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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A131127 Table read by rows: 2*A007318(n,m) - A167374(n,m).

Original entry on oeis.org

1, 3, 1, 2, 5, 1, 2, 6, 7, 1, 2, 8, 12, 9, 1, 2, 10, 20, 20, 11, 1, 2, 12, 30, 40, 30, 13, 1, 2, 14, 42, 70, 70, 42, 15, 1, 2, 16, 56, 112, 140, 112, 56, 17, 1, 2, 18, 72, 168, 252, 252, 168, 72, 19, 1, 2, 20, 90, 240, 420, 504, 420, 240, 90, 21, 1, 2, 22, 110, 330, 660, 924, 924, 660, 330, 110, 23, 1
Offset: 0

Views

Author

Gary W. Adamson, Jun 16 2007

Keywords

Comments

Row sums = A000079(n+1), n>0.
Warning: row sums are not A046055! - N. J. A. Sloane, Jul 08 2009
Row sums = A151821(n+1), n>=0. - Alois P. Heinz, Jul 13 2009
A167374 is a modified version of the pair operator A097806 with (1,1,1,...) in the main diagonal and (-1,-1,-1,...) in the subdiagonal.

Examples

			First few rows of the triangle:
  1;
  3,  1;
  2,  5,  1;
  2,  6,  7,  1;
  2,  8, 12,  9,  1;
  2, 10, 20, 20, 11,  1;
  ...
		

Crossrefs

Programs

  • Maple
    T:= (n, m)-> 2*binomial(n, m) -(-1)^(n+m)*`if`(n=m or n=m+1, 1, 0): seq(seq(T(n,m), m=0..n), n=0..12); # Alois P. Heinz, Jul 13 2009
  • Mathematica
    T[n_, m_] := 2*Binomial[n, m] - (-1)^(n+m)*If[n == m || n == m+1, 1, 0];
    Table[Table[T[n, m], {m, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, May 19 2016, translated from Maple *)

Extensions

Edited by N. J. A. Sloane and R. J. Mathar, Jul 09 2009
Corrected and extended by Alois P. Heinz, Jul 13 2009
Definition simplified by Georg Fischer, Jun 07 2023

A000012 The simplest sequence of positive numbers: the all 1's sequence.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

N. J. A. Sloane, May 16 1994

Keywords

Comments

Number of ways of writing n as a product of primes.
Number of ways of writing n as a sum of distinct powers of 2.
Continued fraction for golden ratio A001622.
Partial sums of A000007 (characteristic function of 0). - Jeremy Gardiner, Sep 08 2002
An example of an infinite sequence of positive integers whose distinct pairwise concatenations are all primes! - Don Reble, Apr 17 2005
Binomial transform of A000007; inverse binomial transform of A000079. - Philippe Deléham, Jul 07 2005
A063524(a(n)) = 1. - Reinhard Zumkeller, Oct 11 2008
For n >= 0, let M(n) be the matrix with first row = (n n+1) and 2nd row = (n+1 n+2). Then a(n) = absolute value of det(M(n)). - K.V.Iyer, Apr 11 2009
The partial sums give the natural numbers (A000027). - Daniel Forgues, May 08 2009
From Enrique Pérez Herrero, Sep 04 2009: (Start)
a(n) is also tau_1(n) where tau_2(n) is A000005.
a(n) is a completely multiplicative arithmetical function.
a(n) is both squarefree and a perfect square. See A005117 and A000290. (End)
Also smallest divisor of n. - Juri-Stepan Gerasimov, Sep 07 2009
Also decimal expansion of 1/9. - Enrique Pérez Herrero, Sep 18 2009; corrected by Klaus Brockhaus, Apr 02 2010
a(n) is also the number of complete graphs on n nodes. - Pablo Chavez (pchavez(AT)cmu.edu), Sep 15 2009
Totally multiplicative sequence with a(p) = 1 for prime p. Totally multiplicative sequence with a(p) = a(p-1) for prime p. - Jaroslav Krizek, Oct 18 2009
n-th prime minus phi(prime(n)); number of divisors of n-th prime minus number of perfect partitions of n-th prime; the number of perfect partitions of n-th prime number; the number of perfect partitions of n-th noncomposite number. - Juri-Stepan Gerasimov, Oct 26 2009
For all n>0, the sequence of limit values for a(n) = n!*Sum_{k>=n} k/(k+1)!. Also, a(n) = n^0. - Harlan J. Brothers, Nov 01 2009
a(n) is also the number of 0-regular graphs on n vertices. - Jason Kimberley, Nov 07 2009
Differences between consecutive n. - Juri-Stepan Gerasimov, Dec 05 2009
From Matthew Vandermast, Oct 31 2010: (Start)
1) When sequence is read as a regular triangular array, T(n,k) is the coefficient of the k-th power in the expansion of (x^(n+1)-1)/(x-1).
2) Sequence can also be read as a uninomial array with rows of length 1, analogous to arrays of binomial, trinomial, etc., coefficients. In a q-nomial array, T(n,k) is the coefficient of the k-th power in the expansion of ((x^q -1)/(x-1))^n, and row n has a sum of q^n and a length of (q-1)*n + 1. (End)
The number of maximal self-avoiding walks from the NW to SW corners of a 2 X n grid.
When considered as a rectangular array, A000012 is a member of the chain of accumulation arrays that includes the multiplication table A003991 of the positive integers. The chain is ... < A185906 < A000007 < A000012 < A003991 < A098358 < A185904 < A185905 < ... (See A144112 for the definition of accumulation array.) - Clark Kimberling, Feb 06 2011
a(n) = A007310(n+1) (Modd 3) := A193680(A007310(n+1)), n>=0. For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the three residue classes Modd 3, called [0], [1], and [2], are shown in the array A088520, if there the third row is taken as class [0] after inclusion of 0. - Wolfdieter Lang, Feb 09 2012
Let M = Pascal's triangle without 1's (A014410) and V = a variant of the Bernoulli numbers A027641 but starting [1/2, 1/6, 0, -1/30, ...]. Then M*V = [1, 1, 1, 1, ...]. - Gary W. Adamson, Mar 05 2012
As a lower triangular array, T is an example of the fundamental generalized factorial matrices of A133314. Multiplying each n-th diagonal by t^n gives M(t) = I/(I-t*S) = I + t*S + (t*S)^2 + ... where S is the shift operator A129184, and T = M(1). The inverse of M(t) is obtained by multiplying the first subdiagonal of T by -t and the other subdiagonals by zero, so A167374 is the inverse of T. Multiplying by t^n/n! gives exp(t*S) with inverse exp(-t*S). - Tom Copeland, Nov 10 2012
The original definition of the meter was one ten-millionth of the distance from the Earth's equator to the North Pole. According to that historical definition, the length of one degree of latitude, that is, 60 nautical miles, would be exactly 111111.111... meters. - Jean-François Alcover, Jun 02 2013
Deficiency of 2^n. - Omar E. Pol, Jan 30 2014
Consider n >= 1 nonintersecting spheres each with surface area S. Define point p on sphere S_i to be a "public point" if and only if there exists a point q on sphere S_j, j != i, such that line segment pq INTERSECT S_i = {p} and pq INTERSECT S_j = {q}; otherwise, p is a "private point". The total surface area composed of exactly all private points on all n spheres is a(n)*S = S. ("The Private Planets Problem" in Zeitz.) - Rick L. Shepherd, May 29 2014
For n>0, digital roots of centered 9-gonal numbers (A060544). - Colin Barker, Jan 30 2015
Product of nonzero digits in base-2 representation of n. - Franklin T. Adams-Watters, May 16 2016
Alternating row sums of triangle A104684. - Wolfdieter Lang, Sep 11 2016
A fixed point of the run length transform. - Chai Wah Wu, Oct 21 2016
Length of period of continued fraction for sqrt(A002522) or sqrt(A002496). - A.H.M. Smeets, Oct 10 2017
a(n) is also the determinant of the (n+1) X (n+1) matrix M defined by M(i,j) = binomial(i,j) for 0 <= i,j <= n, since M is a lower triangular matrix with main diagonal all 1's. - Jianing Song, Jul 17 2018
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j) for 1 <= i,j <= n (see Xavier Merlin reference). - Bernard Schott, Dec 05 2018
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = tau(gcd(i,j)) for 1 <= i,j <= n (see De Koninck & Mercier reference). - Bernard Schott, Dec 08 2020

