A101220 -id:A101220 - cp's OEIS frontend

A022101 Fibonacci sequence beginning 1, 11.

1, 11, 12, 23, 35, 58, 93, 151, 244, 395, 639, 1034, 1673, 2707, 4380, 7087, 11467, 18554, 30021, 48575, 78596, 127171, 205767, 332938, 538705, 871643, 1410348, 2281991, 3692339, 5974330, 9666669, 15640999, 25307668, 40948667, 66256335, 107205002, 173461337, 280666339

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(11;n-1-k,k) with n >= 1, a(-1)=10. These are the SW-NE diagonals in P(11;n,k), the (11,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

Cf. A000032, A000045.

a(n) = A109754(10, n+1) = A101220(10, 0, n+1).

a0:=1; a1:=11; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,11},40] (* Harvey P. Dale, Aug 16 2015 *)

a(n) = 10*fibonacci(n)+fibonacci(n+1) \\ Charles R Greathouse IV, Jun 11 2015

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=11. a(-1)=10.
G.f.: (1+10*x)/(1-x-x^2).
a(n-1) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + 5*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 10*A000045(n) + A000045(n+1). - R. J. Mathar, Apr 07 2011
a(n) = 12*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = A000045(n+4) + A000032(n-4) for n > 0. - Bruno Berselli, Sep 27 2017

A098703 a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.

Original entry on oeis.org

0, 1, 2, 6, 17, 50, 148, 441, 1318, 3946, 11825, 35454, 106328, 318929, 956698, 2869950, 8609617, 25828474, 77484812, 232453449, 697358750, 2092073666, 6276216817, 18828643686, 56485920112, 169457742625, 508373199218, 1525119551286

Offset: 0

Ross La Haye, Oct 27 2004

Sums of antidiagonals of A090888.
Partial sums of A099159.
Form an array with m(0,n) = A000045(n), the Fibonacci numbers, and m(i,j) = Sum_{kJ. M. Bergot, May 27 2013

			a(2) = 2 because 3^2 = 9, Luc(1) = 1 and (9 + 1) / 5 = 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein, Golden Ratio
Eric Weisstein, Lucas Number
Eric Weisstein, Fibonacci Number
Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Cf. A000032, A000045, A000244, A000285, A001622, A001654, A022383, A052964, A084178, A090888, A094688, A099159.

I:=[0,1,2]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-3*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 18 2018

f[n_] := (3^n + Fibonacci[n] + Fibonacci[n - 2])/5; Table[ f[n], {n, 0, 27}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{4, -2, -3}, {0, 1, 2}, 30] (* Jean-François Alcover, Feb 17 2018 *)

def A098703(n): return (3**n + lucas_number2(n-1,1,-1))//5
print([A098703(n) for n in range(21)]) # G. C. Greubel, Jun 02 2025

a(n) = (((1 + sqrt(5))^n - (1 - sqrt(5))^n) / (2^n*sqrt(5))) + ((3^n - (((1 + sqrt(5)) / 2)^(n+1) + ((1 - sqrt(5)) / 2)^(n+1))) / 5).
a(n) = (3^n + (((1 + sqrt(5)) / 2)^(n-1) + ((1 - sqrt(5)) / 2)^(n-1))) / 5.
a(n) = (3^n + A000032(n-1))/5 = A000045(n) + (3^n - A000032(n+1))/5.
a(n) = (3^n + A000045(n) + A000045(n-2))/5.
a(n) = (3^n + 4*A000045(n) - A000045(n+2))/5.
a(n) = Sum_{k=0...n-1} (A000045(k)*3^(n-k-1) - A000045(k-2)*2^(n-k-1)).
a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
a(n) = A000045(n) + A094688(n-1).
a(n) = 3^1 * a(n-1) - A000045(n-3), for n > 2.
a(n) = 3^2 * a(n-2) - A000285(n-4), for n > 3.
a(n) = 3^3 * a(n-3) - A022383(n-5), for n > 4.
Limit_{n -> oo} a(n) / a(n-1) = 3.
From Ross La Haye, Dec 21 2004: (Start)
a(n) = A101220(1,3,n).
Binomial transform of unsigned A084178.
Binomial transform of signed A084178 gives the Fibonacci oblongs (A001654). (End)
G.f.: x*(1-2*x)/((1-3*x)*(1-x-x^2)). - Ross La Haye, Aug 09 2005
a(0) = 0, a(1) = 1, a(n) = a(n-1) + a(n-2) + 3^(n-2) for n > 1. - Ross La Haye, Aug 20 2005
Binomial transform of A052964 beginning {0,1,0,3,1,10,...}. - Ross La Haye, May 31 2006

More terms from Robert G. Wilson v, Nov 04 2004

More terms from Ross La Haye, Dec 21 2004

A022102 Fibonacci sequence beginning 1, 12.

