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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A025168 Expansion of e.g.f.: exp(x/(1-2*x)).

Original entry on oeis.org

1, 1, 5, 37, 361, 4361, 62701, 1044205, 19748177, 417787921, 9770678101, 250194150581, 6959638411705, 208919770666777, 6729933476435261, 231512615111396221, 8469125401589550241, 328241040596380393505, 13434223364220816489637, 578931271898150002093381
Offset: 0

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Author

Keywords

Comments

From Peter Bala, Nov 21 2017: (Start)
The sequence terms have the form 4*m + 1 (follows from the recurrence).
For k = 2,3,4,... the difference a(n+k) - a(n) is divisible by k (proof by induction on n making use of the recurrence - the starting case a(k) == a(0) (mod k) for all k follows from the sum formula for a(k)). Hence for each k, the sequence b(n) == a(n) (mod k) is periodic with the exact period dividing k. (End)
Compound Poisson distribution with parameter 1 and distribution Geometric(1/2) has a probability mass function p_n = a(n)*e^(-1/2)/(4^n*n!). More specifically, let S = Sum_{i=0..N} X_i where X_i's are i.i.d. random variables with Geometric(1/2) distribution (i.e., Pr{X_i = k} = 1/2^(k+1) for k=0,1,2...) and N is a random variable with Poisson(1) distribution independent of all X_i's. Then Pr{S=n} = a(n)*e^(-1/2)/(4^n*n!) = a(n)*e^(-1/2)/A047053(n) for nonnegative integers n. - Xiaohan Zhang, Nov 16 2022

Crossrefs

Programs

  • Maple
    with(combstruct); SetSeqSeqL := [T, {T=Set(S), S=Sequence(U,card >= 1), U=Sequence(Z,card >=1)},labeled];
    f:= gfun:-rectoproc({a(n) = (4*n-3)*a(n-1) - 4*(n-2)*(n-1)*a(n-2),a(0)=1,a(1)=1},a(n),remember):
    map(f, [$0..30]); # Robert Israel, Nov 21 2017
  • Mathematica
    Table[ n! 2^n LaguerreL[ n, 1, -1/2 ], {n, 0, 12} ]
    With[{nn=20},CoefficientList[Series[Exp[x/(1-2x)],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 12 2012 *)
  • Sage
    A025168 = lambda n: hypergeometric([-n,-n+1], [], 2)
    [Integer(A025168(n).n(100)) for n in range(20)] # Peter Luschny, Sep 22 2014

Formula

Second LAH transform of A000012. LAH transform of A000262. a(n) = Sum_{k=0..n} 2^(n-k)*n!/k!*binomial(n-1, k-1). - Vladeta Jovovic, Oct 17 2003
Define f_1(x), f_2(x), ... such that f_1(x) = e^x, f_{n+1}(x) = (d/dx)(x^2*f_n(x)), for n=2,3,.... Then a(n) = e^(-1/2)*4*(n-1)*f_n(1/2). - Milan Janjic, May 30 2008
From Vaclav Kotesovec, Jun 22 2013: (Start)
D-finite with recurrence: a(n) = (4*n-3)*a(n-1) - 4*(n-2)*(n-1)*a(n-2).
a(n) ~ 2^(n-3/4)*n^(n-1/4)*exp(sqrt(2*n)-n-1/4) * (1-1/(3*sqrt(2*n))).
(End)
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - x/(x + (k+1)*(1-2*x)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 09 2013
a(n) = hypergeometric([-n,-n+1],[],2). - Peter Luschny, Sep 22 2014
Sum_{n>=0} a(n)/(4^n*n!) = sqrt(e) = A019774. -Xiaohan Zhang, Nov 16 2022

Extensions

Corrected and extended by Vladeta Jovovic, Sep 08 2002

A051188 Sept-factorial numbers.