Examples

			1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + ...)))) = A001622.
1/9 = 0.11111111111111...
From _Wolfdieter Lang_, Feb 09 2012: (Start)
Modd 7 for nonnegative odd numbers not divisible by 3:
A007310: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, ...
Modd 3:  1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
(End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 186.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 692 pp. 90 and 297, Ellipses, Paris, 2004.
  • Xavier Merlin, Méthodix Algèbre, Exercice 1-a), page 153, Ellipses, Paris, 1995.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 277, 284.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
  • Paul Zeitz, The Art and Craft of Mathematical Problem Solving, The Great Courses, The Teaching Company, 2010 (DVDs and Course Guidebook, Lecture 6: "Pictures, Recasting, and Points of View", pp. 32-34).

Crossrefs

Programs

  • Haskell
    a000012 = const 1
    a000012_list = repeat 1 -- Reinhard Zumkeller, May 07 2012
    
  • Magma
    [1 : n in [0..100]];
    
  • Maple
    seq(1, i=0..150);
  • Mathematica
    Array[1 &, 50] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
  • Maxima
    makelist(1, n, 1, 30); /* Martin Ettl, Nov 07 2012 */
    
  • PARI
    {a(n) = 1};
    
  • Python
    print([1 for n in range(90)]) # Michael S. Branicky, Apr 04 2022

Formula

a(n) = 1.
G.f.: 1/(1-x).
E.g.f.: exp(x).
G.f.: Product_{k>=0} (1 + x^(2^k)). - Zak Seidov, Apr 06 2007
Completely multiplicative with a(p^e) = 1.
Regarded as a square array by antidiagonals, g.f. 1/((1-x)(1-y)), e.g.f. Sum T(n,m) x^n/n! y^m/m! = e^{x+y}, e.g.f. Sum T(n,m) x^n y^m/m! = e^y/(1-x). Regarded as a triangular array, g.f. 1/((1-x)(1-xy)), e.g.f. Sum T(n,m) x^n y^m/m! = e^{xy}/(1-x). - Franklin T. Adams-Watters, Feb 06 2006
Dirichlet g.f.: zeta(s). - Ilya Gutkovskiy, Aug 31 2016
a(n) = Sum_{l=1..n} (-1)^(l+1)*2*cos(Pi*l/(2*n+1)) = 1 identically in n >= 1 (for n=0 one has 0 from the undefined sum). From the Jolley reference, (429) p. 80. Interpretation: consider the n segments between x=0 and the n positive zeros of the Chebyshev polynomials S(2*n, x) (see A049310). Then the sum of the lengths of every other segment starting with the one ending in the largest zero (going from the right to the left) is 1. - Wolfdieter Lang, Sep 01 2016
As a lower triangular matrix, T = M*T^(-1)*M = M*A167374*M, where M(n,k) = (-1)^n A130595(n,k). Note that M = M^(-1). Cf. A118800 and A097805. - Tom Copeland, Nov 15 2016

A097805 Number of compositions of n with k parts, T(n, k) = binomial(n-1, k-1) for n, k >= 1 and T(n, 0) = 0^n, triangle read by rows for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

