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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A288875 Triangle read by rows. The rows give the coefficients of the numerator polynomials for the o.g.f.s of the diagonal sequences of triangle A028338.

Original entry on oeis.org

1, 1, 1, 3, 8, 1, 15, 71, 33, 1, 105, 744, 718, 112, 1, 945, 9129, 14542, 5270, 353, 1, 10395, 129072, 300291, 191384, 33057, 1080, 1, 135135, 2071215, 6524739, 6338915, 2033885, 190125, 3265, 1, 2027025, 37237680, 150895836, 204889344, 103829590, 18990320, 1038780, 9824, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jul 21 2017

Keywords

Comments

The Sheffer triangle A028338 of the type (1/sqrt(1-2*x), -(1/2)*log(1 - 2*x)) is called here |S1hat[2,1]|. The o.g.f. of the sequence of diagonal d, d >= 0 is D(d, t) = Sum_{m=0..d} A028338(d+m, m)*t^m. The e.g.f. of these o.g.f.s is taken as ED(y,t) := Sum_{d >= 0} D(d, t)*y^(d+1)/(d+1)!.
This e.g.f. is found to be ED(y,t) = 1 - sqrt(1 - 2*x(t;y)), where x = x(t;y) is the compositional inverse of y = y(t;x) = x*(1 - t*(-log(1-2*x)/(2*x))) = x + t*log(1-2*x)/2. The o.g.f.s are then D(d, t) = P(d, t)/(1 - t)^(2*d+1), with the row polynomials P(d, t) = Sum_{m=0..d} T(d, m)*t^m, d >= 0.
This computation was inspired by an article of P. Bala (see a link under, e.g., A112007) for Sheffer triangles of the Jabotinsky type (1, F(x)). There Sheffer is called exponential Riordan, and the diagonals are labeled by n = d+1, n >= 1.

Examples

			The triangle T(n, m) begins:
n\m      0        1         2         3         4        5       6    7  8 ...
0:       1
1:       1        1
2:       3        8         1
3:      15       71        33         1
4:     105      744       718       112         1
5:     945     9129     14542      5270       353        1
6:   10395   129072    300291    191384     33057     1080       1
7:  135135  2071215   6524739   6338915   2033885   190125    3265    1
8: 2027025 37237680 150895836 204889344 103829590 18990320 1038780 9824  1
...
		

Crossrefs

Programs

  • Mathematica
    De[d_, t_] := Sum[A028338[d+m, m] t^m, {m, 0, d}]; A028338[n_, k_] := SeriesCoefficient[Times @@ Table[x+i, {i, 1, 2n-1, 2}], {x, 0, k}]; P[n_, x_] := De[n, x] (1-x)^(2n+1); T[n_, m_] := Coefficient[P[n, x], x, m]; Table[T[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 24 2017 *)
    T[n_,m_]:=Sum[(-1)^(i-n+m)*Binomial[2*n+1,n-m-i]*(1/(2^i*i!)*Sum[(-1)^(i-j)*Binomial[i,j]*(2*j+1)^(n+i),{j,0,i}]),{i,0,n-m}];Flatten[Table[T[n,m],{n,0,8},{m,0,n}]] (* Detlef Meya, Dec 18 2023, after Peter Bala from A214406 *)

Formula

T(n, m) = [x^m] P(n, x), with the numerator polynomial of the o.g.f. of the diagonal n (main diagonal n=0) D(n, x) = P(n, x)/(1-x)^(2*n+1). See a comment above.
T(n, m) = Sum_{i=0..n-m} ( (-1)^(i-n+m)*binomial(2*n+1,n-m-i)*(1/(2^i*i!))*Sum_{j=0..i} (-1)^(i-j)*binomial(i,j)*(2*j+1)^(n+i) ). - Detlef Meya, Dec 18 2023, after Peter Bala from A214406.

A109409 Coefficients of polynomials triangular sequence produced by removing primes from the odd numbers in A028338.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 9, 10, 1, 0, 0, 0, 0, 9, 10, 1, 0, 0, 0, 0, 0, 9, 10, 1, 0, 0, 0, 0, 0, 135, 159, 25, 1, 0, 0, 0, 0, 0, 0, 135, 159, 25, 1, 0, 0, 0, 0, 0, 0, 0, 135, 159, 25, 1, 0, 0, 0, 0, 0, 0, 0, 2835, 3474, 684, 46, 1
Offset: 1

Views

Author

Roger L. Bagula, May 19 2007

Keywords

Comments

The row sums also appear to be new: b = Flatten[Join[{{1}}, Table[Apply[Plus, Abs[CoefficientList[Product[x + g[n], {n, 0, m}], x]]], {m, 0, 10}]]] {1, 2, 2, 2, 2, 20, 20, 20, 320, 320, 320, 7040} Since the row sum of A028338 is the double factorial A000165: this result seems to be a factorization of the double factorial numbers by relatively sparse nonprime odd numbers. It might be better to reverse the order of the coefficients to get the higher powers first.

Examples

			{1},
{1, 1},
{0, 1, 1},
{0, 0, 1, 1},
{0, 0, 0, 1, 1},
{0, 0, 0, 9, 10, 1},
{0, 0, 0, 0, 9, 10, 1},
{0, 0, 0, 0, 0, 9, 10, 1}
		

Crossrefs

Programs

  • Mathematica
    a = Join[{{1}}, Table[CoefficientList[Product[x + g[n], {n, 0, m}], x], {m, 0, 10}]]; Flatten[a]

Formula

p[n]=Product[If[PrimeQ[2*n+1]==false,x+(2*n+1),x] a(n) =CoefficientList[p[n],x]

A000290 The squares: a(n) = n^2.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500
Offset: 0

