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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A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

Original entry on oeis.org

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

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Author

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

Keywords

Comments

A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
+1
-1 +1
+1 -2 +1
-1 +3 -3 +1
+1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
From Milan Janjic, May 07 2008: (Start)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
See A193242. - Alexander R. Povolotsky, Feb 05 2013
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k, k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and = return to x axis} binomial(3,0)=1 (Uudududd); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
From Daniel Forgues, Mar 12 2015: (Start)
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
{0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111
(End)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the k-th row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024

Examples

			Triangle T(n,k) begins:
   n\k 0   1   2   3   4   5   6   7   8   9  10  11 ...
   0   1
   1   1   1
   2   1   2   1
   3   1   3   3   1
   4   1   4   6   4   1
   5   1   5  10  10   5   1
   6   1   6  15  20  15   6   1
   7   1   7  21  35  35  21   7   1
   8   1   8  28  56  70  56  28   8   1
   9   1   9  36  84 126 126  84  36   9   1
  10   1  10  45 120 210 252 210 120  45  10   1
  11   1  11  55 165 330 462 462 330 165  55  11   1
  ...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - _Dennis P. Walsh_, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - _Dennis P. Walsh_, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - _Gregory L. Simay_, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - _Wolfdieter Lang_, Nov 12 2018

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 68-74.
  • Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
  • A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
  • William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, pp. 140-141.
  • David Hök, Parvisa mönster i permutationer [Swedish], 2007.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
  • Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
  • Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 271-275.
  • A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
  • John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
  • Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, pages 43-52.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 13, 30-33.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
  • Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.

Links

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Axiom
    -- (start)
)set expose add constructor OutputForm
pascal(0,n) == 1
pascal(n,n) == 1
pascal(i,j | 0 < i and i < j) == pascal(i-1,j-1) + pascal(i,j-1)
pascalRow(n) == [pascal(i,n) for i in 0..n]
displayRow(n) == output center blankSeparate pascalRow(n)
for i in 0..20 repeat displayRow i -- (end)
  • GAP
    Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018
  • Haskell
    a007318 n k = a007318_tabl !! n !! k
a007318_row n = a007318_tabl !! n
a007318_list = concat a007318_tabl
a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
-- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010
  • Magma
    /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
  • Maple
    A007318 := (n,k)->binomial(n,k);
  • Mathematica
    Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)
  • Maxima
    create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
  • PARI
    C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011
  • Python
    # See Hobson link. Further programs:
from math import prod,factorial
def C(n,k): return prod(range(n,n-k,-1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
  • Python
    from math import comb, isqrt
def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024
  • Sage
    def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014

Formula

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n
C(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011: (Start)
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012: (Start)
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.: E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0 P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Extensions

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

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

Views

Author

N. J. A. Sloane

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.

Links

Crossrefs

See A065109 for a signed version.
Columns 0-10 are A000079, A001787, A001788(n-1), A001789, A003472, A054849, A002409(n-6), A054851, A140325(n-8), A140354(n-9), A172242.
T(2n,n) gives A059304.
Cf. A007318, A013609, A013610, A062715, A099089, A004211, A135387.
Cf. A001861, A008277, A027465, A027466, A038231, A038243, A038255, A039683, A112857, A118801, A132440, A133156, A133314, A167374, A218272, A238385, A323939.

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

A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003
T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012
The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015
For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017
For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019
From Gary W. Adamson, Apr 25 2022: (Start)
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

Examples

			The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n-1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
From _Roger L. Bagula_, Feb 20 2009: (Start)
  1
  1
  1   1
  1   2
  1   3   1
  1   4   3
  1   5   6   1
  1   6  10   4
  1   7  15  10   1
  1   8  21  20   5
  1   9  28  35  15   1
  1  10  36  56  35   6
  1  11  45  84  70  21   1
  1  12  55 120 126  56   7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size-4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}. - _Dennis P. Walsh_, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151. - _John Molokach_, Jul 08 2013
In the example, you can see that the n-th row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n-1, n-1), T(2n, n). - _Daniel Forgues_, Jul 07 2018

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 141ff.
  • C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.
  • I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 117.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 182-183.

Links

Crossrefs

Row sums = A000045(n+1) (Fibonacci numbers). - Michael Somos, Apr 02 1999
Cf. A000129, A054123, A052553, A000930, A030528, A053122.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.
Cf. A007318, A025581, A055087.
Cf. A000108, A049310, A055151, A098474, A124644, A134264.
Cf. A054142, A053123, A332602.

Programs

  • Haskell
    a011973 n k = a011973_tabf !! n !! k
a011973_row n = a011973_tabf !! n
a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf
-- Reinhard Zumkeller, Jul 14 2015
  • Maple
    a := proc(n) local k; [ seq(binomial(n-k,k),k=0..floor(n/2)) ]; end;
T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(n-k, k) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013
  • Mathematica
    (* first: sum method *) Table[CoefficientList[Sum[Binomial[n - m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
(* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n - 1] + x*L[x, n - 2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
(* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n - m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* Jean-François Alcover, Oct 21 2013 *)
CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {-1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
CoefficientList[Table[x^((n - 1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k))};
  • Sage
    # Prints the table; cf. A145574.
for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012

Formula

Let F(n, x) be the n-th Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1 - y - x*y^2). - Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m - 1, n) + T(m-2, n-1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4. - Rob Arthan, Sep 22 2003
G.f. for k-th column: x^(2*k-1)/(1-x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2-F(n-1, x)*F(n+1, x) = (-x)^(n-1).
The degree of F(n, x) is floor((n-1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
From Roger L. Bagula, Feb 20 2009: (Start)
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(n-m+1, m)*x^m;
p(x,n) = p(x, n - 1) + x*p(x, n - 2). (End)
T(n, k) = A102541(2*n+2, 2*k+1) + A102541(2*n+1, 2*k) - A102541(2*n+3, 2*k+1), n >= 0 and 0 <= k <= floor(n/2). - Johannes W. Meijer, Aug 26 2013
G.f.: 1/(1-x-y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1-x-tx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = -[1+x-sqrt[(1+x)^2 + 4tx^2]]/(2tx) = x - x^2 + (1-t) x^3 + (-1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = -1, h_2 = -t, and h_n = 0 otherwise. - Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1-x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = -L[Cinv(-tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1-x) and Cinv(x) = x (1-x) is the inverse of C(x) = (1-sqrt(1-4x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = -C[t Linv(-x)]/t = [-1 + sqrt(1+4tx/(1+x))]/2t = x - (1+t) x^2 + (1+2t+2t^2) x^3 - (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644. - Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2-k, 1). - Peter Luschny, May 10 2016

A030528 Triangle read by rows: a(n,k) = binomial(k,n-k).