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

Views

Author

Paul Barry, Aug 25 2004

Keywords

Comments

Previous name was: Riordan array (1, 1/(1-x)) read by rows.
Note this Riordan array would be denoted (1, x/(1-x)) by some authors.
Columns have g.f. (x/(1-x))^k. Reverse of A071919. Row sums are A011782. Antidiagonal sums are Fibonacci(n-1). Inverse as Riordan array is (1, 1/(1+x)). A097805=B*A059260*B^(-1), where B is the binomial matrix.
(0,1)-Pascal triangle. - Philippe Deléham, Nov 21 2006
(n+1) * each term of row n generates triangle A127952: (1; 0, 2; 0, 3, 3; 0, 4, 8, 4; ...). - Gary W. Adamson, Feb 09 2007
Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2008
From Paul Weisenhorn, Feb 09 2011: (Start)
Triangle read by rows: T(r,c) is the number of unordered partitions of n=r*(r+1)/2+c into (r+1) parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2.
Triangle read by rows: T(r,c) is the number of unordered partitions of the number n=r*(r+1)/2+(c-1) into r parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2. (End)
Triangle read by rows: T(r,c) is the number of ordered partitions (compositions) of r into c parts. - Juergen Will, Jan 04 2016
From Tom Copeland, Oct 25 2012: (Start)
Given a basis composed of a sequence of polynomials p_n(x) characterized by ladder (creation / annihilation, or raising / lowering) operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x) with p_0(x)=1, giving the number operator # p_n(x) = RL p_n(x) = n p_n(x), the lower triangular padded Pascal matrix Pd (A097805) serves as a matrix representation of the operator exp(R^2*L) = exp(R#) =
1) exp(x^2D) for the set x^n and
2) D^(-1) exp(t*x)D for the set x^n/n! (see A218234).
(End)
From James East, Apr 11 2014: (Start)
Square array a(m,n) with m,n=0,1,2,... read by off-diagonals.
a(m,n) gives the number of order-preserving functions f:{1,...,m}->{1,...,n}. Order-preserving means that x
a(n,n)=A088218(n) is the size of the semigroup O_n of all order-preserving transformations of {1,...,n}.
Read as a triangle, this sequence may be obtained by augmenting Pascal's triangle by appending the column 1,0,0,0,... on the left.
(End)
A formula based on the partitions of n with largest part k is given as a Sage program below. The 'conjugate' formula leads to A048004. - Peter Luschny, Jul 13 2015
From Wolfdieter Lang, Feb 17 2017: (Start)
The transposed of this lower triangular Riordan matrix of the associated type T provides the transition matrix between the monomial basis {x^n}, n >= 0, and the basis {y^n}, n >= 0, with y = x/(1-x): x^0 = 1 = y^0, x^n = Sum_{m >= n} Ttrans(n,m) y^m, for n >= 1, with Ttrans(n,m) = binomial(m-1,n-1).
Therefore, if a transformation with this Riordan matrix from a sequence {a} to the sequence {b} is given by b(n) = Sum_{m=0..n} T(n, m)*a(m), with T(n, m) = binomial(n-1, m-1), for n >= 1, then Sum_{n >= 0} a(n)*x^n = Sum_{n >= 0} b(n)*y^n, with y = x/(1-x) and vice versa. This is a modified binomial transformation; the usual one belongs to the Pascal Riordan matrix A007318. (End)
From Gus Wiseman, Jan 23 2022: (Start)
Also the number of compositions of n with alternating sum k, with k ranging from -n to n in steps of 2. For example, row n = 6 counts the following compositions (empty column indicated by dot):
. (15) (24) (33) (42) (51) (6)
(141) (132) (123) (114)
(1113) (231) (222) (213)
(1212) (1122) (321) (312)
(1311) (1221) (1131) (411)
(2112) (2121)
(2211) (3111)
(11121) (11112)
(12111) (11211)
(111111) (21111)
The reverse-alternating version is the same. Counting compositions by all three parameters (sum, length, alternating sum) gives A345197. Compositions of 2n with alternating sum 2k with k ranging from -n + 1 to n are A034871. (End)
Also the convolution triangle of A000012. - Peter Luschny, Oct 07 2022
From Sergey Kitaev, Nov 18 2023: (Start)
Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with k right-to-left maxima. A right-to-left maximum in a permutation a(1)a(2)...a(n) is position i such that a(j) < a(i) for all i < j.
Number of permutations of length n avoiding simultaneously the patterns 231 and 312 with k right-to-left minima (resp., left-to-right maxima). A right-to-left minimum (resp., left-to-right maximum) in a permutation a(1)a(2)...a(n) is position i such that a(j) > a(i) for all j > i (resp., a(j) < a(i) for all j < i).
Number of permutations of length n avoiding simultaneously the patterns 213 and 312 with k right-to-left maxima (resp., left-to-right maxima).
Number of permutations of length n avoiding simultaneously the patterns 213 and 231 with k right-to-left maxima (resp., right-to-left minima). (End)

Examples

			G.f. = 1 + x * (x + x^3 * (1 + x) + x^6 * (1 + x)^2 + x^10 * (1 + x)^3 + ...). - _Michael Somos_, Aug 20 2006
The triangle T(n, k) begins:
n\k  0 1 2  3  4   5   6  7  8 9 10 ...
0:   1
1:   0 1
2:   0 1 1
3:   0 1 2  1
4:   0 1 3  3  1
5:   0 1 4  6  4   1
6:   0 1 5 10 10   5   1
7:   0 1 6 15 20  15   6  1
8:   0 1 7 21 35  35  21  7  1
9:   0 1 8 28 56  70  56 28  8 1
10:  0 1 9 36 84 126 126 84 36 9  1
... reformatted _Wolfdieter Lang_, Jul 31 2017
From _Paul Weisenhorn_, Feb 09 2011: (Start)
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+c = 18 with (r+1)=6 summands: (5+5+4+2+1+1), (5+5+3+3+1+1), (5+4+4+3+1+1), (5+5+3+2+2+1), (5+4+4+2+2+1), (5+4+3+3+2+1).
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+(c-1) = 17 with r=5 summands: (5+5+4+2+1), (5+5+3+3+1), (5+5+3+2+2), (5+4+4+3+1), (5+4+4+2+2), (5+4+3+3+2).  (End)
From _James East_, Apr 11 2014: (Start)
a(0,0)=1 since there is a unique (order-preserving) function {}->{}.
a(m,0)=0 for m>0 since there is no function from a nonempty set to the empty set.
a(3,2)=4 because there are four order-preserving functions {1,2,3}->{1,2}: these are [1,1,1], [2,2,2], [1,1,2], [1,2,2]. Here f=[a,b,c] denotes the function defined by f(1)=a, f(2)=b, f(3)=c.
a(2,3)=6 because there are six order-preserving functions {1,2}->{1,2,3}: these are [1,1], [1,2], [1,3], [2,2], [2,3], [3,3].
(End)
		

References

  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Part 1, Section 7.2.1.3, 2011.

Crossrefs

Case m=0 of the polynomials defined in A278073.
Cf. A000012 (diagonal), A011782 (row sums), A088218 (central terms).
The terms just left of center in odd-indexed rows are A001791, even A002054.
The odd-indexed rows are A034871.
Row sums without the center are A058622.
The unordered version is A072233, without zeros A008284.
Right half without center has row sums A027306(n-1).
Right half with center has row sums A116406(n).
Left half without center has row sums A294175(n-1).
Left half with center has row sums A058622(n-1).
A025047 counts alternating compositions.
A098124 counts balanced compositions, unordered A047993.
A106356 counts compositions by number of maximal anti-runs.
A344651 counts partitions by sum and alternating sum.
A345197 counts compositions by sum, length, and alternating sum.