Views

Author

Keywords

Comments

To test if a number is a square, see Cohen, p. 40. - N. J. A. Sloane, Jun 19 2011
Zero followed by partial sums of A005408 (odd numbers). - Jeremy Gardiner, Aug 13 2002
Begin with n, add the next number, subtract the previous number and so on ending with subtracting a 1: a(n) = n + (n+1) - (n-1) + (n+2) - (n-2) + (n+3) - (n-3) + ... + (2n-1) - 1 = n^2. - Amarnath Murthy, Mar 24 2004
Sum of two consecutive triangular numbers A000217. - Lekraj Beedassy, May 14 2004
Numbers with an odd number of divisors: {d(n^2) = A048691(n); for the first occurrence of 2n + 1 divisors, see A071571(n)}. - Lekraj Beedassy, Jun 30 2004
See also A000037.
First sequence ever computed by electronic computer, on EDSAC, May 06 1949 (see Renwick link). - Russ Cox, Apr 20 2006
Numbers k such that the imaginary quadratic field Q(sqrt(-k)) has four units. - Marc LeBrun, Apr 12 2006
For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A006881(k)^(n-1)); a(n) = A000005(A000400(n-1)) = A000005(A011557(n-1)) = A000005(A001023(n-1)) = A000005(A001024(n-1)). - Reinhard Zumkeller, Mar 04 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-2) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
Numbers a such that a^1/2 + b^1/2 = c^1/2 and a^2 + b = c. - Cino Hilliard, Feb 07 2008 (this comment needs clarification, Joerg Arndt, Sep 12 2013)
Numbers k such that the geometric mean of the divisors of k is an integer. - Ctibor O. Zizka, Jun 26 2008
Equals row sums of triangle A143470. Example: 36 = sum of row 6 terms: (23 + 7 + 3 + 1 + 1 + 1). - Gary W. Adamson, Aug 17 2008
Equals row sums of triangles A143595 and A056944. - Gary W. Adamson, Aug 26 2008
Number of divisors of 6^(n-1) for n > 0. - J. Lowell, Aug 30 2008
Denominators of Lyman spectrum of hydrogen atom. Numerators are A005563. A000290-A005563 = A000012. - Paul Curtz, Nov 06 2008
a(n) is the number of all partitions of the sum 2^2 + 2^2 + ... + 2^2, (n-1) times, into powers of 2. - Valentin Bakoev, Mar 03 2009
a(n) is the maximal number of squares that can be 'on' in an n X n board so that all the squares turn 'off' after applying the operation: in any 2 X 2 sub-board, a square turns from 'on' to 'off' if the other three are off. - Srikanth K S, Jun 25 2009
Zero together with the numbers k such that 2 is the number of perfect partitions of k. - Juri-Stepan Gerasimov, Sep 26 2009
Totally multiplicative sequence with a(p) = p^2 for prime p. - Jaroslav Krizek, Nov 01 2009
Satisfies A(x)/A(x^2), A(x) = A173277: (1, 4, 13, 32, 74, ...). - Gary W. Adamson, Feb 14 2010
Positive members are the integers with an odd number of odd divisors and an even number of even divisors. See also A120349, A120359, A181792, A181793, A181795. - Matthew Vandermast, Nov 14 2010
Besides the first term, this sequence is the denominator of Pi^2/6 = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + ... . - Mohammad K. Azarian, Nov 01 2011
Partial sums give A000330. - Omar E. Pol, Jan 12 2013
Drmota, Mauduit, and Rivat proved that the Thue-Morse sequence along the squares is normal; see A228039. - Jonathan Sondow, Sep 03 2013
a(n) can be decomposed into the sum of the four numbers [binomial(n, 1) + binomial(n, 2) + binomial(n-1, 1) + binomial(n-1, 2)] which form a "square" in Pascal's Triangle A007318, or the sum of the two numbers [binomial(n, 2) + binomial(n+1, 2)], or the difference of the two numbers [binomial(n+2, 3) - binomial(n, 3)]. - John Molokach, Sep 26 2013
In terms of triangular tiling, the number of equilateral triangles with side length 1 inside an equilateral triangle with side length n. - K. G. Stier, Oct 30 2013
Number of positive roots in the root systems of type B_n and C_n (when n > 1). - Tom Edgar, Nov 05 2013
Squares of squares (fourth powers) are also called biquadratic numbers: A000583. - M. F. Hasler, Dec 29 2013
For n > 0, a(n) is the largest integer k such that k^2 + n is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n is a multiple of k + n is given by k = n^(2*m). - Derek Orr, Sep 03 2014
For n > 0, a(n) is the number of compositions of n + 5 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
a(n), for n >= 3, is also the number of all connected subtrees of a cycle graph, having n vertices. - Viktar Karatchenia, Mar 02 2016
On every sequence of natural continuous numbers with an even number of elements, the summatory of the second half of the sequence minus the summatory of the first half of the sequence is always a square. Example: Sequence from 61 to 70 has an even number of elements (10). Then 61 + 62 + 63 + 64 + 65 = 315; 66 + 67 + 68 + 69 + 70 = 340; 340 - 315 = 25. (n/2)^2 for n = number of elements. - César Aguilera, Jun 20 2016
On every sequence of natural continuous numbers from n^2 to (n+1)^2, the sum of the differences of pairs of elements of the two halves in every combination possible is always (n+1)^2. - César Aguilera, Jun 24 2016
Suppose two circles with radius 1 are tangent to each other as well as to a line not passing through the point of tangency. Create a third circle tangent to both circles as well as the line. If this process is continued, a(n) for n > 0 is the reciprocals of the radii of the circles, beginning with the largest circle. - Melvin Peralta, Aug 18 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Numerators of the solution to the generalization of the Feynman triangle problem, with an offset of 2. 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 denominators of the ratio of the areas is given by A002061. [Cook & Wood, 2004] - Joe Marasco, Feb 20 2017
Equals row sums of triangle A004737, n >= 1. - Martin Michael Musatov, Nov 07 2017
Right-hand side of the binomial coefficient identity Sum_{k = 0..n} (-1)^(n+k+1)*binomial(n,k)*binomial(n + k,k)*(n - k) = n^2. - Peter Bala, Jan 12 2022
Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)-1} such that |A|=|B|=k and A+B contains {0,1,2,...,a(n)-1}} = n. - Michael Chu, Mar 09 2022
Number of 3-permutations of n elements avoiding the patterns 132, 213, 321. See Bonichon and Sun. - Michel Marcus, Aug 20 2022
Number of intercalates in cyclic Latin squares of order 2n (cyclic Latin squares of odd order do not have intercalates). - Eduard I. Vatutin, Feb 15 2024
a(n) is the number of ternary strings of length n with at most one 0, exactly one 1, and no restriction on the number of 2's. For example, a(3)=9, consisting of the 6 permutations of the string 102 and the 3 permutations of the string 122. - Enrique Navarrete, Mar 12 2025

Examples

			For n = 8, a(8) = 8 * 15 - (1 + 3 + 5 + 7 + 9 + 11 + 13) - 7 = 8 * 15 - 49 - 7 = 64. - _Bruno Berselli_, May 04 2010
G.f. = x + 4*x^2 + 9*x^3 + 16*x^4 + 25*x^5 + 36*x^6 + 49*x^7 + 64*x^8 + 81*x^9 + ...
a(4) = 16. For n = 4 vertices, the cycle graph C4 is A-B-C-D-A. The subtrees are: 4 singles: A, B, C, D; 4 pairs: A-B, BC, C-D, A-D; 4 triples: A-B-C, B-C-D, C-D-A, D-A-B; 4 quads: A-B-C-D, B-C-D-A, C-D-A-B, D-A-B-C; 4 + 4 + 4 + 4 = 16. - _Viktar Karatchenia_, Mar 02 2016
		

References

  • G. L. Alexanderson et al., The William Lowell Putnam Mathematical Competition, Problems and Solutions: 1965-1984, "December 1967 Problem B4(a)", pp. 8(157) MAA Washington DC 1985.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Chapter XV, pp. 135-167.
  • R. P. Burn & A. Chetwynd, A Cascade Of Numbers, "The prison door problem" Problem 4 pp. 5-7; 79-80 Arnold London 1996.
  • H. Cohen, A Course in Computational Algebraic Number Theory, Springer, 1996, p. 40.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 31, 36, 38, 63.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), p. 6.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments, Chapter 6 pp. 71-2, W. H. Freeman NY 1988.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), p. 982.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology and §8.6 Figurate Numbers, pp. 264, 290-291.
  • Alfred S. Posamentier, The Art of Problem Solving, Section 2.4 "The Long Cell Block" pp. 10-1; 12; 156-7 Corwin Press Thousand Oaks CA 1996.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 35, 52-53, 129-132, 244.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • 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).
  • J. K. Strayer, Elementary Number Theory, Exercise Set 3.3 Problems 32, 33, p. 88, PWS Publishing Co. Boston MA 1996.
  • C. W. Trigg, Mathematical Quickies, "The Lucky Prisoners" Problem 141 pp. 40, 141, Dover NY 1985.
  • R. Vakil, A Mathematical Mosaic, "The Painted Lockers" pp. 127;134 Brendan Kelly Burlington Ontario 1996.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Crossrefs

Cf. A092205, A128200, A005408, A128201, A002522, A005563, A008865, A059100, A143051, A143470, A143595, A056944, A001157 (inverse Möbius transform), A001788 (binomial transform), A228039, A001105, A004159, A159918, A173277, A095794, A162395, A186646 (Pisano periods), A028338 (2nd diagonal).
A row or column of A132191.
This sequence is related to partitions of 2^n into powers of 2, as it is shown in A002577. So A002577 connects the squares and A000447. - Valentin Bakoev, Mar 03 2009
Boustrophedon transforms: A000697, A000745.
Cf. A342819.
Cf. A013661.

