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.

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A135278 Triangle read by rows, giving the numbers T(n,m) = binomial(n+1, m+1); or, Pascal's triangle A007318 with its left-hand edge removed.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 12, 66, 220, 495, 792, 924, 792
Offset: 0

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Author

Zerinvary Lajos, Dec 02 2007

Keywords

Comments

T(n,m) is the number of m-faces of a regular n-simplex.
An n-simplex is the n-dimensional analog of a triangle. Specifically, a simplex is the convex hull of a set of (n + 1) affinely independent points in some Euclidean space of dimension n or higher, i.e., a set of points such that no m-plane contains more than (m + 1) of them. Such points are said to be in general position.
Reversing the rows gives A074909, which as a linear sequence is essentially the same as this.
From Tom Copeland, Dec 07 2007: (Start)
T(n,k) * (k+1)! = A068424. The comment on permuted words in A068424 shows that T is related to combinations of letters defined by connectivity of regular polytope simplexes.
If T is the diagonally-shifted Pascal matrix, binomial(n+m, k+m), for m=1, then T is a fundamental type of matrix that is discussed in A133314 and the following hold.
The infinitesimal matrix generator is given by A132681, so T = LM(1) of A132681 with inverse LM(-1).
With a(k) = (-x)^k / k!, T * a = [ Laguerre(n,x,1) ], a vector array with index n for the Laguerre polynomials of order 1. Other formulas for the action of T are given in A132681.
T(n,k) = (1/n!) (D_x)^n (D_t)^k Gf(x,t) evaluated at x=t=0 with Gf(x,t) = exp[ t * x/(1-x) ] / (1-x)^2.
[O.g.f. for T ] = 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 }. [ O.g.f. for row sums ] = 1 / { (1-x) * (1-2x) }, giving A000225 (without a leading zero) for the row sums. Alternating sign row sums are all 1. [Sign correction noted by Vincent J. Matsko, Jul 19 2015]
O.g.f. for row polynomials = [ (1+q)**(n+1) - 1 ] / [ (1+q) -1 ] = A(1,n+1,q) on page 15 of reference on Grassmann cells in A008292. (End)
Given matrices A and B with A(n,k) = T(n,k)*a(n-k) and B(n,k) = T(n,k)*b(n-k), then A*B = C where C(n,k) = T(n,k)*[a(.)+b(.)]^(n-k), umbrally. The e.g.f. for the row polynomials of A is {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. - Tom Copeland, Aug 21 2008
A007318*A097806 as infinite lower triangular matrices. - Philippe Deléham, Feb 08 2009
Riordan array (1/(1-x)^2, x/(1-x)). - Philippe Deléham, Feb 22 2012
The elements of the matrix inverse are T^(-1)(n,k)=(-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Relation to K-theory: T acting on the column vector (-0,d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168 and also A104712, A111492, and A238363. - Tom Copeland, Apr 11 2014
Number of walks of length p>0 between any two distinct vertices of the complete graph K_(n+2) is W(n+2,p)=(-1)^(p-1)*Sum_{k=0..p-1} T(p-1,k)*(-n-2)^k = ((n+1)^p - (-1)^p)/(n+2) = (-1)^(p-1)*Sum_{k=0..p-1} (-n-1)^k. This is equal to (-1)^(p-1)*Phi(p,-n-1), where Phi is the cyclotomic polynomial when p is an odd prime. For K_3, see A001045; for K_4, A015518; for K_5, A015521; for K_6, A015531; for K_7, A015540. - Tom Copeland, Apr 14 2014
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-1)^0 + A_1*(x-1)^1 + A_2*(x-1)^2 + ... + A_n*(x-1)^n. This sequence gives A_0, ..., A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
See A074909 for associations among this array, the Bernoulli polynomials and their umbral compositional inverses, and the face polynomials of permutahedra and their duals (cf. A019538). - Tom Copeland, Nov 14 2014
From Wolfdieter Lang, Dec 10 2015: (Start)
A(r, n) = T(n+r-2, r-1) = risefac(n,r)/r! = binomial(n+r-1, r), for n >= 1 and r >= 1, gives the array with the number of independent components of a symmetric tensors of rank r (number of indices) and dimension n (indices run from 1 to n). Here risefac(n, k) is the rising factorial.
As(r, n) = T(n+1, r+1) = fallfac(n, r)/r! = binomial(n, r), r >= 1 and n >= 1 (with the triangle entries T(n, k) = 0 for n < k) gives the array with the number of independent components of an antisymmetric tensor of rank r and dimension n. Here fallfac is the falling factorial. (End)
The h-vectors associated to these f-vectors are given by A000012 regarded as a lower triangular matrix. Read as bivariate polynomials, the h-polynomials are the complete homogeneous symmetric polynomials in two variables, found in the compositional inverse of an e.g.f. for A008292, the h-vectors of the permutahedra. - Tom Copeland, Jan 10 2017
For a correlation between the states of a quantum system and the combinatorics of the n-simplex, see Boya and Dixit. - Tom Copeland, Jul 24 2017