Programs

  • Maple
    b:= proc(n, i, p) option remember; `if`(n=0, p!, `if`(i<1, 0,
          expand(add(b(n-i*j, i-1, p+j)/j!*x^j, j=0..n/i))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, 0)):
    seq(T(n), n=0..20);  # Alois P. Heinz, May 25 2014
    # Alternatively:
    T := proc(k,n) option remember;
    if k=n then 1 elif k=0 then 0 else
    add(T(k-1,n-i), i=1..n-k+1) fi end:
    A097805 := (n,k) -> T(k,n):
    for n from 0 to 12 do seq(A097805(n,k), k=0..n) od; # Peter Luschny, Mar 12 2016
    # Uses function PMatrix from A357368.
    PMatrix(10, n -> 1);  # Peter Luschny, Oct 07 2022
  • Mathematica
    T[0, 0] = 1; T[n_, k_] := Binomial[n-1, k-1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 03 2014, after Paul Weisenhorn *)
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&]],{n,0,10},{k,0,n}] (* Gus Wiseman, Jan 23 2022 *)
  • PARI
    {a(n) = my(m); if( n<2, n==0, n--; m = (sqrtint(8*n + 1) - 1)\2; binomial(m-1, n - m*(m + 1)/2))}; /* Michael Somos, Aug 20 2006 */
    
  • PARI
    T(n,k) = if (k==0, 0^n, binomial(n-1, k-1)); \\ Michel Marcus, May 06 2022
    
  • PARI
    row(n) = vector(n+1, k, k--; if (k==0, 0^n, binomial(n-1, k-1))); \\ Michel Marcus, May 06 2022
    
  • Python
    from math import comb
    def T(n, k): return comb(n-1, k-1) if k != 0 else k**n  # Peter Luschny, May 06 2022
  • Sage
    # Illustrates a basic partition formula, is not efficient as a program for large n.
    def A097805_row(n):
        r = []
        for k in (0..n):
            s = 0
            for q in Partitions(n, max_part=k, inner=[k]):
                s += mul(binomial(q[j],q[j+1]) for j in range(len(q)-1))
            r.append(s)
        return r
    [A097805_row(n) for n in (0..9)] # Peter Luschny, Jul 13 2015
    

Formula

Number triangle T(n, k) defined by T(n,k) = Sum_{j=0..n} binomial(n, j)*if(k<=j, (-1)^(j-k), 0).
T(r,c) = binomial(r-1,c-1), 0 <= c <= r. - Paul Weisenhorn, Feb 09 2011
G.f.: (-1+x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015
a(0,0) = 1, a(n,k) = binomial(n-1,n-k) = binomial(n-1,k-1) Juergen Will, Jan 04 2016
G.f.: (x^1 + x^2 + x^3 + ...)^k = (x/(1-x))^k. - Juergen Will, Jan 04 2016
From Tom Copeland, Nov 15 2016: (Start)
E.g.f.: 1 + x*[e^((x+1)t)-1]/(x+1).
This padded Pascal matrix with the odd columns negated is NpdP = M*S = S^(-1)*M^(-1) = S^(-1)*M, where M(n,k) = (-1)^n A130595(n,k), the inverse Pascal matrix with the odd rows negated, S is the summation matrix A000012, the lower triangular matrix with all elements unity, and S^(-1) = A167374, a finite difference matrix. NpdP is self-inverse, i.e., (M*S)^2 = the identity matrix, and has the e.g.f. 1 - x*[e^((1-x)t)-1]/(1-x).
M = NpdP*S^(-1) follows from the well-known recursion property of the Pascal matrix, implying NpdP = M*S.
The self-inverse property of -NpdP is implied by the self-inverse relation of its embedded signed Pascal submatrix M (cf. A130595). Also see A118800 for another proof.
Let P^(-1) be A130595, the inverse Pascal matrix. Then T = A200139*P^(-1) and T^(-1) = padded P^(-1) = P*A097808*P^(-1). (End)
The center (n>0) is T(2n+1,n+1) = A000984(n) = 2*A001700(n-1) = 2*A088218(n) = A126869(2n) = 2*A138364(2n-1). - Gus Wiseman, Jan 25 2022

Extensions

Corrected by Philippe Deléham, Oct 05 2005
New name using classical terminology by Peter Luschny, Feb 05 2019

A038207 Triangle whose (i,j)-th entry is binomial(i,j)*2^(i-j).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 8, 12, 6, 1, 16, 32, 24, 8, 1, 32, 80, 80, 40, 10, 1, 64, 192, 240, 160, 60, 12, 1, 128, 448, 672, 560, 280, 84, 14, 1, 256, 1024, 1792, 1792, 1120, 448, 112, 16, 1, 512, 2304, 4608, 5376, 4032, 2016, 672, 144, 18, 1, 1024, 5120, 11520, 15360, 13440, 8064, 3360, 960, 180, 20, 1
Offset: 0

Keywords

Comments

This infinite matrix is the square of the Pascal matrix (A007318) whose rows are [ 1,0,... ], [ 1,1,0,... ], [ 1,2,1,0,... ], ...
As an upper right triangle, table rows give number of points, edges, faces, cubes,
4D hypercubes etc. in hypercubes of increasing dimension by column. - Henry Bottomley, Apr 14 2000. More precisely, the (i,j)-th entry is the number of j-dimensional subspaces of an i-dimensional hypercube (see the Coxeter reference). - Christof Weber, May 08 2009
Number of different partial sums of 1+[1,1,2]+[2,2,3]+[3,3,4]+[4,4,5]+... with entries that are zero removed. - Jon Perry, Jan 01 2004
Row sums are powers of 3 (A000244), antidiagonal sums are Pell numbers (A000129). - Gerald McGarvey, May 17 2005
Riordan array (1/(1-2x), x/(1-2x)). - Paul Barry, Jul 28 2005
T(n,k) is the number of elements of the Coxeter group B_n with descent set contained in {s_k}, 0<=k<=n-1. For T(n,n), we interpret this as the number of elements of B_n with empty descent set (since s_n does not exist). - Elizabeth Morris (epmorris(AT)math.washington.edu), Mar 01 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then T(n,k) = the number of elements (x,y) of S for which y has exactly k more elements than x. - Ross La Haye, Oct 12 2007
T(n,k) is number of paths in the first quadrant going from (0,0) to (n,k) using only steps B=(1,0) colored blue, R=(1,0) colored red and U=(1,1). Example: T(3,2)=6 because we have BUU, RUU, UBU, URU, UUB and UUR. - Emeric Deutsch, Nov 04 2007
T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), and two kinds of step (1,0). - Joerg Arndt, Jul 01 2011
T(i,j) is the number of i-permutations of {1,2,3} containing j 1's. Example: T(2,1)=4 because we have 12, 13, 21 and 31; T(3,2)=6 because we have 112, 113, 121, 131, 211 and 311. - Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (2+x)^n. - N-E. Fahssi, Apr 13 2008
Sum of diagonals are Jacobsthal-numbers: A001045. - Mark Dols, Aug 31 2009
Triangle T(n,k), read by rows, given by [2,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 15 2009
Eigensequence of the triangle = A004211: (1, 3, 11, 49, 257, 1539, ...). - Gary W. Adamson, Feb 07 2010
f-vectors ("face"-vectors) for n-dimensional cubes [see e.g., Hoare]. (This is a restatement of Bottomley's above.) - Tom Copeland, Oct 19 2012
With P = Pascal matrix, the sequence of matrices I, A007318, A038207, A027465, A038231, A038243, A038255, A027466 ... = P^0, P^1, P^2, ... are related by Copeland's formula below to the evolution at integral time steps n= 0, 1, 2, ... of an exponential distribution exp(-x*z) governed by the Fokker-Planck equation as given in the Dattoli et al. ref. below. - Tom Copeland, Oct 26 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Unsigned diagonals of A133156 are rows of this array. - Tom Copeland, Oct 11 2014
Omitting the first row, this is the production matrix for A039683, where an equivalent differential operator can be found. - Tom Copeland, Oct 11 2016
T(n,k) is the number of functions f:[n]->[3] with exactly k elements mapped to 3. Note that there are C(n,k) ways to choose the k elements mapped to 3, and there are 2^(n-k) ways to map the other (n-k) elements to {1,2}. Hence, by summing T(n,k) as k runs from 0 to n, we obtain 3^n = Sum_{k=0..n} T(n,k). - Dennis P. Walsh, Sep 26 2017
Since this array is the square of the Pascal lower triangular matrix, the row polynomials of this array are obtained as the umbral composition of the row polynomials P_n(x) of the Pascal matrix with themselves. E.g., P_3(P.(x)) = 1 P_3(x) + 3 P_2(x) + 3 P_1(x) + 1 = (x^3 + 3 x^2 + 3 x + 1) + 3 (x^2 + 2 x + 1) + 3 (x + 1) + 1 = x^3 + 6 x^2 + 12 x + 8. - Tom Copeland, Nov 12 2018
T(n,k) is the number of 2-compositions of n+1 with some zeros allowed that have k zeros; see the Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020
Also the convolution triangle of A000079. - Peter Luschny, Oct 09 2022

