070997.

Cf. A090317, A158196.

Conjecture: +4*(n+1)*a(n) +(-81*n+23)*a(n-1) +10*(51*n-70)*a(n-2) +500*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Feb 05 2015

Conjecture: +4*(n+1)^2*a(n) +(-41*n^2-58*n+23)*a(n-1) +50*(n+2)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Feb 05 2015

A145608 Numbers a(n)=k such that (1/3)(5(2k+1)^2-2) is A057080(n)^2.

Original entry on oeis.org

0, 3, 27, 216, 1704, 13419, 105651, 831792, 6548688, 51557715, 405913035, 3195746568, 25160059512, 198084729531, 1559517776739, 12278057484384, 96664942098336, 761041479302307, 5991666892320123, 47172293659258680, 371386682381749320, 2923921165394735883, 23019982640776137747

Offset: 0

Richard Choulet, Oct 14 2008

Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A057080, A131571, A145607.

RecurrenceTable[{a[0]==0,a[1]==3,a[n]==8a[n-1]-a[n-2]+3},a,{n,30}] (* or *) LinearRecurrence[{9,-9,1},{0,3,27},30] (* Harvey P. Dale, May 06 2013 *)

a(n+2) = 8*a(n+1) - a(n) + 3.
a(n) = (A070997(n)-1)/2 = 3*A076765(n-1). - R. J. Mathar, Oct 16 2008
G.f.: -3*x / ( (x-1)*(x^2-8*x+1) ). - R. J. Mathar, Nov 27 2011
a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3); a(0)=0, a(1)=3, a(2)=27. - Harvey P. Dale, May 06 2013

Made definition and sequence consistent. Changed offset to 0. - R. J. Mathar, Oct 16 2008

A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 39, 1559, 62321, 2491281, 99588919, 3981065479, 159143030241, 6361740144161, 254310462736199, 10166056769303799, 406387960309415761, 16245352355607326641, 649407706263983649879, 25960062898203738668519, 1037753108221885563090881

Offset: 0

Ilya Gutkovskiy, Feb 18 2016

In general, the ordinary generating function for the recurrence relation b(n) = k*b(n - 1) - b(n - 2) with n>1 and b(0)=1, b(1)=1, is (1 - (k - 1)*x)/(1 - k*x +x^2). This recurrence gives the closed form b(n) = (2^( -n - 1)*((k - 2)*(k - sqrt(k^2 - 4))^n + sqrt(k^2 - 4)*(k - sqrt(k^2 - 4))^n - (k - 2)*(sqrt(k^2 - 4) + k)^n + sqrt(k^2 - 4)*(sqrt(k^2 - 4) + k)^n))/sqrt(k^2 - 4).

Index entries for linear recurrences with constant coefficients, signature (40,-1).

Cf. A001519, A001835, A001653, A049685, A070997, A070998, A072256, A078922, A160682, A007805, A075839, A157014, A159664, A159668, A157877, A238379, A097315.

[n le 2 select 1 else 40*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 19 2016

Table[Cosh[n Log[20 + Sqrt[399]]] - Sqrt[19/21] Sinh[n Log[20 + Sqrt[399]]], {n, 0, 17}]
Table[(2^(-n - 2) (38 (40 - 2 Sqrt[399])^n + 2 Sqrt[399] (40 - 2 Sqrt[399])^n - 38 (40 + 2 Sqrt[399])^n + 2 Sqrt[399] (40 + 2 Sqrt[399])^n))/Sqrt[399], {n, 0, 17}]
LinearRecurrence[{40, -1}, {1, 1}, 17]

G.f.: (1 - 39*x)/(1 - 40*x + x^2).
a(n) = cosh(n*log(20 + sqrt(399))) - sqrt(19/21)*sinh(n*log(20 + sqrt(399))).
a(n) = (2^(-n - 2)*(38*(40 - 2*sqrt(399))^n + 2*sqrt(399)*(40 - 2*sqrt(399))^n - 38*(40 + 2*sqrt(399))^n + 2*sqrt(399)*(40 + 2*sqrt(399))^n))/sqrt(399).
Sum_{n>=0} 1/a(n) = 2.0262989201139499769986...

cp's OEIS Frontend

A131751 Numbers that are both centered triangular and centered pentagonal.

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A145607 Numbers k such that (3*(2*k + 1)^2 + 2)/5 is a square.

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A273053 Numbers n such that 15*n^2 + 16 is a square.

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A145542 Numerators in continued fraction expansion of sqrt(3/5).

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A158197 Expansion of (1-x^2*c(x)^4)/(1-4*x*c(x)^2), c(x) the g.f. of A000108.

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A145608 Numbers a(n)=k such that (1/3)*(5*(2k+1)^2-2) is A057080(n)^2.

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A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

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A145607 Numbers k such that (3(2k + 1)^2 + 2)/5 is a square.

A158197 Expansion of (1-x^2c(x)^4)/(1-4x*c(x)^2), c(x) the g.f. of A000108.

A145608 Numbers a(n)=k such that (1/3)(5(2k+1)^2-2) is A057080(n)^2.