Examples

			The triangle T(n, k) begins:
   n\k  0  1   2   3   4   5   6   7   8  9 10 11 ...
   0:   1
   1:   2  1
   2:   3  3   1
   3:   4  6   4   1
   4:   5 10  10   5   1
   5:   6 15  20  15   6   1
   6:   7 21  35  35  21   7   1
   7:   8 28  56  70  56  28   8   1
   8:   9 36  84 126 126  84  36   9   1
   9:  10 45 120 210 252 210 120  45  10  1
  10:  11 55 165 330 462 462 330 165  55 11  1
  11:  12 66 220 495 792 924 792 495 220 66 12  1
  ... reformatted by _Wolfdieter Lang_, Mar 23 2015
Production matrix begins
   2   1
  -1   1   1
   1   0   1   1
  -1   0   0   1   1
   1   0   0   0   1   1
  -1   0   0   0   0   1   1
   1   0   0   0   0   0   1   1
  -1   0   0   0   0   0   0   1   1
   1   0   0   0   0   0   0   0   1   1
- _Philippe Deléham_, Jan 29 2014
From _Wolfdieter Lang_, Nov 08 2018: (Start)
Recurrence [_Philippe Deléham_]: T(7, 3) = 2*35 + 35 - 15 - 20 = 70.
Recurrence from Riordan A- and Z-sequences: [1,1,repeat(0)] and [2, repeat(-1, +1)]: From Z: T(5, 0) = 2*5 - 10 + 10 - 5 + 1 = 6. From A: T(7, 3) = 35 + 35 = 70.
Boas-Buck column k=3 recurrence: T(7, 3) = (5/4)*(1 + 5 + 15 + 35) = 70. (End)
		

Crossrefs

Programs

  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*1^(i-j), j = 1 .. i) od;
  • Mathematica
    Flatten[Table[CoefficientList[D[1/x ((x + 1) Exp[(x + 1) z] - Exp[z]), {z, k}] /. z -> 0, x], {k, 0, 11}]]
    CoefficientList[CoefficientList[Series[1/((1 - x)*(1 - x - x*y)), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    for(n=0, 20, for(k=0, n, print1(1/k!*sum(i=0, n, (prod(j=0, k-1, i-j))), ", "))) \\ Derek Orr, Oct 14 2014
    
