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.

Showing 1-9 of 9 results.

A362849 Triangle read by rows, T(n, k) = A243664(n) * binomial(n, k).

Original entry on oeis.org

1, 1, 1, 21, 42, 21, 1849, 5547, 5547, 1849, 426405, 1705620, 2558430, 1705620, 426405, 203374081, 1016870405, 2033740810, 2033740810, 1016870405, 203374081, 173959321557, 1043755929342, 2609389823355, 3479186431140, 2609389823355, 1043755929342, 173959321557
Offset: 0

Views

Author

Peter Luschny, May 05 2023

Keywords

Examples

			[0]         1;
[1]         1,          1;
[2]        21,         42,         21;
[3]      1849,       5547,       5547,       1849;
[4]    426405,    1705620,    2558430,    1705620,     426405;
[5] 203374081, 1016870405, 2033740810, 2033740810, 1016870405, 203374081;
		

Crossrefs

Family of triangles: A055372 (m=0, Pascal), A362585 (m=1, Fubini), A362586 (m=2, Joffe), this sequence (m=3, A278073).
Cf. A243664 (column 0 and main diagonal).

Programs

  • SageMath
    # uses[TransOrdPart from A362585]
    def A362849(n) -> list[int]: return TransOrdPart(3, n)
    for n in range(6): print(A362849(n))

A094088 E.g.f. 1/(2-cosh(x)) (even coefficients).

Original entry on oeis.org

1, 1, 7, 121, 3907, 202741, 15430207, 1619195761, 224061282907, 39531606447181, 8661323866026007, 2307185279184885001, 734307168916191403507, 275199311597682485597221, 119956934012963778952439407
Offset: 0

Views

Author

Ralf Stephan, Apr 30 2004

Keywords

Comments

With alternating signs, e.g.f.: 1/(2-cos(x)).
7 divides a(3n+2). Ira Gessel remarks: For any odd prime p, the coefficients of 1/(2-cosh(x)) as e.g.f. are periodic with period dividing p-1.
Consider the sequence defined by a(0) = 1; thereafter a(n) = c*Sum_{k = 1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.
a(n) is the number of ordered set partitions of {1,2,...,2n} into even size blocks. - Geoffrey Critzer, Dec 03 2012
Except for a(0), row sums of A241171. - Peter Bala, Aug 20 2014
Exp( Sum_{n >= 1} a(n)*x^n/n) is the o.g.f. for A255928. - Peter Bala, Mar 13 2015
Also the 2-packed words of degree n; cf. A011782, A000670, A094088, A243664, A243665, A243666 for k-packed words for 0<=k<=5. - Peter Luschny, Jul 06 2015

Crossrefs

Programs

  • Maple
    f:=proc(n,k) option remember;  local i;
    if n=0 then 1
    else k*add(binomial(2*n,2*i)*f(n-i,k),i=1..floor(n)); fi; end;
    g:=k->[seq(f(n,k),n=0..40)];g(1); # N. J. A. Sloane, Mar 28 2012
  • Mathematica
    nn=30;Select[Range[0,nn]!CoefficientList[Series[1/(2-Cosh[x]),{x,0,nn}],x],#>0&]  (* Geoffrey Critzer, Dec 03 2012 *)
    a[0]=1; a[n_] := Sum[1/2*(1+(-1)^(2*n))*Sum[((-1)^(k-j)*Binomial[k, j]*Sum[(j-2*i )^(2*n)*Binomial[j, i], {i, 0, j}])/2^j, {j, 1, k}], {k, 1, n}]; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Apr 03 2015, after Vladimir Kruchinin *)
  • Maxima
    a(n):=b(2*n+2);
    b(n):=sum(((sum(((sum((j-2*i)^n*binomial(j,i),i,0,j))*(-1)^(k-j)*binomial(k,j))/2^(j),j,1,k))*((-1)^n+1))/2,k,1,n/2); /* Vladimir Kruchinin, Apr 23 2011 */
    
  • Maxima
    a(n):=sum(sum((i-k)^(2*n)*binomial(2*k,i)*(-1)^(i),i,0,k-1)/(2^(k-1)),k,1,2*n); /* Vladimir Kruchinin, Oct 05 2012 */
  • PARI
    a(n) = if (n == 0, 1, sum(k=1, n, binomial(2*n, 2*n-2*k)*a(n-k)));
    