Original entry on oeis.org

1, 1, 1, 0, 2, 1, 0, 1, 3, 1, 0, 0, 3, 4, 1, 0, 0, 1, 6, 5, 1, 0, 0, 0, 4, 10, 6, 1, 0, 0, 0, 1, 10, 15, 7, 1, 0, 0, 0, 0, 5, 20, 21, 8, 1, 0, 0, 0, 0, 1, 15, 35, 28, 9, 1, 0, 0, 0, 0, 0, 6, 35, 56, 36, 10, 1, 0, 0, 0, 0, 0, 1, 21, 70, 84, 45, 11, 1, 0, 0, 0, 0, 0, 0, 7, 56, 126, 120, 55, 12, 1
Offset: 1

Views

Author

Wolfdieter Lang

Keywords

Comments

A convolution triangle of numbers obtained from A019590.
a(n,m) := s1(-1; n,m), a member of a sequence of triangles including s1(0; n,m)= A023531(n,m) (unit matrix) and s1(2; n,m)= A007318(n-1,m-1) (Pascal's triangle).
The signed triangular matrix a(n,m)*(-1)^(n-m) is the inverse matrix of the triangular Catalan convolution matrix A033184(n+1,m+1), n >= m >= 0, with A033184(n,m) := 0 if n
Riordan array (1+x, x(1+x)). The signed triangle is the Riordan array (1-x,x(1-x)), inverse to (c(x),xc(x)) with c(x) g.f. for A000108. - Paul Barry, Feb 02 2005 [with offset 0]
Also, a(n,k)=number of compositions of n into k parts of 1's and 2's. Example: a(6,4)=6 because we have 2211, 2121, 2112, 1221, 1212 and 1122. - Emeric Deutsch, Apr 05 2005 [see MacMahon and Riordan. - Wolfdieter Lang, Jul 27 2023]
Subtriangle of A026729. - Philippe Deléham, Aug 31 2006
a(n,k) is the number of length n-1 binary sequences having no two consecutive 0's with exactly k-1 1's. Example: a(6,4)=6 because we have 01011, 01101, 01110, 10101, 10110, 11010. - Geoffrey Critzer, Jul 22 2013
Mirrored, shifted Fibonacci polynomials of A011973. The polynomials (illustrated below) of this entry have the property that p(n,t) = t * [p(n-1,t) + p(n-2,t)]. The additive properties of Pascal's triangle (A007318) are reflected in those of these polynomials, as can be seen in the Example Section below and also when the o.g.f. G(x,t) below is expanded as the series x*(1+x) + t * [x*(1+x)]^2 + t^2 * [x*(1+x)]^3 + ... . See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
Rows of this entry appear as columns of an array for an infinitesimal generator presented in the Copeland link. - Tom Copeland, Dec 23 2015
For n >= 2, the n-th row is also the coefficients of the vertex cover polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
With an additional initial matrix element a_(0,0) = 1 and column of zeros a_(n,0) = 0 for n > 0, these are antidiagonals read from bottom to top of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper, which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Cf. A011973. And A169803. - Tom Copeland, Jul 02 2018

Examples

			Triangle starts:
  [ 1]  1
  [ 2]  1   1
  [ 3]  0   2   1
  [ 4]  0   1   3   1
  [ 5]  0   0   3   4   1
  [ 6]  0   0   1   6   5   1
  [ 7]  0   0   0   4  10   6   1
  [ 8]  0   0   0   1  10  15   7   1
  [ 9]  0   0   0   0   5  20  21   8   1
  [10]  0   0   0   0   1  15  35  28   9   1
  [11]  0   0   0   0   0   6  35  56  36  10   1
  [12]  0   0   0   0   0   1  21  70  84  45  11   1
  [13]  0   0   0   0   0   0   7  56 126 120  55  12   1
  ...
From _Tom Copeland_, Nov 04 2014: (Start)
For quick comparison to other polynomials:
  p(1,t) = 1
  p(2,t) = 1 + 1 t
  p(3,t) = 0 + 2 t + 1 t^2
  p(4,t) = 0 + 1 t + 3 t^2 + 1 t^3
  p(5,t) = 0 + 0   + 3 t^2 + 4 t^3 +  1 t^4
  p(6,t) = 0 + 0   + 1 t^2 + 6 t^3 +  5 t^4 +  1 t^5
  p(7,t) = 0 + 0   + 0     + 4 t^3 + 10 t^4 +  6 t^5 + 1 t^6
  p(8,t) = 0 + 0   + 0     + 1 t^3 + 10 t^4 + 15 t^5 + 7 t^6 + 1 t^7
  ...
Reading along columns gives rows for Pascal's triangle. (End)

References

  • P. A. MacMahon, Combinatory Analysis, Two volumes (bound as one), Chelsea Publishing Company, New York, 1960, Vol. I, nr. 124, p. 151.
  • John Riordan, An Introduction to Combinatorial Analysis, John Wiley & Sons, London, 1958. eq. (35), p.124, 11. p. 154.

Links

Crossrefs

Row sums A000045(n+1) (Fibonacci). a(n, 1)= A019590(n) (Fermat's last theorem). Cf. A049403.
Cf. A104597, A146559, A155020, A125145, A057078, A039717, A001792, A057862, A011973, A115139.

Programs

  • Magma
    /* As triangle */ [[Binomial(k, n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Nov 05 2014
  • Maple
    for n from 1 to 12 do seq(binomial(k,n-k),k=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Apr 05 2005
  • Mathematica
    nn=10;CoefficientList[Series[(1+x)/(1-y x - y x^2),{x,0,nn}],{x,y}]//Grid (* Geoffrey Critzer, Jul 22 2013 *)
Table[Binomial[k, n - k], {n, 13}, {k, n}] // Flatten (* Michael De Vlieger, Dec 23 2015 *)
CoefficientList[Table[x^(n/2 - 1) Fibonacci[n + 1, Sqrt[x]], {n, 10}],
   x] // Flatten (* Eric W. Weisstein, Apr 10 2017 *)

