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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A091030 Partial sums of powers of 13 (A001022).

Original entry on oeis.org

1, 14, 183, 2380, 30941, 402234, 5229043, 67977560, 883708281, 11488207654, 149346699503, 1941507093540, 25239592216021, 328114698808274, 4265491084507563, 55451384098598320, 720867993281778161
Offset: 1

Views

Author

Wolfdieter Lang, Jan 23 2004

Keywords

Comments

13^a(n) is highest power of 13 dividing (13^n)!.
For analogs with primes 2, 3, 5, 7 and 11 see A000225, A003462, A003463, A023000 and A016123 respectively.
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1,A[i,i]:=13, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=14, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n)*charpoly(A,1). - Milan Janjic, Feb 21 2010

Examples

			For n=6, a(6) = 1*6 + 12*15 + 144*20 + 1728*15 + 20736*6 + 248832*1 = 402234. - _Bruno Berselli_, Nov 12 2015
		

Crossrefs

Programs

Formula

G.f.: x/((1-13*x)*(1-x)) = (1/(1-13*x) - 1/(1-x))/12.
a(n) = Sum_{k=0..n-1} 13^k = (13^n-1)/12.
a(n) = 13*a(n-1)+1 for n>1, a(1)=1. - Vincenzo Librandi, Feb 05 2011
a(n) = Sum_{k=0...n-1} 12^k*binomial(n,n-1-k). - Bruno Berselli, Nov 12 2015
E.g.f.: exp(x)*(exp(12*x) - 1)/12. - Stefano Spezia, Mar 11 2023

A135518 Generalized repunits in base 15.

Original entry on oeis.org

1, 16, 241, 3616, 54241, 813616, 12204241, 183063616, 2745954241, 41189313616, 617839704241, 9267595563616, 139013933454241, 2085209001813616, 31278135027204241, 469172025408063616, 7037580381120954241, 105563705716814313616, 1583455585752214704241
Offset: 1

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Author

Julien Peter Benney (jpbenney(AT)gmail.com), Feb 19 2008

Keywords

Comments

Primes in this sequence are given in A006033.
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=15, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Partial sums are in A014898. Also, the sequence is related to A014930 by A014930(n) = n*a(n) - Sum_{i=1..n-1}( a(i) ). - Bruno Berselli, Nov 06 2012

Examples

			For n=4, a(4) = 15^3+15^2+15^1+1 = 3375+225+15+1 = 3616.
For n=6, a(6) = 1*6 + 14*15 + 14^2*20 + 14^3*15 + 14^4*6 + 14^5*1 = 813616. - _Bruno Berselli_, Nov 12 2015
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[PadRight[{},n,1],15],{n,20}] (* or *) LinearRecurrence[{16,-15},{1,16},20] (* Harvey P. Dale, Jul 08 2013 *)
  • Maxima
    A135518(n):=(15^n-1)/14$ makelist(A135518(n),n,1,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=(15^n-1)/14 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Python
    def a(n): return int('1'*n, 15)
    print([a(n) for n in range(1, 20)]) # Michael S. Branicky, Jan 16 2021
  • Sage
    [gaussian_binomial(n,1,15) for n in range(1,15)] # Zerinvary Lajos, May 28 2009
    
  • Sage
    [(15^n-1)/14 for n in (1..30)] # Bruno Berselli, Nov 12 2015
    

Formula

a(n) = (15^n - 1)/14.
a(n) = 15*a(n-1) + 1 with n>1, a(1)=1. - Vincenzo Librandi, Aug 03 2010
G.f.: x/((1-x)*(1-15*x)). - Bruno Berselli, Nov 07 2012
a(1)=1, a(2)=16; for n>2, a(n) = 16*a(n-1) - 15*a(n-2). - Harvey P. Dale, Jul 08 2013
a(n) = Sum_{i=0...n-1} 14^i*binomial(n,n-1-i). - Bruno Berselli, Nov 12 2015
E.g.f.: (1/14)*(exp(15*x) - exp(x)). - G. C. Greubel, Oct 17 2016

A135519 Generalized repunits in base 14.

Original entry on oeis.org

1, 15, 211, 2955, 41371, 579195, 8108731, 113522235, 1589311291, 22250358075, 311505013051, 4361070182715, 61054982558011, 854769755812155, 11966776581370171, 167534872139182395, 2345488209948553531
Offset: 1

Views

Author

Julien Peter Benney (jpbenney(AT)gmail.com), Feb 19 2008

Keywords

Comments

Primes are given in A006032.
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1, A[i,i]:=14, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010

Examples

			a(4) = 2955 because (14^4-1)/13 = 38416/13 = 2955.
For n=6, a(6) = 1*6 + 13*15 + 169*20 + 2197*15 + 28561*6 + 371293*1 = 579195. - _Bruno Berselli_, Nov 12 2015
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[PadRight[{}, n, 1], 14], {n, 20}] (* or *) LinearRecurrence[{15, -14}, {1, 15}, 20] (* Harvey P. Dale, Aug 29 2016 *)
  • Maxima
    A135519(n):=(14^n-1)/13$ makelist(A135519(n),n,1,30); /* Martin Ettl, Nov 05 2012 */
  • Sage
    [gaussian_binomial(n,1,14) for n in range(1,15)] # Zerinvary Lajos, May 28 2009
    