Programs

Formula

G.f.: x*(1 + x) / (1 - x)^3.
E.g.f.: exp(x)*(x + x^2).
Dirichlet g.f.: zeta(s-2).
a(n) = a(-n).
Multiplicative with a(p^e) = p^(2*e). - David W. Wilson, Aug 01 2001
Sum of all matrix elements M(i, j) = 2*i/(i+j) (i, j = 1..n). a(n) = Sum_{i = 1..n} Sum_{j = 1..n} 2*i/(i + j). - Alexander Adamchuk, Oct 24 2004
a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 2. - Miklos Kristof, Mar 09 2005
From Pierre CAMI, Oct 22 2006: (Start)
a(n) is the sum of the odd numbers from 1 to 2*n - 1.
a(0) = 0, a(1) = 1, then a(n) = a(n-1) + 2*n - 1. (End)
For n > 0: a(n) = A130064(n)*A130065(n). - Reinhard Zumkeller, May 05 2007
a(n) = Sum_{k = 1..n} A002024(n, k). - Reinhard Zumkeller, Jun 24 2007
Left edge of the triangle in A132111: a(n) = A132111(n, 0). - Reinhard Zumkeller, Aug 10 2007
Binomial transform of [1, 3, 2, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = binomial(n+1, 2) + binomial(n, 2).
This sequence could be derived from the following general formula (cf. A001286, A000330): n*(n+1)*...*(n+k)*(n + (n+1) + ... + (n+k))/((k+2)!*(k+1)/2) at k = 0. Indeed, using the formula for the sum of the arithmetic progression (n + (n+1) + ... + (n+k)) = (2*n + k)*(k + 1)/2 the general formula could be rewritten as: n*(n+1)*...*(n+k)*(2*n+k)/(k+2)! so for k = 0 above general formula degenerates to n*(2*n + 0)/(0 + 2) = n^2. - Alexander R. Povolotsky, May 18 2008
From a(4) recurrence formula a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) and a(1) = 1, a(2) = 4, a(3) = 9. - Artur Jasinski, Oct 21 2008
The recurrence a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) is satisfied by all k-gonal sequences from a(3), with a(0) = 0, a(1) = 1, a(2) = k. - Jaume Oliver Lafont, Nov 18 2008
a(n) = floor(n*(n+1)*(Sum_{i = 1..n} 1/(n*(n+1)))). - Ctibor O. Zizka, Mar 07 2009
Product_{i >= 2} 1 - 2/a(i) = -sin(A063448)/A063448. - R. J. Mathar, Mar 12 2009
a(n) = A002378(n-1) + n. - Jaroslav Krizek, Jun 14 2009
a(n) = n*A005408(n-1) - (Sum_{i = 1..n-2} A005408(i)) - (n-1) = n*A005408(n-1) - a(n-1) - (n-1). - Bruno Berselli, May 04 2010
a(n) == 1 (mod n+1). - Bruno Berselli, Jun 03 2010
a(n) = a(n-1) + a(n-2) - a(n-3) + 4, n > 2. - Gary Detlefs, Sep 07 2010
a(n+1) = Integral_{x >= 0} exp(-x)/( (Pn(x)*exp(-x)*Ei(x) - Qn(x))^2 +(Pi*exp(-x)*Pn(x))^2 ), with Pn the Laguerre polynomial of order n and Qn the secondary Laguerre polynomial defined by Qn(x) = Integral_{t >= 0} (Pn(x) - Pn(t))*exp(-t)/(x-t). - Groux Roland, Dec 08 2010
Euler transform of length-2 sequence [4, -1]. - Michael Somos, Feb 12 2011
A162395(n) = -(-1)^n * a(n). - Michael Somos, Mar 19 2011
a(n) = A004201(A000217(n)); A007606(a(n)) = A000384(n); A007607(a(n)) = A001105(n). - Reinhard Zumkeller, Feb 12 2011
Sum_{n >= 1} 1/a(n)^k = (2*Pi)^k*B_k/(2*k!) = zeta(2*k) with Bernoulli numbers B_k = -1, 1/6, 1/30, 1/42, ... for k >= 0. See A019673, A195055/10 etc. [Jolley eq 319].
Sum_{n>=1} (-1)^(n+1)/a(n)^k = 2^(k-1)*Pi^k*(1-1/2^(k-1))*B_k/k! [Jolley eq 320] with B_k as above.
A007968(a(n)) = 0. - Reinhard Zumkeller, Jun 18 2011
A071974(a(n)) = n; A071975(a(n)) = 1. - Reinhard Zumkeller, Jul 10 2011
a(n) = A199332(2*n - 1, n). - Reinhard Zumkeller, Nov 23 2011
For n >= 1, a(n) = Sum_{d|n} phi(d)*psi(d), where phi is A000010 and psi is A001615. - Enrique Pérez Herrero, Feb 29 2012
a(n) = A000217(n^2) - A000217(n^2 - 1), for n > 0. - Ivan N. Ianakiev, May 30 2012
a(n) = (A000217(n) + A000326(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = A162610(n, n) = A209297(n, n) for n > 0. - Reinhard Zumkeller, Jan 19 2013
a(A000217(n)) = Sum_{i = 1..n} Sum_{j = 1..n} i*j, for n > 0. - Ivan N. Ianakiev, Apr 20 2013
a(n) = A133280(A000217(n)). - Ivan N. Ianakiev, Aug 13 2013
a(2*a(n)+2*n+1) = a(2*a(n)+2*n) + a(2*n+1). - Vladimir Shevelev, Jan 24 2014
a(n+1) = Sum_{t1+2*t2+...+n*tn = n} (-1)^(n+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*4^(t1)*7^(t2)*8^(t3+...+tn). - Mircea Merca, Feb 27 2014
a(n) = floor(1/(1-cos(1/n)))/2 = floor(1/(1-n*sin(1/n)))/6, n > 0. - Clark Kimberling, Oct 08 2014
a(n) = ceiling(Sum_{k >= 1} log(k)/k^(1+1/n)) = -Zeta'[1+1/n]. Thus any exponent greater than 1 applied to k yields convergence. The fractional portion declines from A073002 = 0.93754... at n = 1 and converges slowly to 0.9271841545163232... for large n. - Richard R. Forberg, Dec 24 2014
a(n) = Sum_{j = 1..n} Sum_{i = 1..n} ceiling((i + j - n + 1)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = Product_{j = 1..n-1} 2 - 2*cos(2*j*Pi/n). - Michel Marcus, Jul 24 2015
From Ilya Gutkovskiy, Jun 21 2016: (Start)
Product_{n >= 1} (1 + 1/a(n)) = sinh(Pi)/Pi = A156648.
Sum_{n >= 0} 1/a(n!) = BesselI(0, 2) = A070910. (End)
a(n) = A028338(n, n-1), n >= 1 (second diagonal). - Wolfdieter Lang, Jul 21 2017
For n >= 1, a(n) = Sum_{d|n} sigma_2(d)*mu(n/d) = Sum_{d|n} A001157(d)*A008683(n/d). - Ridouane Oudra, Apr 15 2021
a(n) = Sum_{i = 1..2*n-1} ceiling(n - i/2). - Stefano Spezia, Apr 16 2021
From Richard L. Ollerton, May 09 2021: (Start) For n >= 1,
a(n) = Sum_{k=1..n} psi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} psi(gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} sigma_2(n/gcd(n,k))*mu(gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} sigma_2(gcd(n,k))*mu(n/gcd(n,k))/phi(n/gcd(n,k)). (End)
a(n) = (A005449(n) + A000326(n))/3. - Klaus Purath, May 13 2021
Let T(n) = A000217(n), then a(T(n)) + a(T(n+1)) = T(a(n+1)). - Charlie Marion, Jun 27 2022
a(n) = Sum_{k=1..n} sigma_1(k) + Sum_{i=1..n} (n mod i). - Vadim Kataev, Dec 07 2022
a(n^2) + a(n^2+1) + ... + a(n^2+n) + 4*A000537(n) = a(n^2+n+1) + ... + a(n^2+2n). In general, if P(k,n) = the n-th k-gonal number, then P(2k,n^2) + P(2k,n^2+1) + ... + P(2k,n^2+n) + 4*(k-1)*A000537(n) = P(2k,n^2+n+1) + ... + P(2k,n^2+2n). - Charlie Marion, Apr 26 2024
Sum_{n>=1} 1/a(n) = A013661. - Alois P. Heinz, Oct 19 2024
a(n) = 1 + 3^3*((n-1)/(n+1))^2 + 5^3*((n-1)*(n-2)/((n+1)*(n+2)))^2 + 7^3*((n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)))^2 + ... for n >= 1. - Peter Bala, Dec 09 2024

Extensions

Incorrect comment and example removed by Joerg Arndt, Mar 11 2010

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

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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

A000165 Double factorial of even numbers: (2n)!! = 2^n*n!.