Formula

a(n, m) = 2*(2*m-n+1)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, n
G.f. for m-th column: (x*(1+x))^m.
As a number triangle with offset 0, this is T(n, k) = Sum_{i=0..n} (-1)^(n+i)*binomial(n, i)*binomial(i+k+1, 2k+1). The antidiagonal sums give the Padovan sequence A000931(n+5). Inverse binomial transform of A078812 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
G.f.: (1 + x)/(1 - y*x - y*x^2). - Geoffrey Critzer, Jul 22 2013 [offset 0] [with offset 1: g.f. of row polynomials in y: x*(1+x)*y/(1 - x*(1+x)*y). - Wolfdieter Lang, Jul 27 2023]
From Tom Copeland, Nov 04 2014: (Start)
O.g.f: G(x,t) = x*(1+x) / [1 - t*x*(1+x)] = -P[Cinv(-x),t], where P(x,t)= x / (1 + t*x) and Cinv(x)= x*(1-x) are the compositional inverses in x of Pinv(x,t) = -P(-x,t) = x / (1 - t*x) and C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108.
Therefore, Ginv(x,t) = -C[Pinv(-x,t)] = {-1 + sqrt[1 + 4*x/(1+t*x)]}/2, which is -A124644(-x,t).
This places this array in a family of arrays related by composition of P and C and their inverses and interpolation by t, such as A091867 and A104597, and associated to the Catalan, Motzkin, Fine, and Fibonacci numbers. Cf. A104597 (polynomials shifted in t) A125145, A146559, A057078, A000045, A155020, A125145, A039717, A001792, A057862, A011973, A115139. (End)

Extensions

More terms from Emeric Deutsch, Apr 05 2005

A027465 Cube of lower triangular normalized binomial matrix.

Original entry on oeis.org

1, 3, 1, 9, 6, 1, 27, 27, 9, 1, 81, 108, 54, 12, 1, 243, 405, 270, 90, 15, 1, 729, 1458, 1215, 540, 135, 18, 1, 2187, 5103, 5103, 2835, 945, 189, 21, 1, 6561, 17496, 20412, 13608, 5670, 1512, 252, 24, 1, 19683, 59049, 78732, 61236, 30618, 10206, 2268
Offset: 0

Views

Author

Olivier Gérard, N. J. A. Sloane

Keywords

Comments

Rows of A013610 reversed. - Michael Somos, Feb 14 2002
Row sums are powers of 4 (A000302), antidiagonal sums are A006190 (a(n) = 3*a(n-1) + a(n-2)). - Gerald McGarvey, May 17 2005
Triangle of coefficients in expansion of (3+x)^n.
Also: Pure Galton board of scheme (3,1). Also: Multiplicity (number) of pairs of n-dimensional binary vectors with dot product (overlap) k. There are 2^n = A000079(n) binary vectors of length n and 2^(2n) = 4^n = A000302(n) different pairs to form dot products k = Sum_{i=1..n} v[i]*u[i] between these, 0 <= k <= n. (Since dot products are symmetric, there are only 2^n*(2^n-1)/2 different non-ordered pairs, actually.) - R. J. Mathar, Mar 17 2006
Mirror image of A013610. - Zerinvary Lajos, Nov 25 2007
T(i,j) is the number of i-permutations of 4 objects a,b,c,d, with repetition allowed, containing j a's. - Zerinvary Lajos, Dec 21 2007
The antidiagonals of the sequence formatted as a square array (see Examples section) and summed with alternating signs gives a bisection of Fibonacci sequence, A001906. Example: 81-(27-1)=55. Similar rule applied to rows gives A000079. - Mark Dols, Sep 01 2009
Triangle T(n,k), read by rows, given by (3,0,0,0,0,0,0,0,...)DELTA (1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 09 2011
T(n,k) = binomial(n,k)*3^(n-k), the number of subsets of [2n] with exactly k symmetric pairs, where elements i and j of [2n] form a symmetric pair if i+j=2n+1. Equivalently, if n couples attend a (ticketed) event that offers door prizes, then the number of possible prize distributions that have exactly k couples as dual winners is T(n,k). - Dennis P. Walsh, Feb 02 2012
T(n,k) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that the intersection of A and B contains exactly k elements. For example, T(2,1) = 6 because we have ({1},{1}); ({1},{1,2}); ({2},{2}); ({2},{1,2}); ({1,2},{1}); ({1,2},{2}). Sum_{k=0..n} T(n,k)*k = A002697(n) (see comment there by Ross La Haye). - Geoffrey Critzer, Sep 04 2013
Also the convolution triangle of A000244. - Peter Luschny, Oct 09 2022

Examples

			Example: n = 3 offers 2^3 = 8 different binary vectors (0,0,0), (0,0,1), ..., (1,1,0), (1,1,1). a(3,2) = 9 of the 2^4 = 64 pairs have overlap k = 2: (0,1,1)*(0,1,1) = (1,0,1)*(1,0,1) = (1,1,0)*(1,1,0) = (1,1,1)*(1,1,0) = (1,1,1)*(1,0,1) = (1,1,1)*(0,1,1) = (0,1,1)*(1,1,1) = (1,0,1)*(1,1,1) = (1,1,0)*(1,1,1) = 2.
For example, T(2,1)=6 since there are 6 subsets of {1,2,3,4} that have exactly 1 symmetric pair, namely, {1,4}, {2,3}, {1,2,3}, {1,2,4}, {1,3,4}, and {2,3,4}.
The present sequence formatted as a triangular array:
     1
     3     1
     9     6     1
    27    27     9     1
    81   108    54    12    1
   243   405   270    90   15    1
   729  1458  1215   540  135   18   1
  2187  5103  5103  2835  945  189  21  1
  6561 17496 20412 13608 5670 1512 252 24 1
  ...
A013610 formatted as a triangular array:
  1
  1  3
  1  6   9
  1  9  27   27
  1 12  54  108   81
  1 15  90  270  405   243
  1 18 135  540 1215  1458   729
  1 21 189  945 2835  5103  5103  2187
  1 24 252 1512 5670 13608 20412 17496 6561
   ...
A099097 formatted as a square array:
      1     0     0    0   0 0 0 0 0 0 0 ...
      3     1     0    0   0 0 0 0 0 0 ...
      9     6     1    0   0 0 0 0 0 ...
     27    27     9    1   0 0 0 0 ...
     81   108    54   12   1 0 0 ...
    243   405   270   90  15 1 ...
    729  1458  1215  540 135 ...
   2187  5103  5103 2835 ...
   6561 17496 20412 ...
  19683 59049 ...
  59049 ...

Links

Crossrefs

Cf. A000244, A007318, A013610, A013610, A099097, A027471, A027472, A036216, A036217, A036219, A036220, A036221, A036222, A036223.

