cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-20 of 27 results. Next

A014994 a(n) = (1 - (-12)^n)/13.

Original entry on oeis.org

1, -11, 133, -1595, 19141, -229691, 2756293, -33075515, 396906181, -4762874171, 57154490053, -685853880635, 8230246567621, -98762958811451, 1185155505737413, -14221866068848955, 170662392826187461
Offset: 1

Views

Author

Keywords

Comments

q-integers for q=-12.

Crossrefs

Programs

  • Magma
    I:=[1,-11]; [n le 2 select I[n] else -11*Self(n-1)+12*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012
    
  • Maple
    a:=n->sum ((-12)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008
  • Mathematica
    LinearRecurrence[{-11, 12}, {1, -11}, 30] (* Vincenzo Librandi, Oct 22 2012 *)
  • PARI
    a(n)=(1-(-12)^n)/13 \\ Charles R Greathouse IV, Sep 24 2015
  • Sage
    [gaussian_binomial(n,1,-12) for n in range(1,18)] # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
G.f.: x/((1 - x)*(1 + 12*x)). - Vincenzo Librandi, Oct 22 2012
a(n) = -11*a(n-1) + 12*a(n-2). - Vincenzo Librandi, Oct 22 2012
E.g.f.: (exp(x) - exp(-12*x))/13. - G. C. Greubel, May 26 2018

Extensions

Better name from Ralf Stephan, Jul 14 2013

A014990 a(n) = (1 - (-8)^n)/9.

Original entry on oeis.org

1, -7, 57, -455, 3641, -29127, 233017, -1864135, 14913081, -119304647, 954437177, -7635497415, 61083979321, -488671834567, 3909374676537, -31274997412295, 250199979298361, -2001599834386887, 16012798675095097
Offset: 1

Views

Author

Keywords

Comments

q-integers for q=-8.

Crossrefs

Programs

  • Magma
    I:=[1, -7]; [n le 2 select I[n] else -7*Self(n-1) +8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012
    
  • Maple
    a:=n->sum ((-8)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008
  • Mathematica
    QBinomial[Range[20],1,-8] (* or *) LinearRecurrence[{-7,8},{1,-7},20] (* Harvey P. Dale, Dec 19 2011 *)
  • PARI
    a(n)=(1-(-8)^n)/9 \\ Charles R Greathouse IV, Oct 07 2015
  • Sage
    [gaussian_binomial(n,1,-8) for n in range(1,20)] # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)}
From Philippe Deléham, Feb 13 2007: (Start)
a(1)=1, a(2)=-7, a(n) = -7*a(n-1) + 8*a(n-2) for n > 2.
a(n) = (-1)^(n+1)*A015565(n).
G.f.: x/(1 + 7*x - 8*x^2). (End)
E.g.f.: (exp(x) - exp(-8*x))/9. - G. C. Greubel, May 26 2018

Extensions

Better name from Ralf Stephan, Jul 14 2013

A014992 a(n) = (1 - (-10)^n)/11.

Original entry on oeis.org

1, -9, 91, -909, 9091, -90909, 909091, -9090909, 90909091, -909090909, 9090909091, -90909090909, 909090909091, -9090909090909, 90909090909091, -909090909090909, 9090909090909091, -90909090909090909
Offset: 1

Views

Author

Keywords

Comments

q-integers for q = -10.

Crossrefs

Programs

  • Magma
    I:=[1, -9]; [n le 2 select I[n] else -9*Self(n-1) +10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012
    
  • Maple
    a:=n->sum ((-10)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008
  • Mathematica
    CoefficientList[Series[1/((1 - x)*(1 + 10*x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 22 2012 *)
  • PARI
    for(n=1, 30, print1((1-(-10)^n)/11, ", ")) \\ G. C. Greubel, May 26 2018
  • Sage
    [gaussian_binomial(n,1,-10) for n in range(1,19)] # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
G.f.: x/((1 - x)*(1 + 10*x)). - Vincenzo Librandi, Oct 22 2012
a(n) = -9*a(n-1) + 10*a(n-2). - Vincenzo Librandi, Oct 22 2012
a(n) = (-1)^(n+1)*A015585(n). - R. J. Mathar, Oct 26 2015
E.g.f.: (exp(x) - exp(-10*x))/11. - G. C. Greubel, May 26 2018

Extensions

Better name from Ralf Stephan, Jul 14 2013

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1
Offset: 0

Views

Author

Tom Copeland, Mar 19 2014

Keywords

Comments

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

Examples

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1
		

Crossrefs

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.

Programs

  • Magma
    [[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
    
  • Maple
    A239473 := proc(n,k)
        add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
    end proc; # R. J. Mathar, Jul 21 2016
  • Mathematica
    Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018
    
  • Sage
    Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Extensions

Inverse array added by Tom Copeland, Mar 26 2014
Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

A014991 a(n) = (1 - (-9)^n)/10.

Original entry on oeis.org

1, -8, 73, -656, 5905, -53144, 478297, -4304672, 38742049, -348678440, 3138105961, -28242953648, 254186582833, -2287679245496, 20589113209465, -185302018885184, 1667718169966657, -15009463529699912
Offset: 1

Views

Author

Keywords

Comments

q-integers for q = -9.