Original entry on oeis.org

1, 2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200, 81749606400, 1961990553600, 51011754393600, 1428329123020800, 42849873690624000, 1371195958099968000, 46620662575398912000, 1678343852714360832000, 63777066403145711616000
Offset: 0

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a(n) is also the size of the automorphism group of the graph (edge graph) of the n-dimensional hypercube and also of the geometric automorphism group of the hypercube (the two groups are isomorphic). This group is an extension of an elementary Abelian group (C_2)^n by S_n. (C_2 is the cyclic group with two elements and S_n is the symmetric group.) - Avi Peretz (njk(AT)netvision.net.il), Feb 21 2001
Then a(n) appears in the power series: sqrt(1+sin(y)) = Sum_{n>=0} (-1)^floor(n/2)*y^(n)/a(n) and sqrt((1+cos(y))/2) = Sum_{n>=0} (-1)^n*y^(2n)/a(2n). - Benoit Cloitre, Feb 02 2002
Appears to be the BinomialMean transform of A001907. See A075271. - John W. Layman, Sep 28 2002
Number of n X n monomial matrices with entries 0, +-1.
Also number of linear signed orders.
Define a "downgrade" to be the permutation d which places the items of a permutation p in descending order. This note concerns those permutations that are equal to their double-downgrades. The number of permutations of order 2n having this property are equinumerous with those of order 2n+1. a(n) = number of double-downgrading permutations of order 2n and 2n+1. - Eugene McDonnell (eemcd(AT)mac.com), Oct 27 2003
a(n) = (Integral_{x=0..Pi/2} cos(x)^(2*n+1) dx) where the denominators are b(n) = (2*n)!/(n!*2^n). - Al Hakanson (hawkuu(AT)excite.com), Mar 02 2004
1 + (1/2)x - (1/8)x^2 - (1/48)x^3 + (1/384)x^4 + ... = sqrt(1+sin(x)).
a(n)*(-1)^n = coefficient of the leading term of the (n+1)-th derivative of arctan(x), see Hildebrand link. - Reinhard Zumkeller, Jan 14 2006
a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is j for iDavid Callan, Sep 25 2006
a(n) is the number of increasing plane trees with n+1 edges. (In a plane tree, each subtree of the root is an ordered tree but the subtrees of the root may be cyclically rotated.) Increasing means the vertices are labeled 0,1,2,...,n+1 and each child has a greater label than its parent. Cf. A001147 for increasing ordered trees, A000142 for increasing unordered trees and A000111 for increasing 0-1-2 trees. - David Callan, Dec 22 2006
Hamed Hatami and Pooya Hatami prove that this is an upper bound on the cardinality of any minimal dominating set in C_{2n+1}^n, the Cartesian product of n copies of the cycle of size 2n+1, where 2n+1 is a prime. - Jonathan Vos Post, Jan 03 2007
This sequence and (1,-2,0,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007
a(n) = number of permutations of the multiset {1,1,2,2,...,n,n,n+1,n+1} such that between the two occurrences of i, there is exactly one entry >i, for i=1,2,...,n. Example: a(2) = 8 counts 121323, 131232, 213123, 231213, 232131, 312132, 321312, 323121. Proof: There is always exactly one entry between the two 1s (when n>=1). Given a permutation p in A(n) (counted by a(n)), record the position i of the first 1, then delete both 1s and subtract 1 from every entry to get a permutation q in A(n-1). The mapping p -> (i,q) is a bijection from A(n) to the Cartesian product [1,2n] X A(n-1). - David Callan, Nov 29 2007
Row sums of A028338. - Paul Barry, Feb 07 2009
a(n) is the number of ways to seat n married couples in a row so that everyone is next to their spouse. Compare A007060. - Geoffrey Critzer, Mar 29 2009
From Gary W. Adamson, Apr 21 2009: (Start)
Equals (-1)^n * (1, 1, 2, 8, 48, ...) dot (1, -3, 5, -7, 9, ...).
Example: a(4) = 384 = (1, 1, 2, 8, 48) dot (1, -3, 5, -7, 9) = (1, -3, 10, -56, 432). (End)
exp(x/2) = Sum_{n>=0} x^n/a(n). - Jaume Oliver Lafont, Sep 07 2009
Assuming n starts at 0, a(n) appears to be the number of Gray codes on n bits. It certainly is the number of Gray codes on n bits isomorphic to the canonical one. Proof: There are 2^n different starting positions for each code. Also, each code has a particular pattern of bit positions that are flipped (for instance, 1 2 1 3 1 2 1 for n=3), and these bit position patterns can be permuted in n! ways. - D. J. Schreffler (ds1404(AT)txstate.edu), Jul 18 2010
E.g.f. of 0,1,2,8,... is x/(1-2x/(2-2x/(3-8x/(4-8x/(5-18x/(6-18x/(7-... (continued fraction). - Paul Barry, Jan 17 2011
Number of increasing 2-colored trees with choice of two colors for each edge. In general, if we replace 2 with k we get the number of increasing k-colored trees. For example, for k=3 we get the triple factorial numbers. - Wenjin Woan, May 31 2011
a(n) = row sums of triangle A193229. - Gary W. Adamson, Jul 18 2011
Also the number of permutations of 2n (or of 2n+1) that are equal to their reverse-complements. (See the Egge reference.) Note that the double-downgrade described in the preceding comment (McDonnell) is equivalent to the reverse-complement. - Justin M. Troyka, Aug 11 2011
The e.g.f. can be used to form a generator, [1/(1-2x)] d/dx, for A000108, so a(n) can be applied to A145271 to generate the Catalan numbers. - Tom Copeland, Oct 01 2011
The e.g.f. of 1/a(n) is BesselI(0,sqrt(2*x)). See Abramowitz-Stegun (reference and link under A008277), p. 375, 9.6.10. - Wolfdieter Lang, Jan 09 2012
a(n) = order of the largest imprimitive group of degree 2n with n systems of imprimitivity (see [Miller], p. 203). - L. Edson Jeffery, Feb 05 2012
Row sums of triangle A208057. - Gary W. Adamson, Feb 22 2012
a(n) is the number of ways to designate a subset of elements in each n-permutation. a(n) = A000142(n) + A001563(n) + A001804(n) + A001805(n) + A001806(n) + A001807(n) + A035038(n) * n!. - Geoffrey Critzer, Nov 08 2012
For n>1, a(n) is the order of the Coxeter groups (also called Weyl groups) of types B_n and C_n. - Tom Edgar, Nov 05 2013
For m>0, k*a(m-1) is the m-th cumulant of the chi-squared probability distribution for k degrees of freedom. - Stanislav Sykora, Jun 27 2014
a(n) with 0 prepended is the binomial transform of A120765. - Vladimir Reshetnikov, Oct 28 2015
Exponential self-convolution of A001147. - Vladimir Reshetnikov, Oct 08 2016
Also the order of the automorphism group of the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017
a(n) is the order of the group O_n(Z) = {A in M_n(Z): A*A^T = I_n}, the group of n X n orthogonal matrices over the integers. - Jianing Song, Mar 29 2021
a(n) is the number of ways to tile a (3n,3n)-benzel or a (3n+1,3n+2)-benzel using left stones and two kinds of bones; see Defant et al., below. - James Propp, Jul 22 2023
a(n) is the number of labeled histories for a labeled topology with the modified lodgepole shape and n+1 cherry nodes. - Noah A Rosenberg, Jan 16 2025

Examples

			The following permutations and their reversals are all of the permutations of order 5 having the double-downgrade property:
  0 1 2 3 4
  0 3 2 1 4
  1 0 2 4 3
  1 4 2 0 3
G.f. = 1 + 2*x + 8*x^2 + 48*x^3 + 384*x^4 + 3840*x^5 + 46080*x^6 + 645120*x^7 + ...
		

References

  • 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

Cf. A000142 (n!), A001147 ((2n-1)!!), A032184 (2^n*(n-1)!).
This sequence gives the row sums in A060187, and (-1)^n*a(n) the alternating row sums in A039757.
Also row sums in A028338.
Column k=2 of A329070.