  • Sage
    def A094088(n) :
        @CachedFunction
        def intern(n) :
            if n == 0 : return 1
            if n % 2 != 0 : return 0
            return add(intern(k)*binomial(n,k) for k in range(n)[::2])
        return intern(2*n)
    [A094088(n) for n in (0..14)]  # Peter Luschny, Jul 14 2012
    

Formula

1/(2-cosh(x)) = Sum_{n>=0} a(n)x^(2n)/(2n)! = 1 + x^2/2 + 7x^4/24 + 121x^6/720 + ...
Recurrence: a(0)=1, a(n) = Sum_{k=1..n} C(2n, 2n-2k)*a(n-k).
a(0)=1 and, for n>=1, a(n)=b(2*n) where b(n) = sum(k=1..n/2,((sum(j=1..k, ((sum(i=0..j,(j-2*i)^n*binomial(j,i)))*(-1)^(k-j)*binomial(k,j))/2^(j)))*((-1)^n+1))/2). - Vladimir Kruchinin, Apr 23 2011
E.g.f.: 1/(2-cosh(x))=8*(1-x^2)/(8 - 12*x^2 + x^4*U(0)) where U(k)= 1 + 4*(k+1)*(k+2)/(2*k+3 - x^2*(2*k+3)/(x^2 + 8*(k+1)*(k+2)*(k+3)/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Sep 30 2012
a(n) = Sum_{k=1..2*n} ( Sum_{i=0..k-1} (i-k)^(2*n) * binomial(2*k,i) * (-1)^i )/2^(k-1), n>0, a(0)=1. - Vladimir Kruchinin, Oct 05 2012
a(n) ~ 2*(2*n)! /(sqrt(3) * (log(2+sqrt(3)))^(2*n+1)). - Vaclav Kotesovec, Oct 19 2013

Extensions

Corrected definition, Joerg Arndt, Apr 26 2011

A278073 Triangle read by rows, coefficients of the polynomials P(m, n) = Sum_{k=1..n} binomial(m*n, m*k)* P(m, n-k)*z with P(m, 0) = 1 and m = 3.

Original entry on oeis.org

1, 0, 1, 0, 1, 20, 0, 1, 168, 1680, 0, 1, 1364, 55440, 369600, 0, 1, 10920, 1561560, 33633600, 168168000, 0, 1, 87380, 42771456, 2385102720, 34306272000, 137225088000, 0, 1, 699048, 1160164320, 158411809920, 5105916816000, 54752810112000, 182509367040000
Offset: 0

Views

Author

Peter Luschny, Jan 22 2017

Keywords

Examples

			Triangle begins:
[1]
[0, 1]
[0, 1,    20]
[0, 1,   168,    1680]
[0, 1,  1364,   55440,   369600]
[0, 1, 10920, 1561560, 33633600, 168168000]
		

Crossrefs

Cf. A014606 (diagonal), A243664 (row sums), A002115 (alternating row sums), A281479 (central coefficients), A327023 (refinement).
Cf. A097805 (m=0), A131689 (m=1), A241171 (m=2), A278074 (m=4).

Programs

  • Maple
    P := proc(m, n) option remember; if n = 0 then 1 else
    add(binomial(m*n, m*k)*P(m, n-k)*x, k=1..n) fi end:
    for n from 0 to 6 do PolynomialTools:-CoefficientList(P(3,n), x) od;
    # Alternatively:
    A278073_row := proc(n)
    1/(1-t*((1/3)*exp(x)+(2/3)*exp(-(1/2)*x)*cos((1/2)*x*sqrt(3))-1));
    expand(series(%,x,3*n+1)); (3*n)!*coeff(%,x,3*n);
    PolynomialTools:-CoefficientList(%,t) end:
    for n from 0 to 6 do A278073_row(n) od;
  • Mathematica
    With[{m = 3}, Table[Expand[j!*SeriesCoefficient[1/(1 - t*(MittagLefflerE[m, x^m] - 1)), {x, 0, j}]], {j, 0, 21, m}]];
    Function[arg, CoefficientList[arg, t]] /@ % // Flatten
  • Sage
    R = PowerSeriesRing(ZZ, 'x')
    x = R.gen().O(30)
    @cached_function
    def P(m, n):
        if n == 0: return R(1)
        return expand(sum(binomial(m*n, m*k)*P(m, n-k)*x for k in (1..n)))
    def A278073_row(n): return list(P(3, n))
    for n in (0..6): print(A278073_row(n)) # Peter Luschny, Mar 24 2020

Formula

E.g.f.: 1/(1-t*((1/3)*exp(x)+(2/3)*exp(-(1/2)*x)*cos((1/2)*x*sqrt(3))-1)), nonzero terms.