  • 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=k..n} binomial(j,k) = binomial(n+1, k+1), n >= k >= 0, else 0. (Partial sum of column k of A007318 (Pascal), or summation on the upper binomial index (Graham et al. (GKP), eq. (5.10). For the GKP reference see A007318.) - Wolfdieter Lang, Aug 22 2012
E.g.f.: 1/x*((1 + x)*exp(t*(1 + x)) - exp(t)) = 1 + (2 + x)*t + (3 + 3*x + x^2)*t^2/2! + .... The infinitesimal generator for this triangle has the sequence [2,3,4,...] on the main subdiagonal and 0's elsewhere. - Peter Bala, Jul 16 2013
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
T(n,k) = A193862(n,k)/2^k. - Philippe Deléham, Jan 29 2014
G.f.: 1/((1-x)*(1-x-x*y)). - Philippe Deléham, Mar 13 2014
From Tom Copeland, Mar 26 2014: (Start)
[From Copeland's 2007 and 2008 comments]
A) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 } (same as Deleham's).
B) The infinitesimal generator for T is given in A132681 with m=1 (same as Bala's), which makes connections to the ubiquitous associated Laguerre polynomials of integer orders, for this case the Laguerre polynomials of order one L(n,-t,1).
C) O.g.f. of row e.g.f.s: Sum_{n>=0} L(n,-t,1) x^n = exp[t*x/(1-x)]/(1-x)^2 = 1 + (2+t)x + (3+3*t+t^2/2!)x^2 + (4+6*t+4*t^2/2!+t^3/3!)x^3+ ... .
D) E.g.f. of row o.g.f.s: ((1+t)*exp((1+t)*x)-exp(x))/t (same as Bala's).
E) E.g.f. for T(n,k)*a(n-k): {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. For example, for a(k)=2^k, the e.g.f. for the row o.g.f.s is {(2+t) exp[(2+t)x] - 2 exp(2x)}/t.
(End)
From Tom Copeland, Apr 28 2014: (Start)
With different indexing
A) O.g.f. by row: [(1+t)^n-1]/t.
B) O.g.f. of row o.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
C) E.g.f. of row o.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
These generating functions are related to row e.g.f.s of A111492. (End)
From Tom Copeland, Sep 17 2014: (Start)
A) U(x,s,t)= x^2/[(1-t*x)(1-(s+t)x)] = Sum_{n >= 0} F(n,s,t)x^(n+2) is a generating function for bivariate row polynomials of T, e.g., F(2,s,t)= s^2 + 3s*t + 3t^2 (Buchstaber, 2008).
B) dU/dt=x^2 dU/dx with U(x,s,0)= x^2/(1-s*x) (Buchstaber, 2008).
C) U(x,s,t) = exp(t*x^2*d/dx)U(x,s,0) = U(x/(1-t*x),s,0).
D) U(x,s,t) = Sum[n >= 0, (t*x)^n L(n,-:xD:,-1)] U(x,s,0), where (:xD:)^k=x^k*(d/dx)^k and L(n,x,-1) are the Laguerre polynomials of order -1, related to normalized Lah numbers. (End)
E.g.f. satisfies the differential equation d/dt(e.g.f.(x,t)) = (x+1)*e.g.f.(x,t) + exp(t). - Vincent J. Matsko, Jul 18 2015
The e.g.f. of the Norlund generalized Bernoulli (Appell) polynomials of order m, NB(n,x;m), is given by exponentiation of the e.g.f. of the Bernoulli numbers, i.e., multiple binomial self-convolutions of the Bernoulli numbers, through the e.g.f. exp[NB(.,x;m)t] = (t/(e^t - 1))^(m+1) * e^(xt). Norlund gave the relation to the factorials (x-1)!/(x-1-n)! = (x-1) ... (x-n) = NB(n,x;n), so T(n,m) = NB(m+1,n+2;m+1)/(m+1)!. - Tom Copeland, Oct 01 2015
From Wolfdieter Lang, Nov 08 2018: (Start)
Recurrences from the A- and Z- sequences for the Riordan triangle (see the W. Lang link under A006232 with references), which are A(n) = A019590(n+1), [1, 1, repeat (0)] and Z(n) = (-1)^(n+1)*A054977(n), [2, repeat(-1, 1)]:
T(0, 0) = 1, T(n, k) = 0 for n < k, and T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and T(n, k) = T(n-1, k-1) + T(n-1, k), for n >= m >= 1.
Boas-Buck recurrence for columns (see the Aug 10 2017 remark in A036521 also for references):
T(n, k) = ((2 + k)/(n - k))*Sum_{j=k..n-1} T(j, k), for n >= 1, k = 0, 1, ..., n-1, and input T(n, n) = 1, for n >= 0, (the BB-sequences are alpha(n) = 2 and beta(n) = 1). (End)
T(n, k) = [x^k] Sum_{j=0..n} (x+1)^j. - Peter Luschny, Jul 09 2019

Extensions

Edited by Tom Copeland and N. J. A. Sloane, Dec 11 2007

A014987 a(n) = (1 - (-6)^n)/7.

Original entry on oeis.org

1, -5, 31, -185, 1111, -6665, 39991, -239945, 1439671, -8638025, 51828151, -310968905, 1865813431, -11194880585, 67169283511, -403015701065, 2418094206391, -14508565238345, 87051391430071, -522308348580425, 3133850091482551
Offset: 1

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Author

Keywords

Comments

q-integers for q=-6.
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^n*charpoly(A,2). - Milan Janjic, Jan 27 2010

Crossrefs

Absolute values are in A015540.