Original entry on oeis.org

1, 7, 98, 2058, 57624, 2016840, 84707280, 4150656720, 232436776320, 14643516908160, 1025046183571200, 78928556134982400, 6629998715338521600, 603329883095805465600, 59126328543388935628800
Offset: 0

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Author

Keywords

Comments

For n >= 1, a(n) is the order of the wreath product of the symmetric group S_n and the Abelian group (C_7)^n. - Ahmed Fares (ahmedfares(AT)my-deja.com), May 07 2001

Crossrefs

Programs

Formula

a(n) = n!*7^n =: (7*n)(!^7).
a(n) = 7*A034834(n) = Product_{k=1..n} 7*k, n >= 1.
E.g.f.: 1/(1 - 7*x).
G.f.: 1/(1 - 7*x/(1 - 7*x/(1 - 14*x/(1 - 14*x/(1 - 21*x/(1 - 21*x/(1 - 28*x/(1 - 28*x/(1 - ... (continued fraction). - Philippe Deléham, Jan 08 2012
From Amiram Eldar, Jun 25 2020: (Start)
Sum_{n>=0} 1/a(n) = e^(1/7) (A092516).
Sum_{n>=0} (-1)^n/a(n) = e^(-1/7) (A092750). (End)

A167546 The ED1 array read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 2, 4, 1, 6, 12, 7, 1, 24, 48, 32, 10, 1, 120, 240, 160, 62, 13, 1, 720, 1440, 960, 384, 102, 16, 1, 5040, 10080, 6720, 2688, 762, 152, 19, 1, 40320, 80640, 53760, 21504, 6144, 1336, 212, 22, 1
Offset: 1

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Author

Johannes W. Meijer, Nov 10 2009

Keywords

Comments

The coefficients in the upper right triangle of the ED1 array (m > n) were found with the a(n,m) formula while the coefficients in the lower left triangle of the ED1 array (m <= n) were found with the recurrence relation, see below. We use for the array rows the letter n (>= 1) and for the array columns the letter m (>= 1).
Our procedure for finding the coefficients in the lower left triangle can be compared with the procedure that De Smit and Lenstra used to fill in the hole in the middle of 'The Print Gallery' by M. C. Escher, see the links. In this lithograph Escher made use of the so-called Droste effect, hence we propose to call this square array of numbers the ED1 array.
For the ED2, ED3 and ED4 arrays see A167560, A167572 and A167584.

Examples

			The ED1 array begins with:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1
1, 4, 7, 10, 13, 16, 19, 22, 25, 28
2, 12, 32, 62, 102, 152, 212, 282, 362, 452
6, 48, 160, 384, 762, 1336, 2148, 3240, 4654, 6432
24, 240, 960, 2688, 6144, 12264, 22200, 37320, 59208, 89664
120, 1440, 6720, 21504, 55296, 122880, 245640, 452880, 783144, 1285536
		

Crossrefs

A000012, A016777, 2*A005891, A167547, A167548 and A167549 equal the first sixth rows of the array.
A000142 equals the first column of the array.
A167550 equals the a(n, n+1) diagonal of the array.
A047053 equals the a(n, n) diagonal of the array.
A167558 equals the a(n+1, n) diagonal of the array.
A167551 equals the row sums of the ED1 array read by antidiagonals.
A167552 is a triangle related to the a(n) formulas of rows of the ED1 array.
A167556 is a triangle related to the GF(z) formulas of the rows of the ED1 array.
A167557 is the lower left triangle of the ED1 array.
Cf. A068424 (the (m-1)!/(m-n-1)! factor), A007680 (the (2*n-1)*(n-1)! factor).
Cf. A167560 (ED2 array), A167572 (ED3 array), A167584 (ED4 array).

Programs

  • Maple
    nmax:=10; mmax:=10; for n from 1 to nmax do for m from 1 to n do a(n,m) := 4^(m-1)*(m-1)!*(n-1+m-1)!/(2*m-2)! od; for m from n+1 to mmax do a(n,m):= (2*n-1)*(n-1)! + sum((-1)^(k-1)*binomial(n-1,k)*a(n,m-k),k=1..n-1) od; od: for n from 1 to nmax do for m from 1 to n do d(n,m):=a(n-m+1,m) od: od: T:=1: for n from 1 to nmax do for m from 1 to n do a(T):= d(n,m): T:=T+1: od: od: seq(a(n),n=1..T-1);
  • Mathematica
    nmax = 10; mmax = 10; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[n, m] = 4^(m - 1)*(m - 1)!*((n - 1 + m - 1)!/(2*m - 2)!)]; For[m = n + 1, m <= mmax, m++, a[n, m] = (2*n - 1)*(n - 1)! + Sum[(-1)^(k - 1)*Binomial[n - 1, k]*a[n, m - k], {k, 1, n - 1}]]; ]; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, d[n, m] = a[n - m + 1, m]]; ]; t = 1; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[t] = d[n, m]; t = t + 1]]; Table[a[n], {n, 1, t - 1}] (* Jean-François Alcover, Dec 20 2011, translated from Maple *)

