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
[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;
Cf.
A243664 (column 0 and main diagonal).
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
-
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
-
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 *)
-
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 */
-
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 */
-
a(n) = if (n == 0, 1, sum(k=1, n, binomial(2*n, 2*n-2*k)*a(n-k)));
-
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
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
Triangle begins:
[1]
[0, 1]
[0, 1, 20]
[0, 1, 168, 1680]
[0, 1, 1364, 55440, 369600]
[0, 1, 10920, 1561560, 33633600, 168168000]
-
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;
-
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
-
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
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
-
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
-
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 *)
-
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
-
# uses[CEN from A243664]
A243665 = lambda len: CEN(4,len)
A243665(13) # Peter Luschny, Jul 06 2015
-
# 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
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
-
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
-
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 *)
-
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
-
# uses[CEN from A243664]
A243666 = lambda len: CEN(5,len)
A243666(12) # Peter Luschny, Jul 06 2015
-
# 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
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
[ 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]
-
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)])
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
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]
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
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.
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
-
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]!
-
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
Showing 1-9 of 9 results.
Comments