Programs

  • Haskell
    a000165 n = product [2, 4 .. 2 * n]  -- Reinhard Zumkeller, Mar 28 2015
    
  • Magma
    [2^n*Factorial(n): n in [0..35]]; // Vincenzo Librandi, Apr 22 2011
    
  • Magma
    I:=[2,8]; [1] cat [n le 2 select I[n]  else (3*n-1)*Self(n-1)-2*(n-1)^2*Self(n-2): n in [1..35] ]; // Vincenzo Librandi, Feb 19 2015
    
  • Maple
    A000165 := proc(n) option remember; if n <= 1 then 1 else n*A000165(n-2); fi; end;
    ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card >= 0)}, labelled]: seq(combstruct[count](ZL, size=n), n=0..17); # Zerinvary Lajos, Mar 26 2008
    G(x):=(1-2*x)^(-1): 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..17); # Zerinvary Lajos, Apr 03 2009
    A000165 := proc(n) doublefactorial(2*n) ; end proc; seq(A000165(n),n=0..10) ; # R. J. Mathar, Oct 20 2009
  • Mathematica
    Table[(2 n)!!, {n, 30}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)
    (2 Range[0, 30])!! (* Harvey P. Dale, Jan 23 2015 *)
    RecurrenceTable[{a[n] == 2 n*a[n-1], a[0] == 1}, a, {n,0,30}] (* Ray Chandler, Jul 30 2015 *)
  • PARI
    a(n)=n!<Charles R Greathouse IV, Feb 11 2011
    
  • PARI
    {a(n) = prod( k=1, n, 2*k)}; /* Michael Somos, Jan 04 2013 */
    
  • Python
    from math import factorial
    def A000165(n): return factorial(n)<Chai Wah Wu, Jan 24 2023
    
  • SageMath
    [2^n*factorial(n) for n in range(31)] # G. C. Greubel, Jul 21 2024

Formula

E.g.f.: 1/(1-2*x).
a(n) = A001044(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (2*i+2) = 2^n*Pochhammer(1,n). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
D-finite with recurrence a(n) = 2*n * a(n-1), n>0, a(0)=1. - Paul Barry, Aug 26 2004
This is the binomial mean transform of A001907. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(n) = Integral_{x>=0} x^n*exp(-x/2)/2 dx. - Paul Barry, Jan 28 2008
G.f.: 1/(1-2x/(1-2x/(1-4x/(1-4x/(1-6x/(1-6x/(1-.... (continued fraction). - Paul Barry, Feb 07 2009
a(n) = A006882(2*n). - R. J. Mathar, Oct 20 2009
From Gary W. Adamson, Jul 18 2011: (Start)
a(n) = upper left term in M^n, M = a production matrix (twice Pascal's triangle deleting the first "2", with the rest zeros; cf. A028326):
2, 2, 0, 0, 0, 0, ...
2, 4, 2, 0, 0, 0, ...
2, 6, 6, 2, 0, 0, ...
2, 8, 12, 8, 2, 0, ...
2, 10, 20, 20, 10, 2, ...
... (End)
From Sergei N. Gladkovskii, Apr 11 2013, May 01 2013, May 24 2013, Sep 30 2013, Oct 27 2013: (Start)
Continued fractions:
G.f.: 1 + x*(Q(0) - 1)/(x+1) where Q(k) = 1 + (2*k+2)/(1-x/(x+1/Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 + 2*k*x - 2*x*(k+1)/Q(k+1).
G.f.: G(0)/2 where G(k) = 1 + 1/(1 - x*(2*k+2)/(x*(2*k+2) + 1/G(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 - x*(4*k+2) - 4*x^2*(k+1)^2/Q(k+1).
G.f.: R(0) where R(k) = 1 - x*(2*k+2)/(x*(2*k+2)-1/(1-x*(2*k+2)/(x*(2*k+2) -1/R(k+1)))). (End)
a(n) = (2n-2)*a(n-2) + (2n-1)*a(n-1), n>1. - Ivan N. Ianakiev, Aug 06 2013
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 1)*a(n-1) - 2*(n - 1)^2*a(n-2) with a(1) = 2 and a(2) = 8.
The sequence b(n) = A068102(n) also satisfies this second-order recurrence. This leads to the generalized continued fraction expansion lim_{n -> oo} b(n)/a(n) = log(2) = 1/(2 - 2/(5 - 8/(8 - 18/(11 - ... - 2*(n - 1)^2/((3*n - 1) - ... ))))). (End)
From Amiram Eldar, Jun 25 2020: (Start)
Sum_{n>=0} 1/a(n) = sqrt(e) (A019774).
Sum_{n>=0} (-1)^n/a(n) = 1/sqrt(e) (A092605). (End)
Limit_{n->oo} a(n)^4 / (n * A134372(n)) = Pi. - Daniel Suteu, Apr 09 2022
a(n) = 1/([x^n] hypergeom([1], [1], x/2)). - Peter Luschny, Sep 13 2024
a(n) = Sum_{k=0..n} k!*(n-k)!*binomial(n,k)^2. - Ridouane Oudra, Jul 13 2025

A047999 Sierpiński's [Sierpinski's] triangle (or gasket): triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 2.

Original entry on oeis.org

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

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Comments

Restored the alternative spelling of Sierpinski to facilitate searching for this triangle using regular-expression matching commands in ASCII. - N. J. A. Sloane, Jan 18 2016
Also triangle giving successive states of cellular automaton generated by "Rule 60" and "Rule 102". - Hans Havermann, May 26 2002
Also triangle formed by reading triangle of Eulerian numbers (A008292) mod 2. - Philippe Deléham, Oct 02 2003
Self-inverse when regarded as an infinite lower triangular matrix over GF(2).
Start with [1], repeatedly apply the map 0 -> [00/00], 1 -> [10/11] [Allouche and Berthe]
Also triangle formed by reading triangles A011117, A028338, A039757, A059438, A085881, A086646, A086872, A087903, A104219 mod 2. - Philippe Deléham, Jun 18 2005
J. H. Conway writes (in Math Forum): at least the first 31 rows give odd-sided constructible polygons (sides 1, 3, 5, 15, 17, ... see A001317). The 1's form a Sierpiński sieve. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005
When regarded as an infinite lower triangular matrix, its inverse is a (-1,0,1)-matrix with zeros undisturbed and the nonzero entries in every column form the Prouhet-Thue-Morse sequence (1,-1,-1,1,-1,1,1,-1,...) A010060 (up to relabeling). - David Callan, Oct 27 2006
Triangle read by rows: antidiagonals of an array formed by successive iterates of running sums mod 2, beginning with (1, 1, 1, ...). - Gary W. Adamson, Jul 10 2008
T(n,k) = A057427(A143333(n,k)). - Reinhard Zumkeller, Oct 24 2010
The triangle sums, see A180662 for their definitions, link Sierpiński’s triangle A047999 with seven sequences, see the crossrefs. The Kn1y(n) and Kn2y(n), y >= 1, triangle sums lead to the Sierpiński-Stern triangle A191372. - Johannes W. Meijer, Jun 05 2011
Used to compute the total Steifel-Whitney cohomology class of the Real Projective space. This was an essential component of the proof that there are no product operations without zero divisors on R^n for n not equal to 1, 2, 4 or 8 (real numbers, complex numbers, quaternions, Cayley numbers), proved by Bott and Milnor. - Marcus Jaiclin, Feb 07 2012
T(n,k) = A134636(n,k) mod 2. - Reinhard Zumkeller, Nov 23 2012
T(n,k) = 1 - A219463(n,k), 0 <= k <= n. - Reinhard Zumkeller, Nov 30 2012
From Vladimir Shevelev, Dec 31 2013: (Start)
Also table of coefficients of polynomials s_n(x) of degree n which are defined by formula s_n(x) = Sum_{i=0..n} (binomial(n,i) mod 2)*x^k. These polynomials we naturally call Sierpiński's polynomials. They also are defined by the recursion: s_0(x)=1, s_(2*n+1)(x) = (x+1)*s_n(x^2), n>=0, and s_(2*n)(x) = s_n(x^2), n>=1.
Note that: s_n(1) = A001316(n),
s_n(2) = A001317(n),
s_n(3) = A100307(n),
s_n(4) = A001317(2*n),
s_n(5) = A100308(n),
s_n(6) = A100309(n),
s_n(7) = A100310(n),
s_n(8) = A100311(n),
s_n(9) = A100307(2*n),
s_n(10) = A006943(n),
s_n(16) = A001317(4*n),
s_n(25) = A100308(2*n), etc.
The equality s_n(10) = A006943(n) means that sequence A047999 is obtained from A006943 by the separation by commas of the digits of its terms. (End)
Comment from N. J. A. Sloane, Jan 18 2016: (Start)
Take a diamond-shaped region with edge length n from the top of the triangle, and rotate it by 45 degrees to get a square S_n. Here is S_6:
[1, 1, 1, 1, 1, 1]
[1, 0, 1, 0, 1, 0]
[1, 1, 0, 0, 1, 1]
[1, 0, 0, 0, 1, 0]
[1, 1, 1, 1, 0, 0]
[1, 0, 1, 0, 0, 0].
Then (i) S_n contains no square (parallel to the axes) with all four corners equal to 1 (cf. A227133); (ii) S_n can be constructed by using the greedy algorithm with the constraint that there is no square with that property; and (iii) S_n contains A064194(n) 1's. Thus A064194(n) is a lower bound on A227133(n). (End)
See A123098 for a multiplicative encoding of the rows, i.e., product of the primes selected by nonzero terms; e.g., 1 0 1 => 2^1 * 3^0 * 5^1. - M. F. Hasler, Sep 18 2016
From Valentin Bakoev, Jul 11 2020: (Start)
The Sierpinski's triangle with 2^n rows is a part of a lower triangular matrix M_n of dimension 2^n X 2^n. M_n is a block matrix defined recursively: M_1= [1, 0], [1, 1], and for n>1, M_n = [M_(n-1), O_(n-1)], [M_(n-1), M_(n-1)], where M_(n-1) is a block matrix of the same type, but of dimension 2^(n-1) X 2^(n-1), and O_(n-1) is the zero matrix of dimension 2^(n-1) X 2^(n-1). Here is how M_1, M_2 and M_3 look like:
1 0 1 0 0 0 1 0 0 0 0 0 0 0
1 1 1 1 0 0 1 1 0 0 0 0 0 0 - It is seen the self-similarity of the
1 0 1 0 1 0 1 0 0 0 0 0 matrices M_1, M_2, ..., M_n, ...,
1 1 1 1 1 1 1 1 0 0 0 0 analogously to the Sierpinski's fractal.
1 0 0 0 1 0 0 0
1 1 0 0 1 1 0 0
1 0 1 0 1 0 1 0
1 1 1 1 1 1 1 1
M_n can also be defined as M_n = M_1 X M_(n-1) where X denotes the Kronecker product. M_n is an important matrix in coding theory, cryptography, Boolean algebra, monotone Boolean functions, etc. It is a transformation matrix used in computing the algebraic normal form of Boolean functions. Some properties and links concerning M_n can be seen in LINKS. (End)
Sierpinski's gasket has fractal (Hausdorff) dimension log(A000217(2))/log(2) = log(3)/log(2) = 1.58496... (and cf. A020857). This gasket is the first of a family of gaskets formed by taking the Pascal triangle (A007318) mod j, j >= 2 (see CROSSREFS). For prime j, the dimension of the gasket is log(A000217(j))/log(j) = log(j(j + 1)/2)/log(j) (see Reiter and Bondarenko references). - Richard L. Ollerton, Dec 14 2021