Programs

  • Magma
    I:=[1,-5]; [n le 2 select I[n] else -5*Self(n-1)+6*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 22 2012
    
  • Maple
    a:=n->sum ((-6)^j, j=0..n): seq(a(n), n=0..25); # Zerinvary Lajos, Dec 16 2008
  • Mathematica
    LinearRecurrence[{-5, 6}, {1, -5}, 30] (* Vincenzo Librandi Oct 22 2012 *)
  • PARI
    a(n)=(1-(-6)^n)/7 \\ Charles R Greathouse IV, Sep 24 2015
  • Sage
    [gaussian_binomial(n,1,-6) for n in range(1,22)] # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = a(n-1) + q^(n-1) = (q^n - 1) / (q - 1).
G.f.: x/((1+6*x)*(1-x)).
a(n) = -5*a(n-1) + 6*a(n-2). - Vincenzo Librandi Oct 22 2012
E.g.f.: (exp(x) - exp(-6*x))/7. - G. C. Greubel, May 26 2018

Extensions

Better name from Ralf Stephan, Jul 14 2013

A015528 a(n) = 3*a(n-1) + 10*a(n-2).

Original entry on oeis.org

0, 1, 3, 19, 87, 451, 2223, 11179, 55767, 279091, 1394943, 6975739, 34876647, 174387331, 871928463, 4359658699, 21798260727, 108991369171, 544956714783, 2724783836059, 13623918656007, 68119594328611, 340597969545903, 1702989851923819, 8514949251230487, 42574746272929651
Offset: 0

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Author

Keywords

Comments

The ratio a(n+1)/a(n) converges to 5 as n approaches infinity. - Felix P. Muga II, Mar 10 2014

Programs

  • Magma
    [5^n/7-(-2)^n/7: n in [0..30]]; // Vincenzo Librandi, Aug 23 2011
    
  • Mathematica
    Join[{a=0,b=1},Table[c=3*b+10*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
    LinearRecurrence[{3,10},{0,1},30] (* or *) CoefficientList[Series[x/(1-x (10x+3)),{x,0,29}],x] (* Harvey P. Dale, Jan 27 2012 *)
  • PARI
    for(n=0,30, print1((5^n - (-2)^n)/7, ", ")) \\ G. C. Greubel, Jan 01 2018
  • Sage
    [lucas_number1(n,3,-10) for n in range(0, 23)]# Zerinvary Lajos, Apr 22 2009
    

Formula

a(n) = 3*a(n-1) + 10*a(n-2).
a(n) = (5^n - (-2)^n)/7. Binomial transform is A015540. - Paul Barry, Feb 07 2004
G.f.: x/(1 - x*(10*x+3)). - Harvey P. Dale, Jan 27 2012
E.g.f.: exp(-2*x)*(exp(7*x) - 1)/7. - Elmo R. Oliveira, Apr 02 2025

A109501 Number of closed walks of length n on the complete graph on 7 nodes from a given node.

Original entry on oeis.org

1, 0, 6, 30, 186, 1110, 6666, 39990, 239946, 1439670, 8638026, 51828150, 310968906, 1865813430, 11194880586, 67169283510, 403015701066, 2418094206390, 14508565238346, 87051391430070, 522308348580426, 3133850091482550, 18803100548895306, 112818603293371830
Offset: 0

Views

Author

Mitch Harris, Jun 30 2005

Keywords

Crossrefs

Cf. A109502.
Cf. sequences with the same recurrence form: A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499, A015531, A109500, A015540. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008

Programs

  • Magma
    [(6^n + 6*(-1)^n)/7: n in [0..30]]; // G. C. Greubel, Dec 30 2017
  • Mathematica
    k=0;lst={k};Do[k=6^n-k;AppendTo[lst, k], {n, 1, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
    CoefficientList[Series[(1 - 5*x)/(1 - 5*x - 6*x^2), {x, 0, 50}], x] (* or *) LinearRecurrence[{5,6}, {1,0}, 30] (* G. C. Greubel, Dec 30 2017 *)
  • PARI
    for(n=0,30, print1((6^n + 6*(-1)^n)/7, ", ")) \\ G. C. Greubel, Dec 30 2017
    

Formula

G.f.: (1 - 5*x)/(1 - 5*x - 6*x^2).
a(n) = (6^n + 6*(-1)^n)/7.
a(n) = 6^(n-1) - a(n-1), a(0) = 1. - Jon E. Schoenfield, Feb 09 2015
a(n) = 5*a(n-1) + 6*a(n-2). - G. C. Greubel, Dec 30 2017
E.g.f.: exp(-x)*(exp(7*x) + 6)/7. - Elmo R. Oliveira, Aug 17 2024

Extensions

Corrected by Franklin T. Adams-Watters, Sep 18 2006
Edited by Jon E. Schoenfield, Feb 09 2015

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

A062160 Square array T(n,k) = (n^k - (-1)^k)/(n+1), n >= 0, k >= 0, read by falling antidiagonals.