  • Sage
    [(14^n-1)/13 for n in (1..30)] # Bruno Berselli, Nov 12 2015
    

Formula

a(n) = (14^n - 1)/13.
a(n) = 14*a(n-1) + 1 for n>1, a(1)=1. - Vincenzo Librandi, Aug 03 2010
a(n) = Sum_{i=0..n-1} 13^i*binomial(n,n-1-i). - Bruno Berselli, Nov 12 2015
From G. C. Greubel, Oct 17 2016: (Start)
G.f.: x/((1-x)*(1-14*x)).
E.g.f.: (1/13)*(exp(14*x) - exp(x)). (End)

A033282 Triangle read by rows: T(n, k) is the number of diagonal dissections of a convex n-gon into k+1 regions.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 1, 9, 21, 14, 1, 14, 56, 84, 42, 1, 20, 120, 300, 330, 132, 1, 27, 225, 825, 1485, 1287, 429, 1, 35, 385, 1925, 5005, 7007, 5005, 1430, 1, 44, 616, 4004, 14014, 28028, 32032, 19448, 4862, 1, 54, 936, 7644, 34398, 91728, 148512, 143208, 75582, 16796
Offset: 3

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Author

Keywords

Comments

T(n+3, k) is also the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's cluster algebra of finite type A_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of the triangle of Narayana numbers A001263. For example, x^2 + 5*x + 5 = y^2 + 3*y + 1. - Paul Boddington, Mar 07 2003
Number of standard Young tableaux of shape (k+1,k+1,1^(n-k-3)), where 1^(n-k-3) denotes a sequence of n-k-3 1's (see the Stanley reference).
Number of k-dimensional 'faces' of the n-dimensional associahedron (see Simion, p. 168). - Mitch Harris, Jan 16 2007
Mirror image of triangle A126216. - Philippe Deléham, Oct 19 2007
For relation to Lagrange inversion or series reversion and the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see A133437. - Tom Copeland, Sep 29 2008
Row generating polynomials 1/(n+1)*Jacobi_P(n,1,1,2*x+1). Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p. 60]. See A001263 for the corresponding array of h-vectors for associahedra of type A_n. See A063007 and A080721 for the f-vectors for associahedra of type B and type D respectively. - Peter Bala, Oct 28 2008
f-vectors of secondary polytopes for Grobner bases for optimization and integer programming (see De Loera et al. and Thomas). - Tom Copeland, Oct 11 2011
From Devadoss and O'Rourke's book: The Fulton-MacPherson compactification of the configuration space of n free particles on a line segment with a fixed particle at each end is the n-Dim Stasheff associahedron whose refined f-vector is given in A133437 which reduces to A033282. - Tom Copeland, Nov 29 2011
Diagonals of A132081 are rows of A033282. - Tom Copeland, May 08 2012
The general results on the convolution of the refined partition polynomials of A133437, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these polynomials. - Tom Copeland, Sep 20 2016
The signed triangle t(n, k) =(-1)^k* T(n+2, k-1), n >= 1, k = 1..n, seems to be obtainable from the partition array A111785 (in Abramowitz-Stegun order) by adding the entries corresponding to the partitions of n with the number of parts k. E.g., triangle t, row n=4: -1, (6+3) = 9, -21, 14. - Wolfdieter Lang, Mar 17 2017
The preceding conjecture by Lang is true. It is implicit in Copeland's 2011 comments in A086810 on the relations among a gf and its compositional inverse for that entry and inversion through A133437 (a differently normalized version of A111785), whose integer partitions are the same as those for A134685. (An inversion pair in Copeland's 2008 formulas below can also be used to prove the conjecture.) In addition, it follows from the relation between the inversion formula of A111785/A133437 and the enumeration of distinct faces of associahedra. See the MathOverflow link concernimg Loday and the Aguiar and Ardila reference in A133437 for proofs of the relations between the partition polynomials for inversion and enumeration of the distinct faces of the A_n associahedra, or Stasheff polytopes. - Tom Copeland, Dec 21 2017
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/(n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 07 2022
Chapoton's observation above is correct: the precise expansion is (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/ (n!*(n+1)!) = Sum_{k = 0..n-1} (-1)^k*T(n+2,n-k-1)*binomial(x+2*n-k,2*n-k), as can be verified using the WZ algorithm. For example, n = 4 gives (x+1)*(x+2)^2*(x+3)^2*(x+4)^2*(x+5)/(4!*5!) = 14*binomial(x+8,8) - 21*binomial(x+7,7) + 9*binomial(x+6,6) - binomial(x+5,5). - Peter Bala, Jun 24 2023