Formula

a(n,m) = (2*(m-1)!/(m-n-1)!)*Integral_{y>=0} sinh(y*(2*n-1))/cosh(y)^(2*m-1) for m > n.
The (n-1)-differences of the n-th array row lead to the recurrence relation
Sum_{k=0..n-1} (-1)^k*binomial(n-1,k)*a(n,m-k) = (2*n-1)*(n-1)!
which in its turn leads to, see also A167557,
a(n,m) = 4^(m-1)*(m-1)!*(n+m-2)!/(2*m-2)! for m <= n.

A144828 Partial products of successive terms of A017113; a(0)=1.

Original entry on oeis.org

1, 4, 48, 960, 26880, 967680, 42577920, 2214051840, 132843110400, 9033331507200, 686533194547200, 57668788341964800, 5305528527460761600, 530552852746076160000, 57299708096576225280000, 6646766139202842132480000, 824199001261152424427520000, 108794268166472120024432640000
Offset: 0

Views

Author

Philippe Deléham, Sep 21 2008

Keywords

Comments

a(n) is the number of signed permutations of length 4n that are equal to their reverse-inverses. Note that the reverse-inverse of a permutation is equivalent to a 90-degree rotation of the permutation's diagram (see the Hardt and Troyka link). - Justin M. Troyka, Aug 11 2011
Define the bar operation as an operation on signed permutation that flips the sign of each entry. Then a(n) is the number of signed permutations of length 2n that are equal to the bar of their inverses and equal to their reverse-complements (see the Hardt and Troyka link). - Justin M. Troyka, Aug 11 2011

Examples

			a(0)=1, a(1)=4, a(2)=4*12=48, a(3)=4*12*20=960, a(4)=4*12*20*28=26880, ...
Since a(1) = 4, there are 4 signed permutations of 4 that are equal to their reverse-inverses.  These are: (+2,+4,+1,+3), (+3,+1,+4,+2), (-2,-4,-1,-3), (-3,-1,-4,-2). - _Justin M. Troyka_, Aug 11 2011
G.f. = 1 + 4*x + 48*x^2 + 960*x^3 + 26880*x^4 + 967680*x^5 + 42577920*x^6 + ...
		

Crossrefs

Essentially the same as A052714. - N. J. A. Sloane, Feb 03 2013
Sequences of the form m^(n-1)*n!*Catalan(n-1): A001813 (m=1), A052714 (or this sequence) (m=2), A221954 (m=3), A052734 (m=4), A221953 (m=5), A221955 (m=6).

Programs

  • Magma
    [2^k *Factorial(2*k) / Factorial(k): k in [0..20]]; // Vincenzo Librandi, Aug 11 2011
    
  • Maple
    A144828:= n-> 2^n*n!*binomial(2*n,n); seq(A144828(n), n=0..30); # G. C. Greubel, Apr 02 2021
  • Mathematica
    Table[4^n (2 n - 1)!!, {n, 0, 15}] (* Vincenzo Librandi, May 14 2015 *)
    Join[{1},FoldList[Times,(8*Range[0,20]+4)]] (* Harvey P. Dale, Dec 01 2015 *)
  • PARI
    a(n)=binomial(2*n,n)*n!<Charles R Greathouse IV, Jan 17 2012
    
  • PARI
    {a(n) = if( n<0, (-1)^n / a(-n), 2^n *(2*n)! / n!)}; /* Michael Somos, Jan 06 2017 */
    
  • Sage
    [2^n*factorial(n+1)*catalan_number(n) for n in (0..30)] # G. C. Greubel, Apr 02 2021