Crossrefs

Programs

  • Magma
    I:=[1,-8]; [n le 2 select I[n] else -8*Self(n-1)+9*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012
    
  • Maple
    a:=n->sum ((-9)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008
  • Mathematica
    ((-9)^Range[30]-1)/-10 (* or *) LinearRecurrence[{-8,9},{1,-8},30] (* Harvey P. Dale, Aug 08 2011 *)
    CoefficientList[Series[1/((1 - x)*(1 + 9*x)), {x, 0, 30}], x]; (* Vincenzo Librandi, Oct 22 2012 *)
  • PARI
    for(n=1,30, print1((1-(-9)^n)/10, ", ")) \\ G. C. Greubel, May 26 2018
  • Sage
    [gaussian_binomial(n,1,-9) for n in range(1,19)] # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
a(0)=1, a(1)=-8, a(n) = -8*a(n-1) + 9*a(n-2). - Harvey P. Dale, Aug 08 2011
G.f.: x/((1 - x)*(1 + 9*x)). - Vincenzo Librandi, Oct 22 2012
E.g.f.: (exp(x) - exp(-9*x))/10. - G. C. Greubel, May 26 2018

Extensions

Better name from Ralf Stephan, Jul 14 2013

A014993 a(n) = (1 - (-11)^n)/12.

Original entry on oeis.org

1, -10, 111, -1220, 13421, -147630, 1623931, -17863240, 196495641, -2161452050, 23775972551, -261535698060, 2876892678661, -31645819465270, 348104014117971, -3829144155297680, 42120585708274481
Offset: 1

Views

Author

Keywords

Comments

q-integers for q = -11.

Crossrefs

Programs

  • Magma
    I:=[1, -10]; [n le 2 select I[n] else -10*Self(n-1) +11*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012
    
  • Maple
    a:=n->sum ((-11)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008
  • Mathematica
    LinearRecurrence[{-10, 11}, {1, -10}, 40] (* Vincenzo Librandi, Oct 22 2012 *)
  • PARI
    for(n=1,30, print1((1-(-11)^n)/12, ", ")) \\ G. C. Greubel, May 26 2018
  • Sage
    [gaussian_binomial(n,1,-11) for n in range(1,18)] # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = a(n-1) + q^{(n-1)} = {(q^n - 1) / (q - 1)}.
G.f.: x/((1 - x)*(1 + 11*x)). - Vincenzo Librandi, Oct 22 2012
a(n) = -10*a(n-1) + 11*a(n-2). - Vincenzo Librandi, Oct 22 2012
E.g.f.: (exp(x) - exp(-11*x))/12. - G. C. Greubel, May 26 2018

Extensions

Better name from Ralf Stephan, Jul 14 2013

A074087 Coefficient of q^1 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=b*nu(n-1)+lambda*(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(2,3).

Original entry on oeis.org

0, 0, 0, 6, 33, 144, 570, 2118, 7587, 26448, 90420, 304470, 1013061, 3338112, 10911150, 35423862, 114342855, 367242336, 1174368360, 3741029094, 11876859369, 37591894320, 118659631650, 373630740966, 1173847761003
Offset: 0

Views

Author

Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 19 2002

Keywords

Comments

The coefficient of q^0 is A014983(n+1).

Examples

			The first 6 nu polynomials are nu(0)=1, nu(1)=2, nu(2)=7, nu(3)=20+6q, nu(4)=61+33q+21q^2, nu(5)=182+144q+120q^2+78q^3+18q^4, so the coefficients of q^1 are 0,0,0,6,33,144.
		

Crossrefs

Coefficients of q^0, q^2 and q^3 are in A014983, A074088 and A074089. Related sequences with other values of b and lambda are in A074082-A074086.

Programs

  • Magma
    I:=[0,0,6,33]; [0] cat [n le 4 select I[n] else 4*Self(n-1) + 2*Self(n-2) -12*Self(n-3) -9*Self(n-4): n in [1..30]]; // G. C. Greubel, May 26 2018
  • Mathematica
    b=2; lambda=3; expon=1; nu[0]=1; nu[1]=b; nu[n_] := nu[n]=Together[b*nu[n-1]+lambda(1-q^(n-1))/(1-q)nu[n-2]]; a[n_] := Coefficient[nu[n], q, expon]
    (* Second program: *)
    Join[{0}, LinearRecurrence[{4,2,-12,-9}, {0,0,6,33}, 50]] (* G. C. Greubel, May 26 2018 *)
  • PARI
    x='x+O('x^30); concat([0,0,0], Vec((6*x^3 +9*x^4)/(1-2*x-3*x^2)^2)) \\ G. C. Greubel, May 26 2018
    

Formula

G.f.: (6*x^3 +9*x^4)/(1-2*x-3*x^2)^2.
a(n) = 4*a(n-1) +2*a(n-2) -12*a(n-3) -9*a(n-4) for n>=5.

