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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A351200 Number of patterns of length n with all distinct runs.

Original entry on oeis.org

1, 1, 3, 11, 53, 305, 2051, 15731, 135697, 1300869, 13726431, 158137851, 1975599321, 26607158781, 384347911211, 5928465081703, 97262304328573, 1691274884085061, 31073791192091251, 601539400910369671, 12238270940611270161, 261071590963047040241
Offset: 0

Views

Author

Gus Wiseman, Feb 09 2022

Keywords

Comments

We define a pattern to be a finite sequence covering an initial interval of positive integers. Patterns are counted by A000670 and ranked by A333217.

Examples

			The a(1) = 1 through a(3) = 11 patterns:
  (1)  (1,1)  (1,1,1)
       (1,2)  (1,1,2)
       (2,1)  (1,2,2)
              (1,2,3)
              (1,3,2)
              (2,1,1)
              (2,1,3)
              (2,2,1)
              (2,3,1)
              (3,1,2)
              (3,2,1)
The complement for n = 3 counts the two patterns (1,2,1) and (2,1,2).

Links

Crossrefs

The version for run-lengths instead of runs is A351292.
A000670 counts patterns, ranked by A333217.
A005649 counts anti-run patterns, complement A069321.
A005811 counts runs in binary expansion.
A032011 counts patterns with distinct multiplicities.
A044813 lists numbers whose binary expansion has distinct run-lengths.
A060223 counts Lyndon patterns, necklaces A019536, aperiodic A296975.
A131689 counts patterns by number of distinct parts.
A238130 and A238279 count compositions by number of runs.
A297770 counts distinct runs in binary expansion.
A345194 counts alternating patterns, up/down A350354.
Counting words with all distinct runs:
- A351013 = compositions, for run-lengths A329739, ranked by A351290.
- A351016 = binary words, for run-lengths A351017.
- A351018 = binary expansions, for run-lengths A032020, ranked by A175413.
- A351202 = permutations of prime factors.
- A351642 = word structures.
Row sums of A351640.
Cf. A003242, A098504, A098859, A106356, A242882, A325545, A328592, A329740, A351014, A351204, A351291.

Programs

  • Mathematica
    allnorm[n_]:=If[n<=0,{{}},Function[s,Array[Count[s,y_/;y<=#]+1&,n]] /@Subsets[Range[n-1]+1]];
Table[Length[Select[Join@@Permutations/@allnorm[n],UnsameQ@@Split[#]&]],{n,0,6}]
  • PARI
    \\ here LahI is A111596 as row polynomials.
LahI(n,y)={sum(k=1, n, y^k*(-1)^(n-k)*(n!/k!)*binomial(n-1, k-1))}
S(n)={my(p=prod(k=1, n, 1 + y*x^k + O(x*x^n))); 1 + sum(i=1, (sqrtint(8*n+1)-1)\2, polcoef(p,i,y)*LahI(i,y))}
R(q)={[subst(serlaplace(p), y, 1) | p<-Vec(q)]}
seq(n)={my(q=S(n)); concat([1], sum(k=1, n, R(q^k-1)*sum(r=k, n, binomial(r, k)*(-1)^(r-k)) ))} \\ Andrew Howroyd, Feb 12 2022

Extensions

Terms a(10) and beyond from Andrew Howroyd, Feb 12 2022

A187535 Central Lah numbers: a(n) = A105278(2*n,n) = A008297(2*n,n).

Original entry on oeis.org

1, 2, 36, 1200, 58800, 3810240, 307359360, 29682132480, 3339239904000, 428906814336000, 61934143990118400, 9931984545324441600, 1751339941492209868800, 336796142594655744000000, 70149825129001153536000000, 15732267448930658699673600000
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Comments

a(n) is the number of Lah partitions of a set of size 2n with n blocks.

Crossrefs

Cf. A008297, A111596, A066667, A187536, A187538, A187539, A187540, A187542 - A187548, A090181, A125558.

