A051921 Duplicate of A023998.
1, 1, 3, 16, 131, 1496, 22482, 426833, 9934563, 277006192, 9085194458
Offset: 0
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.
A(2,2) = 3: 1234, 12|34, 14|23. A(2,3) = 5: 123456, 123|456, 126|345, 135|246, 156|234. A(2,4) = 9: 12345678, 1234|5678, 1238|4567, 1247|3568, 1278|3456, 1346|2578, 1368|2457, 1467|2358, 1678|2345. A(3,2) = 16: 123456, 1234|56, 1236|45, 1245|36, 1256|34, 12|3456, 12|34|56, 12|36|45, 1346|25, 1456|23, 14|2356, 14|23|56, 16|2345, 16|23|45, 14|25|36, 16|25|34. Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, 1, ... 1, 1, 1, 1, 1, 1, 1, ... 1, 2, 3, 5, 9, 17, 33, ... 1, 5, 16, 64, 298, 1540, 8506, ... 1, 15, 131, 1613, 25097, 461105, 9483041, ... 1, 52, 1496, 69026, 4383626, 350813126, 33056715626, ... 1, 203, 22482, 4566992, 1394519922, 573843627152, 293327384637282, ...
A:= proc(n, k) option remember; `if`(k*n=0, 1, add( binomial(n, j)^k*(n-j)*A(j, k), j=0..n-1)/n) end: seq(seq(A(n, d-n), n=0..d), d=0..12);
A[n_, k_] := A[n, k] = If[k*n == 0, 1, Sum[Binomial[n, j]^k*(n-j)*A[j, k], {j, 0, n-1}]/n]; Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Jan 17 2017, translated from Maple *)
For n=3, there are the 3! ordinary permutations (of rank 3), 18 block permutations of rank 2 (2! for each pair of partitions of rank 2) and the single rank 1 one.
Table[Sum[StirlingS2[n,k]^2k!,{k,0,n}],{n,0,100}] (* Emanuele Munarini, Jul 04 2011 *)
makelist(sum(stirling2(n,k)^2*k!,k,0,n),n,0,24); /* Emanuele Munarini, Jul 04 2011 */
a(n) = if (n==0, 1, sum(k=1, n, k!*stirling(n, k, 2)^2)); \\ Michel Marcus, Jun 18 2019
Triangle begins: 1; 1,2; 1,9,6; 1,34,72,24; 1,125,650,600,120; ... T(4,2) = 34: There are 7 partitions of the set {1,2,3,4} into 2 blocks. The four partitions {1,2,3}{4}, {1,2,4}{3}, {1,3,4}{2} and {2,3,4}{1} give rise to 4*4 = 16 uniform block permutations while the remaining 3 partitions {1,2}{3,4}, {1,3}{2,4} and {1,4}{2,3} give 2!*3*3 = 18 uniform block permutations : thus in total there are 16+18 = 34 block permutations between the set partitions of {1,2,3,4} into 2 blocks.
#A061691 #J = sum {n>=0} z^n/n!^2 J := BesselJ(0, 2*i*sqrt(z)): G := exp(x*(J(z)-1)): Gser := simplify(series(G, z = 0, 12)): for n from 1 to 10 do P[n] := n!^2*sort(coeff(Gser, z, n)) od: for n from 1 to 10 do seq(coeff(P[n],x,k), k = 1..n) od; # yields sequence in triangular form # second Maple program: b:= proc(n) option remember; expand(`if`(n=0, 1, add(x*b(n-i)*binomial(n, i)/i!, i=1..n))) end: T:= n-> (p-> seq(coeff(p, x, i)/i!, i=1..n))(b(n)*n!): seq(T(n), n=1..12); # Alois P. Heinz, Sep 10 2019
max = 9; g := Exp[x*(BesselI[0, 2*Sqrt[z]] - 1)]; gser = Series[g, {z, 0, max}, {x, 0, max}]; t[n_, k_] := n!^2*SeriesCoefficient[ gser // Normal, {z, 0, n}, {x, 0, k}]; Flatten[ Table[ t[n, k], {n, 1, max}, {k, 1, n}]] (* Jean-François Alcover, Apr 04 2012, after Maple *)
T(3,2) = 12: 1a|2a3b, 1b|2a3b, 1a3b|2a, 1a3b|2b, 1a2b|3a, 1a2b|3b, 1a|2a|3b, 1a|2b|3a, 1b|2a|3a, 1a|2b|3b, 1b|2a|3b, 1b|2b|3a. Triangle T(n,k) begins: 1; 0, 1; 0, 1, 3; 0, 1, 12, 16; 0, 1, 41, 156, 131; 0, 1, 140, 1155, 2460, 1496; 0, 1, 497, 8020, 32600, 47355, 22482; 0, 1, 1848, 55629, 385420, 1004360, 1098678, 426833; ...