Programs

  • Haskell
    a027465 n k = a027465_tabl !! n !! k
a027465_row n = a027465_tabl !! n
a027465_tabl = iterate (\row ->
   zipWith (+) (map (* 3) (row ++ [0])) (map (* 1) ([0] ++ row))) [1]
-- Reinhard Zumkeller, May 26 2013
  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*3^(i-j), j = 0 .. i) od; # Zerinvary Lajos, Nov 25 2007
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, n -> 3^(n-1)); # Peter Luschny, Oct 09 2022
  • Mathematica
    t[n_, k_] := Binomial[n, k]*3^(n-k); Table[t[n, n-k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 19 2012 *)
  • PARI
    {T(n, k) = polcoeff( (3 + x)^n, k)}; /* Michael Somos, Feb 14 2002 */

Formula

Numerators of lower triangle of (b^2)[ i, j ] where b[ i, j ] = binomial(i-1, j-1)/2^(i-1) if j <= i, 0 if j > i.
Triangle whose (i, j)-th entry is binomial(i, j)*3^(i-j).
a(n, m) = 4^(n-1)*Sum_{j=m..n} b(n, j)*b(j, m) = 3^(n-m)*binomial(n-1, m-1), n >= m >= 1; a(n, m) := 0, n < m. G.f. for m-th column: (x/(1-3*x))^m (m-fold convolution of A000244, powers of 3). - Wolfdieter Lang, Feb 2006
G.f.: 1 / (1 - x(3+y)).
a(n,k) = 3*a(n-1,k) + a(n-1,k-1) - R. J. Mathar, Mar 17 2006
From the formalism of A133314, the e.g.f. for the row polynomials of A027465 is exp(x*t)*exp(3x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-3x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*3x). The results generalize for 3 replaced by any number. - Tom Copeland, Aug 18 2008
T(n,k) = A164942(n,k)*(-1)^k. - Philippe Deléham, Oct 09 2011
Let P and P^T be the Pascal matrix and its transpose and H = P^3 = A027465. Then from the formalism of A132440 and A218272,
exp[x*z/(1-3z)]/(1-3z) = exp(3z D_z z) e^(x*z)= exp(3D_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} (3z)^n L_n(-x/3), where D is the derivative operator and L_n(x) are the regular (not normalized) Laguerre polynomials. - Tom Copeland, Oct 26 2012
E.g.f. for column k: x^k/k! * exp(3x). - Geoffrey Critzer, Sep 04 2013

A034867 Triangle of odd-numbered terms in rows of Pascal's triangle.

Original entry on oeis.org

1, 2, 3, 1, 4, 4, 5, 10, 1, 6, 20, 6, 7, 35, 21, 1, 8, 56, 56, 8, 9, 84, 126, 36, 1, 10, 120, 252, 120, 10, 11, 165, 462, 330, 55, 1, 12, 220, 792, 792, 220, 12, 13, 286, 1287, 1716, 715, 78, 1, 14, 364, 2002, 3432, 2002, 364, 14, 15, 455, 3003, 6435, 5005, 1365, 105, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also triangle of numbers of n-sequences of 0,1 with k subsequences of consecutive 01 because this number is C(n+1,2*k+1). - Roger Cuculiere (cuculier(AT)imaginet.fr), Nov 16 2002
From Gary W. Adamson, Oct 17 2008: (Start)
Received from Herb Conn:
Let T = tan x, then
tan x = T
tan 2x = 2T / (1 - T^2)
tan 3x = (3T - T^3) / (1 - 3T^2)
tan 4x = (4T - 4T^3) / (1 - 6T^2 + T^4)
tan 5x = (5T - 10T^3 + T^5) / (1 - 10T^2 + 5T^4)
tan 6x = (6T - 20T^3 + 6T^5) / (1 - 15T^2 + 15T^4 - T^6)
tan 7x = (7T - 35T^3 + 21T^5 - T^7) / (1 - 21T^2 + 35T^4 - 7T^6)
tan 8x = (8T - 56T^3 + 56T^5 - 8T^7) / (1 - 28T^2 + 70T^4 - 28T^6 + T^8)
tan 9x = (9T - 84T^3 + 126T^5 - 36T^7 + T^9) / (1 - 36 T^2 + 126T^4 - 84T^6 + 9T^8)
... To get the next one in the series, (tan 10x), for the numerator add:
9....84....126....36....1 previous numerator +
1....36....126....84....9 previous denominator =
10..120....252...120...10 = new numerator
For the denominator add:
......9.....84...126...36...1 = previous numerator +
1....36....126....84....9.... = previous denominator =
1....45....210...210...45...1 = new denominator
...where numerators = A034867, denominators = A034839
(End)
Column k is the sum of columns 2k and 2k+1 of A007318. - Philippe Deléham, Nov 12 2008
Triangle, with zeros omitted, given by (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
The row polynomials N(n,x) = Sum_{k=0..floor((n-1)/2)} T(n-1,k)*x^k, and D(n,x) = Sum_{k=0..floor(n/2)} A034839(n,k)*x^k, n >= 1, satisfy the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 = D(1,x). This is due to the Pascal triangle A007318 recurrence. Q(n,x) := tan(n*x)/tan(x) satisfies the recurrence Q(n,x) = (1 + Q(n-1,x))/(1 - v(x)*Q(n-1,x)) with input Q(1,x) = 1 and v = v(x) := (tan(x))^2. This recurrence is obtained from the addition theorem for tan(n*x) using n = 1 + (n-1). Therefore Q(n,x) = N(n,-v(x))/D(n,-v(x)). This proves the Gary W. Adamson contribution from above. See also A220673. This calculation was motivated by an e-mail of Thomas Olsen. The Oliver/Prodinger and Ma references resort to HAKEM Al Memo 239, Item 16, for the tan(n*x) formula in terms of tan(x). - Wolfdieter Lang, Jan 17 2013
The infinitesimal generator (infinigen) for the Narayana polynomials A090181/A001263 can be formed from the row polynomials P(n,y) of this entry. The resulting matrix is an instance of a matrix representation of the analytic infinigens presented in A145271 for general sets of binomial Sheffer polynomials and in A001263 and A119900 specifically for the Narayana polynomials. Given the column vector of row polynomials V = (1, P(1,x) = 2x, P(2,y) = 3x + x^2, P(3,y) = 4x + 4x^2, ...), form the lower triangular matrix M(n,k) = V(n-k,n-k), i.e., diagonally multiply the matrix with all ones on the diagonal and below by the components of V. Form the matrix MD by multiplying A132440^Transpose = A218272 = D (representing derivation of o.g.f.s) by M, i.e., MD = M*D. The non-vanishing component of the first row of (MD)^n * V / (n+1)! is the n-th Narayana polynomial. - Tom Copeland, Dec 09 2015
The diagonals of this entry are A078812 (also shifted A128908 and unsigned A053122, which are embedded in A030528, A102426, A098925, A109466, A092865). Equivalently, the antidiagonals of A078812 are the rows of A034867. - Tom Copeland, Dec 12 2015
Binomial(n,2k+1) is also the number of permutations avoiding both 132 and 213 with k peaks, i.e., positions with w[i]w[i+2]. - Lara Pudwell, Dec 19 2018
Binomial(n,2k+1) is also the number of permutations avoiding both 123 and 132 with k peaks, i.e., positions with w[i]w[i+2]. - Lara Pudwell, Dec 19 2018
The row polynomial P(n, x) = Sum_{0..floor(n/2)} T(n, k)*x^k appears as numerator polynomial of the diagonal sequence m of triangle A104698 as follows. G(m, x) = P(m, x^2)/(1 - x)^(m+1), for m >= 0. - Wolfdieter Lang, May 14 2025
Number of acyclic orientations of the path graph on n+1 vertices, with k-1 sinks. - Per W. Alexandersson, Aug 15 2025

