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.

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.
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
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

A001147 Double factorial of odd numbers: a(n) = (2*n-1)!! = 1*3*5*...*(2*n-1).

Original entry on oeis.org

1, 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075, 13749310575, 316234143225, 7905853580625, 213458046676875, 6190283353629375, 191898783962510625, 6332659870762850625, 221643095476699771875, 8200794532637891559375, 319830986772877770815625
Offset: 0

Keywords

Comments

The solution to Schröder's third problem.
Number of fixed-point-free involutions in symmetric group S_{2n} (cf. A000085).
a(n-2) is the number of full Steiner topologies on n points with n-2 Steiner points. [corrected by Lyle Ramshaw, Jul 20 2022]
a(n) is also the number of perfect matchings in the complete graph K(2n). - Ola Veshta (olaveshta(AT)my-deja.com), Mar 25 2001
Number of ways to choose n disjoint pairs of items from 2*n items. - Ron Zeno (rzeno(AT)hotmail.com), Feb 06 2002
Number of ways to choose n-1 disjoint pairs of items from 2*n-1 items (one item remains unpaired). - Bartosz Zoltak, Oct 16 2012
For n >= 1 a(n) is the number of permutations in the symmetric group S_(2n) whose cycle decomposition is a product of n disjoint transpositions. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001
a(n) is the number of distinct products of n+1 variables with commutative, nonassociative multiplication. - Andrew Walters (awalters3(AT)yahoo.com), Jan 17 2004. For example, a(3)=15 because the product of the four variables w, x, y and z can be constructed in exactly 15 ways, assuming commutativity but not associativity: 1. w(x(yz)) 2. w(y(xz)) 3. w(z(xy)) 4. x(w(yz)) 5. x(y(wz)) 6. x(z(wy)) 7. y(w(xz)) 8. y(x(wz)) 9. y(z(wx)) 10. z(w(xy)) 11. z(x(wy)) 12. z(y(wx)) 13. (wx)(yz) 14. (wy)(xz) 15. (wz)(xy).
a(n) = E(X^(2n)), where X is a standard normal random variable (i.e., X is normal with mean = 0, variance = 1). So for instance a(3) = E(X^6) = 15, etc. See Abramowitz and Stegun or Hoel, Port and Stone. - Jerome Coleman, Apr 06 2004
Second Eulerian transform of 1,1,1,1,1,1,... The second Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} E(n,k)s(k), where E(n,k) is a second-order Eulerian number (A008517). - Ross La Haye, Feb 13 2005
Integral representation as n-th moment of a positive function on the positive axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/2)/sqrt(2*Pi*x) dx, n >= 0. - Karol A. Penson, Oct 10 2005
a(n) is the number of binary total partitions of n+1 (each non-singleton block must be partitioned into exactly two blocks) or, equivalently, the number of unordered full binary trees with n+1 labeled leaves (Stanley, ex 5.2.6). - Mitch Harris, Aug 01 2006
a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is i for iDavid Callan, Sep 25 2006
a(n) is the number of increasing ordered rooted trees on n+1 vertices where "increasing" means the vertices are labeled 0,1,2,...,n so that each path from the root has increasing labels. Increasing unordered rooted trees are counted by the factorial numbers A000142. - David Callan, Oct 26 2006
Number of perfect multi Skolem-type sequences of order n. - Emeric Deutsch, Nov 24 2006
a(n) = total weight of all Dyck n-paths (A000108) when each path is weighted with the product of the heights of the terminal points of its upsteps. For example with n=3, the 5 Dyck 3-paths UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD have weights 1*2*3=6, 1*2*2=4, 1*2*1=2, 1*1*2=2, 1*1*1=1 respectively and 6+4+2+2+1=15. Counting weights by height of last upstep yields A102625. - David Callan, Dec 29 2006
a(n) is the number of increasing ternary trees on n vertices. Increasing binary trees are counted by ordinary factorials (A000142) and increasing quaternary trees by triple factorials (A007559). - David Callan, Mar 30 2007
From Tom Copeland, Nov 13 2007, clarified in first and extended in second paragraph, Jun 12 2021: (Start)
a(n) has the e.g.f. (1-2x)^(-1/2) = 1 + x + 3*x^2/2! + ..., whose reciprocal is (1-2x)^(1/2) = 1 - x - x^2/2! - 3*x^3/3! - ... = b(0) - b(1)*x - b(2)*x^2/2! - ... with b(0) = 1 and b(n+1) = -a(n) otherwise. By the formalism of A133314, Sum_{k=0..n} binomial(n,k)*b(k)*a(n-k) = 0^n where 0^0 := 1. In this sense, the sequence a(n) is essentially self-inverse. See A132382 for an extension of this result. See A094638 for interpretations.
This sequence aerated has the e.g.f. e^(t^2/2) = 1 + t^2/2! + 3*t^4/4! + ... = c(0) + c(1)*t + c(2)*t^2/2! + ... and the reciprocal e^(-t^2/2); therefore, Sum_{k=0..n} cos(Pi k/2)*binomial(n,k)*c(k)*c(n-k) = 0^n; i.e., the aerated sequence is essentially self-inverse. Consequently, Sum_{k=0..n} (-1)^k*binomial(2n,2k)*a(k)*a(n-k) = 0^n. (End)
From Ross Drewe, Mar 16 2008: (Start)
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant but the pairs are not distinguishable, i.e., arrangements which are the same after permutations of the labels are equivalent.
If this sequence and A000680 are denoted by a(n) and b(n) respectively, then a(n) = b(n)/n! where n! = the number of ways of permuting the pair labels.
For example, there are 90 ways of arranging the elements of 3 pairs [1 1], [2 2], [3 3] when the pairs are distinguishable: A = { [112233], [112323], ..., [332211] }.
By applying the 6 relabeling permutations to A, we can partition A into 90/6 = 15 subsets: B = { {[112233], [113322], [221133], [223311], [331122], [332211]}, {[112323], [113232], [221313], [223131], [331212], [332121]}, ....}
Each subset or equivalence class in B represents a unique pattern of pair relationships. For example, subset B1 above represents {3 disjoint pairs} and subset B2 represents {1 disjoint pair + 2 interleaved pairs}, with the order being significant (contrast A132101). (End)
A139541(n) = a(n) * a(2*n). - Reinhard Zumkeller, Apr 25 2008
a(n+1) = Sum_{j=0..n} A074060(n,j) * 2^j. - Tom Copeland, Sep 01 2008
From Emeric Deutsch, Jun 05 2009: (Start)
a(n) is the number of adjacent transpositions in all fixed-point-free involutions of {1,2,...,2n}. Example: a(2)=3 because in 2143=(12)(34), 3412=(13)(24), and 4321=(14)(23) we have 2 + 0 + 1 adjacent transpositions.
a(n) = Sum_{k>=0} k*A079267(n,k).
(End)
Hankel transform is A137592. - Paul Barry, Sep 18 2009
(1, 3, 15, 105, ...) = INVERT transform of A000698 starting (1, 2, 10, 74, ...). - Gary W. Adamson, Oct 21 2009
a(n) = (-1)^(n+1)*H(2*n,0), where H(n,x) is the probabilists' Hermite polynomial. The generating function for the probabilists' Hermite polynomials is as follows: exp(x*t-t^2/2) = Sum_{i>=0} H(i,x)*t^i/i!. - Leonid Bedratyuk, Oct 31 2009
The Hankel transform of a(n+1) is A168467. - Paul Barry, Dec 04 2009
Partial products of odd numbers. - Juri-Stepan Gerasimov, Oct 17 2010
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
a(n) is the number of subsets of {1,...,n^2} that contain exactly k elements from {1,...,k^2} for k=1,...,n. For example, a(3)=15 since there are 15 subsets of {1,2,...,9} that satisfy the conditions, namely, {1,2,5}, {1,2,6}, {1,2,7}, {1,2,8}, {1,2,9}, {1,3,5}, {1,3,6}, {1,3,7}, {1,3,8}, {1,3,9}, {1,4,5}, {1,4,6}, {1,4,7}, {1,4,8}, and {1,4,9}. - Dennis P. Walsh, Dec 02 2011
a(n) is the leading coefficient of the Bessel polynomial y_n(x) (cf. A001498). - Leonid Bedratyuk, Jun 01 2012
For n>0: a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j)^2 for 1 <= i,j <= n. - Enrique Pérez Herrero, Jan 14 2013
a(n) is also the numerator of the mean value from 0 to Pi/2 of sin(x)^(2n). - Jean-François Alcover, Jun 13 2013
a(n) is the size of the Brauer monoid on 2n points (see A227545). - James Mitchell, Jul 28 2013
For n>1: a(n) is the numerator of M(n)/M(1) where the numbers M(i) have the property that M(n+1)/M(n) ~ n-1/2 (for example, large Kendell-Mann numbers, see A000140 or A181609, as n --> infinity). - Mikhail Gaichenkov, Jan 14 2014
a(n) = the number of upper-triangular matrix representations required for the symbolic representation of a first order central moment of the multivariate normal distribution of dimension 2(n-1), i.e., E[X_1*X_2...*X_(2n-2)|mu=0, Sigma]. See vignette for symmoments R package on CRAN and Phillips reference below. - Kem Phillips, Aug 10 2014
For n>1: a(n) is the number of Feynman diagrams of order 2n (number of internal vertices) for the vacuum polarization with one charged loop only, in quantum electrodynamics. - Robert Coquereaux, Sep 15 2014
Aerated with intervening zeros (1,0,1,0,3,...) = a(n) (cf. A123023), the e.g.f. is e^(t^2/2), so this is the base for the Appell sequence A099174 with e.g.f. e^(t^2/2) e^(x*t) = exp(P(.,x),t) = unsigned A066325(x,t), the probabilist's (or normalized) Hermite polynomials. P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for A066325(x,t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), where UP(n,t) are the polynomials of A066325 and, e.g., (P(.,t))^n = P(n,t). - Tom Copeland, Nov 15 2014
a(n) = the number of relaxed compacted binary trees of right height at most one of size n. A relaxed compacted binary tree of size n is a directed acyclic graph consisting of a binary tree with n internal nodes, one leaf, and n pointers. It is constructed from a binary tree of size n, where the first leaf in a post-order traversal is kept and all other leaves are replaced by pointers. These links may point to any node that has already been visited by the post-order traversal. The right height is the maximal number of right-edges (or right children) on all paths from the root to any leaf after deleting all pointers. The number of unbounded relaxed compacted binary trees of size n is A082161(n). See the Genitrini et al. link. - Michael Wallner, Jun 20 2017
Also the number of distinct adjacency matrices in the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017
From Christopher J. Smyth, Jan 26 2018: (Start)
a(n) = the number of essentially different ways of writing a probability distribution taking n+1 values as a sum of products of binary probability distributions. See comment of Mitch Harris above. This is because each such way corresponds to a full binary tree with n+1 leaves, with the leaves labeled by the values. (This comment is due to Niko Brummer.)
Also the number of binary trees with root labeled by an (n+1)-set S, its n+1 leaves by the singleton subsets of S, and other nodes labeled by subsets T of S so that the two daughter nodes of the node labeled by T are labeled by the two parts of a 2-partition of T. This also follows from Mitch Harris' comment above, since the leaf labels determine the labels of the other vertices of the tree.
(End)
a(n) is the n-th moment of the chi-squared distribution with one degree of freedom (equivalent to Coleman's Apr 06 2004 comment). - Bryan R. Gillespie, Mar 07 2021
Let b(n) = 0 for n odd and b(2k) = a(k); i.e., let the sequence b(n) be an aerated version of this entry. After expanding the differential operator (x + D)^n and normal ordering the resulting terms, the integer coefficient of the term x^k D^m is n! b(n-k-m) / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n. E.g., (x+D)^2 = x^2 + 2xD + D^2 + 1 with D = d/dx. The result generalizes to the raising (R) and lowering (L) operators of any Sheffer polynomial sequence by replacing x by R and D by L and follows from the disentangling relation e^{t(L+R)} = e^{t^2/2} e^{tR} e^{tL}. Consequently, these are also the coefficients of the reordered 2^n permutations of the binary symbols L and R under the condition LR = RL + 1. E.g., (L+R)^2 = LL + LR + RL + RR = LL + 2RL + RR + 1. (Cf. A344678.) - Tom Copeland, May 25 2021
From Tom Copeland, Jun 14 2021: (Start)
Lando and Zvonkin present several scenarios in which the double factorials occur in their role of enumerating perfect matchings (pairings) and as the nonzero moments of the Gaussian e^(x^2/2).
Speyer and Sturmfels (p. 6) state that the number of facets of the abstract simplicial complex known as the tropical Grassmannian G'''(2,n), the space of phylogenetic T_n trees (see A134991), or Whitehouse complex is a shifted double factorial.
These are also the unsigned coefficients of the x[2]^m terms in the partition polynomials of A134685 for compositional inversion of e.g.f.s, a refinement of A134991.
a(n)*2^n = A001813(n) and A001813(n)/(n+1)! = A000108(n), the Catalan numbers, the unsigned coefficients of the x[2]^m terms in the partition polynomials A133437 for compositional inversion of o.g.f.s, a refinement of A033282, A126216, and A086810. Then the double factorials inherit a multitude of analytic and combinatoric interpretations from those of the Catalan numbers, associahedra, and the noncrossing partitions of A134264 with the Catalan numbers as unsigned-row sums. (End)
Connections among the Catalan numbers A000108, the odd double factorials, values of the Riemann zeta function and its derivative for integer arguments, and series expansions of the reduced action for the simple harmonic oscillator and the arc length of the spiral of Archimedes are given in the MathOverflow post on the Riemann zeta function. - Tom Copeland, Oct 02 2021
b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b.)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022