Examples

			The triangle T(n, k) begins:
n\k  0  1   2    3     4     5      6      7     8     9
3:   1
4:   1  2
5:   1  5   5
6:   1  9  21   14
7:   1 14  56   84    42
8:   1 20 120  300   330   132
9:   1 27 225  825  1485  1287    429
10:  1 35 385 1925  5005  7007   5005   1430
11:  1 44 616 4004 14014 28028  32032  19448  4862
12:  1 54 936 7644 34398 91728 148512 143208 75582 16796
... reformatted. - _Wolfdieter Lang_, Mar 17 2017
		

References

  • S. Devadoss and J. O'Rourke, Discrete and Computational Geometry, Princeton Univ. Press, 2011 (See p. 241.)
  • Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994. Exercise 7.50, pages 379, 573.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.8.

Crossrefs

Cf. diagonals: A000012, A000096, A033275, A033276, A033277, A033278, A033279; A000108, A002054, A002055, A002056, A007160, A033280, A033281; row sums: A001003 (Schroeder numbers, first term omitted). See A086810 for another version.
A007160 is a diagonal. Cf. A001263.
With leading zero: A086810.
Cf. A019538 'faces' of the permutohedron.
Cf. A063007 (f-vectors type B associahedra), A080721 (f-vectors type D associahedra), A126216 (mirror image).
Cf. A248727 for a relation to f-polynomials of simplices.
Cf. A111785 (contracted partition array, unsigned; see a comment above).
Antidiagonal sums give A005043. - Jordan Tirrell, Jun 01 2017

Programs

  • Magma
    [[Binomial(n-3, k)*Binomial(n+k-1, k)/(k+1): k in [0..(n-3)]]: n in [3..12]];  // G. C. Greubel, Nov 19 2018
    
  • Maple
    T:=(n,k)->binomial(n-3,k)*binomial(n+k-1,k)/(k+1): seq(seq(T(n,k),k=0..n-3),n=3..12); # Muniru A Asiru, Nov 24 2018
  • Mathematica
    t[n_, k_] = Binomial[n-3, k]*Binomial[n+k-1, k]/(k+1);
    Flatten[Table[t[n, k], {n, 3, 12}, {k, 0, n-3}]][[1 ;; 52]] (* Jean-François Alcover, Jun 16 2011 *)
  • PARI
    Q=(1+z-(1-(4*w+2+O(w^20))*z+z^2+O(z^20))^(1/2))/(2*(1+w)*z);for(n=3,12,for(m=1,n-2,print1(polcoef(polcoef(Q,n-2,z),m,w),", "))) \\ Hugo Pfoertner, Nov 19 2018
    
  • PARI
    for(n=3,12, for(k=0,n-3, print1(binomial(n-3,k)*binomial(n+k-1,k)/(k+1), ", "))) \\ G. C. Greubel, Nov 19 2018
    
  • Sage
    [[ binomial(n-3,k)*binomial(n+k-1,k)/(k+1) for k in (0..(n-3))] for n in (3..12)] # G. C. Greubel, Nov 19 2018

Formula

G.f. G = G(t, z) satisfies (1+t)*G^2 - z*(1-z-2*t*z)*G + t*z^4 = 0.
T(n, k) = binomial(n-3, k)*binomial(n+k-1, k)/(k+1) for n >= 3, 0 <= k <= n-3.
From Tom Copeland, Nov 03 2008: (Start)
Two g.f.s (f1 and f2) for A033282 and their inverses (x1 and x2) can be derived from the Drake and Barry references.
1. a: f1(x,t) = y = {1 - (2t+1) x - sqrt[1 - (2t+1) 2x + x^2]}/[2x (t+1)] = t x + (t + 2 t^2) x^2 + (t + 5 t^2 + 5 t^3) x^3 + ...
b: x1 = y/[t + (2t+1)y + (t+1)y^2] = y {1/[t/(t+1) + y] - 1/(1+y)} = (y/t) - (1+2t)(y/t)^2 + (1+ 3t + 3t^2)(y/t)^3 +...
2. a: f2(x,t) = y = {1 - x - sqrt[(1-x)^2 - 4xt]}/[2(t+1)] = (t/(t+1)) x + t x^2 + (t + 2 t^2) x^3 + (t + 5 t^2 + 5 t^3) x^4 + ...
b: x2 = y(t+1) [1- y(t+1)]/[t + y(t+1)] = (t+1) (y/t) - (t+1)^3 (y/t)^2 + (t+1)^4 (y/t)^3 + ...
c: y/x2(y,t) = [t/(t+1) + y] / [1- y(t+1)] = t/(t+1) + (1+t) y + (1+t)^2 y^2 + (1+t)^3 y^3 + ...
x2(y,t) can be used along with the Lagrange inversion for an o.g.f. (A133437) to generate A033282 and show that A133437 is a refinement of A033282, i.e., a refinement of the f-polynomials of the associahedra, the Stasheff polytopes.
y/x2(y,t) can be used along with the indirect Lagrange inversion (A134264) to generate A033282 and show that A134264 is a refinement of A001263, i.e., a refinement of the h-polynomials of the associahedra.
f1[x,t](t+1) gives a generator for A088617.
f1[xt,1/t](t+1) gives a generator for A060693, with inverse y/[1 + t + (2+t) y + y^2].
f1[x(t-1),1/(t-1)]t gives a generator for A001263, with inverse y/[t + (1+t) y + y^2].
The unsigned coefficients of x1(y t,t) are A074909, reverse rows of A135278. (End)
G.f.: 1/(1-x*y-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-.... (continued fraction). - Paul Barry, Feb 06 2009
Let h(t) = (1-t)^2/(1+(u-1)*(1-t)^2) = 1/(u + 2*t + 3*t^2 + 4*t^3 + ...), then a signed (n-1)-th row polynomial of A033282 is given by u^(2n-1)*(1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=2. The power series expansion of h(t) is related to A181289 (cf. A086810). - Tom Copeland, Sep 06 2011
With a different offset, the row polynomials equal 1/(1 + x)*Integral_{0..x} R(n,t) dt, where R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*t^k are the row polynomials of A063007. - Peter Bala, Jun 23 2016
n-th row polynomial = ( LegendreP(n-1,2*x + 1) - LegendreP(n-3,2*x + 1) )/((4*n - 6)*x*(x + 1)), n >= 3. - Peter Bala, Feb 22 2017
n*T(n+1, k) = (4n-6)*T(n, k-1) + (2n-3)*T(n, k) - (n-3)*T(n-1, k) for n >= 4. - Fang Lixing, May 07 2019

