A015249 -id:A015249 - cp's OEIS frontend

A015109 Triangle of Gaussian (or q-binomial) coefficients for q = -2.

1, 1, 1, 1, -1, 1, 1, 3, 3, 1, 1, -5, 15, -5, 1, 1, 11, 55, 55, 11, 1, 1, -21, 231, -385, 231, -21, 1, 1, 43, 903, 3311, 3311, 903, 43, 1, 1, -85, 3655, -25585, 56287, -25585, 3655, -85, 1, 1, 171, 14535, 208335, 875007, 875007, 208335, 14535, 171, 1, 1, -341, 58311

Offset: 0

Olivier Gérard

May be read as a symmetric triangular (T(n,k)=T(n,n-k); k=0,...,n; n=0,1,...) or square array (A(n,r)=A(r,n)=T(n+r,r), read by antidiagonals). The diagonals of the former (or rows/columns of the latter) are A000012 (k=0), A077925 (k=1), A015249 (k=2), A015266 (k=3), A015287 (k=4), A015305 (k=5), A015323 (k=6), A015338 (k=7), A015356 (k=8), A015371 (k=9), A015386 (k=10), A015405 (k=11), A015423 (k=12), ... - M. F. Hasler, Nov 04 2012
The elements of the inverse matrix are apparently T^(-1)(n,k) = (-1)^n*A157785(n,k). - R. J. Mathar, Mar 12 2013
Fu et al. give two combinatorial interpretations of the (unsigned) q-binomial coefficients when q is a negative integer. - Peter Bala, Nov 02 2017

			From _Roger L. Bagula_, Feb 10 2009: (Start)
  1;
  1,   1;
  1,  -1,     1;
  1,   3,     3,      1;
  1,  -5,    15,     -5,      1;
  1,  11,    55,     55,     11,      1;
  1, -21,   231,   -385,    231,    -21,      1;
  1,  43,   903,   3311,   3311,    903,     43,     1;
  1, -85,  3655, -25585,  56287, -25585,   3655,   -85,   1;
  1, 171, 14535, 208335, 875007, 875007, 208335, 14535, 171, 1;  (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
J. A. de Azcarraga and J. A. Macfarlane, Group Theoretical Foundations of Fractional Supersymmetry, arxiv:hep-th/9506177 (1995).
S. Fu, V. Reiner, D. Stanton and N. Thiem, The negative q-binomial, arXiv:1108.4702 [math.CO], 2011.
R. Parthasarathy, q-Fermionic Numbers and Their Roles in Some Physical Problems, arxiv:quant-ph/0403216, 2004.

Cf. A015152 (row sums).
Cf. A022166 (q=2), A022167 (q=3), A022168 (q=4), A022169 (q=5), A022170 (q=6), A022171 (q=7), A022172 (q=8), A022173 (q=9), A022174 (q=10), A022175 (q=11), A022176 (q=12), A022177 (q=13), A022178 (q=14), A022179 (q=15), A022180 (q=16), A022181 (q=17), A022182 (q=18), A022183 (q=19), A022184 (q=20), A022185 (q=21), A022186 (q=22), A022187 (q=23), A022188 (q=24).
Analogous triangles for other q: A015110 (q=-3), A015112 (q=-4), A015113 (q=-5), A015116 (q=-6), A015117 (q=-7), A015118 (q=-8), A015121 (q=-9), A015123 (q=-10), A015124 (q=-11), A015125 (q=-12), A015129 (q=-13), A015132 (q=-14), A015133 (q=-15).

qBinomial:= func< n,k,q | k eq 0 select 1 else (&*[(1-q^(n-j+1))/(1-q^j): j in [1..k]]) >;
[qBinomial(n,k,-2): k in [0..n], n in [0..10]]; // A015109 // G. C. Greubel, Nov 30 2021

A015109 := proc(n, k)
   mul( ((-2)^(1+n-i)-1)/((-2)^i-1) ,i=1..k) ;
end proc: # R. J. Mathar, Mar 12 2013

T[n_, k_, q_]:= Product[(1 - q^(n-j+1))/(1 - q^j), {j, k}];
Table[T[n,k,-2], {n,0,10}, {k,0,n}]//Flatten (* Roger L. Bagula, Feb 10 2009 *)(* modified by G. C. Greubel, Nov 30 2021 *)
Table[QBinomial[n, k, -2], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, Apr 09 2016 *)

T015109(n, k, q=-2)=prod(i=1, k, (q^(1+n-i)-1)/(q^i-1)) \\ (Indexing is that of the triangular array: 0 <= k <= n = 0,1,2,...) \\ M. F. Hasler, Nov 04 2012

flatten([[q_binomial(n,k,-2) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Nov 30 2021

T(n, k) = q-binomial(n, k, -2).