Examples

			a(3) = 1*3*5 = 15.
From _Joerg Arndt_, Sep 10 2013: (Start)
There are a(3)=15 involutions of 6 elements without fixed points:
  #:    permutation           transpositions
  01:  [ 1 0 3 2 5 4 ]      (0, 1) (2, 3) (4, 5)
  02:  [ 1 0 4 5 2 3 ]      (0, 1) (2, 4) (3, 5)
  03:  [ 1 0 5 4 3 2 ]      (0, 1) (2, 5) (3, 4)
  04:  [ 2 3 0 1 5 4 ]      (0, 2) (1, 3) (4, 5)
  05:  [ 2 4 0 5 1 3 ]      (0, 2) (1, 4) (3, 5)
  06:  [ 2 5 0 4 3 1 ]      (0, 2) (1, 5) (3, 4)
  07:  [ 3 2 1 0 5 4 ]      (0, 3) (1, 2) (4, 5)
  08:  [ 3 4 5 0 1 2 ]      (0, 3) (1, 4) (2, 5)
  09:  [ 3 5 4 0 2 1 ]      (0, 3) (1, 5) (2, 4)
  10:  [ 4 2 1 5 0 3 ]      (0, 4) (1, 2) (3, 5)
  11:  [ 4 3 5 1 0 2 ]      (0, 4) (1, 3) (2, 5)
  12:  [ 4 5 3 2 0 1 ]      (0, 4) (1, 5) (2, 3)
  13:  [ 5 2 1 4 3 0 ]      (0, 5) (1, 2) (3, 4)
  14:  [ 5 3 4 1 2 0 ]      (0, 5) (1, 3) (2, 4)
  15:  [ 5 4 3 2 1 0 ]      (0, 5) (1, 4) (2, 3)
(End)
G.f. = 1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + 135135*x^7 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, (26.2.28).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 317.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 228, #19.
  • Hoel, Port and Stone, Introduction to Probability Theory, Section 7.3.
  • F. K. Hwang, D. S. Richards and P. Winter, The Steiner Tree Problem, North-Holland, 1992, see p. 14.
  • C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw-Hill, 1980, pages 466-467.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.6 and also p. 178.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer-Verlag, New York, 1999, p. 73.
  • G. Watson, The Theory of Bessel Functions, Cambridge Univ. Press, 1922.

Crossrefs

Cf. A086677; A055142 (for this sequence, |a(n+1)| + 1 is the number of distinct products which can be formed using commutative, nonassociative multiplication and a nonempty subset of n given variables).
Constant terms of polynomials in A098503. First row of array A099020.
Subsequence of A248652.
Cf. A082161 (relaxed compacted binary trees of unbounded right height).
Cf. A053871 (binomial transform).

Programs

  • GAP
    A001147 := function(n) local i, s, t; t := 1; i := 0; Print(t, ", "); for i in [1 .. n] do t := t*(2*i-1); Print(t, ", "); od; end; A001147(100); # Stefano Spezia, Nov 13 2018
    
  • Haskell
    a001147 n = product [1, 3 .. 2 * n - 1]
    a001147_list = 1 : zipWith (*) [1, 3 ..] a001147_list
    -- Reinhard Zumkeller, Feb 15 2015, Dec 03 2011
    
  • Magma
    A001147:=func< n | n eq 0 select 1 else &*[ k: k in [1..2*n-1 by 2] ] >; [ A001147(n): n in [0..20] ]; // Klaus Brockhaus, Jun 22 2011
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015
    
  • Maple
    f := n->(2*n)!/(n!*2^n);
    A001147 := proc(n) doublefactorial(2*n-1); end: # R. J. Mathar, Jul 04 2009
    A001147 := n -> 2^n*pochhammer(1/2, n); # Peter Luschny, Aug 09 2009
    G(x):=(1-2*x)^(-1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..19); # Zerinvary Lajos, Apr 03 2009; aligned with offset by Johannes W. Meijer, Aug 11 2009
    series(hypergeom([1,1/2],[],2*x),x=0,20); # Mark van Hoeij, Apr 07 2013
  • Mathematica
    Table[(2 n - 1)!!, {n, 0, 19}] (* Robert G. Wilson v, Oct 12 2005 *)
    a[ n_] := 2^n Gamma[n + 1/2] / Gamma[1/2]; (* Michael Somos, Sep 18 2014 *)
    Join[{1}, Range[1, 41, 2]!!] (* Harvey P. Dale, Jan 28 2017 *)
    a[ n_] := If[ n < 0, (-1)^n / a[-n], SeriesCoefficient[ Product[1 - (1 - x)^(2 k - 1), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 27 2017 *)
    (2 Range[0, 20] - 1)!! (* Eric W. Weisstein, Jul 22 2017 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(sum(binomial(n-1,i)*binomial(n-i-1,j)*a(i)*a(j)*a(n-i-j-1),j,0,n-i-1),i,0,n-1); /* Vladimir Kruchinin, May 06 2020 */
  • PARI
    {a(n) = if( n<0, (-1)^n / a(-n), (2*n)! / n! / 2^n)}; /* Michael Somos, Sep 18 2014 */
    
  • PARI
    x='x+O('x^33); Vec(serlaplace((1-2*x)^(-1/2))) \\ Joerg Arndt, Apr 24 2011
    
  • Python
    from sympy import factorial2
    def a(n): return factorial2(2 * n - 1)
    print([a(n) for n in range(101)])  # Indranil Ghosh, Jul 22 2017
    
  • Sage
    [rising_factorial(n+1,n)/2^n for n in (0..15)] # Peter Luschny, Jun 26 2012
    

Formula

E.g.f.: 1 / sqrt(1 - 2*x).
D-finite with recurrence: a(n) = a(n-1)*(2*n-1) = (2*n)!/(n!*2^n) = A010050(n)/A000165(n).
a(n) ~ sqrt(2) * 2^n * (n/e)^n.
Rational part of numerator of Gamma(n+1/2): a(n) * sqrt(Pi) / 2^n = Gamma(n+1/2). - Yuriy Brun, Ewa Dominowska (brun(AT)mit.edu), May 12 2001
With interpolated zeros, the sequence has e.g.f. exp(x^2/2). - Paul Barry, Jun 27 2003
The Ramanujan polynomial psi(n+1, n) has value a(n). - Ralf Stephan, Apr 16 2004
a(n) = Sum_{k=0..n} (-2)^(n-k)*A048994(n, k). - Philippe Deléham, Oct 29 2005
Log(1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + ...) = x + 5/2*x^2 + 37/3*x^3 + 353/4*x^4 + 4081/5*x^5 + 55205/6*x^6 + ..., where [1, 5, 37, 353, 4081, 55205, ...] = A004208. - Philippe Deléham, Jun 20 2006
1/3 + 2/15 + 3/105 + ... = 1/2. [Jolley eq. 216]
Sum_{j=1..n} j/a(j+1) = (1 - 1/a(n+1))/2. [Jolley eq. 216]
1/1 + 1/3 + 2/15 + 6/105 + 24/945 + ... = Pi/2. - Gary W. Adamson, Dec 21 2006
a(n) = (1/sqrt(2*Pi))*Integral_{x>=0} x^n*exp(-x/2)/sqrt(x). - Paul Barry, Jan 28 2008
a(n) = A006882(2n-1). - R. J. Mathar, Jul 04 2009
G.f.: 1/(1-x-2x^2/(1-5x-12x^2/(1-9x-30x^2/(1-13x-56x^2/(1- ... (continued fraction). - Paul Barry, Sep 18 2009
a(n) = (-1)^n*subs({log(e)=1,x=0},coeff(simplify(series(e^(x*t-t^2/2),t,2*n+1)),t^(2*n))*(2*n)!). - Leonid Bedratyuk, Oct 31 2009
a(n) = 2^n*gamma(n+1/2)/gamma(1/2). - Jaume Oliver Lafont, Nov 09 2009
G.f.: 1/(1-x/(1-2x/(1-3x/(1-4x/(1-5x/(1- ...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
The g.f. of a(n+1) is 1/(1-3x/(1-2x/(1-5x/(1-4x/(1-7x/(1-6x/(1-.... (continued fraction). - Paul Barry, Dec 04 2009
a(n) = Sum_{i=1..n} binomial(n,i)*a(i-1)*a(n-i). - Vladimir Shevelev, Sep 30 2010
E.g.f.: A(x) = 1 - sqrt(1-2*x) satisfies the differential equation A'(x) - A'(x)*A(x) - 1 = 0. - Vladimir Kruchinin, Jan 17 2011
a(n) = A123023(2*n). - Michael Somos, Jul 24 2011
a(n) = (1/2)*Sum_{i=1..n} binomial(n+1,i)*a(i-1)*a(n-i). See link above. - Dennis P. Walsh, Dec 02 2011
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*Stirling_1(n+k,k) [Kauers and Ko].
a(n) = A035342(n, 1), n >= 1 (first column of triangle).
a(n) = A001497(n, 0) = A001498(n, n), first column, resp. main diagonal, of Bessel triangle.
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n and sum of top row terms of M^(n-1), where M = a variant of the (1,2) Pascal triangle (Cf. A029635) as the following production matrix:
1, 2, 0, 0, 0, ...
1, 3, 2, 0, 0, ...
1, 4, 5, 2, 0, ...
1, 5, 9, 7, 2, ...
...
For example, a(3) = 15 is the left term in top row of M^3: (15, 46, 36, 8) and a(4) = 105 = (15 + 46 + 36 + 8).
(End)
G.f.: A(x) = 1 + x/(W(0) - x); W(k) = 1 + x + x*2*k - x*(2*k + 3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2011
a(n) = Sum_{i=1..n} binomial(n,i-1)*a(i-1)*a(n-i). - Dennis P. Walsh, Dec 02 2011
a(n) = A009445(n) / A014481(n). - Reinhard Zumkeller, Dec 03 2011
a(n) = (-1)^n*Sum_{k=0..n} 2^(n-k)*s(n+1,k+1), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = (2*n)4! = Gauss_factorial(2*n,4) = Product{j=1..2*n, gcd(j,4)=1} j. - Peter Luschny, Oct 01 2012
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k - 1)/(1 - x*(2*k + 2)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1 + x/Q(0), where Q(k) = 1 + (2*k - 1)*x - 2*x*(k + 1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) - 1 + 2*x*(2*k + 2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(2*k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k + 1)/(4*k + 2 - 2*x*(2*k + 1)*(4*k + 3)/(x*(4*k + 3) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
a(n) = (2*n - 3)*a(n-2) + (2*n - 2)*a(n-1), n > 1. - Ivan N. Ianakiev, Jul 08 2013
G.f.: G(0), where G(k) = 1 - x*(k+1)/(x*(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 04 2013
a(n) = 2*a(n-1) + (2n-3)^2*a(n-2), a(0) = a(1) = 1. - Philippe Deléham, Oct 27 2013
G.f. of reciprocals: Sum_{n>=0} x^n/a(n) = 1F1(1; 1/2; x/2), confluent hypergeometric Function. - R. J. Mathar, Jul 25 2014
0 = a(n)*(+2*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Sep 18 2014
a(n) = (-1)^n / a(-n) = 2*a(n-1) + a(n-1)^2 / a(n-2) for all n in Z. - Michael Somos, Sep 18 2014
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 3.
The sequence b(n) = A087547(n), beginning [1, 4, 52, 608, 12624, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} b(n)/a(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End)
E.g.f of the sequence whose n-th element (n = 1,2,...) equals a(n-1) is 1-sqrt(1-2*x). - Stanislav Sykora, Jan 06 2017
Sum_{n >= 1} a(n)/(2*n-1)! = exp(1/2). - Daniel Suteu, Feb 06 2017
a(n) = A028338(n, 0), n >= 0. - Wolfdieter Lang, May 27 2017
a(n) = (Product_{k=0..n-2} binomial(2*(n-k),2))/n!. - Stefano Spezia, Nov 13 2018
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} C(n-1,i)*C(n-i-1,j)*a(i)*a(j)*a(n-i-j-1), a(0)=1, - Vladimir Kruchinin, May 06 2020
From Amiram Eldar, Jun 29 2020: (Start)
Sum_{n>=1} 1/a(n) = sqrt(e*Pi/2)*erf(1/sqrt(2)), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(Pi/(2*e))*erfi(1/sqrt(2)), where erfi is the imaginary error function. (End)
G.f. of reciprocals: R(x) = Sum_{n>=0} x^n/a(n) satisfies (1 + x)*R(x) = 1 + 2*x*R'(x). - Werner Schulte, Nov 04 2024

Extensions

Removed erroneous comments: neither the number of n X n binary matrices A such that A^2 = 0 nor the number of simple directed graphs on n vertices with no directed path of length two are counted by this sequence (for n = 3, both are 13). - Dan Drake, Jun 02 2009

A002061 Central polygonal numbers: a(n) = n^2 - n + 1.