Extensions

Missing factor of 2 for expansions of f1 and f2 added by Tom Copeland, Apr 12 2009

A015531 Linear 2nd order recurrence: a(n) = 4*a(n-1) + 5*a(n-2).

Original entry on oeis.org

0, 1, 4, 21, 104, 521, 2604, 13021, 65104, 325521, 1627604, 8138021, 40690104, 203450521, 1017252604, 5086263021, 25431315104, 127156575521, 635782877604, 3178914388021, 15894571940104, 79472859700521, 397364298502604, 1986821492513021, 9934107462565104
Offset: 0

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Author

Keywords

Comments

Number of walks of length n between any two distinct vertices of the complete graph K_6. Example: a(2)=4 because the walks of length 2 between the vertices A and B of the complete graph ABCDEF are: ACB, ADB, AEB and AFB. - Emeric Deutsch, Apr 01 2004
General form: k=5^n-k. Also: A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-4, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 27 2010
Pisano period lengths: 1, 2, 6, 2, 2, 6, 6, 4, 18, 2, 10, 6, 4, 6, 6, 8, 16, 18, 18, 2,... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 5 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
For odd n, a(n) is congruent to 1 (mod 10). For even n > 0, a(n) is congruent to 4 (mod 10). - Iain Fox, Dec 30 2017

Crossrefs

A083425 shifted right.
Cf. A033115 (partial sums), A213128.

Programs

Formula

From Paul Barry, Apr 20 2003: (Start)
a(n) = (5^n -(-1)^n)/6.
G.f.: x/((1-5*x)*(1+x)).
E.g.f.(exp(5*x)-exp(-x))/6. (End) (corrected by M. F. Hasler, Jan 29 2012)
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*6^(k-1). - Paul Barry, May 13 2003
a(n) = 5^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
a(n) = ((2+sqrt(9))^n - (2-sqrt(9))^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009
a(n) = round(5^n/6). - Mircea Merca, Dec 28 2010
The logarithmic generating function 1/6*log((1+x)/(1-5*x)) = x + 4*x^2/2 + 21*x^3/3 + 104*x^4/4 + ... has compositional inverse 6/(5+exp(-6*x)) - 1, the e.g.f. for a signed version of A213128. - Peter Bala, Jun 24 2012
a(n) = (-1)^(n-1)*Sum_{k=0..(n-1)} A135278(n-1,k)*(-6)^k = (5^n - (-1)^n)/6 = (-1)^(n-1)*Sum_{k=0..(n-1)} (-5)^k. Equals (-1)^(n-1)*Phi(n,-5) when n is an odd prime, where Phi is the cyclotomic polynomial. - Tom Copeland, Apr 14 2014

A060540 Square array read by antidiagonals downwards: T(n,k) = (n*k)!/(k!^n*n!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 15, 1, 1, 35, 280, 105, 1, 1, 126, 5775, 15400, 945, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1, 6435, 66512160, 96197645544, 5194672859376, 4509264634875, 36212176000, 2027025, 1
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Comments

The Copeland link gives the associations of this entry with the operator calculus of Appell Sheffer polynomials, the combinatorics of simple set partitions encoded in the Faa di Bruno formula for composition of analytic functions (formal Taylor series), the Pascal matrix, and the geometry of the n-dimensional simplices (hypertriangles, or hypertetrahedra). These, in turn, are related to simple instances of the application of the exponential formula / principle / schema giving the number of not-necessarily-connected objects composed from an ensemble of connected objects. - Tom Copeland, Jun 09 2021

Examples

			Array begins:
  1,   1,       1,          1,             1,                 1, ...
  1,   3,      10,         35,           126,               462, ...
  1,  15,     280,       5775,        126126,           2858856, ...
  1, 105,   15400,    2627625,     488864376,       96197645544, ...
  1, 945, 1401400, 2546168625, 5194672859376, 11423951396577720, ...
  ...
		

