A063727 -id:A063727 - cp's OEIS frontend

A016095 Triangular array T(n,k) read by rows, where T(n,k) = coefficient of x^n*y^k in 1/(1-x-y-(x+y)^2).

1, 1, 1, 2, 4, 2, 3, 9, 9, 3, 5, 20, 30, 20, 5, 8, 40, 80, 80, 40, 8, 13, 78, 195, 260, 195, 78, 13, 21, 147, 441, 735, 735, 441, 147, 21, 34, 272, 952, 1904, 2380, 1904, 952, 272, 34, 55, 495, 1980, 4620, 6930, 6930, 4620, 1980, 495, 55

Offset: 0

N. J. A. Sloane, Jan 23 2001

Triangle T(n,k), 0<=k<=n, read by rows, given by [1, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 10 2005

			Triangle begins:
  1;
  1,  1;
  2,  4,  2;
  3,  9,  9,  3;
  5, 20, 30, 20,  5;
  8, 40, 80, 80, 40, 8;
  ...

Columns include A000045, A023607. Central diagonal is A102307. Antidiagonal sums are in A063727.

read transforms; 1/(1-x-y-(x+y)^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);

T[n_, k_] := SeriesCoefficient[1/(1-x-y-(x+y)^2), {x, 0, n}, {y, 0, k}]; Table[T[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 04 2017 *)

G.f.: 1/(1-x-y-(x+y)^2).
T(n,k) = Fibonacci(n+1)*binomial(n,k) = A000045(n+1)*A007318(n,k). - Philippe Deléham, Oct 14 2006
Sum_{k=0..floor(n/2)} T(n-k,k) = A123392(n). - Philippe Deléham, Oct 14 2006
G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x*(1+y)/((2*k+2+ x*(1+y))*x*(1+y)+ 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
T(n,k) = T(n-1,k)+T(n-1,k-1)+T(n-2,k)+2*T(n-2,k-1)+T(n-2,k-2), T(0,0) = T(1,0) = T(1,1) = 1, T(2,0) = T(2,2) = 2, T(2,1) = 4, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 12 2013

A006483 a(n) = Fibonacci(n)*2^n + 1.

Original entry on oeis.org

1, 3, 5, 17, 49, 161, 513, 1665, 5377, 17409, 56321, 182273, 589825, 1908737, 6176769, 19988481, 64684033, 209321985, 677380097, 2192048129, 7093616641, 22955425793, 74285318145, 240392339457, 777925951489, 2517421260801, 8146546327553, 26362777698305

Offset: 0

Dennis S. Kluk (mathemagician(AT)ameritech.net)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. S. Kluk and N. J. A. Sloane, Correspondence, 1979.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (3,2,-4).

Cf. A000045, A063727, A087206.

Equals A103435 + 1.

[Fibonacci(n)*2^n + 1: n in [0..50]]; // Vincenzo Librandi, Jun 09 2013

A006483:=-(-1+6*z**2)/(z-1)/(4*z**2+2*z-1); # Simon Plouffe in his 1992 dissertation

lst={};Do[AppendTo[lst, Fibonacci[n]*2^n+1], {n, 0, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 19 2008 *)
CoefficientList[Series[(-(- 1 + 6 x^2)) / ((1 - x) (1 - 2 x - 4 x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
LinearRecurrence[{3,2,-4},{1,3,5},40] (* Harvey P. Dale, Aug 01 2021 *)

G.f.: -(-1+6*x^2)/((1-x)*(1-2*x-4*x^2)).

G.f. in Formula field corrected by Vincenzo Librandi, Jun 09 2013

A234357 Array T(n,k) by antidiagonals: T(n,k) = n^k * Fibonacci(k).

Original entry on oeis.org

1, 2, 2, 3, 8, 3, 4, 18, 24, 5, 5, 32, 81, 80, 8, 6, 50, 192, 405, 256, 13, 7, 72, 375, 1280, 1944, 832, 21, 8, 98, 648, 3125, 8192, 9477, 2688, 34, 9, 128, 1029, 6480, 25000, 53248, 45927, 8704, 55, 10, 162, 1536, 12005, 62208, 203125, 344064, 223074, 28160, 89, 11, 200, 2187

Offset: 0

Ralf Stephan, Dec 24 2013

			Array starts:
1,  2,   3,    5,     8,     13,    21,   34, 55, 89,...    (A000045)
2,  8,  24,   80,   256,    832,  2688, 8704,...   (A063727, A085449)
3, 18,  81,  405,  1944,   9477, 45927,...         (A122069, A099012)
4, 32, 192, 1280,  8192,  53248,...                         (A099133)
5, 50, 375, 3125, 25000, 203125,...
6, 72, 648, 6480, 62208, 606528,...
...
Columns: A000027, A001105, A117642.