Examples

			Triangle T starts:
  n\k   0   1   2   3   4  5 ...   ----------------------------------------
0:    1
1:    2
2:    3   1
3:    4   4
4:    5  10   1
5:    6  20   6
6:    7  35  21   1
7:    8  56  56   8
8:    9  84 126  36   1
9:   10 120 252 120  10
 10:   11 165 462 330  55  1
 11:   12 220 792 792 220 12
... ... reformatted and extended by - _Wolfdieter Lang_, May 14 2025

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 136.

Links

Crossrefs

Cf. A000129, A007318, A034839, A034867, A084938, A131980, A220673.
Cf. A001263, A090181, A119900, A132440, A145271.
Cf. A030528, A053122, A078812, A092865, A098925, A102426, A109466, A128908, A218272, A104698.
From Wolfdieter Lang, May 14 2025:(Start)
Row length A008619. Row sums A000079. Alternating row sums A009545(n+1).
Column sequences (with certain offsets): A000027, A000292, A000389, A000580, A000582, A001288, ... (End)

Programs

  • Magma
    /* as a triangle */ [[Binomial(n+1,2*k+1): k in [0..Floor(n/2)]]: n in [0..20]]; // G. C. Greubel, Mar 06 2018
  • Maple
    seq(seq(binomial(n+1,2*k+1), k=0..floor(n/2)), n=0..14); # Emeric Deutsch, Apr 01 2005
  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 12;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x]
v[n_, x_] := u[n - 1, x] + v[n - 1, x]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]  (* A034839 as a triangle *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]  (* A034867 as a triangle *)
(* Clark Kimberling, Feb 18 2012 *)
Table[Binomial[n+1, 2*k+1], {n,0,20}, {k,0,Floor[n/2]}]//Flatten (* G. C. Greubel, Mar 06 2018 *)
  • PARI
    for(n=0,20, for(k=0,floor(n/2), print1(binomial(n+1,2*k+1), ", "))) \\ G. C. Greubel, Mar 06 2018

Formula

T(n,k) = C(n+1,2k+1) = Sum_{i=k..n-k} C(i,k) * C(n-i,k).
E.g.f.: 1+(exp(x)*sinh(x*sqrt(y)))/sqrt(y). - Vladeta Jovovic, Mar 20 2005
G.f.: 1/((1-z)^2-t*z^2). - Emeric Deutsch, Apr 01 2005
T(n,k) = Sum_{j = 0..n} A034839(j,k). - Philippe Deléham, May 18 2005
Pell(n+1) = A000129(n+1) = Sum_{k=0..n} T(n,k) * 2^k = (1/n!) Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007
T(n,k) = A007318(n,2k) + A007318(n,2k+1). - Philippe Deléham, Nov 12 2008
O.g.f for column k, k>=0: (1/(1-x)^2)*(x/(1-x))^(2*k). See the G.f. of this array given above by Emeric Deutsch. - Wolfdieter Lang, Jan 18 2013
T(n,k) = (x^(2*k+1))*((1+x)^n-(1-x)^n)/2. - L. Edson Jeffery, Jan 15 2014

Extensions

More terms from Emeric Deutsch, Apr 01 2005

A102426 Triangle read by rows giving coefficients of polynomials defined by F(0,x)=0, F(1,x)=1, F(n,x) = F(n-1,x) + x*F(n-2,x).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 4, 1, 1, 6, 5, 1, 4, 10, 6, 1, 1, 10, 15, 7, 1, 5, 20, 21, 8, 1, 1, 15, 35, 28, 9, 1, 6, 35, 56, 36, 10, 1, 1, 21, 70, 84, 45, 11, 1, 7, 56, 126, 120, 55, 12, 1, 1, 28, 126, 210, 165, 66, 13, 1, 8, 84, 252, 330, 220, 78, 14, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1
Offset: 0

Views

Author

Russell Walsmith, Jan 08 2005

Keywords

Comments

Essentially the same as A098925: a(0)=0 followed by A098925. - R. J. Mathar, Aug 30 2008
F(n) + 2x * F(n-1) gives Lucas polynomials (cf. A034807). - Maxim Krikun (krikun(AT)iecn.u-nancy.fr), Jun 24 2007
After the initial 0, these are the nonzero coefficients of the Fibonacci polynomials; see the Mathematica section. - Clark Kimberling, Oct 10 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
Aside from the initial zeros, these are the antidiagonals read from bottom to top of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper, which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse of A011973. - Tom Copeland, Jul 02 2018

Examples

			The first few polynomials are:
  0
  1
  1
  x + 1
  2*x + 1
  x^2 + 3*x + 1
  3*x^2 + 4*x + 1
------------------
From _Tom Copeland_, Jan 19 2016: (Start)
[n]:
0:  0
1:  1
2:  1
3:  1  1
4:  2  1
5:  1  3  1
6:  3  4  1
7:  1  6  5   1
8:  4 10  6   1
9:  1 10 15   7   1
10: 5 20 21   8   1
11: 1 15 35  28   9  1
12: 6 35 56  36  10  1
13: 1 21 70  84  45 11 1
(End)

References

  • Dominique Foata and Guo-Niu Han, Multivariable tangent and secant q-derivative polynomials, Manuscript, Mar 21 2012.

Links

Crossrefs

Upward diagonals sums are A062200. Downward rows are A102427. Row sums are A000045. Row terms reversed = A011973. Also A102428, A102429.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.
Cf. A078812, A085478.