Crossrefs

Main diagonal is A057599.
Related to A057599, see also A096126 and A246048.
Cf. A060358, A361948 (includes row/col 0).
Cf. A000217, A000292, A000332, A000389, A000579, A000580, A007318, A036040, A099174, A133314, A132440, A135278 (associations in Copeland link).

Programs

  • Mathematica
    T[n_, k_] := (n*k)!/(k!^n*n!);
    Table[T[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 29 2018 *)
  • PARI
    { i=0; for (m=1, 20, for (n=1, m, k=m - n + 1; write("b060540.txt", i++, " ", (n*k)!/(k!^n*n!))); ) } \\ Harry J. Smith, Jul 06 2009

Formula

T(n,k) = (n*k)!/(k!^n*n!) = T(n-1,k)*A060543(n,k) = A060538(n,k)/k!.
T(n,k) = Product_{j=2..n} binomial(j*k-1,k-1). - M. F. Hasler, Aug 22 2014

Extensions

Definition reworded by M. F. Hasler, Aug 23 2014

A068424 Triangle of falling factorials, read by rows: T(n, k) = n*(n-1)*...*(n-k+1), n > 0, 1 <= k <= n.

Original entry on oeis.org

1, 2, 2, 3, 6, 6, 4, 12, 24, 24, 5, 20, 60, 120, 120, 6, 30, 120, 360, 720, 720, 7, 42, 210, 840, 2520, 5040, 5040, 8, 56, 336, 1680, 6720, 20160, 40320, 40320, 9, 72, 504, 3024, 15120, 60480, 181440, 362880, 362880, 10, 90, 720, 5040, 30240, 151200, 604800, 1814400, 3628800, 3628800
Offset: 1

Views

Author

David Wasserman, Mar 13 2003

Keywords

Comments

Triangle in which the n-th row begins with n and the k-th term is obtained by multiplying the (k-1)-th term by (n-k+1) until n-k+1 = 1. - Amarnath Murthy, Nov 11 2002
Has many diagonals in common with A105725. - Zerinvary Lajos, Apr 14 2006
Also: the square array of rising factorials A(n,k) = n*(n+1)*(n+2)*...*(n+k-1) read by antidiagonals downwards. There are no perfect squares in T(n,k) for k >= 2 [see Rigge]. T(n,k) is divisible by a prime exceeding k, if n >= 2*k [see Saradha and Shorey]. - R. J. Mathar, May 02 2007
T(n,k) is the number of injective functions f from {1,...,k} into {1,...,n}, since there are n choices for f(1), then (n-1) choices for f(2), ... and (n-k+1) choices for f(k). E.g., T(3,2)=6 because there are exactly 6 injective functions f:{1,2}->{1,2,3}, namely, f1={(1,1),(2,2)}, f2={(1,1),(2,3)}, f3={(1,2),(2,1)}, f4={(1,2),(2,3)}, f5={(1,3),(2,1)} and f6={(1,3),(2,2)}. - Dennis P. Walsh, Oct 18 2007
Permuted words defined by the connectivity of regular simplices are related to T by T = A135278 * (1!, 2!, 3!, 4!, ...). E.g., for T(4,k) with k-1 = simplex number, label the vertices of a tetrahedron with a, b, c, d, then the 0-simplex, the points, a,b,c,d gives 4 * 1 = 4 words; the 1-simplex, the edges: (ab or ba), (ac or ca), (ad or da), (bc or cb), (bd or db), (cd or dc) gives 6 * 2 = 12 words; the 2-simplex, the faces: (abc or ...), (acd or ...), (adb or ...), (bcd or ...) gives 4 * 6 = 24 words; the 3-simplex, (abcd or ....) gives 1 * 24 = 24 words. - Tom Copeland, Dec 08 2007
Reversal of the triangle by rows = (n+1) * n-th row of triangle A094587. - Gary W. Adamson, May 03 2009
From Geoffrey Critzer, May 06 2009: (Start)
The rectangular array R(n,k), read by diagonals is the number of ways n people can queue up in k (possibly empty) distinct queues. R(n,k) = (n+k-1)!/(k-1)!; R(n,k) = (n+k-1)*R(n-1,k).
Northwest corner:
1, 2, 3, 4, 5, ...;
2, 6, 12, 20, 30, ...;
6, 24, 60, 120, 210, ...;
24, 120, 360, 840, 1680, ...;
120, 720, 2520, 6720, 15120, ...;
...
Example: R(2,2)=6 because there are six ways that two people can get in line at a fast food restaurant that has two order windows open. Let 1 and 2 represent the two people and a | will separate the lines. 12|; 21|; |12; |21; 1|2; 2|1. (End)
Cf. [Hardy and Wright], Theorem 34.
The e.g.f. of the Norlund generalized Bernoulli (Appell) polynomials of order m, NB(n,x;m), is given by exponentiation of the e.g.f. of the Bernoulli numbers, i.e., multiple binomial self-convolutions of the Bernoulli numbers, through the e.g.f. exp[NB(.,x;m)t] = [t/(e^t-1)]^(m+1) * e^(xt). Norlund gave the relation to the factorials (x-1)!/(x-1-k)! = (x-1) ... (x-k) = NB(k,x;k), so T(n,k) = NB(k,n+1;k). - Tom Copeland, Oct 01 2015
T(n,k) is the number of sequences without repetition (partial permutations) of k elements taken from an n-set. For sequences with repetition (k-tuples) cf. A075363. - Manfred Boergens, Jun 18 2023