Original entry on oeis.org

1, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, 241, 273, 307, 343, 381, 421, 463, 507, 553, 601, 651, 703, 757, 813, 871, 931, 993, 1057, 1123, 1191, 1261, 1333, 1407, 1483, 1561, 1641, 1723, 1807, 1893, 1981, 2071, 2163, 2257, 2353, 2451, 2551, 2653
Offset: 0

Keywords

Comments

These are Hogben's central polygonal numbers denoted by the symbol
...2....
....P...
...2.n..
(P with three attachments).
Also the maximal number of 1's that an n X n invertible {0,1} matrix can have. (See Halmos for proof.) - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jul 07 2001
Maximal number of interior regions formed by n intersecting circles, for n >= 1. - Amarnath Murthy, Jul 07 2001
The terms are the smallest of n consecutive odd numbers whose sum is n^3: 1, 3 + 5 = 8 = 2^3, 7 + 9 + 11 = 27 = 3^3, etc. - Amarnath Murthy, May 19 2001
(n*a(n+1)+1)/(n^2+1) is the smallest integer of the form (n*k+1)/(n^2+1). - Benoit Cloitre, May 02 2002
For n >= 3, a(n) is also the number of cycles in the wheel graph W(n) of order n. - Sharon Sela (sharonsela(AT)hotmail.com), May 17 2002
Let b(k) be defined as follows: b(1) = 1 and b(k+1) > b(k) is the smallest integer such that Sum_{i=b(k)..b(k+1)} 1/sqrt(i) > 2; then b(n) = a(n) for n > 0. - Benoit Cloitre, Aug 23 2002
Drop the first three terms. Then n*a(n) + 1 = (n+1)^3. E.g., 7*1 + 1 = 8 = 2^3, 13*2 + 1 = 27 = 3^3, 21*3 + 1 = 64 = 4^3, etc. - Amarnath Murthy, Oct 20 2002
Arithmetic mean of next 2n - 1 numbers. - Amarnath Murthy, Feb 16 2004
The n-th term of an arithmetic progression with first term 1 and common difference n: a(1) = 1 -> 1, 2, 3, 4, 5, ...; a(2) = 3 -> 1, 3, ...; a(3) = 7 -> 1, 4, 7, ...; a(4) = 13 -> 1, 5, 9, 13, ... - Amarnath Murthy, Mar 25 2004
Number of walks of length 3 between any two distinct vertices of the complete graph K_{n+1} (n >= 1). Example: a(2) = 3 because in the complete graph ABC we have the following walks of length 3 between A and B: ABAB, ACAB and ABCB. - Emeric Deutsch, Apr 01 2004
Narayana transform of [1, 2, 0, 0, 0, ...] = [1, 3, 7, 13, 21, ...]. Let M = the infinite lower triangular matrix of A001263 and let V = the Vector [1, 2, 0, 0, 0, ...]. Then A002061 starting (1, 3, 7, ...) = M * V. - Gary W. Adamson, Apr 25 2006
The sequence 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, ... is the trajectory of 3 under repeated application of the map n -> n + 2 * square excess of n, cf. A094765.
Also n^3 mod (n^2+1). - Zak Seidov, Aug 31 2006
Also, omitting the first 1, the main diagonal of A081344. - Zak Seidov, Oct 05 2006
Ignoring the first ones, these are rectangular parallelepipeds with integer dimensions that have integer interior diagonals. Using Pythagoras: sqrt(a^2 + b^2 + c^2) = d, an integer; then this sequence: sqrt(n^2 + (n+1)^2 + (n(n+1))^2) = 2T_n + 1 is the first and most simple example. Problem: Are there any integer diagonals which do not satisfy the following general formula? sqrt((k*n)^2 + (k*(n+(2*m+1)))^2 + (k*(n*(n+(2*m+1)) + 4*T_m))^2) = k*d where m >= 0, k >= 1, and T is a triangular number. - Marco Matosic, Nov 10 2006
Numbers n such that a(n) is prime are listed in A055494. Prime a(n) are listed in A002383. All terms are odd. Prime factors of a(n) are listed in A007645. 3 divides a(3*k-1), 7 divides a(7*k-4) and a(7*k-2), 7^2 divides a(7^2*k-18) and a(7^2*k+19), 7^3 divides a(7^3*k-18) and a(7^3*k+19), 7^4 divides a(7^4*k+1048) and a(7^4*k-1047), 7^5 divides a(7^5*k+1354) and a(7^5*k-1353), 13 divides a(13*k-9) and a(13*k-3), 13^2 divides a(13^2*k+23) and a(13^2*k-22), 13^3 divides a(13^3*k+1037) and a(13^3*k-1036). - Alexander Adamchuk, Jan 25 2007
Complement of A135668. - Kieren MacMillan, Dec 16 2007
From William A. Tedeschi, Feb 29 2008: (Start)
Numbers (sorted) on the main diagonal of a 2n X 2n spiral. For example, when n=2:
.
7---8---9--10
| |
6 1---2 11
| | |
5---4---3 12
|
16--15--14--13
.
Cf. A137928. (End)
a(n) = AlexanderPolynomial[n] defined as Det[Transpose[S]-n S] where S is Seifert matrix {{-1, 1}, {0, -1}}. - Artur Jasinski, Mar 31 2008
Starting (1, 3, 7, 13, 21, ...) = binomial transform of [1, 2, 2, 0, 0, 0]; example: a(4) = 13 = (1, 3, 3, 1) dot (1, 2, 2, 0) = (1 + 6 + 6 + 0). - Gary W. Adamson, May 10 2008
Starting (1, 3, 7, 13, ...) = triangle A158821 * [1, 2, 3, ...]. - Gary W. Adamson, Mar 28 2009
Starting with offset 1 = triangle A128229 * [1,2,3,...]. - Gary W. Adamson, Mar 26 2009
a(n) = k such that floor((1/2)*(1 + sqrt(4*k-3))) + k = (n^2+1), that is A000037(a(n)) = A002522(n) = n^2 + 1, for n >= 1. - Jaroslav Krizek, Jun 21 2009
For n > 0: a(n) = A170950(A002522(n-1)), A170950(a(n)) = A174114(n), A170949(a(n)) = A002522(n-1). - Reinhard Zumkeller, Mar 08 2010
From Emeric Deutsch, Sep 23 2010: (Start)
a(n) is also the Wiener index of the fan graph F(n). The fan graph F(n) is defined as the graph obtained by joining each node of an n-node path graph with an additional node. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. The Wiener polynomial of the graph F(n) is (1/2)t[(n-1)(n-2)t + 2(2n-1)]. Example: a(2)=3 because the corresponding fan graph is a cycle on 3 nodes (a triangle), having distances 1, 1, and 1.
(End)
For all elements k = n^2 - n + 1 of the sequence, sqrt(4*(k-1)+1) is an integer because 4*(k-1) + 1 = (2*n-1)^2 is a perfect square. Building the intersection of this sequence with A000225, k may in addition be of the form k = 2^x - 1, which happens only for k = 1, 3, 7, 31, and 8191. [Proof: Still 4*(k-1)+1 = 2^(x+2) - 7 must be a perfect square, which has the finite number of solutions provided by A060728: x = 1, 2, 3, 5, or 13.] In other words, the sequence A038198 defines all elements of the form 2^x - 1 in this sequence. For example k = 31 = 6*6 - 6 + 1; sqrt((31-1)*4+1) = sqrt(121) = 11 = A038198(4). - Alzhekeyev Ascar M, Jun 01 2011
a(n) such that A002522(n-1) * A002522(n) = A002522(a(n)) where A002522(n) = n^2 + 1. - Michel Lagneau, Feb 10 2012
Left edge of the triangle in A214661: a(n) = A214661(n, 1), for n > 0. - Reinhard Zumkeller, Jul 25 2012
a(n) = A215630(n, 1), for n > 0; a(n) = A215631(n-1, 1), for n > 1. - Reinhard Zumkeller, Nov 11 2012
Sum_{n > 0} arccot(a(n)) = Pi/2. - Franz Vrabec, Dec 02 2012
If you draw a triangle with one side of unit length and one side of length n, with an angle of Pi/3 radians between them, then the length of the third side of the triangle will be the square root of a(n). - Elliott Line, Jan 24 2013
a(n+1) is the number j such that j^2 = j + m + sqrt(j*m), with corresponding number m given by A100019(n). Also: sqrt(j*m) = A027444(n) = n * a(n+1). - Richard R. Forberg, Sep 03 2013
Let p(x) the interpolating polynomial of degree n-1 passing through the n points (n,n) and (1,1), (2,1), ..., (n-1,1). Then p(n+1) = a(n). - Giovanni Resta, Feb 09 2014
The number of square roots >= sqrt(n) and < n+1 (n >= 0) gives essentially the same sequence, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, ... . - Michael G. Kaarhus, May 21 2014
For n > 1: a(n) is the maximum total number of queens that can coexist without attacking each other on an [n+1] X [n+1] chessboard. Specifically, this will be a lone queen of one color placed in any position on the perimeter of the board, facing an opponent's "army" of size a(n)-1 == A002378(n-1). - Bob Selcoe, Feb 07 2015
a(n+1) is, for n >= 1, the number of points as well as the number of lines of a finite projective plane of order n (cf. Hughes and Piper, 1973, Theorem 3.5., pp. 79-80). For n = 3, a(4) = 13, see the 'Finite example' in the Wikipedia link, section 2.3, for the point-line matrix. - Wolfdieter Lang, Nov 20 2015
Denominators of the solution to the generalization of the Feynman triangle problem. If each vertex of a triangle is joined to the point (1/p) along the opposite side (measured say clockwise), then the area of the inner triangle formed by these lines is equal to (p - 2)^2/(p^2 - p + 1) times the area of the original triangle, p > 2. For example, when p = 3, the ratio of the areas is 1/7. The numerators of the ratio of the areas is given by A000290 with an offset of 2. [Cook & Wood, 2004.] - Joe Marasco, Feb 20 2017
n^2 equal triangular tiles with side lengths 1 X 1 X 1 may be put together to form an n X n X n triangle. For n>=2 a(n-1) is the number of different 2 X 2 X 2 triangles being contained. - Heinrich Ludwig, Mar 13 2017
For n >= 0, the continued fraction [n, n+1, n+2] = (n^3 + 3n^2 + 4n + 2)/(n^2 + 3n + 3) = A034262(n+1)/a(n+2) = n + (n+2)/a(n+2); e.g., [2, 3, 4] = A034262(3)/a(4) = 30/13 = 2 + 4/13. - Rick L. Shepherd, Apr 06 2017
Starting with b(1) = 1 and not allowing the digit 0, let b(n) = smallest nonnegative integer not yet in the sequence such that the last digit of b(n-1) plus the first digit of b(n) is equal to k for k = 1, ..., 9. This defines 9 finite sequences, each of length equal to a(k), k = 1, ..., 9. (See A289283-A289287 for the cases k = 5..9.) For k = 10, the sequence is infinite (A289288). For example, for k = 4, b(n) = 1,3,11,31,32,2,21,33,12,22,23,13,14. These terms can be ordered in the following array of size k*(k-1)+1:
1 2 3
21 22 23
31 32 33
11 12 13 14
.
The sequence ends with the term 1k, which lies outside the rectangular array and gives the term +1 (see link).- Enrique Navarrete, Jul 02 2017
The central polygonal numbers are the delimiters (in parenthesis below) when you write the natural numbers in groups of odd size 2*n+1 starting with the group {2} of size 1: (1) 2 (3) 4,5,6 (7) 8,9,10,11,12 (13) 14,15,16,17,18,19,20 (21) 22,23,24,25,26,27,28,29,30 (31) 32,33,34,35,36,37,38,39,40,41,42 (43) ... - Enrique Navarrete, Jul 11 2017
Also the number of (non-null) connected induced subgraphs in the n-cycle graph. - Eric W. Weisstein, Aug 09 2017
Since (n+1)^2 - (n+1) + 1 = n^2 + n + 1 then from 7 onwards these are also exactly the numbers that are represented as 111 in all number bases: 111(2)=7, 111(3)=13, ... - Ron Knott, Nov 14 2017
Number of binary 2 X (n-1) matrices such that each row and column has at most one 1. - Dmitry Kamenetsky, Jan 20 2018
Observed to be the squares visited by bishop moves on a spirally numbered board and moving to the lowest available unvisited square at each step, beginning at the second term (cf. A316667). It should be noted that the bishop will only travel to squares along the first diagonal of the spiral. - Benjamin Knight, Jan 30 2019
From Ed Pegg Jr, May 16 2019: (Start)
Bound for n-subset coverings. Values in A138077 covered by difference sets.
C(7,3,2), {1,2,4}
C(13,4,2), {0,1,3,9}
C(21,5,2), {3,6,7,12,14}
C(31,6,2), {1,5,11,24,25,27}
C(43,7,2), existence unresolved
C(57,8,2), {0,1,6,15,22,26,45,55}
Next unresolved cases are C(111,11,2) and C(157,13,2). (End)
"In the range we explored carefully, the optimal packings were substantially irregular only for n of the form n = k(k+1)+1, k = 3, 4, 5, 6, 7, i.e., for n = 13, 21, 31, 43, and 57." (cited from Lubachevsky, Graham link, Introduction). - Rainer Rosenthal, May 27 2020
From Bernard Schott, Dec 31 2020: (Start)
For n >= 1, a(n) is the number of solutions x in the interval 1 <= x <= n of the equation x^2 - [x^2] = (x - [x])^2, where [x] = floor(x). For n = 3, the a(3) = 7 solutions in the interval [1, 3] are 1, 3/2, 2, 9/4, 5/2, 11/4 and 3.
This sequence is the answer to the 4th problem proposed during the 20th British Mathematical Olympiad in 1984 (see link B.M.O 1984. and Gardiner reference). (End)
Called "Hogben numbers" after the British zoologist, statistician and writer Lancelot Thomas Hogben (1895-1975). - Amiram Eldar, Jun 24 2021
Minimum Wiener index of 2-degenerate graphs with n+1 vertices (n>0). A maximal 2-degenerate graph can be constructed from a 2-clique by iteratively adding a new 2-leaf (vertex of degree 2) adjacent to two existing vertices. The extremal graphs are maximal 2-degenerate graphs with diameter at most 2. - Allan Bickle, Oct 14 2022
a(n) is the number of parking functions of size n avoiding the patterns 123, 213, and 312. - Lara Pudwell, Apr 10 2023
Repeated iteration of a(k) starting with k=2 produces Sylvester's sequence, i.e., A000058(n) = a^n(2), where a^n is the n-th iterate of a(k). - Curtis Bechtel, Apr 04 2024
a(n) is the maximum number of triangles that can be traversed by starting from a triangle and moving to adjacent triangles via an edge, without revisiting any triangle, in an n X n X n equilateral triangular grid made up of n^2 unit equilateral triangles. - Kiran Ananthpur Bacche, Jan 16 2025