T(n, k, q) = Product_{j=1..k} ( (1 - q^(n-j+1))/(1 - q^j) ), for q = -2. - Roger L. Bagula, Feb 10 2009

Edited by M. F. Hasler, Nov 04 2012

A084175 Jacobsthal oblong numbers.

Original entry on oeis.org

0, 1, 3, 15, 55, 231, 903, 3655, 14535, 58311, 232903, 932295, 3727815, 14913991, 59650503, 238612935, 954429895, 3817763271, 15270965703, 61084037575, 244335800775, 977343902151, 3909374210503, 15637499638215, 62549992960455

Offset: 0

Paul Barry, May 18 2003

Inverse binomial transform is A001019 doubled up.
Binomial transform is A084177.
Partial sums of A003683.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

Except for initial terms, same as A015249 and A084152.

Cf. A001654, A001656, A084158, A084159, A099930.

List([0..30], n-> (2^(2*n+1) -(-2)^n -1)/9); # G. C. Greubel, Sep 21 2019

[(2*4^n-(-2)^n-1)/9: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

for n from 1 to 25 do print(round(2^n/3)*round(2^(n+1)/3)) od; # Gary Detlefs, Feb 10 2010

Table[(2*4^n -(-2)^n -1)/9, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011, modified by G. C. Greubel, Sep 21 2019 *)
LinearRecurrence[{3,6,-8}, {0,1,3}, 25] (* Jean-François Alcover, Sep 21 2017 *)

a(n)=(2*4^n-(-2)^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015

def A084175(n): return ((m:=1<Chai Wah Wu, Apr 25 2025

[gaussian_binomial(n, 2, -2) for n in range(1, 26)] # Zerinvary Lajos, May 28 2009

a(n) = A001045(n)*A001045(n+1).
a(n) = (2*4^n - (-2)^n - 1)/9;
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3), a(0)=0, a(1)=1, a(2)=3.
G.f.: x/((1+2*x)*(1-x)*(1-4*x)).
E.g.f.: (2*exp(4*x) - exp(x) - exp(-2*x))/9.
a(n+1) - 4*a(n) = 1, -1, 3, -5, 11, ... = A001045(n+1) signed. - Paul Curtz, May 19 2008
a(n) = round(2^n/3) * round(2^(n+1)/3). - Gary Detlefs, Feb 10 2010
From Peter Bala, Mar 30 2015: (Start)
The shifted o.g.f. A(x) := 1/( (1 + 2*x)*(1 - x)*(1 - 4*x) ) = 1/(1 - 3*x - 6*x^2 + 8*x^3). Hence A(x) == 1/(1 - 3*x + 3*x^2 - x^3) (mod 9) == 1/(1 - x)^3 (mod 9). It follows by Theorem 1 of Heninger et al. that (A(x))^(1/3) = 1 + x + 4*x^2 + 10*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} J(3*n)/J(n)*x^n/n ), where J(n) = A001045(n) are the Jacobsthal numbers. Cf. A001656, A099930. (End)

A015265 Gaussian binomial coefficient [ n,2 ] for q = -13.

Original entry on oeis.org

1, 157, 26690, 4508570, 761974851, 128773405047, 21762709934980, 3677897920745140, 621564749363392901, 105044442632566365137, 17752510805031727164870, 3000174326048697741925710, 507029461102251552321630151, 85687978926280231101185088427

Offset: 2

Olivier Gérard, Dec 11 1999

J. Goldman and G.-C. Rota, The number of subspaces of a vector space, pp. 75-83 of W. T. Tutte, editor, Recent Progress in Combinatorics. Academic Press, NY, 1969.
I. P. Goulden and D. M. Jackson, Combinatorial Enumeration. Wiley, NY, 1983, p. 99.
M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351.

Vincenzo Librandi, Table of n, a(n) for n = 2..200
Index entries for linear recurrences with constant coefficients, signature (157, 2041, -2197).

Cf. Gaussian binomial coefficients [n,2] for q=-2,...,-12: A015249, A015251, A015253, A015255, A015257 A015258, A015259, A015260, A015261, A015262, A015264.