PARI
```
T(n,k)=n^k*fibonacci(k)
```

T(n,k)=polcoeff(Ser(1/(1-n*x-n^2*x^2)),k)

G.f. of n-th row: 1/(1 - n*x - n^2*x^2).

Recurrence: T(n,k) = n*T(n,k-1) + n^2*T(n,k-2), starting n, 2*n^2.

A269991 Decimal expansion of Sum_{n >= 1} 2^(1-n)/Fibonacci(n).

Original entry on oeis.org

1, 6, 8, 4, 8, 1, 3, 1, 4, 4, 4, 8, 9, 5, 7, 6, 0, 9, 6, 3, 1, 6, 5, 5, 4, 3, 3, 7, 3, 8, 0, 0, 7, 8, 2, 3, 0, 2, 3, 7, 0, 6, 3, 8, 8, 2, 4, 5, 7, 0, 8, 6, 8, 2, 0, 9, 4, 3, 1, 7, 6, 1, 8, 8, 5, 9, 5, 0, 5, 6, 0, 0, 2, 8, 0, 4, 9, 4, 5, 4, 9, 8, 9, 1, 0, 8

Offset: 1

Clark Kimberling, Mar 12 2016

			1.684813144489576096316554337380078230...

Cf. A000045, A269992.

Cf. A063727, A085449, A103435, A209084.

x = N[Sum[2^(1 - n)/Fibonacci[n], {n, 1, 500}], 100]
RealDigits[x][[1]]

suminf(n=1, 2^(1-n)/fibonacci(n)) \\ Michel Marcus, Feb 01 2021

Equals Sum_{n>=0} 1/A063727(n) = Sum_{n>=1} 1/A085449(n) = 2 * Sum_{n>=1} 1/A103435(n) = 4 * Sum_{n>=1} 1/A209084(n). - Amiram Eldar, Feb 01 2021

A086344 a(n) = -2a(n-1) + 4a(n-2), a(0) = 1, a(1) = 0.

Original entry on oeis.org

1, 0, 4, -8, 32, -96, 320, -1024, 3328, -10752, 34816, -112640, 364544, -1179648, 3817472, -12353536, 39976960, -129368064, 418643968, -1354760192, 4384096256, -14187233280, 45910851584, -148570636288, 480784678912, -1555851902976, 5034842521600, -16293092655104

Offset: 0

Paul Barry, Jul 17 2003

Inverse binomial transform of (1,1,5,5,25,25,.....).

The absolute values are the constant terms of the reduction by x^2->x+1 of the polynomial p(n,x) given for d=sqrt(x+1) by p(n,x)=((x+d)^n-(x-d)^n)/(2d), for n>=1. The coefficient of x under this reduction is given by A103435. See A192232 for a discussion of reduction. - Clark Kimberling, Jun 29 2011

Index entries for linear recurrences with constant coefficients, signature (-2,4).

seq((-2)^n * combinat:-fibonacci(n-1), n = 0 .. 100); # Robert Israel, Oct 02 2014

LinearRecurrence[{-2,4},{1,0},40] (* Harvey P. Dale, Oct 10 2018 *)

G.f.: (1+2*x)/((1+(1+sqrt(5))*x)(1+(1-sqrt(5))*x)) = ( -1-2*x ) / ( -1-2*x+4*x^2 ).
E.g.f.: exp(-x)*(cosh(sqrt(5)*x)+sinh(sqrt(5)*x)/sqrt(5)).
a(n)=(sqrt(5)-1)^n*(sqrt(5)/10+1/2)+(-sqrt(5)-1)^n*(1/2-sqrt(5)/10).
(-1)^n*a(n) = A063727(n) - 2*A063727(n-1). - R. J. Mathar, Jul 19 2012
(-1)^n*a(n) = sum(k=0..n, binomial(n,k)*(F(n+1)-F(n))), F(n) Fibonacci number A000045. - Peter Luschny, Oct 01 2014
a(n) = (-2)^n *A000045(n-1). - Robert Israel, Oct 02 2014