Examples

			G.f. = 1 + x + 3*x^2 + 7*x^3 + 13*x^4 + 21*x^5 + 31*x^6 + 43*x^7 + ...
		

References

  • Archimedeans Problems Drive, Eureka, 22 (1959), 15.
  • Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 1 of the British Mathematical Olympiad 2007, page 160.
  • Anthony Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Problem 4 pp. 64 and 173 (1984).
  • Paul R. Halmos, Linear Algebra Problem Book, MAA, 1995, pp. 75-6, 242-4.
  • Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 87.
  • Daniel R. Hughes and Frederick Charles Piper, Projective Planes, Springer, 1973.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A010000 (minimum Weiner index of 3-degenerate graphs).

Programs

  • GAP
    List([0..50], n->n^2-n+1); # Muniru A Asiru, May 27 2018
  • Haskell
    a002061 n = n * (n - 1) + 1  -- Reinhard Zumkeller, Dec 18 2013
    
  • Magma
    [ n^2 - n + 1 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 12 2014
    
  • Maple
    A002061 := proc(n)
        numtheory[cyclotomic](6,n) ;
    end proc:
    seq(A002061(n), n=0..20); # R. J. Mathar, Feb 07 2014
  • Mathematica
    FoldList[#1 + #2 &, 1, 2 Range[0, 50]] (* Robert G. Wilson v, Feb 02 2011 *)
    LinearRecurrence[{3, -3, 1}, {1, 1, 3}, 60] (* Harvey P. Dale, May 25 2011 *)
    Table[n^2 - n + 1, {n, 0, 50}] (* Wesley Ivan Hurt, Jun 12 2014 *)
    CoefficientList[Series[(1 - 2x + 3x^2)/(1 - x)^3, {x, 0, 52}], x] (* Robert G. Wilson v, Feb 18 2018 *)
    Cyclotomic[6, Range[0, 100]] (* Paolo Xausa, Feb 09 2024 *)
  • Maxima
    makelist(n^2 - n + 1,n,0,55); /* Martin Ettl, Oct 16 2012 */
    
  • PARI
    a(n) = n^2 - n + 1
    

Formula

G.f.: (1 - 2*x + 3*x^2)/(1-x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = -(n-5)*a(n-1) + (n-2)*a(n-2).
a(n) = Phi_6(n) = Phi_3(n-1), where Phi_k is the k-th cyclotomic polynomial.
a(1-n) = a(n). - Michael Somos, Sep 04 2006
a(n) = a(n-1) + 2*(n-1) = 2*a(n-1) - a(n-2) + 2 = 1+A002378(n-1) = 2*A000124(n-1) - 1. - Henry Bottomley, Oct 02 2000 [Corrected by N. J. A. Sloane, Jul 18 2010]
a(n) = A000217(n) + A000217(n-2) (sum of two triangular numbers).
From Paul Barry, Mar 13 2003: (Start)
x*(1+x^2)/(1-x)^3 is g.f. for 0, 1, 3, 7, 13, ...
a(n) = 2*C(n, 2) + C(n-1, 0).
E.g.f.: (1+x^2)*exp(x). (End)
a(n) = ceiling((n-1/2)^2). - Benoit Cloitre, Apr 16 2003. [Hence the terms are about midway between successive squares and so (except for 1) are not squares. - N. J. A. Sloane, Nov 01 2005]
a(n) = 1 + Sum_{j=0..n-1} (2*j). - Xavier Acloque, Oct 08 2003
a(n) = floor(t(n^2)/t(n)), where t(n) = A000217(n). - Jon Perry, Feb 14 2004
a(n) = leftmost term in M^(n-1) * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 0 1 2 / 0 0 1]. E.g., a(6) = 31 since M^5 * [1 1 1] = [31 11 1]. - Gary W. Adamson, Nov 11 2004
a(n+1) = n^2 + n + 1. a(n+1)*a(n) = (n^6-1)/(n^2-1) = n^4 + n^2 + 1 = a(n^2+1) (a product of two consecutive numbers from this sequence belongs to this sequence). (a(n+1) + a(n))/2 = n^2 + 1. (a(n+1) - a(n))/2 = n. a((a(n+1) + a(n))/2) = a(n+1)*a(n). - Alexander Adamchuk, Apr 13 2006
a(n+1) is the numerator of ((n + 1)! + (n - 1)!)/ n!. - Artur Jasinski, Jan 09 2007
a(n) = A132111(n-1, 1), for n > 1. - Reinhard Zumkeller, Aug 10 2007
a(n) = Det[Transpose[{{-1, 1}, {0, -1}}] - n {{-1, 1}, {0, -1}}]. - Artur Jasinski, Mar 31 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n >= 3. - Jaume Oliver Lafont, Dec 02 2008
a(n) = A176271(n,1) for n > 0. - Reinhard Zumkeller, Apr 13 2010
a(n) == 3 (mod n+1). - Bruno Berselli, Jun 03 2010
a(n) = (n-1)^2 + (n-1) + 1 = 111 read in base n-1 (for n > 2). - Jason Kimberley, Oct 18 2011
a(n) = A228643(n, 1), for n > 0. - Reinhard Zumkeller, Aug 29 2013
a(n) = sqrt(A058031(n)). - Richard R. Forberg, Sep 03 2013
G.f.: 1 / (1 - x / (1 - 2*x / (1 + x / (1 - 2*x / (1 + x))))). - Michael Somos, Apr 03 2014
a(n) = A243201(n - 1) / A003215(n - 1), n > 0. - Mathew Englander, Jun 03 2014
For n >= 2, a(n) = ceiling(4/(Sum_{k = A000217(n-1)..A000217(n) - 1}, 1/k)). - Richard R. Forberg, Aug 17 2014
A256188(a(n)) = 1. - Reinhard Zumkeller, Mar 26 2015
Sum_{n>=0} 1/a(n) = 1 + Pi*tanh(Pi*sqrt(3)/2)/sqrt(3) = 2.79814728056269018... . - Vaclav Kotesovec, Apr 10 2016
a(n) = A101321(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A000217(n-1) + A000124(n-1), n > 0. - Torlach Rush, Aug 06 2018
Sum_{n>=1} arctan(1/a(n)) = Pi/2. - Amiram Eldar, Nov 01 2020
Sum_{n=1..M} arctan(1/a(n)) = arctan(M). - Lee A. Newberg, May 08 2024
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*sech(sqrt(3)*Pi/2).
Product_{n>=2} (1 - 1/a(n)) = Pi*sech(sqrt(3)*Pi/2). (End)
For n > 1, sqrt(a(n)+sqrt(a(n)-sqrt(a(n)+sqrt(a(n)- ...)))) = n. - Diego Rattaggi, Apr 17 2021
a(n) = (1 + (n-1)^4 + n^4) / (1 + (n-1)^2 + n^2) [see link B.M.O. 2007 and Steve Dinh reference]. - Bernard Schott, Dec 27 2021

Extensions

Partially edited by Joerg Arndt, Mar 11 2010
Partially edited by Bruno Berselli, Dec 19 2013

A008290 Triangle T(n,k) of rencontres numbers (number of permutations of n elements with k fixed points).

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 9, 8, 6, 0, 1, 44, 45, 20, 10, 0, 1, 265, 264, 135, 40, 15, 0, 1, 1854, 1855, 924, 315, 70, 21, 0, 1, 14833, 14832, 7420, 2464, 630, 112, 28, 0, 1, 133496, 133497, 66744, 22260, 5544, 1134, 168, 36, 0, 1, 1334961, 1334960, 667485, 222480, 55650, 11088, 1890, 240, 45, 0, 1
Offset: 0

Keywords

Comments

This is a binomial convolution triangle (Sheffer triangle) of the Appell type: (exp(-x)/(1-x),x), i.e., the e.g.f. of column k is (exp(-x)/(1-x))*(x^k/k!). See the e.g.f. given by V. Jovovic below. - Wolfdieter Lang, Jan 21 2008
The formula T(n,k) = binomial(n,k)*A000166(n-k), with the derangements numbers (subfactorials) A000166 (see also the Charalambides reference) shows the Appell type of this triangle. - Wolfdieter Lang, Jan 21 2008
T(n,k) is the number of permutations of {1,2,...,n} having k pairs of consecutive right-to-left minima (0 is considered a right-to-left minimum for each permutation). Example: T(4,2)=6 because we have 1243, 1423, 4123, 1324, 3124 and 2134; for example, 1324 has right-to-left minima in positions 0-1,3-4 and 2134 has right-to-left minima in positions 0,2-3-4, the consecutive ones being joined by "-". - Emeric Deutsch, Mar 29 2008
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for the sequence aC(0,1). - Tom Copeland, Sep 10 2008
A refinement of this triangle is given by A036039. - Tom Copeland, Nov 06 2012
This triangle equals (A211229(2*n,2*k)) n,k >= 0. - Peter Bala, Dec 17 2014

Examples

			exp((y-1)*x)/(1-x) = 1 + y*x + (1/2!)*(1+y^2)*x^2 + (1/3!)*(2 + 3*y + y^3)*x^3 + (1/4!)*(9 + 8*y + 6*y^2 + y^4)*x^4 + (1/5!)*(44 + 45*y + 20*y^2 + 10*y^3 + y^5)*x^5 + ...
Triangle begins:
       1
       0      1
       1      0     1
       2      3     0     1
       9      8     6     0    1
      44     45    20    10    0    1
     265    264   135    40   15    0   1
    1854   1855   924   315   70   21   0  1
   14833  14832  7420  2464  630  112  28  0 1
  133496 133497 66744 22260 5544 1134 168 36 0 1
...
From _Peter Bala_, Feb 13 2017: (Start)
The infinitesimal generator has integer entries given by binomial(n,k)*(n-k-1)! for n >= 2 and 0 <= k <= n-2 and begins
   0
   0  0
   1  0  0
   2  3  0  0
   6  8  6  0 0
  24 30 20 10 0 0
...
It is essentially A238363 (unsigned and omitting the main diagonal), A211603 (with different offset) and appears to be A092271, again without the main diagonal. (End)
		

References

  • Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 173, Table 5.2 (without row n=0 and column k=0).
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 194.
  • Arnold Kaufmann, Introduction à la combinatorique en vue des applications, Dunod, Paris, 1968. See p. 92.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.