A:= proc(n, k) option remember; `if`(n=0, 1, add(A(n-j, k)* binomial(n-1, j-1)*binomial(k, j), j=1..min(k, n))) end: T:= (n, k)-> add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k): seq(seq(T(n, k), k=0..n), n=0..10);
A[n_, k_] := A[n, k] = If[n == 0, 1, Sum[A[n-j, k] Binomial[n-1, j-1]* Binomial[k, j], {j, 1, Min[k, n]}]]; T[n_, k_] := Sum[A[n, k-i] (-1)^i Binomial[k, i], {i, 0, k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 30 2020, after Alois P. Heinz *)
E.g.f.: A(x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 152*x^4/4! + 1944*x^5/5! + ... Related expansions: A(x/(1-x))/(1-x) = 1 + 2*x + 9*x^2/2! + 68*x^3/3! + 760*x^4/4! + ... A(x) + x*A'(x) = 1 + 2*x + 9*x^2/2! + 68*x^3/3! + 760*x^4/4! + ... Also, a(n) appears in the expansion: B(x) = 1 + x + 3*x^2/2!^2 + 17*x^3/3!^2 + 152*x^4/4!^2 + 1944*x^5/5!^2 + ... where log(B(x)) = x + x^2/(2*2!) + x^3/(3*3!) + x^4/(4*4!) + x^5/(5*5!) + ...
b:= proc(n) option remember; `if`(n=0, 1, add(b(n-i)*binomial(n-1, i-1)/i, i=1..n)) end: a:= n-> b(n)*n!: seq(a(n), n=0..25); # Alois P. Heinz, May 11 2016
a[ n_] := If[ n<0, 0, n!^2 Assuming[ x>0, SeriesCoefficient[ Exp[ Integrate[ (Exp[t] - 1)/t, {t, 0, x}]], {x, 0, n}]]]; (* Michael Somos, Dec 28 2012 *) a[ n_] := If[ n<0, 0, n!^2 Assuming[ x>0, SeriesCoefficient[ Exp[ ExpIntegralEi[x] - Log[x] - EulerGamma], {x, 0, n}]]]; (* Michael Somos, Dec 28 2012 *) Table[Sum[BellY[n, k, 1/Range[n]], {k, 0, n}] n!, {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
{a(n)=local(A=1+x,B);for(i=1,n,B=subst(A,x,x/(1-x+x*O(x^n)))/(1-x);A=1+intformal((B-A)/x));n!*polcoeff(A,n)}
{a(n)=if(n<0,0,if(n==0,1,(n-1)!*sum(k=0,n-1,binomial(n,k)*a(k)/k!)))}
{a(n)=n!^2*polcoeff(exp(sum(m=1,n,x^m/(m*m!))+x*O(x^n)),n)}
T(3,2) = 20: 1a2a3b, 1a2b3b, 1a|2a3b, 1a|2b3b, 1b|2a3a, 1b|2a3b, 1a3b|2a, 1b3b|2a, 1a3a|2b, 1a3b|2b, 1a2b|3a, 1b2b|3a, 1a2a|3b, 1a2b|3b, 1a|2a|3b, 1a|2b|3a, 1b|2a|3a, 1a|2b|3b, 1b|2a|3b, 1b|2b|3a. Triangle T(n,k) begins: 1; 0, 1; 0, 2, 3; 0, 5, 20, 16; 0, 15, 122, 237, 131; 0, 52, 774, 2751, 3524, 1496; 0, 203, 5247, 30470, 68000, 65055, 22482; 0, 877, 38198, 341244, 1181900, 1913465, 1462320, 426833; ...
A:= proc(n, k) option remember; `if`(n=0, 1, add(A(n-j, k)* binomial(n-1, j-1)*binomial(k+j-1, j), j=1..n)) end: T:= (n, k)-> add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k): seq(seq(T(n, k), k=0..n), n=0..10);
A[n_, k_] := A[n, k] = If[n == 0, 1, Sum[A[n-j, k] Binomial[n-1, j-1]* Binomial[k + j - 1, j], {j, n}]]; T[n_, k_] := Sum[A[n, k - i] (-1)^i Binomial[k, i], {i, 0, k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 30 2020, after Alois P. Heinz *)
a[0] = 1; a[n_] := a[n] = n Sum[Binomial[n - 1, k]^2 a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 18}] nmax = 18; CoefficientList[Series[Exp[Sqrt[x] BesselI[1, 2 Sqrt[x]]], {x, 0, nmax}], x] Range[0, nmax]!^2
a[0] = 1; a[n_] := a[n] = -(1/n) Sum[Binomial[n, k]^2 (n - k) a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 20}] nmax = 20; CoefficientList[Series[Exp[-Sum[x^k/(k!)^2, {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]!^2
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