Examples

			Triangle begins with T(0,0):
   1;
   2,  1;
   4,  4,  1;
   8, 12,  6,  1;
  16, 32, 24,  8,  1;
  32, 80, 80, 40, 10,  1;
  ... -  corrected by _Clark Kimberling_, Aug 05 2011
Seen as an array read by descending antidiagonals:
[0] 1, 2,  4,   8,    16,    32,    64,     128,     256, ...     [A000079]
[1] 1, 4,  12,  32,   80,    192,   448,    1024,    2304, ...    [A001787]
[2] 1, 6,  24,  80,   240,   672,   1792,   4608,    11520, ...   [A001788]
[3] 1, 8,  40,  160,  560,   1792,  5376,   15360,   42240, ...   [A001789]
[4] 1, 10, 60,  280,  1120,  4032,  13440,  42240,   126720, ...  [A003472]
[5] 1, 12, 84,  448,  2016,  8064,  29568,  101376,  329472, ...  [A054849]
[6] 1, 14, 112, 672,  3360,  14784, 59136,  219648,  768768, ...  [A002409]
[7] 1, 16, 144, 960,  5280,  25344, 109824, 439296,  1647360, ... [A054851]
[8] 1, 18, 180, 1320, 7920,  41184, 192192, 823680,  3294720, ... [A140325]
[9] 1, 20, 220, 1760, 11440, 64064, 320320, 1464320, 6223360, ... [A140354]
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 155.
  • H. S. M. Coxeter, Regular Polytopes, Dover Publications, New York (1973), p. 122.

Programs

  • GAP
    Flat(List([0..15], n->List([0..n], k->Binomial(n, k)*2^(n-k)))); # Stefano Spezia, Nov 21 2018
  • Haskell
    a038207 n = a038207_list !! n
    a038207_list = concat $ iterate ([2,1] *) [1]
    instance Num a => Num [a] where
       fromInteger k = [fromInteger k]
       (p:ps) + (q:qs) = p + q : ps + qs
       ps + qs         = ps ++ qs
       (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
        *                = []
    -- Reinhard Zumkeller, Apr 02 2011
    
  • Haskell
    a038207' n k = a038207_tabl !! n !! k
    a038207_row n = a038207_tabl !! n
    a038207_tabl = iterate f [1] where
       f row = zipWith (+) ([0] ++ row) (map (* 2) row ++ [0])
    -- Reinhard Zumkeller, Feb 27 2013
    
  • Magma
    /* As triangle */ [[(&+[Binomial(n,i)*Binomial(i,k): i in [k..n]]): k in [0..n]]: n in [0..15]]; // Vincenzo Librandi, Nov 16 2018
    
  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*2^(i-j), j = 0 .. i) end do; # yields sequence in triangular form - Emeric Deutsch, Nov 04 2007
    # Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
    PMatrix(10, n -> 2^(n-1)); # Peter Luschny, Oct 09 2022
  • Mathematica
    Table[CoefficientList[Expand[(y + x + x^2)^n], y] /. x -> 1, {n, 0,10}] // TableForm (* Geoffrey Critzer, Nov 20 2011 *)
    Table[Binomial[n,k]2^(n-k),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, May 22 2020 *)
  • PARI
    {T(n, k) = polcoeff((x+2)^n, k)}; /* Michael Somos, Apr 27 2000 */
    
  • Sage
    def A038207_triangle(dim):
        M = matrix(ZZ,dim,dim)
        for n in range(dim): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+2*M[n-1,k]
        return M
    A038207_triangle(9)  # Peter Luschny, Sep 20 2012
    