Formula

a(n) = Sum_{k=0..n} A132393(n,k)*4^k*8^(n-k).
a(n) = A052714(n+1). - R. J. Mathar, Oct 01 2008
a(n) = 2^n *(2*n)! / n!. - Justin M. Troyka, Aug 11 2011
G.f.: 1/(1-4x/(1-8x/(1-12x/(1-16x/(1-20x/(1-24x/(1-28x/(1-32x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2012
a(n) = (-4)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
E.g.f.: 1/sqrt(1-8*x). - Philippe Deléham, May 14 2015
a(n) = 4^n * A001147(n). - Philippe Deléham, May 14 2015
a(n) = 8^n * Gamma(n + 1/2) / sqrt(Pi). - Daniel Suteu, Jan 06 2017
0 = a(n)*(8*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) and a(n) = (-1)^n / a(-n) for all n in Z. - Michael Somos, Jan 06 2017
a(n) = 2^n * (n+1)! * Catalan(n). - G. C. Greubel, Apr 02 2021
Sum_{n>=0} 1/a(n) = 1 + e^(1/8)*sqrt(Pi)*erf(1/(2*sqrt(2)))/(2*sqrt(2)), where erf is the error function. - Amiram Eldar, Dec 20 2022

A075499 Stirling2 triangle with scaled diagonals (powers of 4).

Original entry on oeis.org

1, 4, 1, 16, 12, 1, 64, 112, 24, 1, 256, 960, 400, 40, 1, 1024, 7936, 5760, 1040, 60, 1, 4096, 64512, 77056, 22400, 2240, 84, 1, 16384, 520192, 989184, 435456, 67200, 4256, 112, 1, 65536, 4177920, 12390400, 7956480, 1779456, 169344, 7392, 144, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 02 2002

Keywords

Comments

This is a lower triangular infinite matrix of the Jabotinsky type. See the Knuth reference given in A039692 for exponential convolution arrays.
The row polynomials p(n,x) := Sum_{m=1..n} a(n,m)x^m, n >= 1, have e.g.f. J(x; z)= exp((exp(4*z) - 1)*x/4) - 1
Also the inverse Bell transform of the quadruple factorial numbers 4^n*n! (A047053) adding 1,0,0,0,... as column 0. For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015

Examples

			[1]; [4,1]; [16,12,1]; ...; p(3,x) = x(16 + 12*x + x^2).
From _Andrew Howroyd_, Mar 25 2017: (Start)
Triangle starts
*     1
*     4      1
*    16     12      1
*    64    112     24      1
*   256    960    400     40     1
*  1024   7936   5760   1040    60    1
*  4096  64512  77056  22400  2240   84   1
* 16384 520192 989184 435456 67200 4256 112 1
(End)
		

Crossrefs

Columns 1-7 are A000302, A016152, A019677, A075907-A075910. Row sums are A004213.

Programs

  • Mathematica
    Table[(4^(n - m)) StirlingS2[n, m], {n, 9}, {m, n}] // Flatten (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    for(n=1, 11, for(m=1, n, print1(4^(n - m) * stirling(n, m, 2),", ");); print();) \\ Indranil Ghosh, Mar 25 2017
  • Sage
    # uses[inverse_bell_transform from A265605]
    # Adds a column 1,0,0,... at the left side of the triangle.
    multifact_4_4 = lambda n: prod(4*k + 4 for k in (0..n-1))
    inverse_bell_matrix(multifact_4_4, 9) # Peter Luschny, Dec 31 2015
    

Formula

a(n, m) = (4^(n-m)) * stirling2(n, m).
a(n, m) = (Sum_{p=0..m-1} A075513(m, p)*((p+1)*4)^(n-m))/(m-1)! for n >= m >= 1, else 0.
a(n, m) = 4m*a(n-1, m) + a(n-1, m-1), n >= m >= 1, else 0, with a(n, 0) := 0 and a(1, 1)=1.
G.f. for m-th column: (x^m)/Product_{k=1..m}(1-4k*x), m >= 1.
E.g.f. for m-th column: (((exp(4x)-1)/4)^m)/m!, m >= 1.