Original entry on oeis.org

1, 12, 13, 25, 38, 63, 101, 164, 265, 429, 694, 1123, 1817, 2940, 4757, 7697, 12454, 20151, 32605, 52756, 85361, 138117, 223478, 361595, 585073, 946668, 1531741, 2478409, 4010150, 6488559, 10498709

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(12;n-1-k,k) with n>=1, a(-1)=11. These are the SW-NE diagonals in P(12;n,k), the (12,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

Cf. A000045, A001175, A093645, A101220, A109754.

a0:=1; a1:=12; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,12},40] (* Harvey P. Dale, Jan 23 2012 *)

a(n) = if(n==0, 1, if(n==1, 12, a(n-1)+a(n-2))) \\ Felix Fröhlich, Jun 09 2022

Vec((1+11*x)/(1-x-x^2) + O(x^20)) \\ Felix Fröhlich, Jun 09 2022

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=12. a(-1):=11.
G.f.: (1+11*x)/(1-x-x^2).
a(n) = A109754(11, n+1) = A101220(11, 0, n+1).
a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (11/2)*((1+sqrt(5))^(n-1)-(1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=13. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 11*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 13*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = F(n+5) + F(n-1) - F(n-5) for F(n) the Fibonacci number A000045(n). - Greg Dresden and Aamen Muharram, Jun 09 2022

A022110 Fibonacci sequence beginning 1, 20.

Original entry on oeis.org

1, 20, 21, 41, 62, 103, 165, 268, 433, 701, 1134, 1835, 2969, 4804, 7773, 12577, 20350, 32927, 53277, 86204, 139481, 225685, 365166, 590851, 956017, 1546868, 2502885, 4049753, 6552638, 10602391, 17155029, 27757420, 44912449, 72669869, 117582318, 190252187

Offset: 0

N. J. A. Sloane

a(n-1) = Sum(P(20;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1) = 19. These are the SW-NE diagonals in P(20;n,k), the (20,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

a(n) = A109754(19, n+1) = A101220(19, 0, n+1).

a0:=1; a1:=20; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

a={};b=1;c=20;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
LinearRecurrence[{1, 1}, {1, 20}, 35] (* Paolo Xausa, Feb 22 2024 *)

a(n) = a(n-1)+a(n-2), n >= 2, a(0) = 1, a(1) = 20.

G.f.: (1+19*x)/(1-x-x^2).

A053311 Partial sums of A000285.

Original entry on oeis.org

1, 5, 10, 19, 33, 56, 93, 153, 250, 407, 661, 1072, 1737, 2813, 4554, 7371, 11929, 19304, 31237, 50545, 81786, 132335, 214125, 346464, 560593, 907061, 1467658, 2374723, 3842385, 6217112, 10059501, 16276617, 26336122

Offset: 0

Barry E. Williams, Mar 06 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., pp. 189, 194-196.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000285.

a(n) = A101220(4, 1, n+1).

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x)/((x-1)*(x^2+x-1)))); // G. C. Greubel, May 24 2018

CoefficientList[Series[(1+3*x)/((x-1)*(x^2+x-1)), {x, 0, 50}], x] (* G. C. Greubel, May 24 2018 *)

x='x+O('x^30); Vec((1+3*x)/((x-1)*(x^2+x-1))) \\ G. C. Greubel, May 24 2018

a(n) = a(n-1) + a(n-2) + 4; a(0)=1, a(1)=5; n >= 1.
a(n) = 4*F(n+2) + F(n+1) - 4, where F(k) is A000045(k).
From R. J. Mathar, Apr 29 2013: (Start)
G.f.: ( 1+3*x ) / ( (x-1)*(x^2+x-1) ).
a(n) = A000071(n+3) + 3*A000071(n+2) = A000285(n+2) - 4. (End)

A099159 a(n) = (L(n-2)+2*3^n)/5.