Crossrefs

Mirror of triangle A098825.
Cf. A080955.
Cf. A000012, A000142 (row sums), A000354.
Cf. A170942. Sub-triangle of A211229.
T(2n,n) gives A281262.

Programs

  • Haskell
    a008290 n k = a008290_tabl !! n !! k
    a008290_row n = a008290_tabl !! n
    a008290_tabl = map reverse a098825_tabl
    -- Reinhard Zumkeller, Dec 16 2013
  • Maple
    T:= proc(n,k) T(n, k):= `if`(k=0, `if`(n<2, 1-n, (n-1)*
          (T(n-1, 0)+T(n-2, 0))), binomial(n, k)*T(n-k, 0))
        end:
    seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Mar 15 2013
  • Mathematica
    a[0] = 1; a[1] = 0; a[n_] := Round[n!/E] /; n >= 1 size = 8; Table[Binomial[n, k]a[n - k], {n, 0, size}, {k, 0, n}] // TableForm (* Harlan J. Brothers, Mar 19 2007 *)
    T[n_, k_] := Subfactorial[n-k]*Binomial[n, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2017 *)
    T[n_, k_] := If[n<1, Boole[n==0 && k==0], T[n, k] = T[n-1, k-1] + T[n-1, k]*(n-1-k) + T[n-1, k+1]*(k+1)]; (* Michael Somos, Sep 13 2024 *)
    T[0, 0]:=1; T[n_, 0]:=T[n, 0]=n  T[n-1, 0]+(-1)^n; T[n_, k_]:=T[n, k]=n/k T[n-1, k-1];
    Flatten@Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Oliver Seipel, Nov 26 2024 *)
  • PARI
    {T(n, k) = if(k<0 || k>n, 0, n!/k! * sum(i=0, n-k, (-1)^i/i!))}; /* Michael Somos, Apr 26 2000 */
    

Formula

T(n, k) = T(n-1, k)*n + binomial(n, k)*(-1)^(n-k) = T(n, k-1)/k + binomial(n, k)*(-1)^(n-k)/(n-k+1) = T(n-1, k-1)*n/k = T(n-k, 0)*binomial(n, k) = A000166(n-k)*binomial(n,k) [with T(0, 0) = 1]; so T(n, n) = 1, T(n, n-1) = 0, T(n, n-2) = n*(n-1)/2 for n >= 0.
Sum_{k=0..n} T(n, k) = Sum_{k=0..n} k * T(n, k) = n! for all n > 0, n, k integers. - Wouter Meeussen, May 29 2001
From Vladeta Jovovic, Aug 12 2002: (Start)
O.g.f. for k-th column: (1/k!)*Sum_{i>=k} i!*x^i/(1+x)^(i+1).
O.g.f. for k-th row: k!*Sum_{i=0..k} (-1)^i/i!*(1-x)^i. (End)
E.g.f.: exp((y-1)*x)/(1-x). - Vladeta Jovovic, Aug 18 2002
E.g.f. for number of permutations with exactly k fixed points is x^k/(k!*exp(x)*(1-x)). - Vladeta Jovovic, Aug 25 2002
Sum_{k=0..n} T(n, k)*x^k is the permanent of the n X n matrix with x's on the diagonal and 1's elsewhere; for x = 0, 1, 2, 3, 4, 5, 6 see A000166, A000142, A000522, A010842, A053486, A053487, A080954. - Philippe Deléham, Dec 12 2003; for x = 1+i see A009551 and A009102. - John M. Campbell, Oct 11 2011
T(n, k) = Sum_{j=0..n} A008290(n, j)*k^(n-j) is the permanent of the n X n matrix with 1's on the diagonal and k's elsewhere; for k = 0, 1, 2 see A000012, A000142, A000354. - Philippe Deléham, Dec 13 2003
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*binomial(j,k)*n!/j!. - Paul Barry, May 25 2006
T(n,k) = (n!/k!)*Sum_{j=0..n-k} ((-1)^j)/j!, 0 <= k <= n. From the Appell type of the triangle and the subfactorial formula.
T(n,0) = n*Sum_{j=0..n-1} (j/(j+1))*T(n-1,j), T(0,0)=1. From the z-sequence of this Sheffer triangle z(j)=j/(j+1) with e.g.f. (1-exp(x)*(1-x))/x. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
T(n,k) = (n/k)*T(n-1,k-1) for k >= 1. See above. From the a-sequence of this Sheffer triangle a(0)=1, a(n)=0, n >= 1 with e.g.f. 1. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
From Henk P. J. van Wijk, Oct 29 2012: (Start)
T(n,k) = T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k=0 and
T(n,k) = T(n-1,k-1) + T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k>=1.
(End)
T(n,k) = A098825(n,n-k). - Reinhard Zumkeller, Dec 16 2013
Sum_{k=0..n} k^2 * T(n, k) = 2*n! if n > 1. - Michael Somos, Jun 06 2017
From Tom Copeland, Jul 26 2017: (Start)
The lowering and raising operators of this Appell sequence of polynomials P(n,x) are L = d/dx and R = x + d/dL log[exp(-L)/(1-L)] = x-1 + 1/(1-L) = x + L + L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L)^(-1) exp(-L) x^n = (1+L+L^2+...)(x-1)^n = n! Sum_{k=0..n} (x-1)^k / k!.
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
For more on the infinitesimal generator, noted by Bala below, see A238385. (End)
Sum_{k=0..n} k^m * T(n,k) = A000110(m)*n! if n >= m. - Zhujun Zhang, May 24 2019
Sum_{k=0..n} (k+1) * T(n,k) = A098558(n). - Alois P. Heinz, Mar 11 2022
From Alois P. Heinz, May 20 2023: (Start)
Sum_{k=0..n} (-1)^k * T(n,k) = A000023(n).
Sum_{k=0..n} (-1)^k * k * T(n,k) = A335111(n). (End)
T(n,k) = A145224(n,k)+A145225(n,k), refined by even and odd perms. - R. J. Mathar, Jul 06 2023

Extensions

Comments and more terms from Michael Somos, Apr 26 2000 and Christian G. Bower, Apr 26 2000

A094587 Triangle of permutation coefficients arranged with 1's on the diagonal. Also, triangle of permutations on n letters with exactly k+1 cycles and with the first k+1 letters in separate cycles.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24, 12, 4, 1, 120, 120, 60, 20, 5, 1, 720, 720, 360, 120, 30, 6, 1, 5040, 5040, 2520, 840, 210, 42, 7, 1, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 3628800
Offset: 0

Author

Paul Barry, May 13 2004

Keywords

Comments

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016
Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007
From Tom Copeland, Nov 01 2007: (Start)
T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = Sum_{j=0..n} (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x), formally
5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)
6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)
7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
From Peter Bala, Jul 10 2008: (Start)
This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0.....................1...............2.......3......4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.
Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.
When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).
We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).
An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).
Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)
(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009
Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019
The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009
T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011
T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011
T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012
The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013
For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016
See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016
Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016
The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017
Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017
If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

Examples

			Rows begin {1}, {1,1}, {2,2,1}, {6,6,3,1}, ...
For n=3 and k=1, T(3,1)=6 since there are exactly 6 permutations of {1,2,3,4} with exactly 2 cycles and with 1 and 2 in separate cycles. The permutations are (1)(2 3 4), (1)(2 4 3), (1 3)(2 4), (1 4)(2 3), (1 3 4)(2), and (1 4 3)(2). - _Dennis P. Walsh_, Jan 24 2011
Triangle begins:
     1,
     1,    1,
     2,    2,    1,
     6,    6,    3,    1,
    24,   24,   12,    4,    1,
   120,  120,   60,   20,    5,    1,
   720,  720,  360,  120,   30,    6,    1,
  5040, 5040, 2520,  840,  210,   42,    7,    1
The production matrix is:
      1,     1,
      1,     1,     1,
      2,     2,     1,    1,
      6,     6,     3,    1,    1,
     24,    24,    12,    4,    1,   1,
    120,   120,    60,   20,    5,   1,   1,
    720,   720,   360,  120,   30,   6,   1,   1,
   5040,  5040,  2520,  840,  210,  42,   7,   1,   1,
  40320, 40320, 20160, 6720, 1680, 336,  56,   8,   1,   1
which is the exponential Riordan array A094587, or [1/(1-x),x], with an extra superdiagonal of 1's.
Inverse begins:
   1,
  -1,  1,
   0, -2,  1,
   0,  0, -3,  1,
   0,  0,  0, -4,  1,
   0,  0,  0,  0, -5,  1,
   0,  0,  0,  0,  0, -6,  1,
   0,  0,  0,  0,  0,  0, -7,  1
		

Programs

  • Haskell
    a094587 n k = a094587_tabl !! n !! k
    a094587_row n = a094587_tabl !! n
    a094587_tabl = map fst $ iterate f ([1], 1)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
    
  • Maple
    T := proc(n, m): n!/m! end: seq(seq(T(n, m), m=0..n), n=0..9);  # Johannes W. Meijer, Oct 07 2009, revised Nov 25 2012
    # Alternative: Note that if you leave out 'abs' you get A021009.
    T := proc(n, k) option remember; if n = 0 and k = 0 then 1 elif k < 0 or k > n then 0 else abs((n + k)*T(n-1, k) - T(n-1, k-1)) fi end: #  Peter Luschny, Dec 30 2021
  • Mathematica
    Flatten[Table[Table[n!/k!, {k,0,n}], {n,0,10}]] (* Geoffrey Critzer, Dec 11 2011 *)
  • Sage
    def A094587_row(n): return (factorial(n)*exp(x).taylor(x,0,n)).list()
    for n in (0..7): print(A094587_row(n)) # Peter Luschny, Sep 28 2017