Examples

			Triangle begins:
  1;
  2,  2;
  3,  6,   6;
  4, 12,  24,  24;
  5, 20,  60, 120, 120;
  6, 30, 120, 360, 720, 720;
Square begins:
    1,   2,    3,    4,     5, ...
    2,   6,   12,   20,    30, ...
    6,  24,   60,  120,   210, ...
   24, 120,  360,  840,  1680, ...
  120, 720, 2520, 6720, 15120, ...
		

References

  • G. H. Hardy, E. M. Wright, An Introduction to the Theory of Numbers, Clarendon Press, Fifth edition, 1979, p. 64.
  • O. Rigge, 9th Congr. Math. Scan., Helsingfors, 1938, Mercator, 1939, pp. 155-160.

Crossrefs

Same as A008279 for k>0.
Cf. A094587. - Gary W. Adamson, May 03 2009
Appears in A167546. - Johannes W. Meijer, Nov 12 2009

Programs

  • Mathematica
    Flatten[Table[n!/(n-k)!, {n, 10}, {k, n}]] (* or, from version 7: *)
    Flatten[Table[FactorialPower[n, k], {n, 10}, {k, n}]]  (* Jean-François Alcover, Jun 17 2011, updated Sep 29 2016 *)
  • PARI
    T(n,k)=if(k<1 || k>n,0,n!/(n-k)!)

Formula

As a triangle: T(n,k) = k!*binomial(n,k) = n!/(n-k)!, 1 <= k <= n. - Michael Somos, Apr 05 2003
E.g.f.: exp(x)*x*y/(1-x*y). - Michael Somos, Apr 05 2003
As a square: A(n,k) = (n+k-1)!/(k-1)!, 1 <= k <= n. - Ron L.J. van den Burg, Nov 28 2021

A103371 Number triangle T(n,k) = C(n,n-k)*C(n+1,n-k).

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 12, 1, 5, 40, 60, 20, 1, 6, 75, 200, 150, 30, 1, 7, 126, 525, 700, 315, 42, 1, 8, 196, 1176, 2450, 1960, 588, 56, 1, 9, 288, 2352, 7056, 8820, 4704, 1008, 72, 1, 10, 405, 4320, 17640, 31752, 26460, 10080, 1620, 90, 1, 11, 550, 7425, 39600, 97020
Offset: 0

Views

Author

Paul Barry, Feb 03 2005

Keywords

Comments

Columns include A000027, A002411, A004302, A108647, A134287. Row sums are C(2n+1,n+1) or A001700.
T(n-1,k-1) is the number of ways to put n identical objects into k of altogether n distinguishable boxes. See the partition array A035206 from which this triangle arises after summing over all entries related to partitions with fixed part number k.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008
The o.g.f. of the (n+1)-th diagonal is given by G(n, x) = (n+1)*Sum_{k=1..n} A001263(n, k)*x^(k-1) / (1 - x)^(2*n+1), for n >= 1 and for n = 0 it is G(0, x) = 1/(1-x). - Wolfdieter Lang, Jul 31 2017

Examples

			The triangle T(n, k) begins:
n\k  0   1    2     3     4     5     6    7  8 9 ...
0:   1
1:   2   1
2:   3   6    1
3:   4  18   12     1
4:   5  40   60    20     1
5:   6  75  200   150    30     1
6:   7 126  525   700   315    42     1
7:   8 196 1176  2450  1960   588    56    1
8:   9 288 2352  7056  8820  4704  1008   72  1
9:  10 405 4320 17640 31752 26460 10080 1620 90 1
...  reformatted. - _Wolfdieter Lang_, Jul 31 2017
From _R. J. Mathar_, Mar 29 2013: (Start)
The matrix inverse starts
       1;
      -2,       1;
       9,      -6,      1;
     -76,      54,    -12,      1;
    1055,    -760,    180,    -20,   1;
  -21906,   15825,  -3800,    450, -30,   1;
  636447, -460026, 110775, -13300, 945, -42, 1; (End)
O.g.f. of 4th diagonal [4, 40,200, ...] is G(3, x) = 4*(1 + 3*x + x^2)/(1 - x)^7, from the n = 3 row [1, 3, 1] of A001263. See a comment above. - _Wolfdieter Lang_, Jul 31 2017
		

Crossrefs

Cf. A007318, A000894 (central terms), A132813 (mirrored).