A087205 a(n) = -2a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 0, 8, -16, 64, -192, 640, -2048, 6656, -21504, 69632, -225280, 729088, -2359296, 7634944, -24707072, 79953920, -258736128, 837287936, -2709520384, 8768192512, -28374466560, 91821703168, -297141272576, 961569357824, -3111703805952

Offset: 0

Paul Barry, Aug 25 2003

Inverse binomial transform of A087204.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,4)

Cf. A000045, A063727, A084222, A085449, A087204.

[(-1)^(n+1)*2^n*Fibonacci(n-2): n in [0..50]]; // G. C. Greubel, Oct 08 2018

Table[-(-2)^n*Fibonacci[n - 2], {n, 0, 50}] (* G. C. Greubel, Oct 08 2018 *)
LinearRecurrence[{-2,4},{1,2},30] (* Harvey P. Dale, Jan 24 2022 *)

Vec((4*x+1)/(-4*x^2+2*x+1)+O(x^66)) \\ Joerg Arndt, Jul 14 2013

vector(50, n, n--; (-1)^(n+1)*2^n*fibonacci(n-2)) \\ G. C. Greubel, Oct 08 2018

a(n) = (-1-sqrt(5))^n * (1/2-3*sqrt(5)/10) + (-1+sqrt(5))^n * (1/2+3*sqrt(5)/10).
G.f.: (4*x +1)/(-4*x^2 +2*x +1). - Joerg Arndt, Jul 14 2013
a(n+2) = A085449(n)*(-1)^(n+1); a(n+3) = A063727(n)*(-1)^n.
a(n) = -(-2)^n*F(n-2) for n >= 0, with F = A000045, and F(-1) = 1, F(-2) = -1. - Wolfdieter Lang, Oct 08 2018

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 44, 312, 2224, 15840, 112832, 803712, 5724928, 40779264, 290475008, 2069084160, 14738305024, 104982503424, 747801460736, 5326668791808, 37942424436736, 270267896954880, 1925146777223168, 13713023838978048, 97679317251653632, 695780094221746176

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,8).

Sequences of the form a(n) = c*a(n-1) + d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A000045,A001045,A006130,A006131,A015440,A015441,A015442,A015443,A015445,A015446
.A000129,A002605,A015518,A063727,A002532,A083099,A015519,A003683,A002534,A083102
.A006190,A007482,A030195,A015521,A015523,A083858,A015524,A015525,A099012,A015528
.A001076,A090017,A015530,A057087,A015531,A085939,A015532,A180222,A015533,A180226
.A052918,A015535,A015536,A015537,A057088,A015540,A015541,A015544,A015545,A180250
.A005668,A135030,A090018,A135032,A015551,A057089,A015552,A189800,A189801,A190005
.A054413,A015555,A015559,A015561,A015562,A015564,A057090,A015565,A015566,A015568
.A041025,A190331,A015574,A190510,A015575,A190560,A015576,A057091,A015577,A190953
.A099371,A015579,A181353,A015580,A015581,A153191,A015583,A015584,A057092,A015585
A041041,A191014,A015588,A190954,A190955,A190956,A015589,A190957,A015591,A057093

I:=[0,1]; [n le 2 select I[n] else 6*Self(n-1)+8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2011

LinearRecurrence[{6, 8}, {0, 1}, 50]
CoefficientList[Series[-(x/(-1+6 x+8 x^2)),{x,0,50}],x] (* Harvey P. Dale, Jul 26 2011 *)

a(n)=([0,1; 8,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

G.f.: x/(1 - 2*x*(3+4*x)). - Harvey P. Dale, Jul 26 2011

A099173 Array, A(k,n), read by diagonals: g.f. of k-th row x/(1-2x-(k-1)x^2).