Examples

			Triangle begins:
              1,
             1,1,
            1,0,1,
           1,1,1,1,
          1,0,0,0,1,
         1,1,0,0,1,1,
        1,0,1,0,1,0,1,
       1,1,1,1,1,1,1,1,
      1,0,0,0,0,0,0,0,1,
     1,1,0,0,0,0,0,0,1,1,
    1,0,1,0,0,0,0,0,1,0,1,
   1,1,1,1,0,0,0,0,1,1,1,1,
  1,0,0,0,1,0,0,0,1,0,0,0,1,
  ...
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Brand, Neal; Das, Sajal; Jacob, Tom. The number of nonzero entries in recursively defined tables modulo primes. Proceedings of the Twenty-first Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1990). Congr. Numer. 78 (1990), 47--59. MR1140469 (92h:05004).
  • John W. Milnor and James D. Stasheff, Characteristic Classes, Princeton University Press, 1974, pp. 43-49 (sequence appears on p. 46).
  • H.-O. Peitgen, H. Juergens and D. Saupe: Chaos and Fractals (Springer-Verlag 1992), p. 408.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

Crossrefs

Sequences based on the triangles formed by reading Pascal's triangle mod m: (this sequence) (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).
Other versions: A090971, A038183.
From Johannes W. Meijer, Jun 05 2011: (Start)
A106344 is a skew version of this triangle.
Triangle sums (see the comments): A001316 (Row1; Related to Row2), A002487 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A007306 (Kn3, Kn4), A060632 (Fi1, Fi2), A120562 (Ca1, Ca2), A112970 (Gi1, Gi2), A127830 (Ze3, Ze4). (End)

Programs

  • Haskell
    import Data.Bits (xor)
    a047999 :: Int -> Int -> Int
    a047999 n k = a047999_tabl !! n !! k
    a047999_row n = a047999_tabl !! n
    a047999_tabl = iterate (\row -> zipWith xor ([0] ++ row) (row ++ [0])) [1]
    -- Reinhard Zumkeller, Dec 11 2011, Oct 24 2010
    
  • Magma
    A047999:= func< n,k | BitwiseAnd(n-k, k) eq 0 select 1 else 0 >;
    [A047999(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 03 2024
  • Maple
    # Maple code for first M rows (here M=10) - N. J. A. Sloane, Feb 03 2016
    ST:=[1,1,1]; a:=1; b:=2; M:=10;
    for n from 2 to M do ST:=[op(ST),1];
    for i from a to b-1 do ST:=[op(ST), (ST[i+1]+ST[i+2]) mod 2 ]; od:
    ST:=[op(ST),1];
    a:=a+n; b:=a+n; od:
    ST; # N. J. A. Sloane
    # alternative
    A047999 := proc(n,k)
        modp(binomial(n,k),2) ;
    end proc:
    seq(seq(A047999(n,k),k=0..n),n=0..12) ; # R. J. Mathar, May 06 2016
  • Mathematica
    Mod[ Flatten[ NestList[ Prepend[ #, 0] + Append[ #, 0] &, {1}, 13]], 2] (* Robert G. Wilson v, May 26 2004 *)
    rows = 14; ca = CellularAutomaton[60, {{1}, 0}, rows-1]; Flatten[ Table[ca[[k, 1 ;; k]], {k, 1, rows}]] (* Jean-François Alcover, May 24 2012 *)
    Mod[#,2]&/@Flatten[Table[Binomial[n,k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 26 2019 *)
  • PARI
    \\ Recurrence for Pascal's triangle mod p, here p = 2.
    p = 2; s=13; T=matrix(s,s); T[1,1]=1;
    for(n=2,s, T[n,1]=1; for(k=2,n, T[n,k] = (T[n-1,k-1] + T[n-1,k])%p ));
    for(n=1,s,for(k=1,n,print1(T[n,k],", "))) \\ Gerald McGarvey, Oct 10 2009
    
  • PARI
    A011371(n)=my(s);while(n>>=1,s+=n);s
    T(n,k)=A011371(n)==A011371(k)+A011371(n-k) \\ Charles R Greathouse IV, Aug 09 2013
    
  • PARI
    T(n,k)=bitand(n-k,k)==0 \\ Charles R Greathouse IV, Aug 11 2016
    
  • Python
    def A047999_T(n,k):
        return int(not ~n & k) # Chai Wah Wu, Feb 09 2016
    

Formula

Lucas's Theorem is that T(n,k) = 1 if and only if the 1's in the binary expansion of k are a subset of the 1's in the binary expansion of n; or equivalently, k AND NOT n is zero, where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 09 2016 and N. J. A. Sloane, Feb 10 2016
Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).
T(n,k) = T(n-1,k-1) XOR T(n-1,k), 0 < k < n; T(n,0) = T(n,n) = 1. - Reinhard Zumkeller, Dec 13 2009
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 2 = |T(n-1,k-1) - T(n-1,k)|, 0 < k < n; T(n,0) = T(n,n) = 1. - Rick L. Shepherd, Feb 23 2018
From Vladimir Shevelev, Dec 31 2013: (Start)
For polynomial {s_n(x)} we have
s_0(x)=1; for n>=1, s_n(x) = Product_{i=1..A000120(n)} (x^(2^k_i) + 1),
if the binary expansion of n is n = Sum_{i=1..A000120(n)} 2^k_i;
G.f. Sum_{n>=0} s_n(x)*z^n = Product_{k>=0} (1 + (x^(2^k)+1)*z^(2^k)) (0
Let x>1, t>0 be real numbers. Then
Sum_{n>=0} 1/s_n(x)^t = Product_{k>=0} (1 + 1/(x^(2^k)+1)^t);
Sum_{n>=0} (-1)^A000120(n)/s_n(x)^t = Product_{k>=0} (1 - 1/(x^(2^k)+1)^t).
In particular, for t=1, x>1, we have
Sum_{n>=0} (-1)^A000120(n)/s_n(x) = 1 - 1/x. (End)
From Valentin Bakoev, Jul 11 2020: (Start)
(See my comment about the matrix M_n.) Denote by T(i,j) the number in the i-th row and j-th column of M_n (0 <= i, j < 2^n). When i>=j, T(i,j) is the j-th number in the i-th row of the Sierpinski's triangle. For given i and j, we denote by k the largest integer of the type k=2^m and k
T(i,0) = T(i,i) = 1, or
T(i,j) = 0 if i < j, or
T(i,j) = T(i-k,j), if j < k, or
T(i,j) = T(i-k,j-k), if j >= k.
Thus, for given i and j, T(i,j) can be computed in O(log_2(i)) steps. (End)