Original entry on oeis.org

1, 1, 4, 11, 33, 98, 293, 877, 2628, 7879, 23629, 70874, 212601, 637769, 1913252, 5739667, 17218857, 51656338, 154968637, 464905301, 1394714916, 4184143151, 12552426869, 37657276426, 112971822513, 338915456593, 1016746352068, 3050239027547, 9150717036273

Offset: 0

Paul Barry, Oct 01 2004

Binomial transform of A052964.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Amya Luo, Pattern Avoidance in Nonnesting Permutations, Undergraduate Thesis, Dartmouth College (2024). See p. 16.
Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Cf. A000032, A052964, A098703, A101220,

A099159[n_] := (LucasL[n-2] + 2*3^n)/5; Array[A099159, 30, 0] (* or *)
LinearRecurrence[{4, -2, -3}, {1, 1, 4}, 30] (* Paolo Xausa, Jun 20 2024 *)

G.f.: (1-3*x+2*x^2)/((1-3*x)*(1-x-x^2)).
a(n) = ((1+sqrt(5))/2)^n*(3/10-sqrt(5)/10) + ((1-sqrt(5))/2)^n*(3/10+sqrt(5)/10) + 3^n*2/5.
a(n) = Sum_{k=0..n} (-2*0^k-Fib(k-4)) * 3^(n-k).
a(n) = A098703(n+1) - A098703(n). - Ross La Haye, Sep 11 2005

A104004 Expansion of (1-x) * (1+x) / ((1-2x)(1-x-x^2)).

Original entry on oeis.org

1, 3, 7, 16, 35, 75, 158, 329, 679, 1392, 2839, 5767, 11678, 23589, 47555, 95720, 192427, 386451, 775486, 1555153, 3117071, 6245088, 12507887, 25044431, 50135230, 100345485, 200812363, 401821144, 803960099, 1608434427, 3217700894, 6436748057

Offset: 0

Creighton Dement, Feb 24 2005

A floretion-generated sequence relating to Fibonacci numbers and powers of 2. The sequence results from a particular transform of the sequence A000079*(-1)^n (powers of 2).

Floretion Algebra Multiplication Program, FAMP Code: 1jesforseq[ ( 5'i + .5i' + .5'ii' + .5e)*( + .5j' + .5'kk' + .5'ki' + .5e ) ], 1vesforseq = A000079(n+1)*(-1)^(n+1), ForType: 1A. Identity used: jesfor = jesrightfor + jesleftfor

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2).

Cf. A000045, A000079, A001911, A008466, A016777, A022958, A027934, A042950, A078024, A099036, A221719.

[3*2^n-Fibonacci(n+3): n in [0..40]]; // Vincenzo Librandi, Aug 18 2017

with (combinat):a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=fibonacci(n-1)+2*a[n-1] od: seq(a[n], n=1..26); # Zerinvary Lajos, Mar 17 2008

LinearRecurrence[{3, -1, -2}, {1, 3, 7}, 80] (* Vincenzo Librandi, Aug 18 2017 *)
CoefficientList[Series[(1-x)(1+x)/((2x-1)(x^2+x-1)),{x,0,40}],x] (* Harvey P. Dale, Oct 12 2024 *)
A104004[n_]:= 3*2^n -Fibonacci[n+3]; (* G. C. Greubel, Jun 05 2025 *)

def A104004(n): return 3*2**n - fibonacci(n+3) # G. C. Greubel, Jun 05 2025

4*a(n) = A008466(n+3) + A027973(n) (FAMP result).
Suggestions made by Superseeker: (Start)
a(n+2) - a(n+1) - a(n) = A042950(n+1).
Coefficients of g.f.*(1-x)/(1+x) match A099036.
Coefficients of g.f./(1+x) match A027934.
Coefficients of g.f./(1-x^2) match A008466. (End)
a(n) = A101220(3, 2, n+1) - A101220(3, 2, n). - Ross La Haye, Aug 05 2005
a(n) = 3*2^n - Fibonacci(n+3) = A221719(n) + 1. - Ralf Stephan, May 20 2007, Hugo Pfoertner, Mar 06 2024
a(n) = (3*2^n - (2^(-n)*((1-sqrt(5))^n*(-2+sqrt(5)) + (1+sqrt(5))^n*(2+sqrt(5)))) / sqrt(5)). - Colin Barker, Aug 18 2017
From G. C. Greubel, Jun 05 2025: (Start)
Sum_{k=0..n} A022958(k+1)*a(n-k) = A001911(n+1).
Sum_{k=0..n} (-1)^k*A016777(k)*a(n-k) = A078024(n).
E.g.f.: 3*exp(2*x) - (2/sqrt(5))*exp(x/2)*( 2*sinh(sqrt(5)*x/2) + sqrt(5)*cosh(sqrt(5)*x/2) ). (End)

A106517 Convolution of Fibonacci(n-1) and 3^n.