Formula

T(n, k) = Sum_{i=0..n} binomial(n,i)*binomial(i,k).
T(n, k) = (-1)^k*A065109(n,k).
G.f.: 1/(1-2*z-t*z). - Emeric Deutsch, Nov 04 2007
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
From the formalism of A133314, the e.g.f. for the row polynomials of A038207 is exp(x*t)*exp(2x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-2x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*2x). The results generalize for 2 replaced by any number. - Tom Copeland, Aug 18 2008
Sum_{k=0..n} T(n,k)*x^k = (2+x)^n. - Philippe Deléham, Dec 15 2009
n-th row is obtained by taking pairwise sums of triangle A112857 terms starting from the right. - Gary W. Adamson, Feb 06 2012
T(n,n) = 1 and T(n,k) = T(n-1,k-1) + 2*T(n-1,k) for kJon Perry, Oct 11 2012
The e.g.f. for the n-th row is given by umbral composition of the normalized Laguerre polynomials A021009 as p(n,x) = L(n, -L(.,-x))/n! = 2^n L(n, -x/2)/n!. E.g., L(2,x) = 2 -4*x +x^2, so p(2,x)= (1/2)*L(2, -L(.,-x)) = (1/2)*(2*L(0,-x) + 4*L(1,-x) + L(2,-x)) = (1/2)*(2 + 4*(1+x) + (2+4*x+x^2)) = 4 + 4*x + x^2/2. - Tom Copeland, Oct 20 2012
From Tom Copeland, Oct 26 2012: (Start)
From the formalism of A132440 and A218272:
Let P and P^T be the Pascal matrix and its transpose and H= P^2= A038207.
Then with D the derivative operator,
exp(x*z/(1-2*z))/(1-2*z)= exp(2*z D_z z) e^(x*z)= exp(2*D_x (x D_x)) e^(z*x)
= (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
= (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T
= Sum_{n>=0} z^n * 2^n Lag_n(-x/2)= exp[z*EF(.,x)], an o.g.f. for the f-vectors (rows) of A038207 where EF(n,x) is an e.g.f. for the n-th f-vector. (Lag_n(x) are the un-normalized Laguerre polynomials.)
Conversely,
exp(z*(2+x))= exp(2D_x) exp(x*z)= exp(2x) exp(x*z)
= (1 x x^2 x^3 ...) H^T (1 z z^2/2! z^3/3! ...)^T
= (1 z z^2/2! z^3/3! ...) H (1 x x^2 x^3 ...)^T
= exp(z*OF(.,x)), an e.g.f for the f-vectors of A038207 where
OF(n,x)= (2+x)^n is an o.g.f. for the n-th f-vector.
(End)
G.f.: R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ (1+y))*x/((2*k+2+ (1+y))*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
A038207 = exp[M*B(.,2)] where M = A238385-I and (B(.,x))^n = B(n,x) are the Bell polynomials (cf. A008277). B(n,2) = A001861(n). - Tom Copeland, Apr 17 2014
T = (A007318)^2 = A112857*|A167374| = |A118801|*|A167374| = |A118801*A167374| = |P*A167374*P^(-1)*A167374| = |P*NpdP*A167374|. Cf. A118801. - Tom Copeland, Nov 17 2016
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial 2^n*Sum_{k = 0..n} binomial(n,k)*x^k/k!. For example, the e.g.f. for the third subdiagonal is exp(x)*(8 + 24*x + 12*x^2 + 4*x^3/3) = 8 + 32*x + 80*x^2/2! + 160*x^3/3! + .... - Peter Bala, Mar 05 2017
T(3*k+2,k) = T(3*k+2,k+1), T(2*k+1,k) = 2*T(2*k+1,k+1). - Yuchun Ji, May 26 2020
From Robert A. Russell, Aug 05 2020: (Start)
G.f. for column k: x^k / (1-2*x)^(k+1).
E.g.f. for column k: exp(2*x) * x^k / k!. (End)
Also the array A(n, k) read by descending antidiagonals, where A(n, k) = (-1)^n*Sum_{j= 0..n+k} binomial(n + k, j)*hypergeom([-n, j+1], [1], 1). - Peter Luschny, Nov 09 2021

A128064 Triangle T with T(n,n)=n, T(n,n-1)=-(n-1) and otherwise T(n,k)=0; 0

Original entry on oeis.org

1, -1, 2, 0, -2, 3, 0, 0, -3, 4, 0, 0, 0, -4, 5, 0, 0, 0, 0, -5, 6, 0, 0, 0, 0, 0, -6, 7, 0, 0, 0, 0, 0, 0, -7, 8, 0, 0, 0, 0, 0, 0, 0, -8, 9, 0, 0, 0, 0, 0, 0, 0, 0, -9, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -11, 12
Offset: 1

Author

Gary W. Adamson, Feb 14 2007

Keywords

Comments

The matrix inverse = (1/1; 1/2, 1/2; 1/3, 1/3, 1/3;...). Binomial transform of A128064 = A128065. A128064 * A007318 = A103406.
The positive version with row sums 2n+1 is given by T(n,k)=sum{j=k..n, C(n,j)*C(j,k)*(-1)^(n-j)*(j+1)}. - Paul Barry, May 26 2007
Binomial transform of unsigned sequence is A003506. - Gary W. Adamson, Aug 29 2007
Table T(n,k) read by antidiagonals. T(n,1) = n (for n>1), T(n,2) = -n, T(n,k) = 0, k > 2. - Boris Putievskiy, Feb 07 2013

Examples

			First few rows of the triangle are:
1;
-1,2;
0,-2,3;
0,0,-3,4;
0,0,0,-4,5;
0,0,0,0,-5,6;
0,0,0,0,0,-6,7;
...
From _Boris Putievskiy_, Feb 07 2013: (Start)
The start of the sequence as table:
1..-1..0..0..0..0..0...
2..-2..0..0..0..0..0...
3..-3..0..0..0..0..0...
4..-4..0..0..0..0..0...
5..-5..0..0..0..0..0...
6..-6..0..0..0..0..0...
7..-7..0..0..0..0..0...
. . .
(End)
		

Crossrefs

Programs

  • Mathematica
    row[1] = {1}; row[2] = {-1, 2}; row[n_] := Join[Array[0&, n-2], {-n+1, n}]; Table[row[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, Jan 12 2015 *)

Formula

Number triangle T(n,k)=sum{j=k..n, C(n,j)*C(j,k)*(-1)^(j-k)*(j+1)}. - Paul Barry, May 26 2007
a(n) = A002260(n)*A167374(n); a(n) = i*floor((i+2)/(t+2))*(-1)^(i+t+1), where i=n-t*(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Feb 07 2013
G.f.: (-1)^k*[x^k*exp(k*x)]'/exp(k*x)=sum(n>=k, (-1)^n*T(n,k)*x^n). - Vladimir Kruchinin, Oct 18 2013

A118800 Triangle read by rows: T satisfies the matrix products: C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 2, -3, 1, 4, -8, 5, -1, 8, -20, 18, -7, 1, 16, -48, 56, -32, 9, -1, 32, -112, 160, -120, 50, -11, 1, 64, -256, 432, -400, 220, -72, 13, -1, 128, -576, 1120, -1232, 840, -364, 98, -15, 1, 256, -1280, 2816, -3584, 2912, -1568, 560, -128, 17, -1, 512, -2816, 6912, -9984, 9408, -6048, 2688, -816, 162, -19, 1
Offset: 0