A114799 Septuple factorial, 7-factorial, n!7, n!!!!!!!, a(n) = n*a(n-7) if n > 1, else 1.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 6, 7, 8, 18, 30, 44, 60, 78, 98, 120, 288, 510, 792, 1140, 1560, 2058, 2640, 6624, 12240, 19800, 29640, 42120, 57624, 76560, 198720, 379440, 633600, 978120, 1432080, 2016840, 2756160, 7352640, 14418720, 24710400, 39124800
Offset: 0

Views

Author

Jonathan Vos Post, Feb 18 2006

Keywords

Comments

Many of the terms yield multifactorial primes a(n) + 1, e.g.: a(2) + 1 = 3, a(4) + 1 = 5, a(6) + 1 = 7, a(9) + 1 = 19, a(10) + 1 = 31, a(12) + 1 = 61, a(13) + 1 = 79, a(24) + 1 = 12241, a(25) + 1 = 19801, a(26) + 1 = 29641, a(29) + 1 = 76561, a(31) + 1 = 379441, a(35) + 1 = 2016841, a(36) + 1 = 2756161, ...
Equivalently, product of all positive integers <= n congruent to n (mod 7). - M. F. Hasler, Feb 23 2018

Examples

			a(40) = 40 * a(40-7) = 40 * a(33) = 40 * (33*a(26)) = 40 * 33 * (26*a(19)) = 40 * 33 * 26 * (19*a(12)) = 40 * 33 * 26 * 19 * (12*a(5)) = 40 * 33 * 26 * 19 * 12 5 = 39124800.
		

Crossrefs

Programs

  • GAP
    a:= function(n)
        if n<1 then return 1;
        else return n*a(n-7);
        fi;
      end;
    List([0..40], n-> a(n) ); # G. C. Greubel, Aug 20 2019
  • Magma
    b:= func< n | (n lt 8) select n else n*Self(n-7) >;
    [1] cat [b(n): n in [1..40]]; // G. C. Greubel, Aug 20 2019
    
  • Maple
    A114799 := proc(n)
        option remember;
        if n < 1 then
            1;
        else
            n*procname(n-7) ;
        end if;
    end proc:
    seq(A114799(n),n=0..40) ; # R. J. Mathar, Jun 23 2014
    A114799 := n -> product(n-7*k,k=0..(n-1)/7); # M. F. Hasler, Feb 23 2018
  • Mathematica
    a[n_]:= If[n<1, 1, n*a[n-7]]; Table[a[n], {n,0,40}] (* G. C. Greubel, Aug 20 2019 *)
  • PARI
    A114799(n,k=7)=prod(j=0,(n-1)\k,n-j*k) \\ M. F. Hasler, Feb 23 2018
    
  • Sage
    def a(n):
        if (n<1): return 1
        else: return n*a(n-7)
    [a(n) for n in (0..40)] # G. C. Greubel, Aug 20 2019
    

Formula

a(n) = 1 for n <= 1, else a(n) = n*a(n-7).
Sum_{n>=0} 1/a(n) = A288094. - Amiram Eldar, Nov 10 2020

Extensions

Edited by M. F. Hasler, Feb 23 2018

A167560 The ED2 array read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 24, 16, 6, 1, 120, 80, 32, 8, 1, 720, 480, 192, 54, 10, 1, 5040, 3360, 1344, 384, 82, 12, 1, 40320, 26880, 10752, 3072, 680, 116, 14, 1, 362880, 241920, 96768, 27648, 6144, 1104, 156, 16, 1
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Comments

The coefficients in the upper right triangle of the ED2 array (m>n) were found with the a(n,m) formula while the coefficients in the lower left triangle of the ED2 array (m<=n) were found with the recurrence relation, see below. We use for the array rows the letter n (>=1) and for the array columns the letter m (>=1).
The ED2 array is related to the EG1 matrix, see A162005, because sum(EG1(2*m-1,n) * z^(2*m-1), m=1..infinity) = ((2*n-1)!/(4^(n-1)*(n-1)!^2))*int(sinh(y*(2*z))/cosh(y)^(2*n), y=0..infinity).
For the ED1, ED3 and ED4 arrays see A167546, A167572 and A167584.