Programs

  • Maple
    A187535:= n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(A187535(n),n=0..12);
  • Mathematica
    a[n_]:=If[n==0,1,Binomial[2n-1,n-1](2n)!/n!]
Table[a[n],{n,0,12}]
(* Alternative: *)
a[n_] := Binomial[2*n, n] FactorialPower[2*n - 1, n];
Table[a[n], {n, 0, 15}] (* Peter Luschny, Jun 15 2022 *)
  • Maxima
    a(n) := if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(a(n),n,0,12);
  • Sage
    [catalan_number(n)*binomial(2*n-1,n)*factorial(n+1) for n in range(15)] # Peter Luschny, Oct 07 2014

Formula

a(n) = binomial(2n-1,n-1)*(2n)!/n! (for n>0).
D-finite with recurrence (n+1)*a(n+1) = 4*(2n+1)^2*a(n) - delta(n,0).
a(n) ~ 2^(4*n)*n^n*exp(-n)/sqrt(2*n*Pi).
a(n)*a(n+2) - a(n+1)^2 is >= 0 and is a multiple of 2^(n+3) for all nonnegative n.
a(n) == 0 (mod 10) for n>3.
E.g.f.: 1/2 + K(16x)/Pi, where K(z) is the complete elliptic integral of the first kind, which can also be written as a Legendre function of the second kind.
a(n) = Catalan(n)*C(2*n-1,n)*(n+1)!. - Peter Luschny, Oct 07 2014
a(n) = A125558(n)*(n+1)! = A090181(2*n,n)*(n+1)!. - Peter Luschny, Oct 07 2014
a(n) = (2/n)*(Gamma(2*n)^2/Gamma(n)^3) for n>0. - Peter Luschny, Oct 17 2014

A001754 Lah numbers: a(n) = n!*binomial(n-1,2)/6.

Original entry on oeis.org

0, 0, 1, 12, 120, 1200, 12600, 141120, 1693440, 21772800, 299376000, 4390848000, 68497228800, 1133317785600, 19833061248000, 366148823040000, 7113748561920000, 145120470663168000, 3101950060425216000, 69337707233034240000, 1617879835437465600000
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n+1) = Sum_{pi in Symm(n)} Sum_{i=1..n} max(pi(i)-i,0)^2, i.e., the sum of the squares of the positive displacement of all letters in all permutations on n letters. - Franklin T. Adams-Watters, Oct 25 2006

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 156.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 44.
  • 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).

Links

Crossrefs

Column 3 of A008297.
Column m=3 of unsigned triangle A111596.
Cf. A005990, A053495, A260665.
Cf. A001113, A001620, A091725, A099285.

Programs

  • Magma
    [Factorial(n)*Binomial(n-1, 2)/6: n in [1..25]]; // Vincenzo Librandi, Oct 11 2011
  • Maple
    [seq(n!*binomial(n-1,2)/6, n=1..40)];
  • Mathematica
    Table[(n-2)*(n-1)*n!/12, {n, 21}] (* Arkadiusz Wesolowski, Nov 26 2012 *)
With[{nn=30},CoefficientList[Series[(x/(1-x))^3/6,{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Oct 04 2017 *)
  • Sage
    [factorial(n-1)*binomial(n,3)/2 for n in (1..30)] # G. C. Greubel, May 10 2021

Formula

E.g.f.: ((x/(1-x))^3)/3!.
If we define f(n,i,x) = Sum_{k=i..n} (Sum_{j=i..k} binomial(k,j) * Stirling1(n,k) * Stirling2(j,i)*x^(k-j)) then a(n+1) = (-1)^n*f(n,2,-4), n >= 2. - Milan Janjic, Mar 01 2009
a(n) = Sum_{k>=1} k * A260665(n,k). - Alois P. Heinz, Nov 14 2015
D-finite with recurrence (-n+5)*a(n) + (n-2)*(n-3)*a(n-1) = 0, n >= 4. - R. J. Mathar, Jan 06 2021
From Amiram Eldar, May 02 2022: (Start)
Sum_{n>=3} 1/a(n) = 6*(gamma - Ei(1)) + 9, where gamma = A001620 and Ei(1) = A091725.
Sum_{n>=3} (-1)^(n+1)/a(n) = 18*(gamma - Ei(-1)) - 12/e - 9, where Ei(-1) = -A099285 and e = A001113. (End)

A111595 Triangle of coefficients of square of Hermite polynomials divided by 2^n with argument sqrt(x/2).

