A090248 -id:A090248 - cp's OEIS frontend

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127

Offset: 0

Paul Barry, Aug 16 2003

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).
The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005
a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence L_{5,n}.
Tanya Khovanova, Recursive Sequences
Wikipedia, Metallic mean
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A006497, A014448, A085447.

Cf. A086902, A000032, A052918.

I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)

{a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */

[lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.
a(n) = A100236(n) + 1.
E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007
a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008
A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010
a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010
a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020
From Peter Bala, Jul 09 2025 : (Start)
The following series telescope (Cf. A000032):
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A099365 Squares of A052918(n-1) (generalized Fibonacci).

Original entry on oeis.org

0, 1, 25, 676, 18225, 491401, 13249600, 357247801, 9632441025, 259718659876, 7002771375625, 188815108482001, 5091005157638400, 137268324147754801, 3701153746831741225, 99793882840309258276, 2690733682941518232225

Offset: 0

Wolfdieter Lang, Oct 18 2004

See the comment in A099279. This is example a=5.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 5 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 5 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 30 2023

G. C. Greubel, Table of n, a(n) for n = 0..650
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (26,26,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, this sequence, A099366, A099367, A099369, A099372, A099374.

[(2/29)*(Evaluate(ChebyshevFirst(n), 27/2) -(-1)^n): n in [0..30]]; // G. C. Greubel, Aug 21 2022

with (combinat):seq(fibonacci(n,5)^2,n=0..16); # Zerinvary Lajos, Apr 09 2008

LinearRecurrence[{26,26,-1},{0,1,25},30] (* Harvey P. Dale, Sep 25 2019 *)

def A099365(n): return (2/29)*(chebyshev_T(n, 27/2) - (-1)^n)
[A099365(n) for n in (0..30)] # G. C. Greubel, Aug 21 2022

a(n) = A052918(n-1)^2, n >= 1, a(0) = 0.
a(n) = 26*a(n-1) + 26*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=25.
a(n) = 27*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 27/2) - (-1)^n)/29 with twice the Chebyshev's T(n, x) polynomials of the first kind. 2*T(n, 27/2) = A090248(n).
G.f.: x*(1-x)/((1-27*x+x^2)*(1+x)) = x*(1-x)/(1-26*x-26*x^2+x^3).
a(n) = (1 - (-1)^n)/2 + 25*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 30 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (5 + sqrt(29))/10 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A097781 Chebyshev polynomials S(n,27) with Diophantine property.

Original entry on oeis.org

1, 27, 728, 19629, 529255, 14270256, 384767657, 10374456483, 279725557384, 7542215592885, 203360095450511, 5483180361570912, 147842509666964113, 3986264580646460139, 107481301167787459640, 2898008866949614950141

Offset: 0

Wolfdieter Lang, Aug 31 2004

All positive integer solutions of Pell equation b(n)^2 - 725*a(n)^2 = +4 together with b(n)=A090248(n+1), n>=0. Note that D=725=29*5^2 is not squarefree.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 27's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,26}. - Milan Janjic, Jan 26 2015

			(x,y) = (27;1), (727;27), (19602;728), ... give the positive integer solutions to x^2 - 29*(5*y)^2 =+4.

Vincenzo Librandi, Table of n, a(n) for n = 0..700
R. Flórez, R. A. Higuita, A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (27,-1).

Cf. A078362, A078366.

I:=[1, 27, 728]; [n le 3 select I[n] else 27*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

with (combinat):seq(fibonacci(2*n, 5)/5, n=1..16); # Zerinvary Lajos, Apr 20 2008

Join[{a=1,b=27},Table[c=27*b-a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2011 *)
CoefficientList[Series[1/(1 - 27 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 24 2012 *)

[lucas_number1(n,27,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = S(n, 27) = U(n, 27/2) = S(2*n+1, sqrt(29))/sqrt(29) with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x).
a(n) = 27*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=27; a(-1)=0.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (27+5*sqrt(29))/2 and am = (27-5*sqrt(29))/2.
G.f.: 1/(1-27*x+x^2).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*26^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/5*(5 + sqrt(29)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 5/54*(5 + sqrt(29)). - Peter Bala, Dec 23 2012

A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 4, 7, 2, 2, 5, 14, 18, 2, 2, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 2, 8, 47, 198, 527, 724, 322, 2, 2, 9, 62, 322, 1154, 2525, 2702, 843, 2, 2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, 2, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2

Offset: 0

William W. Collier, Feb 18 2018

Note the similarity in form of the recursive steps in the array definition above and the polynomial definition under FORMULA.

			i\j |0  1   2    3      4       5        6          7           8            9
----+-------------------------------------------------------------------------
   0|2  2   2    2      2       2        2          2           2            2
   1|2  3   7   18     47     123      322        843        2207         5778
   2|2  4  14   52    194     724     2702      10084       37634       140452
   3|2  5  23  110    527    2525    12098      57965      277727      1330670
   4|2  6  34  198   1154    6726    39202     228486     1331714      7761798
   5|2  7  47  322   2207   15127   103682     710647     4870847     33385282
   6|2  8  62  488   3842   30248   238142    1874888    14760962    116212808
   7|2  9  79  702   6239   55449   492802    4379769    38925119    345946302
   8|2 10  98  970   9602   95050   940898    9313930    92198402    912670090
   9|2 11 119 1298  14159  154451  1684802   18378371   200477279   2186871698
  10|2 12 142 1692  20162  240252  2862862   34114092   406506242   4843960812
  11|2 13 167 2158  27887  360373  4656962   60180133   777684767  10049721838
  12|2 14 194 2702  37634  524174  7300802  101687054  1416317954  19726764302
  13|2 15 223 3330  49727  742575 11088898  165590895  2472774527  36926027010
  14|2 16 254 4048  64514 1028176 16386302  261152656  4162056194  66331746448
  15|2 17 287 4862  82367 1395377 23639042  400468337  6784322687 114933017342
  16|2 18 322 5778 103682 1860498 33385282  599074578 10749957122 192900153618
  17|2 19 359 6802 128879 2441899 46267202  876634939 16609796639 314709501202
  18|2 20 398 7940 158402 3160100 63043598 1257711860 25091193602 500566160180
  19|2 21 439 9198 192719 4037901 84603202 1772629341 37140612959 778180242798