A243665 Number of 4-packed words of degree n.

Original entry on oeis.org

1, 1, 71, 35641, 65782211, 323213457781, 3482943541940351, 72319852680213967921, 2637329566270689344838491, 157544683317273333844553610061, 14601235867276343036803577794300631, 2010110081536549910297353731858747088201, 396647963186245408341324212422008625649510771
Offset: 0

Views

Author

N. J. A. Sloane, Jun 14 2014

Keywords

Comments

See Novelli-Thibon (2014) for precise definition.

Crossrefs

Cf. A011782, A000670, A094088, A243664, A243665, A243666 for k-packed words of degree n for 0<=k<=5.

Programs

  • Maple
    1/(2-(cos(t^(1/4))+cosh(t^(1/4)))/2): series(%,t,14): seq((4*n)!*coeff(%,t,n),n=0..12); # Peter Luschny, Jul 07 2015
  • Mathematica
    g[t_] := (Cos[t] + Cosh[t])/2;
    a[n_] := (4n)! SeriesCoefficient[1/(2 - g[t^(1/4)]), {t, 0, n}];
    Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jul 14 2018, after Peter Luschny *)
  • PARI
    seq(n)={my(a=vector(n+1)); a[1]=1; for(n=1, n, a[1+n]=sum(k=1, n, binomial(4*n, 4*k) * a[1+n-k])); a} \\ Andrew Howroyd, Jan 21 2020
  • Sage
    # uses[CEN from A243664]
    A243665 = lambda len: CEN(4,len)
    A243665(13) # Peter Luschny, Jul 06 2015
    
  • Sage
    # Alternatively:
    def PackedWords4(n):
        shapes = ([x*4 for x in p] for p in Partitions(n))
        return sum(factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in shapes)
    [PackedWords4(n) for n in (0..12)] # Peter Luschny, Aug 02 2015
    

Formula

a(n) = (4*n)! * [t^n] 1/(2-g(t^(1/4))) with g(t) = (cos(t) + cosh(t))/2. - Peter Luschny, Jul 07 2015
a(0) = 1; a(n) = Sum_{k=1..n} binomial(4*n,4*k) * a(n-k). - Ilya Gutkovskiy, Jan 21 2020

Extensions

a(0)=1 prepended, more terms from Peter Luschny, Jul 06 2015

A243666 Number of 5-packed words of degree n.

Original entry on oeis.org

1, 1, 253, 762763, 11872636325, 633287284180541, 90604069581412784683, 29529277377602939454694793, 19507327717978242212109900308085, 23927488379043876045061553841299192011, 50897056444296458534155179226333868898628813, 177758773838827813873239281786548960244155096117573
Offset: 0

Views

Author

N. J. A. Sloane, Jun 14 2014

Keywords

Comments

See Novelli-Thibon (2014) for precise definition.

Crossrefs

Cf. A011782, A000670, A094088, A243664, A243665, A243666 for k-packed words of degree n for 0<=k<=5.

Programs

  • Maple
    a := (5+sqrt(5))/4: b := (5-sqrt(5))/4: g := t -> (exp(t)+2*exp(t-a*t)*cos(t*sqrt(b/2))+2*exp(t-b*t)*cos(t*sqrt(a/2)))/5: series(1/(2-g(t)),t,56): seq((5*n)!*(coeff(simplify(%),t,5*n)),n=0..11); # Peter Luschny, Jul 07 2015
  • Mathematica
    b = (5 - Sqrt[5])/4; c = (5 + Sqrt[5])/4;
    g[t_] := (Exp[t] + 2*Exp[t - c*t]*Cos[t*Sqrt[b/2]] + 2*Exp[t - b*t]* Cos[t*Sqrt[c/2]])/5;
    a[n_] := (5n)! SeriesCoefficient[1/(2 - g[t]), { t, 0, 5 n}] // Simplify;
    Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 12}] (* Jean-François Alcover, Jul 14 2018, after Peter Luschny *)
  • PARI
    seq(n)={my(a=vector(n+1)); a[1]=1; for(n=1, n, a[1+n]=sum(k=1, n, binomial(5*n, 5*k) * a[1+n-k])); a} \\ Andrew Howroyd, Jan 21 2020
  • Sage
    # uses[CEN from A243664]
    A243666 = lambda len: CEN(5,len)
    A243666(12) # Peter Luschny, Jul 06 2015
    