Programs

  • Haskell
    a103371 n k = a103371_tabl !! n !! k
    a103371_row n = a103371_tabl !! n
    a103371_tabl = map reverse a132813_tabl
    -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    /* As triangle */ [[Binomial(n,n-k)*Binomial(n+1,n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 01 2017
    
  • Maple
    A103371 := (n,k) -> binomial(n,k)^2*(n+1)/(k+1);
    seq(print(seq(A103371(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011
  • Mathematica
    Flatten[Table[Binomial[n,n-k]Binomial[n+1,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 26 2014 *)
    CoefficientList[Series[Series[E^(x(1+y))(BesselI[0,2*x*Sqrt[y]]+BesselI[1,2*x*Sqrt[y]]/Sqrt[y]),{x,0,8}],{y,0,8}],{x,y}]*Range[0,8]! (* Natalia L. Skirrow, Apr 14 2025 *)
  • Maxima
    create_list(binomial(n,k)*binomial(n+1,k+1),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n+1,k+1), ", "))) \\ G. C. Greubel, Nov 09 2018

Formula

Number triangle T(n, k) = C(n, n-k)*C(n+1, n-k) = C(n, k)*C(n+1, k+1); Column k of this triangle has g.f. Sum_{j=0..k} (C(k, j)*C(k+1, j) * x^(k+j))/(1-x)^(2*k+2); coefficients of the numerators are the rows of the reverse triangle C(n, k)*C(n+1, k).
T(n,k) = C(n, k)*Sum_{j=0..(n-k)} C(n-j, k). - Paul Barry, Jan 12 2006
T(n,k) = (n+1-k)*N(n+1,k+1), with N(n,k):=A001263(n,k), the Narayana triangle (with offset [1,1]).
O.g.f.: ((1-(1-y)*x)/sqrt((1-(1+y)*x)^2-4*x^2*y) -1)/2, (from o.g.f. of A001263, Narayana triangle). - Wolfdieter Lang, Nov 13 2007
From Peter Bala, Jan 24 2008: (Start)
Matrix product of A007318 and A122899.
O.g.f. for row n: (1-x)^n*P(n,1,0,(1+x)/(1-x)) = 1/(2*x)*(1-x)^(n+1)*( Legendre_P(n+1,(1+x)/(1-x)) - Legendre_P(n,(1+x)/(1-x)) ), where P(n,a,b,x) denotes the Jacobi polynomial.
O.g.f. for column k: x^k/(1-x)^(k+2)*P(k,0,1,(1+x)/(1-x)). Compare with A008459. (End)
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,1). (Cf. A194595, A197653, A197654). - Peter Luschny, Oct 20 2011
T(n,k) = A003056(n+1,k+1)*C(n,k)^2/(k+1). - Peter Luschny, Oct 29 2011
T(n,k) = A007318(n, k)*A135278(n, k), n >= k >= 0. - Wolfdieter Lang, Jul 31 2017
From Natalia L. Skirrow, Apr 14 2025: (Start)
T(n,k) = A008459(n,k) + n*N(n,k+1).
E.g.f.: e^(x*(1+y))*(I_0(2*x*sqrt(y)) + I_1(2*x*sqrt(y))/sqrt(y)), where I_n is the modified Bessel function of the first kind. (The I_0 contributes A008459(n,k), the I_1 contributes n*N(n,k+1))
O.g.f. for row n: (n+1)*2F1(-n,-n;2;y) = (n+1)*2F1(2+n,2+n;2;y)*(1-y)^(2*(n+1)) (by Euler's hypergeometric transformation); (n+1)*2F1(2+n,2+n;2;y) is the o.g.f. for row n of (k+n+1)!^2/(k!*(k+1)!*n!*(n+1)!), which is column n+1 of A132812.
O.g.f. for column k: 2F1(1+k,2+k;1;x)*x^k = 2F1(-k,-1-k;1;x)*x^k/(1-x)^(2+2*k). 2F1(-k,-1-k;1;x) is the kth row of A132813, the reflection of the kth row of this triangle.
O.g.f. for diagonal d (beginning at a(d,0)): (d+1)*x^d*2F1(d+1,d+2;2;x*y). 2F1(d+1,d+2;2;x) = 2F1(1-d,-d;2;x)/(1-x)^(2*d+1), numerator being the o.g.f. of row d of the Narayana triangle.
These respectively yield:
T(n,k) = Sum_{i=0..n+k} C(2*(n+1),i)*(-1)^i*A132812(n+1+k-i,n+1),
T(d+k,k) = Sum_{i=0..k} C(d-i+1+2*k,d-i)*T(k,k-i),
T(d+k,k) = Sum_{i=0..d} C(k-i + 2*d,k-i)*N(d,i+1)*(d+1).
E.g.f. for column k: 1F1(2+k;1;x)*x^k/k!.
E.g.f. for diagonal d: (d+1)*x^d*1F1(d+2;2;x*y)/d!. (End)