Extensions

Additional links from Lekraj Beedassy, Jan 22 2004

A161198 Triangle of polynomial coefficients related to the series expansions of (1-x)^((-1-2*n)/2).

Original entry on oeis.org

1, 1, 2, 3, 8, 4, 15, 46, 36, 8, 105, 352, 344, 128, 16, 945, 3378, 3800, 1840, 400, 32, 10395, 39048, 48556, 27840, 8080, 1152, 64, 135135, 528414, 709324, 459032, 160720, 31136, 3136, 128
Offset: 0

Author

Johannes W. Meijer, Jun 08 2009, Jul 22 2011

Keywords

Comments

The series expansion of (1-x)^((-1-2*n)/2) = sum(b(p)*x^p, p=0..infinity) for n = 0, 1, 2, .. can be described with b(p) = (F(p,n)/ (2*n-1)!!)*(binomial(2*p,p)/4^(p)) with F(x,n) = 2^n * product( x+(2*k-1)/2, k=1..n). The roots of the F(x,n) polynomials can be found at p = (1-2*k)/2 with k from 1 to n for n = 0, 1, 2, .. . The coefficients of the F(x,n) polynomials lead to the triangle given above. The triangle row sums lead to A001147.
Quite surprisingly we discovered that sum(b(p)*x^p, p=0..infinity) = (1-x)^(-1-2*n)/2, for n = -1, -2, .. . We assume that if m = n+1 then the value returned for product(f(k), k = m..n) is 1 and if m> n+1 then 1/product(f(k), k=n+1..m-1) is the value returned. Furthermore (1-2*n)!! = (-1)^(n+1)/(2*n-3)!! for n = 1, 2, 3 .. . This leads to b(p) = ((-1-2*n)!!/ G(p,n))*(binomial(2*p,p) /4^(p)) for n = -1, -2, .. . For the G(p,n) polynomials we found that G(p,n) = F(-p,-n). The roots of the G(p,n) polynomials can be found at p=(2*k-1)/2 with k from 1 to (-n) for n = -1, -2, .. . The coefficients of the G(p,n) polynomials lead to a second triangle that stands with its head on top of the first one. It is remarkable that the row sums lead once again to A001147.
These two triangles together look like an hourglass so we propose to call the F(p,n) and the G(p,n) polynomials the hourglass polynomials.
Triangle T(n,k), read by rows, given by (1, 2, 3, 4, 5, 6, 7, 8, 9, ...) DELTA (2, 0, 2, 0, 2, 0, 2, 0, 2, ...) where DELTA is the operator defined in A084938. Philippe Deléham, May 14 2015.

Examples

			From _Gary W. Adamson_, Jul 19 2011: (Start)
The first few rows of matrix M are:
  1, 2,  0,  0, 0, ...
  1, 3,  2,  0, 0, ...
  1, 4,  5,  2, 0, ...
  1, 5,  9,  7, 2, ...
  1, 6, 14, 16, 9, ... (End)
The first few G(p,n) polynomials are:
  G(p,-3) = 15 - 46*p + 36*p^2 - 8*p^3
  G(p,-2) = 3 - 8*p + 4*p^2
  G(p,-1) = 1 - 2*p
The first few F(p,n) polynomials are:
  F(p,0) = 1
  F(p,1) = 1 + 2*p
  F(p,2) = 3 + 8*p + 4*p^2
  F(p,3) = 15 + 46*p + 36*p^2 + 8*p^3
The first few rows of the upper and lower hourglass triangles are:
  [15, -46, 36, -8]
  [3, -8, 4]
  [1, -2]
  [1]
  [1, 2]
  [3, 8, 4]
  [15, 46, 36, 8]
		

Crossrefs

Cf. A001790 [(1-x)^(-1/2)], A001803 [(1-x)^(-3/2)], A161199 [(1-x)^(-5/2)] and A161201 [(1-x)^(-7/2)].
Cf. A002596 [(1-x)^(1/2)], A161200 [(1-x)^(3/2)] and A161202 [(1-x)^(5/2)].
A046161 gives the denominators of the series expansions of all (1-x)^((-1-2*n)/2).
A028338 is a scaled triangle version, A039757 is a scaled signed triangle version and A109692 is a transposed scaled triangle version.
A001147 is the first left hand column and equals the row sums.
A004041 is the second left hand column divided by 2, A028339 is the third left hand column divided by 4, A028340 is the fourth left hand column divided by 8, A028341 is the fifth left hand column divided by 16.
A000012, A000290, A024196, A024197 and A024198 are the first (n-m=0), second (n-m=1), third (n-m=2), fourth (n-m=3) and fifth (n-m=4) right hand columns divided by 2^m.
A074599 * A025549 is not always equals the second left hand column.
Cf. A029635. [Gary W. Adamson, Jul 19 2011]

Programs

  • Maple
    nmax:=7; for n from 0 to nmax do a(n,n):=2^n: a(n,0):=doublefactorial(2*n-1) od: for n from 2 to nmax do for m from 1 to n-1 do a(n,m) := 2*a(n-1,m-1)+(2*n-1)*a(n-1,m) od: od: seq(seq(a(n,k), k=0..n), n=0..nmax);
    nmax:=7: M := Matrix(1..nmax+1,1..nmax+1): A029635 := proc(n,k): binomial(n,k) + binomial(n-1,k-1) end: for i from 1 to nmax do for j from 1 to i+1 do M[i,j] := A029635(i,j-1) od: od: for n from 0 to nmax do B := M^n: for m from 0 to n do a(n,m):= B[1,m+1] od: od: seq(seq(a(n,m), m=0..n), n=0..nmax);
    A161198 := proc(n,k) option remember; if k > n or k < 0 then 0 elif n = 0 and k = 0 then 1 else 2*A161198(n-1, k-1) + (2*n-1)*A161198(n-1, k) fi end:
    seq(print(seq(A161198(n,k), k = 0..n)), n = 0..6);  # Peter Luschny, May 09 2013
  • Mathematica
    nmax = 7; a[n_, 0] := (2*n-1)!!; a[n_, n_] := 2^n; a[n_, m_] := a[n, m] = 2*a[n-1, m-1]+(2*n-1)*a[n-1, m]; Table[a[n, m], {n, 0, nmax}, {m, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014, after Maple *)
  • PARI
    for(n=0,9, print(Vec(Ser( 2^n*prod( k=1,n, x+(2*k-1)/2 ),,n+1))))  \\ M. F. Hasler, Jul 23 2011
    
  • Sage
    @CachedFunction
    def A161198(n,k):
        if k > n or k < 0 : return 0
        if n == 0 and k == 0: return 1
        return 2*A161198(n-1,k-1)+(2*n-1)*A161198(n-1,k)
    for n in (0..6): [A161198(n,k) for k in (0..n)]  # Peter Luschny, May 09 2013

Formula

a(n,m) := coeff(2^(n)*product((x+(2*k-1)/2),k=1..n), x, m) for n = 0, 1, .. ; m = 0, 1, .. .
a(n, m) = 2*a(n-1,m-1)+(2*n-1)*a(n-1,m) with a(n, n) = 2^n and a(n, 0) = (2*n-1)!!.
a(n,m) = the (m+1)-th term in the top row of M^n, where M is an infinite square production matrix; M[i,j] = A029635(i,j-1) = binomial(i, j-1) + binomial(i-1, j-2) with A029635 the (1.2)-Pascal triangle, see the examples and second Maple program. [Gary W. Adamson, Jul 19 2011]
T(n,k) = 2^k * A028338(n,k). - Philippe Deléham, May 14 2015

A004041 Scaled sums of odd reciprocals: a(n) = (2*n + 1)!!*(Sum_{k=0..n} 1/(2*k + 1)).