Original entry on oeis.org

0, 1, 0, -1, 1, 0, 1, 0, 1, 0, -1, 1, 1, 1, 0, 1, 0, 3, 2, 1, 0, -1, 1, 5, 7, 3, 1, 0, 1, 0, 11, 20, 13, 4, 1, 0, -1, 1, 21, 61, 51, 21, 5, 1, 0, 1, 0, 43, 182, 205, 104, 31, 6, 1, 0, -1, 1, 85, 547, 819, 521, 185, 43, 7, 1, 0, 1, 0, 171, 1640, 3277, 2604, 1111, 300, 57, 8, 1, 0, -1, 1, 341, 4921, 13107, 13021, 6665, 2101, 455, 73, 9, 1, 0
Offset: 0

Views

Author

Henry Bottomley, Jun 08 2001

Keywords

Comments

For n >= 1, T(n, k) equals the number of walks of length k between any two distinct vertices of the complete graph K_(n+1). - Peter Bala, May 30 2024

Examples

			From _Seiichi Manyama_, Apr 12 2019: (Start)
Square array begins:
   0, 1, -1,  1,  -1,    1,    -1,      1, ...
   0, 1,  0,  1,   0,    1,     0,      1, ...
   0, 1,  1,  3,   5,   11,    21,     43, ...
   0, 1,  2,  7,  20,   61,   182,    547, ...
   0, 1,  3, 13,  51,  205,   819,   3277, ...
   0, 1,  4, 21, 104,  521,  2604,  13021, ...
   0, 1,  5, 31, 185, 1111,  6665,  39991, ...
   0, 1,  6, 43, 300, 2101, 14706, 102943, ... (End)
		

Crossrefs

Related to repunits in negative bases (cf. A055129 for positive bases).
Main diagonal gives A081216.
Cf. A109502.

Programs

  • Maple
    seq(print(seq((n^k - (-1)^k)/(n+1), k = 0..10)), n = 0..10); # Peter Bala, May 31 2024
  • Mathematica
    T[n_,k_]:=(n^k - (-1)^k)/(n+1); Join[{0},Table[Reverse[Table[T[n-k,k],{k,0,n}]],{n,12}]]//Flatten (* Stefano Spezia, Feb 20 2024 *)

Formula

T(n, k) = n^(k-1) - n^(k-2) + n^(k-3) - ... + (-1)^(k-1) = n^(k-1) - T(n, k-1) = n*T(n, k-1) - (-1)^k = (n - 1)*T(n, k-1) + n*T(n, k-2) = round[n^k/(n+1)] for n > 1.
T(n, k) = (-1)^(k+1) * resultant( n*x + 1, (x^k-1)/(x-1) ). - Max Alekseyev, Sep 28 2021
G.f. of row n: x/((1+x) * (1-n*x)). - Seiichi Manyama, Apr 12 2019
E.g.f. of row n: (exp(n*x) - exp(-x))/(n+1). - Stefano Spezia, Feb 20 2024
From Peter Bala, May 31 2024: (Start)
Binomial transform of the m-th row: Sum_{k = 0..n} binomial(n, k)*T(m, k) = (m + 1)^(n-1) for n >= 1.
Let R(m, x) denote the g.f. of the m-th row of the square array. Then R(m_1, x) o R(m_2, x) = R(m_1 + m_2 + m_1*m_2, x), where o denotes the black diamond product of power series as defined by Dukes and White. Cf. A109502.
T(m_1 + m_2 + m_1*m_2, k) = Sum_{i = 0..k} Sum_{j = i..k} binomial(k, i)* binomial(k-i, j-i)*T(m_1, j)*T(m_2, k-i). (End)

A033116 Base-6 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

Original entry on oeis.org

1, 6, 37, 222, 1333, 7998, 47989, 287934, 1727605, 10365630, 62193781, 373162686, 2238976117, 13433856702, 80603140213, 483618841278, 2901713047669, 17410278286014, 104461669716085, 626770018296510, 3760620109779061
Offset: 1