  • Sage
    # Alternatively:
    def PackedWords5(n):
        shapes = ([x*5 for x in p] for p in Partitions(n))
        return sum(factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in shapes)
    [PackedWords5(n) for n in (0..11)] # Peter Luschny, Aug 02 2015
    

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(5*n,5*k) * a(n-k). - Ilya Gutkovskiy, Jan 21 2020

Extensions

a(0)=1 prepended, more terms from Peter Luschny, Jul 06 2015

A260883 Number of m-shape ordered set partitions, square array read by ascending antidiagonals, A(m, n) for m, n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 3, 9, 1, 1, 7, 13, 35, 1, 1, 21, 121, 75, 161, 1, 1, 71, 1849, 3907, 541, 913, 1, 1, 253, 35641, 426405, 202741, 4683, 6103, 1, 1, 925, 762763, 65782211, 203374081, 15430207, 47293, 47319, 1, 1, 3433, 17190265, 11872636325, 323213457781, 173959321557
Offset: 1

Views

Author

Peter Luschny, Aug 02 2015

Keywords

Comments

A set partition of m-shape is a partition of a set with cardinality m*n for some n >= 0 such that the sizes of the blocks are m times the parts of the integer partitions of n. It is ordered if the positions of the blocks are taken into account.
If m = 0, all possible sizes are zero. Thus the number of ordered set partitions of 0-shape is the number of ordered partitions of n (partition numbers A101880).
If m = 1, the set is {1, 2, ..., n} and the set of all possible sizes are the integer partitions of n. Thus the number of ordered set partitions of 1-shape is a Fubini number (sequence A000670).
If m = 2, the set is {1, 2, ..., 2n} and the number of ordered set partitions of 2-shape is also the number of 2-packed words of degree n (sequence A094088).

Examples

			[ n ] [0  1   2      3         4            5                  6]
[ m ] -----------------------------------------------------------
[ 0 ] [1, 1,  3,     9,       35,          161,              913]  A101880
[ 1 ] [1, 1,  3,    13,       75,          541,             4683]  A000670
[ 2 ] [1, 1,  7,   121,     3907,       202741,         15430207]  A094088
[ 3 ] [1, 1, 21,  1849,   426405,    203374081,     173959321557]  A243664
[ 4 ] [1, 1, 71, 35641, 65782211, 323213457781, 3482943541940351]  A243665
        A244174
For example the number of ordered set partitions of {1,2,...,9} with sizes in [9], [6,3] and [3,3,3] is 1, 168 and 1680 respectively. Thus A(3,3) = 1849.
Formatted as a triangle:
[1]
[1, 1]
[1, 1, 3]
[1, 1, 3, 9]
[1, 1, 7, 13, 35]
[1, 1, 21, 121, 75, 161]
[1, 1, 71, 1849, 3907, 541, 913]
[1, 1, 253, 35641, 426405, 202741, 4683, 6103]
		

Crossrefs

Programs

  • Sage
    def A260883(m, n):
        shapes = ([x*m for x in p] for p in Partitions(n))
        return sum(factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in shapes)
    for m in (0..4): print([A260883(m, n) for n in (0..6)])

Formula

From Petros Hadjicostas, Aug 02 2019: (Start)
Conjecture: For n >= 0, let P be the set of all possible lists (a_1, ..., a_n) of nonnegative integers such that a_1*1 + a_2*2 + ... + a_n*n = n. Consider terms of the form multinomial(n*m, m*[1,..., 1, 2,..., 2,..., n,..., n]) * multinomial(a_1 + ... + a_n, [a_1,..., a_n]), where in the list [1,..., 1, 2,..., 2,..., n,..., n] the number 1 occurs a_1 times, 2 occurs a_2 times, ..., and n occurs a_n times. (Here a_n = 0 or 1.) Summing these terms over P we get A(m, n) provided m >= 1. (End)