Author

Paul D. Hanna, May 02 2006

Keywords

Comments

The matrix square, T^2, consists of columns that are all the same.
Matrix inverse is triangle A118801. Row sums form {0^n, n>=0}.
Unsigned row sums equal A025192(n) = 2*3^(n-1), n>=1.
Row squared sums equal A051708.
Antidiagonal sums equals all 1's.
Unsigned antidiagonal sums form A078057 (with offset).
Antidiagonal squared sums form A002002(n) = Sum_{k=0..n-1} C(n,k+1)*C(n+k,k), n>=1.
From Paul Barry, Nov 10 2008: (Start)
T is [1,1,0,0,0,...] DELTA [ -1,0,0,0,0,...] or C(1,n) DELTA -C(0,n). (DELTA defined in A084938).
The positive matrix T_p is [1,1,0,0,0,...] DELTA [1,0,0,0,0,...]. T_p*C^-1 is
[0,1,0,0,0,....] DELTA [1,0,0,0,0,...] which is C(n-1,k-1) for n,k>=1. (End)
The triangle formed by deleting the minus signs is the mirror of the self-fusion of Pascal's triangle; see Comments at A081277 and A193722. - Clark Kimberling, Aug 04 2011
Riordan array ( (1 - x)/(1 - 2*x), -x/(1 - 2*x) ). Cf. A209149. The matrix square is the Riordan array ( (1 - x)^2/(1 - 2*x), x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013
From Peter Bala, Feb 23 2019: (Start)
There is a 1-parameter family of solutions to the simultaneous equations C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle. Let T(k) denote the Riordan array ( (1 - k*x)/(1 - (k + 1)*x), -x/(1 - (k + 1)*x) ) so that T(1) = T. Then C*T(k)*C = T(k)^-1 and T(k)*C*T(k) = C^-1, for arbitrary k. For arbitrary m, the Riordan arrays (T(k)*C^m)^2 and (C^m*T(k))^2 both belong to the Appell subgroup of the Riordan group.
More generally, given a fixed m, we can ask for a lower triangular array X solving the simultaneous equations (C^m)*X*(C^m) = X^-1 and X*(C^m)*X = C^(-m). A 1-parameter family of solutions is given by the Riordan arrays X = ( (1 - m*k*x)/(1 - m*(k + 1)*x), -x/(1 - m*(k + 1)*x) ). The Riordan arrays X^2 , (X*C^n)^2 and (C^n*X)^2, for arbitrary n, all belong to the Appell subgroup of the Riordan group. (End)

Examples

			Triangle begins:
     1;
     1,    -1;
     2,    -3,     1;
     4,    -8,     5,     -1;
     8,   -20,    18,     -7,     1;
    16,   -48,    56,    -32,     9,     -1;
    32,  -112,   160,   -120,    50,    -11,     1;
    64,  -256,   432,   -400,   220,    -72,    13,    -1;
   128,  -576,  1120,  -1232,   840,   -364,    98,   -15,    1;
   256, -1280,  2816,  -3584,  2912,  -1568,   560,  -128,   17,   -1;
   512, -2816,  6912,  -9984,  9408,  -6048,  2688,  -816,  162,  -19,  1;
  1024, -6144, 16640, -26880, 28800, -21504, 11424, -4320, 1140, -200, 21, -1;
  ...
The matrix square, T^2, equals:
   1;
   0,  1;
   1,  0,  1;
   2,  1,  0,  1;
   4,  2,  1,  0,  1;
   8,  4,  2,  1,  0,  1;
  16,  8,  4,  2,  1,  0,  1;
  32, 16,  8,  4,  2,  1,  0,  1;
  64, 32, 16,  8,  4,  2,  1,  0,  1; ...
where all columns are the same.
		

Crossrefs

Cf. A118801 (inverse), A025192 (unsigned row sums), A051708 (row squared sums), A078057 (unsigned antidiagonal sums), A002002 (antidiagonal squared sums).

Programs

  • Mathematica
    (* This program generates A118800 as the mirror of the self-fusion of Pascal's triangle. *)
    z = 8; a = 1; b = 1; c = 1; d = 1;
    p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n;
    t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, 0];
    w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, _] = 1;
    g[n_] := CoefficientList[w[n, -x], x];
    TableForm[Table[Reverse[Abs@g[n]], {n, -1, z}]]
    Flatten[Table[Reverse[Abs@g[n]], {n, -1, z}]] (* A081277 *)
    TableForm[Table[g[n], {n, -1, z}]]
    Flatten[Table[g[n], {n, -1, z}]] (* A118800 *)
    (* Clark Kimberling, Aug 04 2011 *)
    T[ n_, k_] := If[ n<0 || k<0, 0, (-1)^k 2^(n-k) (Binomial[ n, k] + Binomial[ n-1, n-k]) / 2]; (* Michael Somos, Nov 25 2016 *)
  • PARI
    {T(n,k)=if(n==0&k==0,1,(-1)^k*2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2)}
    for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • PARI
    /* Chebyshev Polynomials as Antidiagonals: */
    {T(n,k)=local(Ox=x*O(x^(2*k))); polcoeff(((1+sqrt(1-x^2+Ox))^(n+k)+(1-sqrt(1-x^2+Ox))^(n+k))/2,2*k,x)}
    for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • Sage
    # uses[riordan_square from A321620]
    # Computes the unsigned triangle.
    riordan_square((1-x)/(1-2*x), 8) # Peter Luschny, Jan 03 2019

Formula

T(n,k) = (-1)^k * 2^(n-k) * ( C(n,k) + C(n-1,k-1) )/2 for n>=k>=0 with T(0,0) = 1. Antidiagonals form the coefficients of Chebyshev polynomials: T(n,k) = [x^(2*n)] [(1+sqrt(1-x^2))^(n+k) + (1-sqrt(1-x^2))^(n+k)]/2.
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
O.g.f.: (1 - t)/(1 + t*(x - 2)) = 1 + (1 - x)*t + (2 - 3*x + x^2)^t^2 + (4 - 8*x + 5*x^2 - x^3)*t^3 + .... Row polynomial R(n,x) = (1 - x)*(2 - x)^(n-1) for n >= 1. - Peter Bala, Jul 17 2013
T(n,k)=2*T(n-1,k)-T(n-1,k-1) with T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 25 2013
G.f. for row n (n>=1): Sum_{k=0..n} T(n,k)*x^k = (1-x)*(2-x)^(n-1). - Philippe Deléham, Nov 25 2013
From Tom Copeland, Nov 15 2016: (Start)
E.g.f. is [1 + (1-x)e^((2-x)t)]/(2-x), so the row polynomials are p_n(x) = (1-q,(x))^n, umbrally, where (q.(x))^k = q_k(x) are the row polynomials of A239473, or, equivalently, T = M*A239473, where M is the inverse Pascal matrix C^(-1) = A130595 with the odd rows negated, i.e., M(n,k) = (-1)^n C^(-1)(n,k) with e.g.f. exp[(1-x)t]. Cf. A200139: A200139(n,k) = (-1)^k* A118800(n,k).
TCT = C^(-1) = A130595 and A239473 = A000012*C^(-1) = S*C^(-1) imply (M*S)^2 = Identity matrix, i.e., M*S = (M*S)^(-1) = S^(-1)*M^(-1) = A167374*M^(-1). Note that M = M^(-1). Cf. A097805. (End)

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1
Offset: 0

Author

Abdullahi Umar, Aug 25 2008

Keywords

Comments

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

Examples

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)
		

Programs

  • Maple
    A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011
  • Mathematica
    Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