Examples

			The ED2 array begins with:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
6, 16, 32, 54, 82, 116, 156, 202, 254, 312
24, 80, 192, 384, 680, 1104, 1680, 2432, 3384, 4560
120, 480, 1344, 3072, 6144, 11160, 18840, 30024, 45672, 66864
720, 3360, 10752, 27648, 61440, 122880, 226800, 392832, 646128, 1018080
		

Crossrefs

A000012, A005843 (n>=1), 2*A104249 (n>=1), A167561, A167562 and A167563 equal the first sixth rows of the array.
A000142 equals the first column of the array.
A047053 equals the a(n, n) diagonal of the array.
2*A034177 equals the a(n+1, n) diagonal of the array.
A167570 equals the a(n+2, n) diagonal of the array,
A167564 equals the row sums of the ED2 array read by antidiagonals.
A167565 is a triangle related to the a(n) formulas of the rows of the ED2 array.
A167568 is a triangle related to the GF(z) formulas of the rows of the ED2 array.
A167569 is the lower left triangle of the ED2 array.
Cf. A162005 (EG1 triangle).
Cf. A167546 (ED1 array), A167572 (ED3 array), A167584 (ED4 array).

Programs

  • Maple
    nmax:=10; mmax:=10; for n from 1 to nmax do for m from 1 to n do a(n,m) := 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! od; for m from n+1 to mmax do a(n,m):= n! + sum((-1)^(k-1)*binomial(n-1,k)*a(n,m-k),k=1..n-1) od; od: for n from 1 to nmax do for m from 1 to n do d(n,m):=a(n-m+1,m) od: od: T:=1: for n from 1 to nmax do for m from 1 to n do a(T):= d(n,m): T:=T+1: od: od: seq(a(n),n=1..T-1);
    # alternative
    A167560 := proc(n,m)
        option remember ;
        if m > n then
            n!+add( (-1)^(k-1)*binomial(n-1,k)*procname(n,m-k),k=1..n-1) ;
        else
            4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! ;
        end if;
    end proc:
    seq( seq(A167560(d-m,m),m=1..d-1),d=2..12) ; # R. J. Mathar, Jun 28 2024
  • Mathematica
    nmax = 10; mmax = 10; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[n, m] = 4^(m - 1)*(m - 1)!*((n + m - 1)!/(2*m - 1)!)]; For[m = n + 1, m <= mmax, m++, a[n, m] = n! + Sum[(-1)^(k - 1)*Binomial[n - 1, k]*a[n, m - k], {k, 1, n - 1}]]; ]; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, d[n, m] = a[n - m + 1, m]]; ]; t = 1; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[t] = d[n, m]; t = t + 1]]; Table[a[n], {n, 1, t - 1}] (* Jean-François Alcover, Dec 20 2011, translated from Maple *)

Formula

a(n,m) = ((m-1)!/((m-n-1)!))*int(sinh(y*(2*n))/(cosh(y))^(2*m),y=0..infinity) for m>n.
The (n-1)-differences of the n-th array row lead to the recurrence relation
sum((-1)^k*binomial(n-1,k)*a(n-1,m-k),k=0..n-1) = n!
which in its turn leads to, see A167569,
a(n,m) = 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! if m<=n.

A144827 Partial products of successive terms of A017029; a(0)=1.

Original entry on oeis.org

1, 4, 44, 792, 19800, 633600, 24710400, 1136678400, 60243955200, 3614637312000, 242180699904000, 17921371792896000, 1451631115224576000, 127743538139762688000, 12135636123277455360000, 1237834884574300446720000, 134924002418598748692480000, 15651184280557454848327680000
Offset: 0

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Author

Philippe Deléham, Sep 21 2008

Keywords

Examples

			a(0)=1, a(1)=4, a(2)=4*11=44, a(3)=4*11*18=792, a(4)=4*11*18*25=19800, ...
		