Programs

  • Magma
    [0] cat [Binomial(Floor(n/2)+k, Floor((n-1)/2-k) ): k in [0..Floor((n-1)/2)], n in [0..17]]; // G. C. Greubel, Oct 13 2019
  • Mathematica
    Join[{0}, Table[ Select[ CoefficientList[ Fibonacci[n, x], x], 0 < # &], {n, 0, 17}]//Flatten] (* Clark Kimberling, Oct 10 2013 and slightly modified by Robert G. Wilson v, May 03 2017 *)
  • PARI
    F(n) = if (n==0, 0, if (n==1, 1, F(n-1) + x*F(n-2)));
tabf(nn) = for (n=0, nn, print(Vec(F(n)))); \\ Michel Marcus, Feb 10 2020

Formula

Alternatively, as n is even or odd: T(n-2, k) + T(n-1, k-1) = T(n, k), T(n-2, k) + T(n-1, k) = T(n, k)
T(n, k) = binomial(floor(n/2)+k, floor((n-1)/2-k) ). - Paul Barry, Jun 22 2005
Beginning with the second polynomial in the example and offset=0, P(n,t)= Sum_{j=0..n}, binomial(n-j,j)*x^j with the convention that 1/k! is zero for k=-1,-2,..., i.e., 1/k! = lim_{c->0} 1/(k+c)!. - Tom Copeland, Oct 11 2014
From Tom Copeland, Jan 19 2016: (Start)
O.g.f.: (x + x^2 - x^3) / (1 - (2+t)*x^2 + x^4) = (x^2 (even part) + x*(1-x^2) (odd)) / (1 - (2+t)*x^2 + x^4).
Recursion relations:
A) p(n,t) = p(n-1,t) + p(n-2,t) for n=2,4,6,8,...
B) p(n,t) = t*p(n-1,t) + p(n-2,t) for n=3,5,7,...
C) a(n,k) = a(n-2,k) + a(n-1,k) for n=4,6,8,...
D) a(n,k) = a(n-2,k) + a(n-1,k-1) for n=3,5,7,...
Relation A generalized to MV(n,t;r) = P(2n+1,t) + r R(2n,t) for n=1,2,3,... (cf. A078812 and A085478) is the generating relation on p. 229 of Andre-Jeannine for the generalized Morgan-Voyce polynomials, e.g., MV(2,t;r) = p(5,t) + r*p(4,t) = (1 + 3t + t^2) + r*(2 + t) = (1 + 2r) + (3 + r)*t + t^2, so P(n,t) = MV(n-4,t;1) for n=4,6,8,... .
The even and odd polynomials are also presented in Trzaska and Ferri.
Dropping the initial 0 and re-indexing with initial m=0 gives the row polynomials Fb(m,t) = p(n+1,t) below with o.g.f. G(t,x)/x, starting with Fb(0,t) = 1, Fb(1,t) = 1, Fb(2,t) = 1 + t, and Fb(3,t) = 2 + t.
The o.g.f. x/G(x,t) = (1 - (2+t)*x^2 + x^4) / (1 + x - x^2) then generates a sequence of polynomials IFb(t) such that the convolution Sum_{k=0..n} IFb(n-k,t) Fb(k,t) vanishes for n>1 and is one for n=0. These linear polynomials have the basic Fibonacci numbers A000045 as an overall factor:
IFb(0,t) = 1
IFb(1,t) = -1
IFb(2,t) = -t
IFb(3,t) = -1 (1-t)
IFb(4,t) = 2 (1-t)
IFb(5,t) = -3 (1-t)
IFb(6,t) = 5 (1-t)
IFb(7,t) = -8 (1-t)
IFb(8,t) = 13 (1-t)
... .
(End)

Extensions

Name corrected by John K. Sikora, Feb 10 2020

A181289 Triangle read by rows: T(n,k) is the number of 2-compositions of n having length k (0 <= k <= n).

Original entry on oeis.org

1, 0, 2, 0, 3, 4, 0, 4, 12, 8, 0, 5, 25, 36, 16, 0, 6, 44, 102, 96, 32, 0, 7, 70, 231, 344, 240, 64, 0, 8, 104, 456, 952, 1040, 576, 128, 0, 9, 147, 819, 2241, 3400, 2928, 1344, 256, 0, 10, 200, 1372, 4712, 9290, 11040, 7840, 3072, 512, 0, 11, 264, 2178, 9108, 22363
Offset: 0

Views

Author

Emeric Deutsch, Oct 12 2010

Keywords

Comments

A 2-composition of n is a nonnegative matrix with two rows, such that each column has at least one nonzero entry and whose entries sum up to n. The length of the 2-composition is the number of columns.
From Tom Copeland, Sep 06 2011: (Start)
R(t,z) = (1-z)^2 / ((1+t)*(1-z)^2-1) = 1/(t - (2*z + 3*z^2 + 4*z^3 + 5*z^4 + ...)) = 1/t + (1/t)^2*2*z + (1/t)^3*(4+3t)*z^2 + (1/t)^4*(8+12*t+4*t^2)*z^3 + ... gives row reversed polynomials of A181289 with G(t,z) = R(1/t,z)/t.
R(t,z) is related to generators for A033282 and A001003 (t=1) and can be umbrally extended to give a partition generator for A133437. (End)
A refined, reverse version of this array is given in A253722. - Tom Copeland, May 02 2015
The infinitesimal generator (infinigen) for the face polynomials of associahedra A086810/A033282, read as decreasing powers, (and for the dual simplicial complex read as increasing powers) can be formed from the row polynomials P(n,t) of this entry. This type of infinigen is presented in A145271 for general sets of binomial Sheffer polynomials. This specific infinigen is presented in analytic form in A086810. Given the column vector of row polynomials V = (P(0,t) = 1, P(1,y) = 2 t, P(2,y) = 3 t + 4 t^2, P(3,y) = 4 t + 12 t^2 + 8 t^3, ...), form the lower triangular matrix M(n,k) = V(n-k,n-k), i.e., diagonally multiply the matrix with all ones on the diagonal and below by the components of V. Form the matrix MD by multiplying A132440^Transpose = A218272 = D (representing derivation of o.g.f.s) by M, i.e., MD = M*D. The non-vanishing component of the first row of (MD)^n * V / (n+1)! is the n-th face polynomial. - Tom Copeland, Dec 11 2015
T is the convolution triangle of the positive integers starting at 2 (see A357368). - Peter Luschny, Oct 19 2022

Examples

			Triangle starts:
  1;
  0,  2;
  0,  3,   4;
  0,  4,  12,    8;
  0,  5,  25,   36,   16;
  0,  6,  44,  102,   96,    32;
  0,  7,  70,  231,  344,   240,    64;
  0,  8, 104,  456,  952,  1040,   576,   128;
  0,  9, 147,  819, 2241,  3400,  2928,  1344,   256;
  0, 10, 200, 1372, 4712,  9290, 11040,  7840,  3072,  512;
  0, 11, 264, 2178, 9108, 22363, 34332, 33488, 20224, 6912, 1024;

Links

Crossrefs

Cf. A003480 (row sums), A181290.
Cf. A253722, A033282, A001003, A133437.
Cf. A086810, A132440, A145271, A218272.
Cf. A000297 (column 3), A006636 (column 4), A006637 (column 5).