William W. Collier, a(i,j) = f(i+2,j)
William W. Collier, Experimental Mathematics on Wisteria Tables, Talk to Poughkeepsie ACM Chapter.
OEIS Wiki, The (1,2) Pascal Triangle.

The array first appeared in A298675.
Rows 1 through 29 of the array appear in these OEIS entries: A005248, A003500, A003501, A003499, A056854, A086903, A056918, A087799, A057076, A087800, A078363, A067902, A078365, A090727, A078367, A087215, A078369, A090728, A090729, A090730, A090731, A090732, A090733, A090247, A090248, A090249, A090251. Also entries occur for rows 45, 121, and 320:  A087265, A065705, A089775. Each of these entries asserts that a(i,j)=f(i+2,j) is true for that row.
A few of the columns appear in the OEIS: A008865 (for column 2), A058794 and A007754 (for column 3), and A230586 (for column 5).
Main diagonal gives A343261.

A:= proc(i, j) option remember; `if`(min(i, j)=0, 2,
      `if`(j=1, i+2, (i+2)*A(i, j-1)-A(i, j-2)))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Mar 05 2019

a[, 0] = a[0, ] = 2; a[i_, 1] := i + 2;
a[i_, j_] := a[i, j] =(i + 2) a[i, j - 1] - a[i, j - 2];
Table[a[i - j, j], {i, 0, 10}, {j, 0, i}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Let k be an integer, and let r1 and r2 be the roots of x + 1/x = k. Then f(k,n) = r1^n + r2^n is an integer, for integer n >= 0. Theorem: a(i,j) = f(i+2,j), for i,j >= 0. Proof: See the Collier link.
Define polynomials recursively by:
    p[0](n) = 2, for n >= 0 ( [ and ] demark subscripts).
    p[1](n) = n + 2, for n >= 0.
    p[j](n) = p[j-1](n) * p[1](n) - p[j-2](n), for j > 1, n >= 0. The coefficients of these polynomials occur as the even numbered, upward diagonals in the OEIS Wiki link. Conjecture: a(i,j) = p[j](i), i,j >= 0.

Edited by N. J. A. Sloane, Apr 04 2018

A090251 a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29.

Original entry on oeis.org

2, 29, 839, 24302, 703919, 20389349, 590587202, 17106639509, 495501958559, 14352450158702, 415725552643799, 12041688576511469, 348793243166188802, 10102962363242963789, 292637115290879761079, 8476373381072270107502

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004

a(n+1)/a(n) converges to ((29+sqrt(837))/2) =28.9654761... Lim a(n)/a(n+1) as n approaches infinity = 0.0345238... =2/(29+sqrt(837)) =(29-sqrt(837))/2. Lim a(n+1)/a(n) as n approaches infinity = 28.9654761... = (29+sqrt(837))/2=2/(29-sqrt(837)). Lim a(n)/a(n+1) = 29 - Lim a(n+1)/a(n).
A Chebyshev T-sequence with a Diophantine property.
a(n) gives the general (nonnegative integer) solution of the Pell equation a^2 - 93*(3*b)^2 =+4 with companion sequence b(n)=A097782(n+1), n>=0.

			a(4) =703919 = 29a(3) - a(2) = 29*24302 - 839= ((29+sqrt(837))/2)^4 + ((29-sqrt(837))/2)^4 = 703918.99999857 + 0.00000142 =703919.
(x,y) = (2;0), (29;1), (839;29), (24302,840), ..., give the
nonnegative integer solutions to x^2 - 93*(3*y)^2 =+4.

O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (29, -1).

Cf. A083148, A007610.
a(n)=sqrt(4 + 93*(3*A097782(n-1))^2), n>=1.
Cf. A077428, A078355 (Pell +4 equations).
Cf. A090248 for 2*T(n, 27/2).

a[0] = 2; a[1] = 29; a[n_] := 29a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{29,-1},{2,29},30] (* Harvey P. Dale, May 28 2013 *)

[lucas_number2(n,29,1) for n in range(0,16)] # Zerinvary Lajos, Jun 27 2008

a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29. a(n) = ((29+sqrt(837))/2)^n + ((29-sqrt(837))/2)^n, (a(n))^2 =a(2n)+2.
a(n) = S(n, 29) - S(n-2, 29) = 2*T(n, 29/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 27)=A097781(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (29+3*sqrt(93))/2 and am := (29-3*sqrt(93))/2.
G.f.: (2-29*x)/(1-29*x+x^2).

More terms from Robert G. Wilson v, Jan 30 2004

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

cp's OEIS Frontend

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

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A099365 Squares of A052918(n-1) (generalized Fibonacci).

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A097781 Chebyshev polynomials S(n,27) with Diophantine property.

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A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

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A090251 a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29.

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