Original entry on oeis.org

1, 3, 10, 31, 95, 288, 869, 2615, 7858, 23595, 70819, 212512, 637625, 1913019, 5739290, 17218247, 51655351, 154967040, 464902717, 1394710735, 4184136386, 12552415923, 37657258715, 112971793856, 338915410225, 1016746277043

Offset: 0

Paul Barry, May 05 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Cf. A000032, A000045, A000244, A094688, A101220.

Diagonal sums of number triangle A106516.

I:=[1,3,10]; [n le 3 select I[n] else 4*Self(n-1) -2*Self(n-2) -3*Self(n-3): n in [1..41]]; // G. C. Greubel, Aug 05 2021

LinearRecurrence[{4,-2,-3},{1,3,10},30] (* Harvey P. Dale, Oct 08 2014 *)

a(n) = sum(k=0, n, fibonacci(n-k-1) * 3^k); \\ Michel Marcus, Aug 06 2021

[(2*3^(n+1) - lucas_number2(n+1, 1, -1))/5 for n in (0..40)] # G. C. Greubel, Aug 05 2021

G.f.: (1-x)/((1-x-x^2)*(1-3*x)).
a(n) = Sum_{k=0..n} Fibonacci(n-k-1) * 3^k.
a(n) = A101220(2, 3, n+1). - Ross La Haye, Jul 25 2005
a(n) = A101220(3, 3, n+1) - A101220(3, 3, n). - Ross La Haye, May 31 2006
a(n) = (1/5)*(6*3^n - Lucas(n+1)). - Ralf Stephan, Nov 16 2010
Sum_{k=0..n} a(k) = A094688(n+1). - G. C. Greubel, Aug 05 2021

A022104 Fibonacci sequence beginning 1, 14.

Original entry on oeis.org

1, 14, 15, 29, 44, 73, 117, 190, 307, 497, 804, 1301, 2105, 3406, 5511, 8917, 14428, 23345, 37773, 61118, 98891, 160009, 258900, 418909, 677809, 1096718, 1774527, 2871245, 4645772, 7517017, 12162789

Offset: 0

N. J. A. Sloane

a(n-1)=sum(P(14;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=13. These are the SW-NE diagonals in P(14;n,k), the (14,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

a(n) = A109754(13, n+1) = A101220(13, 0, n+1).

a0:=1; a1:=14; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

a={};b=1;c=14;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,9,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 22 2008 *)
LinearRecurrence[{1,1},{1,14},40] (* Harvey P. Dale, Jun 12 2017 *)

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=14. a(-1):=13.

G.f.: (1+13*x)/(1-x-x^2).

A022106 Fibonacci sequence beginning 1, 16.

Original entry on oeis.org

1, 16, 17, 33, 50, 83, 133, 216, 349, 565, 914, 1479, 2393, 3872, 6265, 10137, 16402, 26539, 42941, 69480, 112421, 181901, 294322, 476223, 770545, 1246768, 2017313, 3264081, 5281394, 8545475, 13826869

Offset: 0

N. J. A. Sloane

a(n-1)=sum(P(16;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=15. These are the SW-NE diagonals in P(16;n,k), the (16,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

a(n) = A109754(15, n+1) = A101220(15, 0, n+1).

a0:=1; a1:=16; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

a={};b=1;c=16;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
LinearRecurrence[{1,1},{1,16},40] (* Harvey P. Dale, Jun 22 2016 *)

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=16. a(-1):=15.

G.f.: (1+15*x)/(1-x-x^2).

cp's OEIS Frontend

A022101 Fibonacci sequence beginning 1, 11.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A098703 a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A022102 Fibonacci sequence beginning 1, 12.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

A022110 Fibonacci sequence beginning 1, 20.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A053311 Partial sums of A000285.

Views

Author

Keywords

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A099159 a(n) = (L(n-2)+2*3^n)/5.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A104004 Expansion of (1-x) * (1+x) / ((1-2*x)*(1-x-x^2)).

Views

Author

Keywords

Comments

Links

Crossrefs

A104004 Expansion of (1-x) * (1+x) / ((1-2x)(1-x-x^2)).