Programs

  • Maple
    T := proc (n, k) if k <= n then sum((-1)^j*2^(k-j)*binomial(k, j)*binomial(n+k-j-1, 2*k-1), j = 0 .. k) else 0 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
# Uses function PMatrix from A357368.
PMatrix(10, n -> n + 1); # Peter Luschny, Oct 19 2022
  • Mathematica
    Table[Sum[(-1)^j*2^(k - j) Binomial[k, j] Binomial[n + k - j - 1, 2 k - 1], {j, 0, k}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 11 2015 *)
  • PARI
    T_xt(max_row) = {my(N=max_row+1, x='x+O('x^N), h=(1-x)^2/((1-x)^2 - t*x*(2-x))); vector(N, n, Vecrev(polcoeff(h, n-1)))}
T_xt(10) \\ John Tyler Rascoe, Apr 05 2025

Formula

T(n,k) = Sum_{j=0..k} (-1)^j*2^(k-j)*binomial(k,j)*binomial(n+k-j-1, 2*k-1) (0 <= k <= n).
G.f.: G(t,x) = (1-x)^2/((1-x)^2 - t*x*(2-x)).
G.f. of column k = x^k*(2-x)^k/(1-x)^{2k} (k>=1) (we have a Riordan array).
Recurrences satisfied by the numbers u_{n,k}=T(n,k) can be found in the Castiglione et al. reference.
Sum_{k=0..n} k*T(n,k) = A181290(n).
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=0, T(1,1)=2, T(2,0)=0, T(1,1)=3, T(2,2)=4, T(n,k)=0, if k < 0 or if k > n. - Philippe Deléham, Nov 29 2013

A169803 Triangle read by rows: T(n,k) = binomial(n+1-k,k) (n >= 0, 0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 2, 0, 1, 3, 1, 0, 1, 4, 3, 0, 0, 1, 5, 6, 1, 0, 0, 1, 6, 10, 4, 0, 0, 0, 1, 7, 15, 10, 1, 0, 0, 0, 1, 8, 21, 20, 5, 0, 0, 0, 0, 1, 9, 28, 35, 15, 1, 0, 0, 0, 0, 1, 10, 36, 56, 35, 6, 0, 0, 0, 0, 0, 1, 11, 45, 84, 70, 21, 1, 0, 0, 0, 0, 0, 1, 12, 55, 120, 126, 56, 7, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Nadia Heninger and N. J. A. Sloane, May 21 2010

Keywords

Comments

T(n,k) = 0 if k <0 or k > n+1-k.
T(n,k) is the number of binary vectors of length n and weight k containing no pair of adjacent 1's.
Take Pascal's triangle A007318 and push the k-th column downwards by 2k-1 places (k>=1).
Row sums are A000045.
From Emanuele Munarini, May 24 2011: (Start)
Diagonal sums are A000930(n+1).
A sparse subset (or scattered subset) of {1,2,...,n} is a subset never containing two consecutive elements. T(n,k) is the number of sparse subsets of {1,2,...,n} having size k. For instance, for n=4 and k=2 we have the 3 sparse 2-subsets of {1,2,3,4}: 13, 14, 24. (End)
As a triangle, row 2*n-1 consists of the coefficients of Morgan-Voyce polynomial B(n,x), A172431, and row 2*n to the coefficients of Morgan-Voyce polynomial b(n,x), A054142.
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
Antidiagonals of the Pascal matrix A007318 read bottom to top, omitting the first antidiagonal. These are also the antidiagonals (omitting the first antidiagonal) read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper, which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is embedded in A102426. - Tom Copeland, Jul 02 2018

Examples

			Triangle begins:
  [1]
  [1, 1]
  [1, 2, 0]
  [1, 3, 1, 0]
  [1, 4, 3, 0, 0]
  [1, 5, 6, 1, 0, 0]
  [1, 6, 10, 4, 0, 0, 0]
  [1, 7, 15, 10, 1, 0, 0, 0]
  [1, 8, 21, 20, 5, 0, 0, 0, 0]
  [1, 9, 28, 35, 15, 1, 0, 0, 0, 0]
  [1, 10, 36, 56, 35, 6, 0, 0, 0, 0, 0]
  [1, 11, 45, 84, 70, 21, 1, 0, 0, 0, 0, 0]
  [1, 12, 55, 120, 126, 56, 7, 0, 0, 0, 0, 0, 0]
  [1, 13, 66, 165, 210, 126, 28, 1, 0, 0, 0, 0, 0, 0]
  [1, 14, 78, 220, 330, 252, 84, 8, 0, 0, 0, 0, 0, 0, 0]
  [1, 15, 91, 286, 495, 462, 210, 36, 1, 0, 0, 0, 0, 0, 0, 0]
  [1, 16, 105, 364, 715, 792, 462, 120, 9, 0, 0, 0, 0, 0, 0, 0, 0]
  [1, 17, 120, 455, 1001, 1287, 924, 330, 45, 1, 0, 0, 0, 0, 0, 0, 0, 0]
  [1, 18, 136, 560, 1365, 2002, 1716, 792, 165, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0]
  [1, 19, 153, 680, 1820, 3003, 3003, 1716, 495, 55, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0]
  ...

Links

Crossrefs

Cf. A000045, A000930, A007318, A011973 (another version), A218272.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways. - N. J. A. Sloane, May 29 2011
A172431 and A054142 describe the odd and even lines of the triangle.

Programs

  • Mathematica
    T[n_,k_]:= Binomial[n+1-k,k]; Table[T[n,k],{n,0,12},{k,0,n}]//Flatten (* Stefano Spezia, Sep 16 2024 *)
  • Maxima
    create_list(binomial(n-k+1,k),n,0,20,k,0,n); /* Emanuele Munarini, May 24 2011 */
  • PARI
    T(n,k)=binomial(n+1-k,k) \\ Charles R Greathouse IV, Oct 24 2012

A121448 Triangle read by rows: T(n,k) is the number of binary trees with n edges and having k vertices of outdegree 1 (n>=0, k>=0). A binary tree is a rooted tree in which each vertex has at most two children and each child of a vertex is designated as its left or right child.