Crossrefs

Programs

  • Magma
    [ 1 ] cat [ &*[ (7*k+4): k in [0..n] ]: n in [0..14] ]; // Klaus Brockhaus, Nov 10 2008
    
  • Mathematica
    FoldList[Times,1,Range[4,150,7]] (* Harvey P. Dale, Apr 25 2014 *)
  • SageMath
    [1]+[4*7^(n-1)*rising_factorial(11/7, n-1) for n in (1..30)] # G. C. Greubel, Feb 22 2022

Formula

a(n) = Sum_{k=0..n} A132393(n,k)*4^k*7^(n-k).
G.f.: 1/(1-4*x/(1-7*x/(1-11*x/(1-14*x/(1-18*x/(1-21*x/(1-25*x/(1-... (continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-3)^n*Sum_{k=0..n} (7/3)^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
From Ilya Gutkovskiy, Mar 23 2017: (Start)
E.g.f.: 1/(1 - 7*x)^(4/7).
a(n) ~ sqrt(2*Pi)*7^n*n^(n+1/14)/(exp(n)*Gamma(4/7)). (End)
a(n) = 4*7^(n-1)*Pochhammer(n-1, 11/7) with a(0) = 1. - G. C. Greubel, Feb 22 2022
Sum_{n>=0} 1/a(n) = 1 + (e/7^3)^(1/7)*(Gamma(4/7) - Gamma(4/7, 1/7)). - Amiram Eldar, Dec 19 2022

Extensions

Corrected a(9) by Vincenzo Librandi, Jul 14 2011

A051142 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -4, 1, 32, -12, 1, -384, 176, -24, 1, 6144, -3200, 560, -40, 1, -122880, 70144, -14400, 1360, -60, 1, 2949120, -1806336, 415744, -47040, 2800, -84, 1, -82575360, 53526528, -13447168, 1732864, -125440, 5152, -112, 1, 2642411520, -1795424256, 483835904
Offset: 1

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Author

Keywords

Comments

a(n,m) = R_n^m(a=0, b=4) in the notation of the given 1961 and 1962 references.
a(n,m) is a Jabotinsky matrix, i.e., the monic row polynomials E(n,x) := Sum_{m=1..n} a(n,m)*x^m = Product_{j=0..n-1} (x - 4*j), n >= 1, and E(0,x) := 1 are exponential convolution polynomials (see A039692 for the definition and a Knuth reference).
This is the signed Stirling1 triangle with diagonal d >= 0 (main diagonal d = 0) scaled with 4^d.
Also the Bell transform of the quadruple factorial numbers Product_{k=0..n-1} (4*k+4) (A047053) giving unsigned values and adding 1, 0, 0, 0, ... as column 0. For the definition of the Bell transform, see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015

Examples

			Triangle a(n,m) (with rows n >= 1 and columns m = 1..n) begins:
        1;
       -4,     1;
       32,   -12,      1;
     -384,   176,    -24,    1;
     6144, -3200,    560,  -40,   1,
  -122880, 70144, -14400, 1360, -60, 1;
  ...
3rd row o.g.f.: E(3,x) = 32*x - 12*x^2 + x^3.
		

Crossrefs

First (m=1) column sequence is: A047053(n-1).
Row sums (signed triangle): A008545(n-1)*(-1)^(n-1).
Row sums (unsigned triangle): A007696(n).
Cf. A008275 (Stirling1 triangle, b=1), A039683 (b=2), A051141 (b=3).

Programs

  • Mathematica
    Table[StirlingS1[n, m] 4^(n - m), {n, 9}, {m, n}] // Flatten (* Michael De Vlieger, Dec 31 2015 *)
  • Sage
    # uses[bell_transform from A264428]
    # Unsigned values and an additional first column (1,0,0,0, ...).
    def A051142_row(n):
        multifact_4_4 = lambda n: prod(4*k + 4 for k in (0..n-1))
        mfact = [multifact_4_4(k) for k in (0..n)]
        return bell_transform(n, mfact)
    [A051142_row(n) for n in (0..9)] # Peter Luschny, Dec 31 2015

Formula

a(n, m) = a(n-1, m-1) - 4*(n-1)*a(n-1, m) for n >= m >= 1; a(n, m) := 0 for n < m; a(n, 0) := 0 for n >= 1; a(0, 0) = 1.
E.g.f. for the m-th column of the signed triangle: (log(1 + 4*x)/4)^m/m!.
a(n, m) = S1(n, m)*4^(n-m), with S1(n, m) := A008275(n, m) (signed Stirling1 triangle).