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 4, 4, 0, 1, 2, 5, 8, 5, 0, 1, 2, 6, 12, 16, 6, 0, 1, 2, 7, 16, 29, 32, 7, 0, 1, 2, 8, 20, 44, 70, 64, 8, 0, 1, 2, 9, 24, 61, 120, 169, 128, 9, 0, 1, 2, 10, 28, 80, 182, 328, 408, 256, 10, 0, 1, 2, 11, 32, 101, 256, 547, 896, 985, 512, 11

Offset: 0

Ralf Stephan, Oct 13 2004

			Square array, A(n, k), begins as:
  0, 1, 2,  3,  4,   5,    6,    7,     8, ... A001477;
  0, 1, 2,  4,  8,  16,   32,   64,   128, ... A000079;
  0, 1, 2,  5, 12,  29,   70,  169,   408, ... A000129;
  0, 1, 2,  6, 16,  44,  120,  328,   896, ... A002605;
  0, 1, 2,  7, 20,  61,  182,  547,  1640, ... A015518;
  0, 1, 2,  8, 24,  80,  256,  832,  2688, ... A063727;
  0, 1, 2,  9, 28, 101,  342, 1189,  4088, ... A002532;
  0, 1, 2, 10, 32, 124,  440, 1624,  5888, ... A083099;
  0, 1, 2, 11, 36, 149,  550, 2143,  8136, ... A015519;
  0, 1, 2, 12, 40, 176,  672, 2752, 10880, ... A003683;
  0, 1, 2, 13, 44, 205,  806, 3457, 14168, ... A002534;
  0, 1, 2, 14, 48, 236,  952, 4264, 18048, ... A083102;
  0, 1, 2, 15, 52, 269, 1110, 5179, 22568, ... A015520;
  0, 1, 2, 16, 56, 304, 1280, 6208, 27776, ... A091914;
Antidiagonal triangle, T(n, k), begins as:
  0;
  0,  1;
  0,  1,  2;
  0,  1,  2,  3;
  0,  1,  2,  4,  4;
  0,  1,  2,  5,  8,  5;
  0,  1,  2,  6, 12, 16,   6;
  0,  1,  2,  7, 16, 29,  32,   7;
  0,  1,  2,  8, 20, 44,  70,  64,   8;
  0,  1,  2,  9, 24, 61, 120, 169, 128,   9;
  0,  1,  2, 10, 28, 80, 182, 328, 408, 256,  10;

G. C. Greubel, Antidiagonals n = 0..50, flattened
Ralf Stephan, Prove or disprove. 100 Conjectures from the OEIS, #16, arXiv:math/0409509 [math.CO], 2004.

Rows m: A001477 (m=0), A000079 (m=1), A000129 (m=2), A002605 (m=3), A015518 (m=4), A063727 (m=5), A002532 (m=6), A083099 (m=7), A015519 (m=8), A003683 (m=9), A002534 (m=10), A083102 (m=11), A015520 (m=12), A091914 (m=13).
Columns q: A000004 (q=0), A000012 (q=1), A009056 (q=2), A008586 (q=3).
Main diagonal gives A357502.

A099173:= func< n,k | (&+[n^j*Binomial(k,2*j+1): j in [0..Floor(k/2)]]) >;
[A099173(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2023

A[k_, n_]:= Which[k==0, n, n==0, 0, True, ((1+Sqrt[k])^n - (1-Sqrt[k])^n)/(2 Sqrt[k])]; Table[A[k-n, n]//Simplify, {k, 0, 12}, {n, 0, k}]//Flatten (* Jean-François Alcover, Jan 21 2019 *)

A(k,n)=sum(i=0,n\2,k^i*binomial(n,2*i+1))

def A099173(n,k): return sum( n^j*binomial(k, 2*j+1) for j in range((k//2)+1) )
flatten([[A099173(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 17 2023

A(n, k) = Sum_{i=0..floor(k/2)} n^i * C(k, 2*i+1) (array).
Recurrence: A(n, k) = 2*A(n, k-1) + (n-1)*A(n, k-2), with A(n, 0) = 0, A(n, 1) = 1.
T(n, k) = A(n-k, k) (antidiagonal triangle).
T(2*n, n) = A357502(n).
A(n, k) = ((1+sqrt(n))^k - (1-sqrt(n))^k)/(2*sqrt(n)). - Jean-François Alcover, Jan 21 2019

A163762 Triangle of coefficients of polynomials H(n,x)=(U^n+L^n)/2+(U^n-L^n)/(2d), where U=x+d, L=x-d, d=(x+4)^(1/2).