Original entry on oeis.org

1, 0, 2, 1, 0, 4, 0, 6, 0, 8, 2, 0, 24, 0, 16, 0, 20, 0, 80, 0, 32, 5, 0, 120, 0, 240, 0, 64, 0, 70, 0, 560, 0, 672, 0, 128, 14, 0, 560, 0, 2240, 0, 1792, 0, 256, 0, 252, 0, 3360, 0, 8064, 0, 4608, 0, 512, 42, 0, 2520, 0, 16800, 0, 26880, 0, 11520, 0, 1024, 0, 924, 0, 18480, 0
Offset: 0

Views

Author

Emeric Deutsch, Jul 31 2006

Keywords

Comments

T(2n,0) = binomial(2n,n)/(n+1) (the Catalan numbers; A000108); T(2n+1,0)=0. T(n,n)=2^n (A000079). Sum(k*T(n,k),k=0..n)=2*binomial(2n,n-1)=2*A001791(n). After deleting the zeros, reflection of A091894.
From Tom Copeland, Feb 07 2016: (Start)
A shifted o.g.f. is OG(x,t) = [1 - 2tx - sqrt[(1-2tx)^2-4x^2]] / (2x) = x + 2t x^2 + (1+4t^2) x^3 + ... with compositional inverse OGinv(x,t) = x / (1 + 2tx + x^2), the shifted o.g.f. for A053117 (mod signs).
For x > 0 and choosing the positive square root, OG(x^2,t) = H(x,t) = x^2 + 2t x^4 + (1+4t^2) x^6 + ... has the compositional inverse Hinv(x,t) = sqrt[x / (1 + 2tx + x^2)] , which satisfies Hinv(H(x, t), t) = x, and which is the generating function for the Legendre polynomials (mod signs, cf. A008316) times sqrt(x).
In general, GB(x,t,b) = [x / (1 - 2tx + x^2)]^b is a generator for the Gegenbauer polynomials times x^b for positive roots with compositional inverse about the origin GBinv(x,t,b) = OG(x^(1/b),-t) for x>0. Cf. A097610.
(End)
From Tom Copeland, Feb 09 2016: (Start)
z1 = OG(x,t) is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,t),z2(x,t)) =(z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-2tx)/x] z + 1, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.
The other zero is given by z2(x,t) = [1 - 2tx + sqrt[(1-2tx)^2-4x^2]] / (2x) = (1 - 2tx)/x - z1(x,t).
The two are zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2). (Added Feb 13 2016. See Landweber et al., p 14. Cf. A097610.)
(End)

Examples

			T(2,2)=4 because, denoting by L (R) an edge going from a vertex to a left (right) child, we have the paths: LL, LR, RL and RR.
Triangle starts:
  1;
  0,2;
  1,0,4;
  0,6,0,8;
  2,0,24,0,16;

Links

Crossrefs

Cf. A000108, A000079, A001791, A091894.
Cf. A038207, A053117, A097610, A126120, A145271, A180874, A218272.
Cf. A008316.

Programs

  • Maple
    T:=proc(n,k) if n-k mod 2 = 0 then 2^k*binomial(n+1,k)*binomial(n+1-k,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    nn=10;Drop[CoefficientList[Series[(1-2x y - ((-4x^2+(1-2x y)^2))^(1/2))/(2 x),{x,0,nn}],{x,y}],1]//Grid  (* Geoffrey Critzer, Feb 20 2013 *)

Formula

T(n,k) = 2^k*binomial(n+1,k)binomial(n+1-k,(n-k)/2)/(n+1) if n-k is even; otherwise, T(n,k) = 0. G.f. G=G(t,z) satisfies G=1+2tzG+z^2*G^2.
T(n,k) = 2^k*A097610(n,k). - Philippe Deléham, Aug 17 2006
From Tom Copeland, Feb 09 2016: (Start)
The following is from the formalism in A097610 with h1 = 2t, h2 = 1, and MT(n,h1,h2) = MT(n,2t,1) and with OG(x,t) defined above.
E.g.f.: M(x,t) = e^(2tx) AC(x) = exp[x MT(.,2t,1)] = exp[x P(.,t)], where AC(x) = I_1(2x)/x = Sum_{n>=0} x^(2n)/(n!(n+1)!) = exp(c.x) is the e.g.f. of A126120.
P(n,t) = MT(n,2t,1) = (c. + 2t)^n = Sum_{k=0..n} binomial(n,k) c(n-k) (2t)^k with c(k) = A126120(k). P(n,t+s) = (c. + 2t + 2s)^n = (P(.,t) + 2s)^n.
P(n,t) = t^n FC(n,c./t) = t^n (2 + c./t)^n, where FC(n,t) = (2 + t)^n are the face polynomials (vectors) of the hypercubes of A038207, i.e., the row polynomials of this entry can be obtained as the umbral composition of the reverse face polynomials with the aerated Catalan numbers of A000108.
The lowering and raising operators for the row polynomials P(n,t) of this entry are L = (1/2) d/dt = (1/2) D and R = 2t + dlog{AC(L)}/dL = 2t + Sum_{n>=0} b(n) L^(2n+1)/(2n+1)! = 2t + L - L^3/3! + 5 L^5/5! - ... with b(n) = (-1)^n A180874(n+1).
Let CP(n,t) = P(n+1,t) with CP(0,t) = 0. Then the infinitesimal generator for CP(n,t) is g(x) d/dx with g(x) = 1 /[dOGinv(x,t)/dx] = x^2 / [(OGinv(x,t))^2 (1 - x^2)] = (1 + 2t x + x^2)^2 / (1 - x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomial CP(n,t), i.e., exp[x g(u)d/du] u |_(u=0) = OG(x,t) = 1 /[1 - x P(.,t)]. Cf. A145271.
g(x) = 1 + 4t x + (3+4t) x^2 + 8t x^3 + 4(1+t^2) x^4 + 8t x^5 + 4(1+t^2) x^6 + 8t x^7 + ... has the repeating coefficients of the vector V = (1, 4t, 3+4t, 8t, 4(1+t^2), 8t, 4(1+t^2), 8t, ...). Form the lower triangular matrix U with all ones on the diagonal and below. Multiply the n-th diagonal of U by V(n), giving the matrix VU with VU(n,k) = V(n-k). Then (1,0,0,0,..) [VU * DM]^n/n! (0,1,0,0,..)^T = CP(n,t) = P(n-1,t) for n>0 with DM being the matrix A218272 representing differentiation of a power series.
(End)
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