Formula

T(n, k) = n!/k! if n >= k >= 0, otherwise 0.
T(n, k) = Sum_{i=k..n} |S1(n+1, i+1)*S2(i, k)| * (-1)^i, with S1, S2 the Stirling numbers.
T(n,k) = (n-k)*T(n-1,k) + T(n-1,k-1). E.g.f.: exp(x*y)/(1-y) = 1 + (1+x)*y + (2+2*x+x^2)*y^2/2! + (6+6*x+3*x^2+x^3)*y^3/3!+ ... . - Peter Bala, Jul 10 2008
A094587 = 1 / ((-1)*A129184 * A127648 + I), I = Identity matrix. - Gary W. Adamson, May 03 2009
From Johannes W. Meijer, Oct 07 2009: (Start)
The o.g.f. of right hand column k is Gf(z;k) = (k-1)!/(1-z)^k, k => 1.
The recurrence relations of the right hand columns lead to Pascal's triangle A007318. (End)
Let f(x) = (1/x)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+1)*R(n,x)-x*R'(n,x). Cf. A132159. - Peter Bala, Oct 28 2011
A padded shifted version of this lower triangular matrix with zeros in the first column and row except for a one in the diagonal position is given by integral(t=0 to t=infinity) exp[-t(I-P)] = 1/(I-P) = I + P^2 + P^3 + ... where P is the infinitesimal generator matrix A218234 and I the identity matrix. The non-padded version is given by P replaced by A132440. - Tom Copeland, Oct 25 2012
From Peter Bala, Aug 28 2013: (Start)
The row polynomials R(n,x) form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x).
Let P(n,x) = Product_{k=0..n-1} (x + k) denote the rising factorial polynomial sequence with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (6, 6, 3, 1) so P(3,x + 1) = (x + 1)*(x + 2)*(x + 3) = 6 + 6*x + 3*x*(x + 1) + x*(x + 1)*(x + 2). (End)
From Tom Copeland, Apr 21 & 26, and Aug 13 2014: (Start)
T-I = M = -A021009*A132440*A021009 with e.g.f. y*exp(x*y)/(1-y). Cf. A132440. Dividing the n-th row of M by n generates the (n-1)th row of T.
T = 1/(I - A132440) = {2*I - exp[(A238385-I)]}^(-1) = unsigned exp[(I-A238385)] = exp[A000670(.)*(A238385-I)] = , umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)/(1-y), so the row polynomials form an Appell sequence with lowering operator d/dx and raising operator x + 1/(1-D).
With L(n,m,x)= Laguerre polynomials of order m, the row polynomials are (-1)^n*n!*L(n,-1-n,x) = (-1)^n*(-1!/(-1-n)!)*K(-n,-1-n+1,x) = n!* K(-n,-n,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-1-n,-:xD:) = (-1)^n*x^(n+1)*:Dx:^n*x^(-1-n) = (-1)^n*x*:xD:^n*x^(-1) = (-1)^n*n!*binomial(xD-1,n) = n!*K(-n,-n,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706 and A132159.
The n-th row of signed M has the coefficients of d[(-:xD:)^n]/d(:Dx:)= f[d/d(-:xD:)](-:xD:)^n with f(y)=y/(y-1), :Dx:^n= n!L(n,0,-:xD:), and (-:xD:)^n = n!L(n,0,:Dx:). M has the coefficients of [D/(1-D)]x^n. (End)
From Tom Copeland, Nov 18 2015: (Start)
Coefficients of the row polynomials of the e.g.f. Sum_{n>=0} P_n(b1,b2,..,bn;t) x^n/n! = e^(P.(..;t) x) = e^(xt) / (1-b.x) = (1 + b1 x + b2 x^2 + b3 x^3 + ...) e^(xt) = 1 + (b1 + t) x + (2 b2 + 2 b1 t + t^2) x^2/2! + (6 b3 + 6 b2 t + 3 b1 t^2 + t^3) x^3/3! + ... , with lowering operator L = d/dt, i.e., L P_n(..;t) = n * P_(n-1)(..;t), and raising operator R = t + d[log(1 + b1 D + b2 D^2 + ...)]/dD = t - Sum_{n>=1} F(n,b1,..,bn) D^(n-1), i.e., R P_n(..,;t) = P_(n+1)(..;t), where D = d/dt and F(n,b1,..,bn) are the Faber polynomials of A263916.
Also P_n(b1,..,bn;t) = CIP_n(t-F(1,b1),-F(2,b1,b2),..,-F(n,b1,..,bn)), the cycle index polynomials A036039.
(End)
The raising operator R = x + 1/(1-D) = x + 1 + D + D^2 + ... in matrix form acting on an o.g.f. (formal power series) is the transpose of the production matrix M below. The linear term x is the diagonal of ones after transposition. The other transposed diagonals come from D^m x^n = n! / (n-m)! x^(n-m). Then P(n,x) = (1,x,x^2,..) M^n (1,0,0,..)^T is a matrix representation of R P(n-1,x) = P(n,x). - Tom Copeland, Aug 17 2016
The row polynomials have e.g.f. e^(xt)/(1-t) = exp(t*q.(x)), umbrally. With p_n(x) the row polynomials of A132013, q_n(x) = v_n(p.(u.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form [T] = [q] = [v]*[p]*[u]. Conversely, p_n(x) = u_n (q.(v.(x))). - Tom Copeland, Nov 10 2016
From the Appell sequence formalism, 1/(1-b.D) t^n = P_n(b1,b2,..,bn;t), the generalized row polynomials noted in the Nov 18 2015 formulas, consistent with the 2007 comments. - Tom Copeland, Nov 22 2016
From Peter Bala, Feb 18 2017: (Start)
G.f.: Sum_{n >= 1} (n*x)^(n-1)/(1 + (n - t)*x)^n = 1 + (1 + t)*x + (2 + 2*t + t^2)*x^2 + ....
n-th row polynomial R(n,t) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^k*(x + k - t)^(n-k) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^(n-k)*(x + k + t)^k, for arbitrary x. The particular case of the latter sum when x = 0 and t = 1 is identity 10.35 in Gould, Vol.4. (End)
Rodrigues-type formula for the row polynomials: R(n, x) = -exp(x)*Int(exp(-x)* x^n, x), for n >= 0. Recurrence: R(n, x) = x^n + n*R(n-1, x), for n >= 1, and R(0, x) = 1. d/dx(R(n, x)) = R(n, x) - x^n, for n >= 0 (compare with the formula from Peter Bala, Aug 28 2013). - Wolfdieter Lang, Dec 23 2019
T(n, k) = Sum_{i=0..n-k} A048994(n-k, i) * n^i for 0 <= k <= n. - Werner Schulte, Jul 26 2022

Extensions

Edited by Johannes W. Meijer, Oct 07 2009
New description from Dennis P. Walsh, Jan 24 2011

A132440 Infinitesimal Pascal matrix: generator (lower triangular matrix representation) of the Pascal matrix, the classical operator xDx, iterated Laguerre transforms, associated matrices of the list partition transform and general Euler transformation for sequences.

Original entry on oeis.org

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

Author

Tom Copeland, Nov 13 2007, Nov 15 2007, Nov 22 2007, Dec 02 2007

Keywords

Comments

Let M(t) = exp(t*T) = lim_{n->oo} (1 + t*T/n)^n.
Pascal matrix = [ binomial(n,k) ] = M(1) = exp(T), truncating the series gives the n X n submatrices.
Inverse Pascal matrix = M(-1) = exp(-T) = matrix for inverse binomial transform.
A(j) = T^j / j! equals the matrix [binomial(n,k) * delta(n-k-j)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e., A(j) is a matrix with all the terms 0 except for the j-th lower (or main for j=0) diagonal, which equals that of the Pascal triangle. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n,k) * d(n-k)] which include as a subset the invertible associated matrices of the list partition transform (LPT) of A133314.
For sequences with b(0) = 1, umbrally,
M[b(.)] = exp(b(.)*T) = [ binomial(n,k) * b(n-k) ] = matrices associated to b by LPT.
[M[b(.)]]^(-1) = exp(c(.)*T) = [ binomial(n,k) * c(n-k) ] = matrices associated to c, where c = LPT(b) . Or,
[M[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[M(b(.))] = M[LPT(b(.))]= M[c(.)].
This is related to xDx, the iterated Laguerre transform and the general Euler transformation of a sequence through the comments in A132013 and A132014 and the relation [Sum_{k=0..n} binomial(n,k) * b(n-k) * d(k)] = M(b)*d, (n-th term). See also A132382.
If b(n,x) is a binomial type Sheffer sequence, then M[b(.,x)]*s(y) = s(x+y) when s(y) = (s(0,y),s(1,y),s(2,y),...) is an array for a Sheffer sequence with the same delta operator as b(n,x) and [M[b(.,x)]]^(-1) is given by the formulas above with b(n) replaced by b(n,x) as b(0,x)=1 for a binomial-type Sheffer sequence.
T = I - A132013 and conversely A132013 = I - T, which is the matrix representation for the iterated mixed order Laguerre transform characterized in A132013 (and A132014).
(I-T)^m generates the group [A132013]^m for m = 0,1,2,... discussed in A132014.
The inverse is 1/(I-T) = I + T + T^2 + T^3 + ... = [A132013]^(-1) = A094587 with the associated sequence (0!,1!,2!,3!,...) under the LPT.
And 1/(I-T)^2 = I + 2*T + 3*T^2 + 4*T^3 + ... = [A132013]^(-2) = A132159 with the associated sequence (1!,2!,3!,4!,...) under the LPT.
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or e.g.f.'s EA(x) and EB(x).
1) b(0) = 0, b(n) = n * a(n-1),
2) B(x) = xDx A(x)
3) B(x) = x * Lag(1,-:xD:) A(x)
4) EB(x) = x * EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x) is the Laguerre polynomial.
So the exponentiated operator can be characterized as
5) exp(t*T) A(x) = exp(t*xDx) A(x) = [Sum_{n=0,1,...} (t*x)^n * Lag(n,-:xD:)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x), a generalized Euler transformation for an o.g.f.,
6) exp(t*T) EA(x) = exp(t*x)*EA(x) = exp[(t+a(.))*x], gen. Euler trf. for an e.g.f.
7) exp(t*T) * a = M(t) * a = [Sum_{k=0..n} binomial(n,k) * t^(n-k) * a(k)].
The umbral extension of formulas 5, 6 and 7 gives formally
8) exp[c(.)*T] A(x) = exp(c(.)*xDx) A(x) = [Sum_{n>=0} (c(.)*x)^n * Lag(n,-:xD:)] A(x) = [exp{[c(.)*u/(1-c(.)*u)]*:xD:} / (1-c(.)*u) ] A(x) (eval. at u=x) = A[x/(1-c(.)*x)]/(1-c(.)*x), where the umbral evaluation should be applied only after a power series in c is obtained,
9) exp[c(.)*T] EA(x) = exp(c(.)*x)*EA(x) = exp[(c(.)+a(.))*x]
10) exp[c(.)*T] * a = M[c(.)] * a = [Sum_{k=0..n} binomial(n,k) * c(n-k) * a(k)] .
The n X n principal submatrix of T is nilpotent, in particular, [Tsub_n]^(n+1) = 0, n=0,1,2,3,....
Note (xDx)^n = x^n D^n x^n = x^n n! (:Dx:)^n/n! = x^n n! Lag(n,-:xD:).
The operator xDx is an important, classical operator explored by among others Dattoli, Al-Salam, Carlitz and Stokes and even earlier investigators.
For a recent treatment of xDx, DxD and more general operators see the paper "Laguerre-type derivatives: Dobinski relations and combinatorial identities". - Karol A. Penson, Sep 15 2009
See Copeland's link for generalized Laguerre functions and connection to fractional differ-integrals in exercises through (:Dx:)^a/a!=(D^a x^a)/a!. - Tom Copeland, Nov 17 2011
From Tom Copeland, Apr 25 2014: (Start)
Conjugation or "similarity" transformations of [dP]=A132440 have an operator interpretation (cf. A074909 and A238363):
In general, select two operators A and B such that A^n = F1(n,B) and B^n = F2(n,A); then A^n = F1(n,F2(.,A)) and B^n = F2(n,F1(.,B)), evaluated umbrally, i.e., F1(n,F2(.,x))=F2(n,F1(.,x))=x^n, implying the polynomials F1 and F2 are an umbral compositional inverse pair.
One such pair are the Bell polynomials Bell(n,x) and falling factorials (x)_n with Bell(n,:xD:)=(xD)^n and (xD)_n=:xD:^n (cf. A074909). Another are the Laguerre polynomials LN(n,x)= n!*Lag(n,x) (A021009), which are umbrally self-inverse, with LN(n,-:xD:)=:Dx:^n and LN(n,:Dx:)= (-:xD:)^n with :Dx:^n=D^n*x^n.
Evaluating, for n>=0, the operator derivative d(B^n)/dA = d(F2(n,A))/dA in the basis B^n, i.e., with A^n finally replaced by F1(n,B), or A^n=F1(.,B)^n=F1(n,B), is equivalent to the matrix conjugation
A) [F2]*[dP]*[F1]
B) = [F2]*[dP]*[F2]^(-1)
C) = [F1]^(-1)*[dP]*[F1],
where [F1] is the lower triangular matrix with the n-th row the coefficients of F1(n,x) and analogously for [F2].
So, given the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose, form the power series V(x)=Rv*Cv(x).
D) dV(B)/dA = Rv * [F2]*[dP]*[F1] * Cv(B).
E) With A=D and B=D, F1(n,x)=F2(n,x)=x^n and [F1]=[F2]=I. Then d(B^n)/dA = d(D^n)/dD = n * D^(n-1); therefore, consistently [F2]*[dP]*[F1] = [dP] and dV(D)/dD = Rv * [dP] * Cv(D). (End)

Examples

			Matrix T begins
  0;
  1,0;
  0,2,0;
  0,0,3,0;
  0,0,0,4,0;
  ...
		

References

  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015, (x^n D^n x^n on p. 187).

Programs

Formula

T = log(P) with the Pascal matrix P:=A007318. This should be read as T_N = log(P_N) with P_N the N X N matrix P, N>=2. Because P_N is lower triangular with all diagonal elements 1, the series log(1_N-(1_N-P_N)) stops after N-1 terms because (1_N-P_N)^N is the 0_N-matrix. - Wolfdieter Lang, Oct 14 2010
Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L p_n(x) = n * p_(n-1)(x) and R p_n(x) = p_(n+1)(x), the matrix T represents the action of R*L*R in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R = D^(-1). - Tom Copeland, Oct 25 2012
From Tom Copeland, Apr 26 2014: (Start)
A) T = exp(A238385-I) - I
B) = [St1]*P*[St2] - I
C) = [St1]*P*[St1]^(-1) - I
D) = [St2]^(-1)*P*[St2] - I
E) = [St2]^(-1)*P*[St1]^(-1) - I
where P=A007318, [St1]=padded A008275 just as [St2]=A048993=padded A008277, and I=identity matrix. (End)
From Robert Israel, Oct 02 2015: (Start)
G.f. Sum_{k >= 1} k x^((k+3/2)^2/2 - 17/8) is related to Jacobi theta functions.
If 8*n+17 = y^2 is a square, then a(n) = (y-3)/2, otherwise a(n) = 0. (End)

Extensions

Missing zero added in table by Tom Copeland, Feb 25 2014

A165900 a(n) = n^2 - n - 1.