Original entry on oeis.org

1, 0, 1, 1, -2, 1, 0, 9, -6, 1, 9, -36, 42, -12, 1, 0, 225, -300, 130, -20, 1, 225, -1350, 2475, -1380, 315, -30, 1, 0, 11025, -22050, 15435, -4620, 651, -42, 1, 11025, -88200, 220500, -182280, 67830, -12600, 1204, -56, 1, 0, 893025, -2381400, 2302020, -1020600, 235494, -29736, 2052, -72
Offset: 0

Views

Author

Wolfdieter Lang, Aug 23 2005

Keywords

Comments

This is a Sheffer triangle (lower triangular exponential convolution matrix). For Sheffer row polynomials see the S. Roman reference and explanations under A048854.
In the umbral notation of the S. Roman reference this would be called Sheffer for ((sqrt(1-2*t))/(1-t), t/(1-t)).
The associated Sheffer triangle is A111596.
Matrix logarithm equals A112239. - Paul D. Hanna, Aug 29 2005
The row polynomials (1/2^n)* H(n,sqrt(x/2))^2, with the Hermite polynomials H(n,x), have e.g.f. (1/sqrt(1-y^2))*exp(x*y/(1+y)).
The row polynomials s(n,x):=sum(a(n,m)*x^m,m=0..n), together with the associated row polynomials p(n,x) of A111596, satisfy the exponential (or binomial) convolution identity s(n,x+y) = sum(binomial(n,k)*s(k,x)*p(n-k,y),k=0..n), n>=0.
The unsigned column sequences are: A111601, A111602, A111777-A111784, for m=1..10.

Examples

			The triangle a(n, m) begins:
n\m       0         1         2          3         4         5       6       7     8    9  10 ...
0:        1
1:        0         1
2:        1        -2         1
3:        0         9        -6          1
4:        9       -36        42        -12         1
5:        0       225      -300        130       -20         1
6:      225     -1350      2475      -1380       315       -30       1
7:        0     11025    -22050      15435     -4620       651     -42       1
8:    11025    -88200    220500    -182280     67830    -12600    1204     -56     1
9:        0    893025  -2381400    2302020  -1020600    235494  -29736    2052   -72    1
10:  893025  -8930250  28279125  -30958200  15961050  -4396140  689850  -63000  3285  -90   1
-------------------------------------------------------------------------------------------------

References

  • R. P. Boas and R. C. Buck, Polynomial Expansions of Analytic Functions, Springer, 1958, p. 41
  • S. Roman, The Umbral Calculus, Academic Press, New York, 1984, p. 128.

Links

Crossrefs

Row sums: A111882. Unsigned row sums: A111883.
Cf. A112239 (matrix log).

Programs

  • Mathematica
    row[n_] := CoefficientList[ 1/2^n*HermiteH[n, Sqrt[x/2]]^2, x]; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)
  • Python
    from sympy import hermite, Poly, sqrt, symbols
x = symbols('x')
def a(n): return Poly(1/2**n*hermite(n, sqrt(x/2))**2, x).all_coeffs()[::-1]
for n in range(11): print(a(n)) # Indranil Ghosh, May 26 2017

Formula

E.g.f. for column m>=0: (1/sqrt(1-x^2))*((x/(1+x))^m)/m!.
a(n, m)=((-1)^(n-m))*(n!/m!)*sum(binomial(2*k, k)*binomial(n-2*k-1, m-1)/(4^k), k=0..floor((n-m)/2)), n>=m>=1. a(2*k, 0)= ((2*k)!/(k!*2^k))^2 = A001818(k), a(2*k+1) = 0, k>=0. a(n, m)=0 if n

A129062 T(n, k) = [x^k] Sum_{k=0..n} Stirling2(n, k)*RisingFactorial(x, k), triangle read by rows, for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 6, 1, 0, 26, 36, 12, 1, 0, 150, 250, 120, 20, 1, 0, 1082, 2040, 1230, 300, 30, 1, 0, 9366, 19334, 13650, 4270, 630, 42, 1, 0, 94586, 209580, 166376, 62160, 11900, 1176, 56, 1, 0, 1091670, 2562354, 2229444, 952728, 220500, 28476, 2016, 72, 1
Offset: 0

Views

Author

Wolfdieter Lang, May 04 2007

Keywords

Comments

Matrix product of Stirling2 with unsigned Stirling1 triangle.
For the subtriangle without column no. m=0 and row no. n=0 see A079641.
The reversed matrix product |S1|. S2 is given in A111596.
As a product of lower triangular Jabotinsky matrices this is a lower triangular Jabotinsky matrix. See the D. E. Knuth references given in A039692 for Jabotinsky type matrices.
E.g.f. for row polynomials P(n,x):=sum(a(n,m)*x^m,m=0..n) is 1/(2-exp(z))^x. See the e.g.f. for the columns given below.
A048993*A132393 as infinite lower triangular matrices. - Philippe Deléham, Nov 01 2009
Triangle T(n,k), read by rows, given by (0,2,1,4,2,6,3,8,4,10,5,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 19 2011.
Also the Bell transform of A000629. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle begins:
  1;
  0,    1;
  0,    2,    1;
  0,    6,    6,    1;
  0,   26,   36,   12,   1;
  0,  150,  250,  120,  20,  1;
  0, 1082, 2040, 1230, 300, 30,  1;

Links

Crossrefs

Columns k=0..3 give A000007, A000629(n-1), A129063, A129064.
Cf. A000629, A000670, A005649, A079641, A129062, A325872, A325873, A383140.