A326587 Coefficients of polynomials related to ordered set partitions. Triangle read by rows, T_{m}(n, k) for m = 3 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 11, 10, 0, 645, 924, 280, 0, 111563, 197802, 101640, 15400, 0, 42567981, 86271640, 57717660, 15415400, 1401400, 0, 30342678923, 67630651098, 53492240256, 19158419280, 3144741600, 190590400
Offset: 0

Views

Author

Peter Luschny, Jul 20 2019

Keywords

Examples

			Triangle starts:
0 [1]
1 [0, 1]
2 [0, 11, 10]
3 [0, 645, 924, 280]
4 [0, 111563, 197802, 101640, 15400]
5 [0, 42567981, 86271640, 57717660, 15415400, 1401400]
6 [0, 30342678923, 67630651098, 53492240256, 19158419280, 3144741600, 190590400]
		

Crossrefs

Row sums A243664. Main diagonal A025035.
A129062 (m=1, associated with A131689), A326477 (m=2, associated with A241171), this sequence (m=3, associated with A278073), A326585 (m=4, associated with A278074).

Programs

Formula

T(n, k) = T_{3}(n, k) where T_{m}(n, k) is defined in A326477.

A327023 Ordered set partitions of the set {1, 2, ..., 3*n} with all block sizes divisible by 3, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

Original entry on oeis.org

1, 1, 1, 20, 1, 168, 1680, 1, 440, 924, 55440, 369600, 1, 910, 10010, 300300, 1261260, 33633600, 168168000, 1, 1632, 37128, 48620, 1113840, 24504480, 17153136, 326726400, 2058376320, 34306272000, 137225088000
Offset: 0

Views

Author

Peter Luschny, Aug 27 2019

Keywords

Comments

T_{m}(n, k) gives the number of ordered set partitions of the set {1, 2, ..., m*n} into sized blocks of shape m*P(n, k), where P(n, k) is the k-th integer partition of n in the 'canonical' order A080577. Here we assume the rows of A080577 to be 0-based and m*[a, b, c,..., h] = [m*a, m*b, m*c,..., m*h]. Here is case m = 3. For instance 3*P(4, .) = [[12], [9, 3], [6, 6], [6, 3, 3], [3, 3, 3, 3]].

Examples

			Triangle starts (note the subdivisions by ';' (A072233)):
[0] [1]
[1] [1]
[2] [1;   20]
[3] [1;  168;  1680]
[4] [1;  440,   924;  55440;  369600]
[5] [1;  910, 10010; 300300, 1261260; 33633600; 168168000]
[6] [1; 1632, 37128,  48620; 1113840, 24504480,  17153136; 326726400, 2058376320;
     34306272000; 137225088000]
.
T(4, 1) = 440 because [9, 3] is the integer partition 3*P(4, 1) in the canonical order and there are 220 set partitions which have the shape [9, 3]. Finally, since the order of the sets is taken into account, one gets 2!*220 = 440.
		

Crossrefs

Row sums: A243664, alternating row sums: A002115, main diagonal: A014606, central column A281479, by length: A278073.
Cf. A178803 (m=0), A133314 (m=1), A327022 (m=2), this sequence (m=3), A327024 (m=4).

Programs

  • Sage
    # uses[GenOrdSetPart from A327022]
    def A327023row(n): return GenOrdSetPart(3, n)
    for n in (0..6): print(A327023row(n))

A352428 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/3)} binomial(n,3*k+1) * a(n-3*k-1).

Original entry on oeis.org

1, 1, 2, 6, 25, 130, 810, 5881, 48806, 455706, 4727881, 53955682, 671730246, 9059714665, 131588822822, 2047796305470, 33992509701721, 599526848094850, 11195864285933682, 220692569175568729, 4579248276057441926, 99767702172338210898, 2277136869014579978473, 54336724559407913237122
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 16 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, 3 k + 1] a[n - 3 k - 1], {k, 0, Floor[(n - 1)/3]}]; Table[a[n], {n, 0, 23}]
    nmax = 23; CoefficientList[Series[1/(1 - Sum[x^(3 k + 1)/(3 k + 1)!, {k, 0, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    my(N=40, x='x+O('x^N)); Vec(serlaplace(1/(1-sum(k=0, N\3, x^(3*k+1)/(3*k+1)!)))) \\ Seiichi Manyama, Mar 23 2022

Formula

E.g.f.: 1 / (1 - Sum_{k>=0} x^(3*k+1) / (3*k+1)!).
Showing 1-9 of 9 results.