A104712 Pascal's triangle, with the first two columns removed.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715
Offset: 2

Views

Author

Gary W. Adamson, Mar 19 2005

Keywords

Comments

A000295 (Eulerian numbers) gives the row sums.
Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle.
From Peter Luschny, Apr 10 2011: (Start)
A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:
(a) Each partition has at least one singleton block;
(c) k is the size of the largest block of the partition;
(b) m = n - k + 1 is the number of parts of the partition.
Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)).
For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11.
This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts
1;
1, 3;
1, 6, 4;
1, 10, 10, 5;
1, 15, 20, 15, 6;
...
(End)
Diagonal sums are A001924(n+1). - Philippe Deléham, Jan 11 2014
Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014

Examples

			The triangle a(n, k) begins:
  n\k  2   3   4    5    6    7    8   9  10 11 12 13
  2:   1
  3:   3   1
  4:   6   4   1
  5:  10  10   5    1
  6:  15  20  15    6    1
  7:  21  35  35   21    7    1
  8:  28  56  70   56   28    8    1
  9:  36  84 126  126   84   36    9   1
  10: 45 120 210  252  210  120   45  10   1
  11: 55 165 330  462  462  330  165  55  11  1
  12: 66 220 495  792  924  792  495 220  66 12  1
  13: 78 286 715 1287 1716 1716 1287 715 286 78 13  1
... reformatted. - _Wolfdieter Lang_, Mar 20 2015
		

Crossrefs

Cf. A000295, A007318, A008292, A104713, A027641/A027642 (first Bernoulli numbers B-), A164555/A027642 (second Bernoulli numbers B+), A176327/A176289.

Programs

  • Magma
    /* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018
  • Mathematica
    t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)
  • PARI
    for(n=2, 10, for(k=2,n, print1(binomial(n,k), ", "))) \\ G. C. Greubel, May 15 2018
    

Formula

T(n,k) = binomial(n,k), for 2 <= k <= n.
From Peter Bala, Jul 16 2013: (Start)
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^3, x/(1 - x)).
O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t)) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)
As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
From Tom Copeland, Apr 11 2014: (Start)
A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!.
B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }
C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....
D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)
O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014
Reverse T = [St2]*dP*[St1]- dP = [St2]*(exp(x*M)-I)*[St1]-(exp(x*M)-I) with top two rows of zeros removed, [St1]=padded A008275 just as [St2]=A048993=padded A008277, dP= A132440, M=A238385-I, and I=identity matrix. Cf. A238363. - Tom Copeland, Apr 26 2014
O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015

Extensions

Edited and extended by David Wasserman, Jul 03 2007

A319225 Number of acyclic spanning subgraphs of a cycle graph, where the sizes of the connected components are given by the prime indices of n.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 4, 1, 2, 4, 5, 4, 6, 5, 5, 1, 7, 5, 8, 5, 6, 6, 9, 5, 3, 7, 2, 6, 10, 12, 11, 1, 7, 8, 7, 9, 12, 9, 8, 6, 13, 14, 14, 7, 7, 10, 15, 6, 4, 7, 9, 8, 16, 7, 8, 7, 10, 11, 17, 21, 18, 12, 8, 1, 9, 16, 19, 9, 11, 16, 20, 14, 21, 13, 8, 10, 9, 18
Offset: 1

Views

Author

Gus Wiseman, Sep 13 2018

Keywords

Comments

a(1) = 1 by convention.
A prime index of n is a number m such that prime(m) divides n.

Examples

			Of the cycle ({1,2,3}, {(1,2),(2,3),(3,1)}) the spanning subgraphs where the sizes of connected components are (2,1) are: ({1,2,3}, {(1,2)}), ({1,2,3}, {(2,3)}), ({1,2,3}, {(3,1)}). Since the prime indices of 6 are (2,1), we conclude a(6) = 3.
		

Crossrefs

Programs

  • Mathematica
    csm[s_]:=With[{c=Select[Tuples[Range[Length[s]],2],And[OrderedQ[#],UnsameQ@@#,Length[Intersection@@s[[#]]]>0]&]},If[c=={},s,csm[Union[Append[Delete[s,List/@c[[1]]],Union@@s[[c[[1]]]]]]]]];
    Table[Length[With[{m=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]},Select[Subsets[Partition[Range[Total[m]],2,1,1],{Total[m]-PrimeOmega[n]}],Sort[Length/@csm[Union[#,List/@Range[Total[m]]]]]==m&]]],{n,30}]

Formula

a(n) = A056239(n) * (Omega(n) - 1)! / Product c_i! where c_i is the multiplicity of prime(i) in the prime factorization of n.
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