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
BellMatrix(n -> polylog(-n,1/2), 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    rows = 9;
t = Table[PolyLog[-n, 1/2], {n, 0, rows}]; T[n_, k_] := BellY[n, k, t];
Table[T[n, k], {n, 0, rows}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)
p[n_] := Sum[StirlingS2[n, k] Pochhammer[x, k], {k, 0, n}];
Table[CoefficientList[FunctionExpand[p[n]], x], {n, 0, 9}] // Flatten (* Peter Luschny, Jun 27 2019 *)
  • Sage
    def a_row(n):
    s = sum(stirling_number2(n,k)*rising_factorial(x,k) for k in (0..n))
    return expand(s).list()
[a_row(n) for n in (0..9)] # Peter Luschny, Jun 28 2019

Formula

a(n,m) = Sum_{k=m..n} S2(n,k) * |S1(k,m)|, n>=0; S2=A048993, S1=A048994.
E.g.f. of column k (with leading zeros): (f(x)^k)/k! with f(x):= -log(1-(exp(x)-1)) = -log(2-exp(x)).
Sum_{0<=k<=n} T(n,k)*x^k = A153881(n+1), A000007(n), A000670(n), A005649(n) for x = -1,0,1,2 respectively. - Philippe Deléham, Nov 19 2011

Extensions

New name by Peter Luschny, Jun 27 2019

A187536 Partial sums of the central Lah numbers (A187535).

Original entry on oeis.org

1, 3, 39, 1239, 60039, 3870279, 311229639, 29993362119, 3369233266119, 432276047602119, 62366420037720519, 9994350965362162119, 1761334292457572030919, 338557476887113316030919, 70488382605888266852030919, 15802755831536546966525630919
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Crossrefs

Cf. A187535, A008297, A111596, A187538 - A187540, A187542 - A187548.

Programs

  • Maple
    A187536 := proc(n) add(A187535(i),i=0..n) ; end proc:
seq(A187536(n),n=0..10) ; # R. J. Mathar, Mar 20 2011
  • Mathematica
    Table[1 + Sum[Binomial[2k-1,k-1](2k)!/k!, {k, 1, n}], {n, 0, 20}]
  • Maxima
    makelist(1+sum(binomial(2*k-1,k-1)*(2*k)!/k!,k,1,n),n,0,12);

Formula

a(n) = 1 + Sum_{k=0..n} binomial(2k-1,k-1)*(2k)!/k!.
(n+2)*a(n+2) - (16n^2 + 49n +3 8)*a(n+1) + 4 *(2n+3)^2*a(n) = 0.
Asymptotically a(n) ~ 2^(4n)n^n exp(-n)/sqrt(2n*pi).

A187538 Alternating partial sums of the central Lah numbers (A187535).

Original entry on oeis.org

1, 1, 35, 1165, 57635, 3752605, 303606755, 29378525725, 3309861378275, 425596952957725, 61508547037160675, 9870475998287280925, 1741469465493922587875, 335054673129161821412125, 69814770455871991714587875, 15662452678474786707959012125, 3764014801927115965888623387875
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187539, A187540, A187542 - A187548.

Programs

  • Maple
    A187538 := proc(n) add( (-1)^(n+k)*A187535(k),k=0..n) ; end proc:
seq(A187538(n),n=0..10) ; # R. J. Mathar, Mar 21 2011
  • Mathematica
    Table[(-1)^n + Sum[(-1)^(n-k)Binomial[2k-1,k-1](2k)!/k!, {k, 1, n}], {n, 0, 20}]
  • Maxima
    makelist((-1)^n+sum((-1)^(n-k)*binomial(2*k-1,k-1)*(2*k)!/k!,k,1,n),n,0,12);

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*A187535(k).
(n+2)*a(n+2) - (16*n^2 + 47*n + 34)*a(n+1) - 4*(2*n+3)^2*a(n) = 0.
a(n) ~ 2^(4*n - 1/2) * n^(n - 1/2) / (sqrt(Pi) * exp(n)). - Vaclav Kotesovec, Mar 30 2018

A187540 Binomial partial sums of the central Lah numbers.