Formula

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

A118801 Triangle T that satisfies the matrix products: C*[T^-1]*C = T and T*[C^-1]*T = C, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -7, 5, -1, 1, -15, 17, -7, 1, 1, -31, 49, -31, 9, -1, 1, -63, 129, -111, 49, -11, 1, 1, -127, 321, -351, 209, -71, 13, -1, 1, -255, 769, -1023, 769, -351, 97, -15, 1, 1, -511, 1793, -2815, 2561, -1471, 545, -127, 17, -1, 1, -1023, 4097, -7423, 7937, -5503, 2561, -799, 161, -19, 1
Offset: 0

Author

Paul D. Hanna, May 02 2006

Keywords

Comments

Matrix inverse is triangle A118800. Row sums are: (1-n). Unsigned row sums equal A007051(n) = (3^n + 1)/2. Row squared sums equal A118802. Antidiagonal sums equal A080956(n) = (n+1)(2-n)/2. Unsigned antidiagonal sums form A024537 (with offset).
T = C^2*D^-1 where matrix product D = C^-1*T*C = T^-1*C^2 has only 2 nonzero diagonals: D(n,n)=-D(n+1,n)=(-1)^n, with zeros elsewhere. Also, [B^-1]*T*[B^-1] = B*[T^-1]*B forms a self-inverse matrix, where B^2 = C and B(n,k) = C(n,k)/2^(n-k). - Paul D. Hanna, May 04 2006
Riordan array ( 1/(1 - x), -x/(1 - 2*x) ) The matrix square is the Riordan array ( (1 - 2*x)/(1 - x)^2, x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013

Examples

			Formulas for initial columns are, for n>=0:
T(n+1,1) = 1 - 2^(n+1);
T(n+2,2) = 1 + 2^(n+1)*n;
T(n+3,3) = 1 - 2^(n+1)*(n*(n+1)/2 + 1);
T(n+4,4) = 1 + 2^(n+1)*(n*(n+1)*(n+2)/6 + n);
T(n+5,5) = 1 - 2^(n+1)*(n*(n+1)*(n+2)*(n+3)/24 + n*(n+1)/2 + 1).
Triangle begins:
1;
1,-1;
1,-3,1;
1,-7,5,-1;
1,-15,17,-7,1;
1,-31,49,-31,9,-1;
1,-63,129,-111,49,-11,1;
1,-127,321,-351,209,-71,13,-1;
1,-255,769,-1023,769,-351,97,-15,1;
1,-511,1793,-2815,2561,-1471,545,-127,17,-1;
1,-1023,4097,-7423,7937,-5503,2561,-799,161,-19,1; ...
The matrix square, T^2, starts:
1;
0,1;
-1,0,1;
-2,-1,0,1;
-3,-2,-1,0,1;
-4,-3,-2,-1,0,1; ...
where all columns are the same.
The matrix product C^-1*T*C = T^-1*C^2 is:
1;
-1,-1;
0, 1, 1;
0, 0,-1,-1;
0, 0, 0, 1, 1; ...
where C(n,k) = n!/(n-k)!/k!.
		

Crossrefs

Cf. A118800 (inverse), A007051 (unsigned row sums), A118802 (Row squared sums), A080956 (antidiagonal sums), A024537 (unsigned antidiagonal sums).
A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011

Programs

  • Mathematica
    Table[(1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]) - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)
  • PARI
    {T(n,k)=if(n==0&k==0,1,1+(-1)^k*2^(n-k+1)*sum(j=0,k\2,binomial(n-2*j-2,k-2*j-1)))}

Formula

T(n,k) = 1 + (-1)^k*2^(n-k+1)*Sum_{j=0..[k/2]} C(n-2j-2,k-2j-1) for n>=k>=0 with T(0,0) = 1.
For k>0, T(n,k) = -T(n-1,k-1) + 2*T(n-1,k). - Gerald McGarvey, Aug 05 2006
O.g.f.: (1 - 2*t)/(1 - t) * 1/(1 + t*(x - 2)) = 1 + (1 - x)*t + (1 - 3*x + x^2)*t^2 + (1 - 7*x + 5*x^2 - x^3)*t^3 + .... - Peter Bala, Jul 17 2013
From Tom Copeland, Nov 17 2016: (Start)
Let M = A200139^(-1) = (unsigned A118800)^(-1) and NpdP be the signed padded Pascal matrix defined in A097805. Then T(n,k) = (-1)^n* M(n,k) and T = P*NpdP = (A239473)^(-1)*P^(-1) = P*A167374*P^(-1) = A156644*P^(-1), where P is the Pascal matrix A007318 with inverse A130595. Cf. A112857.
Signed P^2 = signed A032807 = T*A167374. (End)

A156644 Mirror image of triangle A080233.

Original entry on oeis.org

1, 0, 1, -1, 1, 1, -2, 0, 2, 1, -3, -2, 2, 3, 1, -4, -5, 0, 5, 4, 1, -5, -9, -5, 5, 9, 5, 1, -6, -14, -14, 0, 14, 14, 6, 1, -7, -20, -28, -14, 14, 28, 20, 7, 1, -8, -27, -48, -42, 0, 42, 48, 27, 8, 1, -9, -35, -75, -90, -42, 42, 90, 75, 35, 9, 1
Offset: 0

Author

Philippe Deléham, Feb 12 2009

Keywords

Comments

Inverse of A239473. Equals A007318*A167374. - Tom Copeland, Nov 14 2016

Examples

			Triangle begins as:
   1;
   0,  1;
  -1,  1, 1;
  -2,  0, 2, 1;
  -3, -2, 2, 3, 1;
  -4, -5, 0, 5, 4, 1; ...
		

Crossrefs

Programs

  • Magma
    A156644:= func< n,k | ((2*k-n+1)/(k+1))*Binomial(n,k) >;
    [A156644(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 28 2021
  • Mathematica
    Table[Binomial[n, k] -Binomial[n, k+1], {n,0,10}, {k,0,n}]//Flatten (* Michael De Vlieger, Nov 24 2016 *)
  • Sage
    def A156644(n,k): return ((2*k-n+1)/(k+1))*binomial(n,k)
    flatten([[A156644(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 28 2021
    

Formula

T(n,k) = A080233(n,n-k) = (-1)^(n-k)*A097808(n,k).
T(n,k) = ((2*k-n+1)/(k+1))*binomial(n,k).
T(n,k) = T(n-1,k-1) + T(n-1,k), k>0, with T(n,0) = 1-n = A024000(n), T(n,n) = 1.
T(n,k) = binomial(n,k) - binomial(n,k+1) = Sum_{i=-k-1..k+1} (-1)^(i+1) * binomial(n,k+1+i) * binomial(n+2,k+1-i). - Mircea Merca, Apr 28 2012
Sum_{k=0..n} T(n, k) = A000012(n) = 1^n. - G. C. Greubel, Feb 28 2021

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

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
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