Cf. Gaussian binomial coefficients [n,r] for q=-13: A015286 (r=3), A015303 (r=4), A015321 (r=5), A015337 (r=6), A015355 (r=7), A015370 (r=8), A015385 (r=9), A015402 (r=10), A015422 (r=11), A015438 (r=12). - M. F. Hasler, Nov 03 2012

I:=[1,157,26690]; [n le 3 select I[n] else 157*Self(n-1)+2041*Self(n-2)-2197*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Oct 28 2012

Table[QBinomial[n, 2, -13], {n, 2, 20}] (* Vincenzo Librandi, Oct 28 2012 *)

A015265(n,q=-13)=(1-q^n)*(q^(n-1)-1)/2352 \\ M. F. Hasler, Nov 03 2012

[gaussian_binomial(n,2,-13) for n in range(2,14)] # Zerinvary Lajos, May 27 2009

G.f.: x^2/((1-x)*(1+13*x)*(1-169*x)). - Ralf Stephan, Apr 01 2004
a(2) = 1, a(3) = 157, a(4) = 26690, a(n) = 157*a(n-1) + 2041*a(n-2) - 2197*a(n-3). - Vincenzo Librandi, Oct 28 2012
a(n) = (1/2352)*( (1 - (-13)^n)*((-13)^(n-1) - 1) ). - M. F. Hasler, Nov 03 2012

A084152 Exponential self-convolution of Jacobsthal numbers (divided by 2).

Original entry on oeis.org

0, 0, 1, 3, 15, 55, 231, 903, 3655, 14535, 58311, 232903, 932295, 3727815, 14913991, 59650503, 238612935, 954429895, 3817763271, 15270965703, 61084037575, 244335800775, 977343902151, 3909374210503, 15637499638215, 62549992960455

Offset: 0

Paul Barry, May 16 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

Cf. A001045, A078008, A084153.

Except for initial terms, same as A015249 and A084175.

[(4^n-2+(-2)^n)/18: n in [0..35]]; // Vincenzo Librandi, Jul 05 2011

Join[{a=0,b=0},Table[c=2*b+8*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3,6,-8},{0,0,1},30] (* Harvey P. Dale, Nov 11 2011 *)

def A084152(n): return ((m:=1<>1|1)//3) # Chai Wah Wu, Apr 25 2025

[(4^n-2+(-2)^n)/18 for n in range(41)] # G. C. Greubel, Oct 11 2022

a(n) = (4^n - 2 + (-2)^n)/18.
G.f.: x^2/((1-x)*(1+2*x)*(1-4*x)).
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3).
E.g.f.: (exp(2*x) - exp(-x))^2/18 = (exp(4*x) - 2*exp(x) + exp(-x))/18.
Binomial transform of 0, 0, 1, 0, 9, 0, 81, ... .
a(n) = A001045(n)*A078008(n)/2.
a(n) = floor(2^n/3)ceiling(2^n/3)/2. - Paul Barry, Apr 28 2004

A249993 Expansion of 1/((1+x)(1+2x)(1-4x)).

Original entry on oeis.org

1, 1, 11, 29, 147, 525, 2227, 8653, 35123, 139469, 559923, 2235597, 8950579, 35785933, 143176499, 572640461, 2290692915, 9162509517, 36650562355, 146601200845, 586406900531, 2345623407821, 9382502019891, 37529991302349, 150119998763827, 600479927946445

Offset: 0

Alex Ratushnyak, Dec 27 2014

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,10,8).

Cf. A249992.
Cf. A006095, A171477 for g.f. 1/((1-x)*(1-2*x)*(1-4*x)).
Cf. A015249, A084152, A084175 for g.f. 1/((1-x)*(1+2*x)*(1-4*x)).
Cf. A109765 for g.f. 1/((1+x)*(1-2*x)*(1-4*x)).

[(2^(2*n+3) +(-1)^n*(5*2^(n+1)-3))/15: n in [0..40]]; // G. C. Greubel, Oct 10 2022

CoefficientList[Series[1/((1+x)(1+2x)(1-4x)),{x,0,30}],x] (* or *) LinearRecurrence[{1,10,8},{1,1,11},30] (* Harvey P. Dale, Dec 13 2018 *)

Vec(1/((1+x)*(1+2*x)*(1-4*x)) + O(x^40)) \\ Michel Marcus, Dec 28 2014

[(2^(2*n+3) +(-1)^n*(5*2^(n+1)-3))/15 for n in range(41)] # G. C. Greubel, Oct 10 2022

G.f.: 1/((1+x)*(1+2*x)*(1-4*x)).
a(n) = ( 2^(3+2*n) + (5*2^(1+n) - 3)*(-1)^n )/15. Colin Barker, Dec 28 2014
a(n) = a(n-1) + 10*a(n-2) + 8*a(n-3). - Colin Barker, Dec 28 2014
E.g.f.: (1/15)*(10*exp(-2*x) - 3*exp(-x) + 8*exp(4*x)). - G. C. Greubel, Oct 10 2022

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A015109 Triangle of Gaussian (or q-binomial) coefficients for q = -2.

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A084175 Jacobsthal oblong numbers.

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A015265 Gaussian binomial coefficient [ n,2 ] for q = -13.

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A084152 Exponential self-convolution of Jacobsthal numbers (divided by 2).

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A249993 Expansion of 1/((1+x)*(1+2*x)*(1-4*x)).

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A249993 Expansion of 1/((1+x)(1+2x)(1-4x)).