Original entry on oeis.org

1, 3, 41, 1315, 63825, 4116611, 331127353, 31915763811, 3585520583585, 460054836028675, 66377105303195721, 10637410917472061603, 1874707445757653437681, 360356280811211873453955, 75028021167256736753934425
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187538, A187539, A187542, A187543, A187544, A187545, A187546, A187547, A187548.

Programs

  • Maple
    seq(1+add(binomial(n,k)*binomial(2*k-1,k-1)*(2*k)!/k!, k=1..n), n=0..20);
  • Mathematica
    Table[1 + Sum[Binomial[n, k]Binomial[2k-1,k-1](2k)!/k!, {k, 1, n}], {n, 0, 20}]
  • Maxima
    makelist(1+sum(binomial(n,k)*binomial(2*k-1,k-1)*(2*k)!/k!, k,1,n), n,0,12);
  • PARI
    a(n) = 1+sum(k=0,n, binomial(n,k)*binomial(2*k-1,k-1)*(2*k)!/k!) \\ Charles R Greathouse IV, Feb 07 2017

Formula

Formula: a(n) = 1+sum(binomial(n,k)binomial(2k-1,k-1)(2k)!/k!,k=0..n).
Recurrence: for n>=3, a(n) = 1/n*(-2 +(32 - 48*n + 16*n^2)*a(n-3) + (-31 + 63*n - 32*n^2)*a(n-2) + (3 - 14*n + 16*n^2)*a(n-1) )
E.g.f.: exp(x) (1/2 + 1/Pi K(16x) ), where K(z) is the elliptic integral of the first kind (defined as in Mathematica).
a(n) ~ 16^n*n^(n-1/2)*exp(1/16-n)/sqrt(2*Pi). - Vaclav Kotesovec, Aug 09 2013

A187542 Convolutions of the central Lah numbers (A187535).

Original entry on oeis.org

1, 4, 76, 2544, 123696, 7942080, 635633280, 61009159680, 6831940227840, 874493448514560, 125946241018214400, 20156433977646489600, 3548609812373223628800, 681555522002874494976000, 141810253720479017017344000
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187539, A187540, A187543, A187544, A187545, A187546, A187547, A187548.

Programs

  • Maple
    a := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(sum(a(k)*a(n-k), k=0..n),n=0..12);
  • Mathematica
    a[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[a[k]a[n - k], {k, 0, n}], {n, 0, 20}]
  • Maxima
    a(n) := if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(a(k)*a(n-k),k,0,n),n,0,12);

Formula

a(n) = sum(L(k)L(n-k),k=0..n), where L(n) is a central Lah number.
a(n) ~ n! * 16^n / (Pi*n). - Vaclav Kotesovec, Oct 06 2019

A187539 Alternated binomial partial sums of central Lah numbers (A187535).

Original entry on oeis.org

1, 1, 33, 1097, 54209, 3527889, 285356449, 27608615257, 3110179582593, 399896866564001, 57791843384031521, 9273757516482276201, 1636151050649025202753, 314786007405793614831217, 65590496972310741712688289, 14714600180590751334321307769
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187540, A187542, A187543, A187544, A187545, A187546, A187547, A187548.

Programs

  • Maple
    seq((-1)^n+add((-1)^(n-k)*binomial(n,k)*binomial(2*k-1,k-1)*(2*k)!/k!, k=1..n), n=0..20);
  • Mathematica
    Table[(-1)^n + Sum[(-1)^(n-k)Binomial[n,k]Binomial[2k-1,k-1](2k)!/k!, {k, 1, n}], {n, 0, 20}]
  • Maxima
    makelist((-1)^n+sum((-1)^(n-k)*binomial(n,k)*binomial(2*k-1,k-1) *(2*k)!/k!, k,1,n), n,0,12);

Formula

a(n) = 1+sum((-1)^(n-k)*C(n,k)*C(2k-1,k-1)*(2k)!/k!, k=0..n).
Recurrence: n>=3, a(n) = (2*(-1)^n + (32 - 48*n + 16*n^2)*a(n-3) + (33 - 65*n + 32*n^2)*a(n-2) + (5 - 18*n + 16*n^2)*a(n-1))/n
E.g.f.: exp(-x) (1/2 + 1/pi K(16x) ), where K(z) is the elliptic integral of the first kind (defined as in Mathematica).
a(n) ~ 16^n*n^(n-1/2)/(sqrt(2*Pi)*exp(n+1/16)). - Vaclav Kotesovec, Aug 10 2013
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