Extensions

Edited by Dean Hickerson, Aug 21 2002

A302747 Triangle read by rows: T(0,0) = 1; T(n,k) = -2*T(n-1,k) + 3*T(n-2,k-1) for 0 <= k <= floor(n/2); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, -2, 4, 3, -8, -12, 16, 36, 9, -32, -96, -54, 64, 240, 216, 27, -128, -576, -720, -216, 256, 1344, 2160, 1080, 81, -512, -3072, -6048, -4320, -810, 1024, 6912, 16128, 15120, 4860, 243, -2048, -15360, -41472, -48384, -22680, -2916, 4096, 33792, 103680, 145152, 90720, 20412, 729, -8192, -73728, -253440
Offset: 0

Views

Author

Zagros Lalo, May 04 2018

Keywords

Comments

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle in A303901 ((3-2x)^n).
The coefficients in the expansion of 1/(1-3x+2x^2) are given by the sequence generated by the row sums.

Examples

			Triangle begins:
.
   n | k = 0      1       2       3       4       5       6
  ---+-----------------------------------------------------
   0 |     1
   1 |    -2
   2 |     4      3
   3 |    -8    -12
   4 |    16     36       9
   5 |   -32    -96     -54
   6 |    64    240     216      27
   7 |  -128   -576    -720    -216
   8 |   256   1344    2160    1080      81
   9 |  -512  -3072   -6048   -4320    -810
  10 |  1024   6912   16128   15120    4860     243
  11 | -2048 -15360  -41472  -48384  -22680   -2916
  12 |  4096  33792  103680  145152   90720   20412     729
  13 | -8192 -73728 -253440 -414720 -326592 -108864  -10206
		

References

  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 394-396.

Crossrefs

Row sums give A014983.

Programs

  • Mathematica
    t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, -2 t[n - 1, k] + 3 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten
  • PARI
    T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, -2*T(n-1,k) + 3*T(n-2,k-1)));
    tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018

A015000 q-integers for q=-13.

Original entry on oeis.org

1, -12, 157, -2040, 26521, -344772, 4482037, -58266480, 757464241, -9847035132, 128011456717, -1664148937320, 21633936185161, -281241170407092, 3656135215292197, -47529757798798560, 617886851384381281
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Magma
    I:=[1,-12]; [n le 2 select I[n] else -12*Self(n-1)+13*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012
    
  • Maple
    a:=n->sum ((-13)^j, j=0..n-1): seq(a(n), n=0..20); # Zerinvary Lajos, Dec 16 2008
  • Mathematica
    QBinomial[Range[20],1,-13] (* Harvey P. Dale, May 02 2012 *)
    LinearRecurrence[{-12, 13}, {1, -12}, 30] (* Vincenzo Librandi, Oct 22 2012 *)
  • PARI
    for(n=1, 30, print1((1-(-13)^n)/14, ", ")) \\ G. C. Greubel, May 26 2018
  • Sage
    [gaussian_binomial(n,1,-13) for n in range(1,18)] # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1), with q=-13.
a(n) = Sum_{j=0..n-1} (-13)^j. - Zerinvary Lajos, Dec 16 2008
G.f.: x/((1 - x)*(1 + 13*x)). - Vincenzo Librandi, Oct 22 2012
a(n) = -12*a(n-1) + 13*a(n-2). - Vincenzo Librandi, Oct 22 2012
From G. C. Greubel, May 26 2018: (Start)
a(n) = (1 - (-13)^n)/14.
E.g.f.: (exp(x) - exp(-13*x))/14. (End)

Extensions

Edited by N. J. A. Sloane, Jan 13 2009 at the suggestion of R. J. Mathar

A051068 Partial sums of A014578.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 5, 6, 7, 8, 9, 9, 10, 11, 11, 12, 13, 14, 15, 16, 16, 17, 18, 18, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 26, 27, 28, 29, 29, 30, 31, 31, 32, 33, 34, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 44, 44, 45, 46, 47, 48, 49, 49
Offset: 0

Views

Author

Keywords

Comments

Duplicate of A050294? [Joerg Arndt, Apr 27 2013]
From Michel Dekking, Feb 10 2019: (Start)
The answer to Joerg Arndt's question is: yes (modulo an offset). To see this, it suffices to prove that the two sequences of first differences Da and Db of a= A051068 and b:=A050294 are equal. Clearly the sequence Da of first differences of a is the sequence A014578. According to Philippe Deleham (2004), Da equals 0x = 0110110111110..., where x is the fixed point of the morphism 0->111, 1->110.
From Vladimir Shevelev (2011) we know a formula for b=A050294: b(n) = n-b(floor(n/3)). This gives that the sequence of first differences Db:=(b(n+1)-b(n)) of b satisfies
Db(3m+1) = Db(3m+2) = 1, and Db(3m+3) = 1 - Db(m).
This implies that Db = x, the fixed point of 0->111, 1->110.
(End)

Crossrefs

Formula

a(3^n) = A015518(n+1) = -(-1)^n*A014983(n+1). - Philippe Deléham, Mar 31 2004
Previous Showing 11-20 of 27 results. Next