A245244 Triangle of coefficients of the Pbar polynomials, read by rows.

Original entry on oeis.org

1, -3, 4, 25, -56, 32, -427, 1228, -1184, 384, 12465, -41840, 52416, -29184, 6144, -555731, 2079892, -3076288, 2258688, -829440, 122880, 35135945, -142843304, 237829600, -208562688, 102279168, -26787840, 2949120, -2990414715, 12987478876, -23672564832, 23581133952, -13947525120, 4929576960, -970260480
Offset: 0

Views

Author

Richard P. Brent, Jul 14 2014

Keywords

Comments

Pbar(r,n) is a polynomial of degree r defined by the recurrence
Pbar(r+1,n) = (2*n-1)^2 * Pbar(r,n) - 4*(n-1)^2 * Pbar(r,n-1)
with initial condition Pbar(0,n) = 1.

Examples

			Pbar(1,n) = 4*n-3, Pbar(2,n) = 32*n^2 - 56*n + 25.
Triangle begins:
1,
-3, 4,
25, -56, 32,
-427, 1228, -1184, 384,
12465, -41840, 52416, -29184, 6144,
...
From _Peter Bala_, Jan 22 2018: (Start)
The polynomials Pbar(r,n) as hypergeometric series:
r = 0: n*Pbar(0,n) = n = 1 + 3*(n-1)/(n+1) + 5*(n-1)*(n-2)/((n+1)*(n+2)) + 7*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 1/2.
r = 1: n*Pbar(1,n) = n*(4*n - 3) = 1 + 3^3*(n-1)/(n+1) + 5^3*(n-1)*(n-2)/((n+1)*(n+2)) + 7^3*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 3/2.
r = 2: n*Pbar(2,n) = n*(32*n^2 - 56*n + 25) = 1 + 3^5*(n-1)/(n+1) + 5^5*(n-1)*(n-2)/((n+1)*(n+2)) + 7^5*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 5/2.
The above identities when r = 0 and r = 1 were found by Ramanujan. See Example 5 and Example 13 in Chapter 10 of Berndt. (End)
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 20 and p. 23.

Crossrefs

(-1)^r Pbar(r,0) is sequence A009843. The leading coefficient of Pbar(r,n) is sequence A047053. Cf. also A036970, A083061, A160485 for analogous moments of Bernoulli random walks.

Programs

  • PARI
    N=10; P=vector(N+2); P[1]=1;
    Pbar(r)=P[r+1];
    for (r=0, N, P[r+2] = (2*n-1)^2 * Pbar(r) - 4*(n-1)^2 * subst(Pbar(r),n,n-1) );
    seq=[];  for(r=1,N, seq=concat(seq, Vecrev(P[r])); );  seq
    \\ Joerg Arndt, Jan 27 2015

Formula

In terms of Dumont-Foata polynomials F(r,x,y,z),
Pbar(r,n) = (-4)^r F(r+1,1/2-n,1/2,1/2).
In terms of odd absolute moments of a symmetric Bernoulli random walk with an odd number of steps,
n*C(2*n,n)*Pbar(r,n) = Sum_{k} C(2*n-1,k) * |2*n-1-2*k|^(2*r+1).
In terms of the Pochhammer symbol or ascending factorial (x)_k,
Pbar(r,n) = Sum_{1 <= j <= k <= r+1} (-1)^(j+1)*(1-n)_{k-1}*(2j-1)^(2r+1)/((k-j)!(k)_j).
n*Pbar(r,n) = 1 + 3^(2*r+1)*(n-1)/(n+1) + 5^(2*r+1)*(n-1)*(n-2)/((n+1)*(n+2)) + 7^(2*r+1)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... = Sum_{k = 0..n-1} binomial(n-1,k)/binomial(n+k,k)*(2*k + 1)^(2*r+1); this follows easily from the above recurrence. Examples are given below. - Peter Bala, Jan 22 2018

Extensions

More terms from Joerg Arndt, Jan 27 2015
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