Original entry on oeis.org

-1, -1, 1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255
Offset: 0

Author

Philippe Deléham, Sep 29 2009

Keywords

Comments

Previous name was: Values of Fibonacci polynomial n^2 - n - 1.
Shifted version of the array denoted rB(0,2) in A132382, whose e.g.f. is exp(x)(1-x)^2. Taking the derivative gives the e.g.f. of this sequence. - Tom Copeland, Dec 02 2013
The Fibonacci numbers are generated by the series x/(1 - x - x^2). - T. D. Noe, Dec 04 2013
Absolute value of expression f(k)*f(k+1) - f(k-1)*f(k+2) where f(1)=1, f(2)=n. Sign is alternately +1 and -1. - Carmine Suriano, Jan 28 2014 [Can anybody clarify what is meant here? - Joerg Arndt, Nov 24 2014]
Carmine's formula is a special case related to 4 consecutive terms of a Fibonacci sequence. A generalization of this formula is |a(n)| = |f(k+i)*f(k+j) - f(k)*f(k+i+j)|/F(i)*F(j), where f denotes a Fibonacci sequence with the initial values 1 and n, and F denotes the original Fibonacci sequence A000045. The same results can be obtained with the simpler formula |a(n)| = |f(k+1)^2 - f(k)^2 - f(k+1)*f(k)|. Everything said so far is also valid for Fibonacci sequences f with the initial values f(1) = n - 2, f(2) = 2*n - 3. - Klaus Purath, Jun 27 2022
a(n) is the total number of dollars won when using the Martingale method (bet $1, if win then continue to bet $1, if lose then double next bet) for n trials of a wager with exactly one loss, n-1 wins. For the case with exactly one win, n-1 losses, see A070313. - Max Winnick, Jun 28 2022
Numbers m such that 4*m+5 is a square b^2, where b = 2*n -1, for m = a(n). - Klaus Purath, Jul 23 2022

Examples

			G.f. = -1 - x + x^2 + 5*x^3 + 11*x^4 + 19*x^5 + 29*x^6 + 41*x^7 + ... - _Michael Somos_, Mar 23 2023
		

Crossrefs

A028387 and A110331 are very similar sequences.

Programs

Formula

a(n+2) = (n+1)*a(n+1) - (n+2)*a(n).
G.f.: (x^2+2*x-1)/(1-x)^3.
E.g.f.: exp(x)*(x^2-1).
a(n) = - A188652(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011
a(n) = A214803(A015614(n+1)) for n > 0. - Reinhard Zumkeller, Jul 29 2012
a(n+1) = a(n) + A005843(n) = A002378(n) - 1. - Ivan N. Ianakiev, Feb 18 2013
a(n+2) = A028387(n). - Michael B. Porter, Sep 26 2018
From Klaus Purath, Aug 25 2022: (Start)
a(2*n) = n*(a(n+1) - a(n-1)) -1.
a(2*n+1) = (2*n+1)*(a(n+1) - a(n)) - 1.
a(n+2) = a(n) + 4*n + 2.
a(n) = A014206(n-1) - 3 = A002061(n-1) - 2.
a(n) = A028552(n-2) + 1 = A014209(n-2) + 2 = 2* A034856(n-2) + 3.
a(n) = A008865(n-1) + n = A005563(n-1) - n.
a(n) = A014209(n-3) + 2*n = A028387(n-1) - 2*n.
a(n) = A152015(n)/n, n != 0.
(a(n+k) - a(n-k))/(2*k) = 2*n-1, for any k.
(End)
For n > 1, 1/a(n) = Sum_{k>=1} F(k)/n^(k+1), where F(n) = A000045(n). - Diego Rattaggi, Nov 01 2022
a(n) = a(1-n) for all n in Z. - Michael Somos, Mar 23 2023
For n > 1, 1/a(n) = Sum_{k>=1} F(2k)/((n+1)^(k+1)), where F(2n) = A001906(n). - Diego Rattaggi, Jan 20 2025
From Amiram Eldar, May 11 2025: (Start)
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*Pi/sqrt(5).
Product_{n>=3} 1 - 1/a(n) = -sec(sqrt(5)*Pi/2)*Pi/6.
Product_{n>=2} 1 + 1/a(n) = -sec(sqrt(5)*Pi/2)*Pi. (End)

Extensions

a(22) corrected by Reinhard Zumkeller, Apr 13 2011
Better name from Joerg Arndt, Oct 26 2024

A132159 Lower triangular matrix T(n,j) for double application of an iterated mixed order Laguerre transform inverse to A132014. Coefficients of Laguerre polynomials (-1)^n * n! * L(n,-2-n,x).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 24, 18, 6, 1, 120, 96, 36, 8, 1, 720, 600, 240, 60, 10, 1, 5040, 4320, 1800, 480, 90, 12, 1, 40320, 35280, 15120, 4200, 840, 126, 14, 1, 362880, 322560, 141120, 40320, 8400, 1344, 168, 16, 1, 3628800, 3265920, 1451520, 423360, 90720, 15120
Offset: 0

Author

Tom Copeland, Nov 01 2007

Keywords

Comments

The matrix operation b = T*a can be characterized several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or their e.g.f.'s EA(x) and EB(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a1(.),-1],0], umbrally,
where a1(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0]
2) b(n) = (-1)^n * n! * Lag(n,a(.),-2-n)
3) b(n) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(-2,j) * j! * a(n-j)
4) b(n) = Sum_{j=0..n} binomial(n,j) * (j+1)! * a(n-j)
5) B(x) = (1-xDx))^(-2) A(x), formally
6) B(x) = Sum_{j>=0} (-1)^j * binomial(-2,j) * (xDx)^j A(x)
= Sum_{j>=0} (j+1) * (xDx)^j A(x)
7) B(x) = Sum_{j>=0} (j+1) * x^j * D^j * x^j A(x)
8) B(x) = Sum_{j>=0} (j+1)! * x^j * Lag(j,-:xD:,0) A(x)
9) EB(x) = Sum_{j>=0} x^j * Lag[j,(.)! * Lag[.,a1(.),-1],0]
10) EB(x) = Sum_{j>=0} Lag[j,a1(.),-1] * (-x)^j / (1-x)^(j+1)
11) EB(x) = Sum_{j>=0} x^n * Sum_{j=0..n} (j+1)!/j! * a(n-j) / (n-j)!
12) EB(x) = Sum_{j>=0} (-x)^j * Lag[j,a(.),-2-j]
13) EB(x) = exp(a(.)*x) / (1-x)^2 = (1-x)^(-2) * EA(x)
14) T = A094587^2 = A132013^(-2) = A132014^(-1)
where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the D operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. Thus T(n,k) = binomial(n,k)*c(n-k) . c are also the coefficients in formulas 4 and 8.
The reciprocal sequence to c is d = (1,-2,2,0,0,0,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132014. (A121757 is the reverse of T.)
These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),-1],0] by replacing 2 by m in formulas 2, 3, 5, 6, 12, 13 and 14, or (j+1)! by (m-1+j)!/(m-1)! in 4, 8 and 11. For further discussion of repeated applications of T, see A132014.
The row sums of T = [formula 4 with a(n) all 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A001339. Therefore the e.g.f. of A001339 = [formula 13 with a(n) all 1] = exp(x)*(1-x)^(-2) = exp(x)*exp[c(.)*x)] = exp[(1+c(.))*x].
Note the reciprocal is 1/{exp[(1+c(.))*x]} = exp(-x)*(1-x)^2 = e.g.f. of signed A002061 with leading 1 removed], which makes A001339 and the signed, shifted A002061 reciprocal arrays under the list partition transform of A133314.
The e.g.f. for the row polynomials (see A132382) implies they form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
As noted in item 12 above and reiterated in the Bala formula below, the e.g.f. is e^(x*t)/(1-x)^2, and the Poisson-Charlier polynomials P_n(t,y) have the e.g.f. (1+x)^y e^(-xt) (Feinsilver, p. 5), so the row polynomials R_n(t) of this entry are (-1)^n P_n(t,-2). The associated Appell sequence IR_n(t) that is the umbral compositional inverse of this entry's polynomials has the e.g.f. (1-x)^2 e^(xt), i.e., the e.g.f. of A132014 (noted above), and, therefore, the row polynomials (-1)^n PC(t,2). As umbral compositional inverses, R_n(IR.(t)) = t^n = IR_n(R.(t)), where, by definition, P.(t)^n = P_n(t), is the umbral evaluation. - Tom Copeland, Jan 15 2016
T(n,k) is the number of ways to place (n-k) rooks in a 2 x (n-1) Ferrers board (or diagram) under the Goldman-Haglund i-row creation rook mode for i=2. Triangular recurrence relation is given by T(n,k) = T(n-1,k-1) + (n+1-k)*T(n-1,k). - Ken Joffaniel M. Gonzales, Jan 21 2016

Examples

			First few rows of the triangle are
    1;
    2,  1;
    6,  4,  1;
   24, 18,  6, 1;
  120, 96, 36, 8, 1;
		

Crossrefs

Columns: A000142 (k=0), A001563 (k=1), A001286 (k=2), A005990 (k=3), A061206 (k=4), A062199 (k=5), A062148 (k=6).

Programs

  • Haskell
    a132159 n k = a132159_tabl !! n !! k
    a132159_row n = a132159_tabl !! n
    a132159_tabl = map reverse a121757_tabl
    -- Reinhard Zumkeller, Mar 06 2014
    
  • Magma
    /* As triangle */ [[Binomial(n,k)*Factorial(n-k+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 10 2016
    
  • Maple
    T := proc(n,k) return binomial(n,k)*factorial(n-k+1): end: seq(seq(T(n,k),k=0..n),n=0..10); # Nathaniel Johnston, Sep 28 2011
  • Mathematica
    nn=10;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[y x]/(1-x)^2,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Feb 15 2013 *)
  • Sage
    flatten([[binomial(n,k)*factorial(n-k+1) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, May 19 2021

Formula

T(n,k) = binomial(n,k)*c(n-k).
From Peter Bala, Jul 10 2008: (Start)
T(n,k) = binomial(n,k)*(n-k+1)!.
T(n,k) = (n-k+1)*T(n-1,k) + T(n-1,k-1).
E.g.f.: exp(x*y)/(1-y)^2 = 1 + (2+x)*y + (6+4*x+x^2)*y^2/2! + ... .
This array is the particular case P(2,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0....................1...............2.........3.....4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
See A094587 for some general properties of these arrays.
Other cases recorded in the database include: P(1,0) = Pascal's triangle A007318, P(1,1) = A094587, P(2,0) = A038207, P(3,0) = A027465, P(1,3) = A136215 and P(2,3) = A136216. (End)
Let f(x) = (1/x^2)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+2)*R(n,x)-x*R'(n,x). Cf. A094587. - Peter Bala, Oct 28 2011
Exponential Riordan array [1/(1 - y)^2, y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x). Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = Product_{k = 0..n-1} (2*x + k) with the convention that P(0,x) = 1. Then the present triangle is the triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (24, 18, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 3)*(2*x + 4) = 24 + 18*(2*x) + 6*(2*x)*(2*x + 1) + (2*x)*(2*x + 1)*(2*x + 2). Matrix square of triangle A094587. - Peter Bala, Aug 29 2013
From Tom Copeland, Apr 21 2014: (Start)
T = (I-A132440)^(-2) = {2*I - exp[(A238385-I)]}^(-2) = unsigned exp[2*(I-A238385)] = exp[A005649(.)*(A238385-I)], umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)*(1-y)^(-2), so the row polynomials form an Appell sequence with lowering operator D=d/dx and raising operator x+2/(1-D).
With L(n,m,x) = Laguerre polynomials of order m, the row polynomials are (-1)^n * n! * L(n,-2-n,x) = (-1)^n*(-2!/(-2-n)!)*K(-n,-2-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-2-n,-:xD:) = (-1)^n*x^(n+2)*:Dx:^n*x^(-2-n) = (-1)^n*x^2*:xD:^n*x^(-2) = (-1)^n*n!*binomial(xD-2,n) = (-1)^n*n!*binomial(-2,n)*K(-n,-2-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706.
The generalized Pascal triangle Bala mentions is a special case of the fundamental generalized factorial matrices in A133314. (End)
From Peter Bala, Jul 26 2021: (Start)
O.g.f: 1/y * Sum_{k >= 0} k!*( y/(1 - x*y) )^k = 1 + (2 + x)*y + (6 + 4*x + x^2)*y^2 + ....
First-order recurrence for the row polynomials: (n - x)*R(n,x) = n*(n - x + 1)*R(n-1,x) - x^(n+1) with R(0,x) = 1.
R(n,x) = (x + n + 1)*R(n-1,x) - (n - 1)*x*R(n-2,x) with R(0,x) = 1 and R(1,x) = 2 + x.
R(n,x) = A087981 (x = -2), A000255 (x = -1), A000142 (x = 0), A001339 (x = 1), A081923 (x = 2) and A081924 (x = 3). (End)

Extensions

Formula 3) in comments corrected by Tom Copeland, Apr 20 2014
Title modified by Tom Copeland, Apr 23 2014

A055137 Regard triangle of rencontres numbers (see A008290) as infinite matrix, compute inverse, read by rows.