Original entry on oeis.org

1, 1, 1, 1, 3, 4, 1, 6, 13, 4, 1, 10, 29, 24, 16, 1, 15, 55, 81, 88, 16, 1, 21, 95, 207, 300, 144, 64, 1, 28, 154, 448, 813, 684, 496, 64, 1, 36, 238, 868, 1913, 2352, 2272, 768, 256, 1, 45, 354, 1554, 4077, 6625, 7984, 4704, 2560, 256, 1, 55, 510, 2622, 8061, 16283

Offset: 1

Clark Kimberling, Aug 04 2009

H(n,x)=P(n,x)+Q(n,x), where P and Q are given by A162516, A162517.
H(n,0)=4^Floor(n/2) for n=0,1,2,...
H(n,1)=A063727(n); row sums
(Column 2)=A000217 (triangular numbers)

			First six rows:
1
1...1
1...3...4
1...6..13...4
1..10..29..24..16
1..15..55..81..88..16
Row 6 represents x^5+15*x^4+55*x^3+81*x^2+88*x+16.

A000217, A063727, A162516, A162517.

H(n,x)=2*x*H(n-1,x)-(x^2-x-4)*H(n-2,x), where H(0,x)=1, H(1,x)=x+1.

H(n,x)=(1+1/d)*U^n+(1-1/d)*L^n, where U=x+d, L=x-d, d=(x+4)^(1/2).

A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.

Original entry on oeis.org

0, 4, 8, 32, 96, 320, 1024, 3328, 10752, 34816, 112640, 364544, 1179648, 3817472, 12353536, 39976960, 129368064, 418643968, 1354760192, 4384096256, 14187233280, 45910851584, 148570636288, 480784678912, 1555851902976, 5034842521600, 16293092655104

Offset: 0

Seiichi Kirikami, Mar 06 2012

a(n)/A063727(n) are convergents for A134972.

Abs(Sum_{i=0..n} C(n,n-i)*a(i)-(sqrt(5)-1)* A033887(n))->0. - Seiichi Kirikami, Jan 20 2016

E. W. Cheney, Introduction to Approximation Theory, McGraw-Hill, Inc., 1966.

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A063727, A033887, A085449, A103435, A134972, A269991.

Cf. A086344 (this sequence with signs).

I:=[0,4]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 16 2016

RecurrenceTable[{a[n]==2*a[n-1]+4*a[n-2], a[0]==0, a[1]==4}, a, {n, 30}]
LinearRecurrence[{2, 4}, {0, 4}, 40] (* Vincenzo Librandi, Jan 16 2016 *)

concat(0, Vec(4*x/(1-2*x-4*x^2) + O(x^40))) \\ Michel Marcus, Jan 16 2016

a(n) = (2/sqrt(5))*((1+sqrt(5))^n-(1-sqrt(5))^n).
G.f.: 4*x/(1-2*x-4*x^2). - Bruno Berselli, Mar 08 2012
a(n) = 4*A085449(n) = 2*A103435(n). - Bruno Berselli, Mar 08 2012
Sum_{n>=1} 1/a(n) = (1/4) * A269991. - Amiram Eldar, Feb 01 2021

cp's OEIS Frontend

A016095 Triangular array T(n,k) read by rows, where T(n,k) = coefficient of x^n*y^k in 1/(1-x-y-(x+y)^2).

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A006483 a(n) = Fibonacci(n)*2^n + 1.

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A234357 Array T(n,k) by antidiagonals: T(n,k) = n^k * Fibonacci(k).

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A269991 Decimal expansion of Sum_{n >= 1} 2^(1-n)/Fibonacci(n).

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A086344 a(n) = -2*a(n-1) + 4*a(n-2), a(0) = 1, a(1) = 0.

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A087205 a(n) = -2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.

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A189800 a(n) = 6*a(n-1) + 8*a(n-2), with a(0)=0, a(1)=1.

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A099173 Array, A(k,n), read by diagonals: g.f. of k-th row x/(1-2*x-(k-1)*x^2).

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A086344 a(n) = -2a(n-1) + 4a(n-2), a(0) = 1, a(1) = 0.

A087205 a(n) = -2a(n-1) + 4a(n-2), a(0)=1, a(1)=2.

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

A099173 Array, A(k,n), read by diagonals: g.f. of k-th row x/(1-2x-(k-1)x^2).

A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.