Views

Author

Keywords

Comments

Partial sums of A015540. - Mircea Merca, Dec 28 2010

Crossrefs

Programs

Formula

From R. J. Mathar, Jan 08 2011: (Start)
G.f.: x / ( (1-x)*(1-6*x)*(1+x) ).
a(n) = 6^(n+1)/35 -1/10 -(-1)^n/14. (End)
a(n)=floor(6^(n+1)/35). a(n+1)=sum{k=0..floor(n/2)} 6^(n-2*k). a(n+1)=sum{k=0..n} sum{j=0..k} (-1)^(j+k)*6^j. - Paul Barry, Nov 12 2003, index corrected R. J. Mathar, Jan 08 2011
a(n) = 5*a(n-1) +6*a(n-2)+1. - Zerinvary Lajos, Dec 14 2008
a(n) = floor(6^(n+1)/7)/5 = floor((6*6^n-1)/35) = round((12*6^n-7)/70) = round((6*6^n-6)/35) = ceiling((6*6^n-6)/35). a(n)=a(n-2)+6^(n-1), n>2. - Mircea Merca, Dec 28 2010

A189800 a(n) = 6*a(n-1) + 8*a(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

Views

Author

Keywords

Crossrefs

Programs

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

Formula

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

A092165 Let M = 2 X 2 matrix [ 1 2 / 5 4 ]; a(n) = (1,2) entry of M^n.

Original entry on oeis.org

2, 10, 62, 370, 2222, 13330, 79982, 479890, 2879342, 17276050, 103656302, 621937810, 3731626862, 22389761170, 134338567022, 806031402130, 4836188412782, 29017130476690, 174102782860142, 1044616697160850, 6267700182965102, 37606201097790610, 225637206586743662
Offset: 1

Views

Author

Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Apr 01 2004

Keywords

Crossrefs

Programs

  • Magma
    [(2*6^n - 2*(-1)^n)/7: n in [1..30]]; // Vincenzo Librandi, Jul 21 2015
  • Mathematica
    Table[ MatrixPower[{{1, 2}, {5, 4}}, n][[1, 2]], {n, 20}] (* Robert G. Wilson v, Apr 22 2004 *)
    LinearRecurrence[{5, 6}, {2, 10}, 25] (* Vincenzo Librandi, Jul 21 2015 *)

Formula

a(n) = (2*6^n - 2*(-1)^n)/7.
a(n) = A092164(n) -(-1)^n.
From R. J. Mathar, Apr 20 2009: (Start)
a(n) = 5*a(n-1) + 6*a(n-2) = 2*A015540(n).
G.f.: 2*x/((1+x)*(1-6*x)). (End)

Extensions

Edited by Robert G. Wilson v, Apr 22 2004
More terms from Vincenzo Librandi, Jul 21 2015

A179897 a(n) = (n^(2*n+1) + 1) / (n+1).

Original entry on oeis.org

1, 1, 11, 547, 52429, 8138021, 1865813431, 593445188743, 250199979298361, 135085171767299209, 90909090909090909091, 74619186937936447687211, 73381705110822317661638341, 85180949465178001182799643437, 115244915978498073437814463065839, 179766618030828831251710653305053711
Offset: 0

Views

Author

Martin Saturka (martin(AT)saturka.net), Jul 31 2010

Keywords

Comments

a(n) is the arithmetic mean of the multiset consisting of n lots of 1/n and one lot of n^(2*n+1). This multiset also has an integer valued geometric mean which is equal to n for n > 0.
According to search at OEIS for particular sequence members, a(n) is also: (1+2*n)-th q-integer for q=-n, (2*(n+1))-th cyclotomic polynomial at q=-n, Gaussian binomial coefficient [2*n+1, 2*n] for q=-n, number of walks of length 1+2*n between any two distinct vertices of the complete graph K_(n+1).

Examples

			For n = 2, a(2) = 11 which is the arithmetic mean of {1/2, 1/2, 2^5} = 33 / 3 = 11. The geometric mean is 8^(1/3) = 2, i.e. both are integral.
		

Crossrefs

Main diagonal of A362783.
Values for n = 5, 6 via other ways. Q-integers: A014986, A014987, K_n paths: A015531, A015540, Cyclotomic polynomials: A020504, A020505, Gaussian binomial coefficients: A015391, A015429.

Programs

  • PARI
    a(n) = (n^(2*n + 1) + 1)/(n + 1) \\ Andrew Howroyd, May 03 2023
  • Python
    [(n**(2*n+1)+1)//(n+1) for n in range(1,11)]
    

Formula

a(n) = Sum_{i=0..2*n} (-n)^i.

Extensions

Edited, a(0)=1 prepended and more terms from Andrew Howroyd, May 03 2023
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