Original entry on oeis.org

1, 0, 1, -1, 0, 1, -2, -3, 0, 1, -3, -8, -6, 0, 1, -4, -15, -20, -10, 0, 1, -5, -24, -45, -40, -15, 0, 1, -6, -35, -84, -105, -70, -21, 0, 1, -7, -48, -140, -224, -210, -112, -28, 0, 1, -8, -63, -216, -420, -504, -378, -168, -36, 0, 1, -9, -80, -315, -720
Offset: 0

Author

Christian G. Bower, Apr 25 2000

Keywords

Comments

The n-th row consists of coefficients of the characteristic polynomial of the adjacency matrix of the complete n-graph.
Triangle of coefficients of det(M(n)) where M(n) is the n X n matrix m(i,j)=x if i=j, m(i,j)=i/j otherwise. - Benoit Cloitre, Feb 01 2003
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for rB(0,1). The e.g.f. for the row polynomials is exp(x*t) * exp(x) *(1-x). T(n,k) = Binomial(n,k)* s(n-k) where s = (1,0,-1,-2,-3,...) with an e.g.f. of exp(x)*(1-x) which is the reciprocal of the e.g.f. of A000166. - Tom Copeland, Sep 10 2008
Row sums are: {1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...}. - Roger L. Bagula, Feb 20 2009
T is related to an operational calculus connecting an infinitesimal generator for fractional integro-derivatives with the values of the Riemann zeta function at positive integers (see MathOverflow links). - Tom Copeland, Nov 02 2012
The submatrix below the null subdiagonal is signed and row reversed A127717. The submatrix below the diagonal is A074909(n,k)s(n-k) where s(n)= -n, i.e., multiply the n-th diagonal by -n. A074909 and its reverse A135278 have several combinatorial interpretations. - Tom Copeland, Nov 04 2012
T(n,k) is the difference between the number of even (A145224) and odd (A145225) permutations (of an n-set) with exactly k fixed points. - Julian Hatfield Iacoponi, Aug 08 2024

Examples

			1; 0,1; -1,0,1; -2,-3,0,1; -3,-8,-6,0,1; ...
(Bagula's matrix has a different sign convention from the list.)
From _Roger L. Bagula_, Feb 20 2009: (Start)
  { 1},
  { 0,   1},
  {-1,   0,    1},
  { 2,  -3,    0,    1},
  {-3,   8,   -6,    0,     1},
  { 4, -15,   20,  -10,     0,    1},
  {-5,  24,  -45,   40,   -15,    0,    1},
  { 6, -35,   84, -105,    70,  -21,    0,   1},
  {-7,  48, -140,  224,  -210,  112,  -28,   0,   1},
  { 8, -63,  216, -420,   504, -378,  168, -36,   0, 1},
  {-9,  80, -315,  720, -1050, 1008, -630, 240, -45, 0, 1}
(End)
R(3,x) = (-1)^3*Sum_{permutations p in S_3} sign(p)*(-x)^(fix(p)).
    p   | fix(p) | sign(p) | (-1)^3*sign(p)*(-x)^fix(p)
========+========+=========+===========================
  (123) |    3   |    +1   |      x^3
  (132) |    1   |    -1   |       -x
  (213) |    1   |    -1   |       -x
  (231) |    0   |    +1   |       -1
  (312) |    0   |    +1   |       -1
  (321) |    1   |    -1   |       -x
========+========+=========+===========================
                           | R(3,x) = x^3 - 3*x - 2
- _Peter Bala_, Aug 08 2011
		

References

  • Norman Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University Press, 1993. p. 17.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p.184 problem 3.

Crossrefs

Cf. A005563, A005564 (absolute values of columns 1, 2).
Cf. A000312.

Programs

  • Mathematica
    M[n_] := Table[If[i == j, x, 1], {i, 1, n}, {j, 1, n}]; a = Join[{{1}}, Flatten[Table[CoefficientList[Det[M[n]], x], {n, 1, 10}]]] (* Roger L. Bagula, Feb 20 2009 *)
    t[n_, k_] := (k-n+1)*Binomial[n, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 29 2013, after Pari *)
  • PARI
    T(n,k)=(1-n+k)*if(k<0 || k>n,0,n!/k!/(n-k)!)

Formula

G.f.: (x-n+1)*(x+1)^(n-1) = Sum_(k=0..n) T(n,k) x^k.
T(n, k) = (1-n+k)*binomial(n, k).
k-th column has o.g.f. x^k(1-(k+2)x)/(1-x)^(k+2). k-th row gives coefficients of (x-k)(x+1)^k. - Paul Barry, Jan 25 2004
T(n,k) = Coefficientslist[Det[Table[If[i == j, x, 1], {i, 1, n}, {k, 1, n}],x]. - Roger L. Bagula, Feb 20 2009
From Peter Bala, Aug 08 2011: (Start)
Given a permutation p belonging to the symmetric group S_n, let fix(p) be the number of fixed points of p and sign(p) its parity. The row polynomials R(n,x) have a combinatorial interpretation as R(n,x) = (-1)^n*Sum_{permutations p in S_n} sign(p)*(-x)^(fix(p)). An example is given below.
Note: The polynomials P(n,x) = Sum_{permutations p in S_n} x^(fix(p)) are the row polynomials of the rencontres numbers A008290. The integral results Integral_{x = 0..n} R(n,x) dx = n/(n+1) = Integral_{x = 0..-1} R(n,x) dx lead to the identities Sum_{p in S_n} sign(p)*(-n)^(1 + fix(p))/(1 + fix(p)) = (-1)^(n+1)*n/(n+1) = Sum_{p in S_n} sign(p)/(1 + fix(p)). The latter equality was Problem B6 in the 66th William Lowell Putnam Mathematical Competition 2005. (End)
From Tom Copeland, Jul 26 2017: (Start)
The e.g.f. in Copeland's 2008 comment implies this entry is an Appell sequence of polynomials P(n,x) with lowering and raising operators L = d/dx and R = x + d/dL log[exp(L)(1-L)] = x+1 - 1/(1-L) = x - L - L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L) exp(L) x^n = (1-L) (x+1)^n = (x+1)^n - n (x+1)^(n-1) = (x+1-n)(x+1)^(n-1) = (x+s.)^n umbrally, where (s.)^n = s_n = P(n,0).
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
The exponential infinitesimal generator (infinigen) of this entry is the negated infinigen of A008290, the matrix (M) noted by Bala, related to A238363. Then e^M = [the lower triangular A008290], and e^(-M) = [the lower triangular A055137]. For more on the infinigens, see A238385. (End)
From the row g.f.s corresponding to Bagula's matrix example below, the n-th row polynomial has a zero of multiplicity n-1 at x = 1 and a zero at x = -n+1. Since this is an Appell sequence dP_n(x)/dx = n P_{n-1}(x), the critical points of P_n(x) have the same abscissas as the zeros of P_{n-1}(x); therefore, x = 1 is an inflection point for the polynomials of degree > 2 with P_n(1) = 0, and the one local extremum of P_n has the abscissa x = -n + 2 with the value (-n+1)^{n-1}, signed values of A000312. - Tom Copeland, Nov 15 2019
From Julian Hatfield Iacoponi, Aug 08 2024: (Start)
T(n,k) = A145224(n,k) - A145225(n,k).
T(n,k) = binomial(n,k)*(A003221(n-k)-A000387(n-k)). (End)

Extensions

Additional comments from Michael Somos, Jul 04 2002

A132013 T(n,j) for an iterated mixed order Laguerre transform. Coefficients of the normalized generalized Laguerre polynomials (-1)^n*n!*L(n,1-n,x).

Original entry on oeis.org

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

Author

Tom Copeland, Oct 30 2007

Keywords

Comments

The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x).
1) b(0) = a(0), b(n) = a(n) - n*a(n-1) for n > 0
2) b(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}, umbrally
3) b(n) = n! Sum_{j=0..min(1,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),1-n)
5) B(x) = (1-xDx) A(x) = [1-x*Lag(1,-xD:,0)] A(x)
6) EB(x) = (1-x) EA(x),
where D is the derivative w.r.t. x and Lag(n,x,m) is the associated Laguerre polynomial of order m. These formulas are easily generalized for repeated applications of the operator.
c = (1,-1,0,0,0,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314. The reciprocal sequence is d = (0!,1!,2!,3!,4!,...).
Consequently, the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A094587, which has the property that the terms at and below TI(m,m) are the associated sequence under the list partition transform for the inverse for T^(m+1) for m=0,1,2,3,... .
Row sums of T = [formula 3 with all a(n) = 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A024000 = (1,0,-1,-2,-3,...), with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x) * exp(x) = exp(x)*exp(c(.)*x) = exp[(1+c(.))*x].
Alternating row sums of T = [formula 3 with all a(n) = (-1)^n] = [finite differences of c] = [coefficients of B(x) with A(x) = 1/(1+x)] = (1,-2,3,-4,...), with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x) * exp(-x) = exp(-x)*exp(c(.)*x) = exp[-(1-c(.))*x].
An e.g.f. for the o.g.f.s for repeated applications of T on A(x) is given by
exp[t*(1-xDx)] A(x) = e^t * Sum_{n=0,1,...} (-t*x)^n * Lag(n,-:xD:,0) A(x)
= e^t * exp{[-t*u/(1+t*u)]*:xD:} / (1+t*u) A(x) (eval. at u=x)
= e^t * A[x/(1+t*x)]/(1+t*x) .
See A132014 for more notes on repeated applications.

Examples

			First few rows of the triangle are
   1;
  -1,  1;
   0, -2,  1;
   0,  0, -3,  1;
   0,  0,  0, -4,  1;
   0,  0,  0,  0, -5,  1;
   0,  0,  0,  0,  0, -6,  1;
   0,  0,  0,  0,  0,  0, -7,  1;
		

Programs

  • Maple
    c := n -> `if`(n=0,1,`if`(n=1,-1,0)):
    T := (n,k) -> binomial(n,k)*c(n-k); # Peter Luschny, Nov 14 2016
  • Mathematica
    Table[PadLeft[{-n, 1}, n+1], {n, 0, 13}] // Flatten (* Jean-François Alcover, Apr 29 2014 *)
  • PARI
    row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 1-n)); \\ Michel Marcus, Jul 26 2021

Formula

T(n,k) = binomial(n,k)*c(n-k), with the sequence c defined in the comments.
E.g.f.: exp(x*y)(1-x), which implies the row polynomials form an Appell sequence. More relations can be found in A132382. - Tom Copeland, Dec 03 2013
From Tom Copeland, Apr 21 2014: (Start)
Change notation letting L(n,m,x) = Lag(n,x,m).
Row polynomials: (-1)^n*n!*L(n,1-n,x) = -x^(n-1)*L(1,n-1,x) =
(-1)^n*(1/(1-n)!)*K(-n,1-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
For the row polynomials, the lowering operator = d/dx and the raising operator = x - 1/(1-D).
T = I - A132440 = 2*I - exp[A238385-I] = signed exp[A238385-I], where I = identity matrix.
Operationally, (-1)^n*n!*L(n,1-n,-:xD:) = -x^(n-1)*:Dx:^n*x^(1-n) = (-1)^n*x^(-1)*:xD:^n*x = (-1)^n*n!*binomial(xD+1,n) = (-1)^n*n!*binomial(1,n)*K(-n,1-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
The unsigned row polynomials have e.g.f. (1+t)e^(xt) = exp(t*p.(x)), umbrally, and p_n(x) = (1+D) x^n. With q_n(x) the row polynomials of A094587, p_n(x) = u_n(q.(v.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form unsigned [T] = [p] = [u]*[q]*[v]. Conversely, q_n(x) = v_n (p.(u.(x))). - Tom Copeland, Nov 10 2016
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^k*binomial(n,k)*Lag(k,1,x). - Peter Bala, Jul 25 2021

Extensions

Title modified by Tom Copeland, Apr 21 2014
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