Original entry on oeis.org

1, 4, 23, 176, 1689, 19524, 264207, 4098240, 71697105, 1396704420, 29985521895, 703416314160, 17901641997225, 491250187505700, 14459713484342175, 454441401368236800, 15188465029114325025, 537928935889764226500
Offset: 0

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

n-th elementary symmetric function of the first n+1 odd positive integers.
Also the determinant of the n X n matrix given by m(i,j) = 2*i + 2 = if i = j and otherwise 1. For example, Det[{{4, 1, 1, 1, 1, 1}, {1, 6, 1, 1, 1, 1}, {1, 1, 8, 1, 1, 1}, {1, 1, 1, 10, 1, 1}, {1, 1, 1, 1, 12, 1}, {1, 1, 1, 1, 1, 14}}] = 264207 = a(6). - John M. Campbell, May 20 2011

Examples

			(arctanh(x))^2 = x^2 + 2/3*x^4 + 23/45*x^6 + 44/105*x^8 + ...
		

Crossrefs

Cf. A002428.
From Johannes W. Meijer, Jun 08 2009: (Start)
Equals second left hand column of A028338 triangle.
Equals second right hand column of A109692 triangle.
Equals second left hand column of A161198 triangle divided by 2.
(End)

Programs

  • Mathematica
    Table[(-1)^(n + 1)* Sum[(-2)^(n - k) k (-1)^(n - k) StirlingS1[n + 1, k + 1], {k, 0, n}], {n, 1, 18}] (* Zerinvary Lajos, Jul 08 2009 *)
    FunctionExpand@Table[(2 n + 1)!! (Log[4] + HarmonicNumber[n + 1/2])/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 13 2016 *)

Formula

a(n) = (2*n + 1)!!*(Sum_{k=0..n} 1/(2*k + 1)).
a(n) is coefficient of x^(2*n+2) in (arctanh x)^2, multiplied by (n + 1)*(2*n + 1)!!.
a(n) = Sum_{i=k+1..n} (-1)^(k+1-i)*2^(n-1)*binomial(i-1, k)*s1(n, i) with k = 1, where s1(n, i) are unsigned Stirling numbers of the first kind. - Victor Adamchik (adamchik(AT)ux10.sp.cs.cmu.edu), Jan 23 2001
a(n) ~ 2^(1/2)*log(n)*n*(2n/e)^n. - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
E.g.f.: 1/2*(1 - 2*x)^(-3/2)*(2 - log(1 - 2*x)). - Vladeta Jovovic, Feb 19 2003
Sum_{n>=1} a(n-1)/(n!*n*2^n) = (Pi/2)^2. - Philippe Deléham, Aug 12 2003
For n >= 1, a(n-1) = 2^(n-1)*n!*(Sum_{k=0..n-1} (-1)^k*binomial(1/2, k)/(n - k)). - Milan Janjic, Dec 14 2008
Recurrence: a(n) = 4*n*a(n-1) - (2*n - 1)^2*a(n-2). - Vladimir Reshetnikov, Oct 13 2016

A286718 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
Offset: 0

Author

Wolfdieter Lang, May 18 2017

Keywords

Comments

This is a generalization of the unsigned Stirling1 triangle A132393.
In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/A002209(k+1).
This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018

Examples

			The triangle T(n, k) begins:
n\k        0        1        2       3      4     5    6  7 8 ...
O:         1
1:         1        1
2:         4        5        1
3:        28       39       12       1
4:       280      418      159      22      1
5:      3640     5714     2485     445     35     1
6:     58240    95064    45474    9605   1005    51    1
7:   1106560  1864456   959070  227969  28700  1974   70  1
8:  24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
From _Wolfdieter Lang_, Aug 09 2017: (Start)
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Crossrefs

S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
Column sequences for k = 0..4: A007559, A024216(n-1), A286721(n-2), A382984, A382985.
Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
Row sums: A008544. Alternating row sums: A000007.
Beta sequence: A002208(n+1)/A002209(n+1).

Programs

  • Mathematica
    T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[, -1] = 0; T[0, 0] = 1; T[n, k_] /; nJean-François Alcover, Jun 20 2018 *)

Formula

Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;x,n) with the rising factorial
risefac(d,a;x,n) := Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}{n-k}(a_0,a_1,...,a{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017

A039757 Triangle of coefficients in expansion of (x-1)*(x-3)*(x-5)*...*(x-(2*n-1)).

Original entry on oeis.org

1, -1, 1, 3, -4, 1, -15, 23, -9, 1, 105, -176, 86, -16, 1, -945, 1689, -950, 230, -25, 1, 10395, -19524, 12139, -3480, 505, -36, 1, -135135, 264207, -177331, 57379, -10045, 973, -49, 1, 2027025, -4098240, 2924172, -1038016, 208054, -24640, 1708, -64, 1, -34459425, 71697105, -53809164, 20570444, -4574934, 626934, -53676, 2796, -81, 1
Offset: 0

Author

Ruedi Suter (suter(AT)math.ethz.ch)

Keywords

Comments

Triangle of B-analogs of Stirling numbers of first kind.

Examples

			The triangle T(n, k) begins:
n\k        0        1         2        3        4      5      6    7   8  9
0:         1
1:        -1        1
2:         3       -4         1
3:       -15       23        -9        1
4:       105     -176        86      -16        1
5:      -945     1689      -950      230      -25      1
6:     10395   -19524     12139    -3480      505    -36      1
7:   -135135   264207   -177331    57379   -10045    973    -49    1
8:   2027025 -4098240   2924172 -1038016   208054 -24640   1708  -64   1
9: -34459425 71697105 -53809164 20570444 -4574934 626934 -53676 2796 -81  1
...
row n = 10 :654729075 -1396704420 1094071221 -444647600 107494190 -16486680 1646778 -106800 4335 -100 1
... reformatted and extended. - _Wolfdieter Lang_, May 09 2017
		

Crossrefs

A028338 is unsigned version.
From Johannes W. Meijer, Jun 08 2009: (Start)
A161198 is an unsigned scaled triangle version.
A109692 is an unsigned transposed triangle version.
A000007 equals the row sums. (End)
A000165(n)*(-1)^n (alternating row sums).

Programs

  • Maple
    nmax:=8; mmax:=nmax: for n from 0 to nmax do a(n, 0) := (-1)^n*doublefactorial(2*n-1) od: for n from 0 to nmax do a(n, n) := 1 od: for n from 2 to nmax do for m from 1 to n-1 do a(n, m) := a(n-1, m-1)-(2*n-1)*a(n-1, m) od; od: seq(seq(a(n, m), m=0..n), n=0..nmax); # Johannes W. Meijer, Jun 08 2009, revised Nov 29 2012
  • Mathematica
    a[n_, m_] := a[n, m] = a[n-1, m-1] - (2*n-1)*a[n-1, m]; a[n_, 0] := (-1)^n*(2*n-1)!!; a[n_, n_] = 1; Table[a[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 16 2012, after Johannes W. Meijer *)
  • PARI
    row(n)=Vecrev(prod(i=1,n,'x-2*i+1)) \\ Charles R Greathouse IV, Feb 09 2017

Formula

Triangle T(n, k), read by rows, given by [ -1, -2, -3, -4, -5, -6, -7, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...], where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 20 2005
a(n,m) = a(n-1,m-1) - (2*n-1)*a(n-1,m) with a(n,0) = (-1)^n*(2*n-1)!! and a(n,n) = 1. - Johannes W. Meijer, Jun 08 2009
Exponential Riordan array [1/sqrt(1 + 2*x), 1/2*log(1 + 2*x)] with e.g.f. (1 + 2*x)^((t - 1)/2) = 1 + (t-1)*x + (t-1)*(t-3)*x^2/2! + .... - Peter Bala, Jun 23 2014
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