A192723 Erroneous version of A102221.
1, 5, 55, 1077, 32951, 1451723, 87054773, 6818444405, 675900963271, 82717196780955, 12248810651651332, 2158585005685222202, 446445657799551722026, 107087164031952012004954
Offset: 1
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a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k]^3 a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 14}] nmax = 14; CoefficientList[Series[1/(1 - Sum[x^k/(k!)^3, {k, 1, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!^3
A(2,2) = M(2; 2)^2 + M(2; 1,1)^2 = 1 + 4 = 5. Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, ... 1, 1, 1, 1, 1, 1, ... 2, 3, 5, 9, 17, 33, ... 4, 13, 55, 271, 1459, 8263, ... 8, 75, 1077, 19353, 395793, 8718945, ... 16, 541, 32951, 2699251, 262131251, 28076306251, ...
b:= proc(n, k) option remember; `if`(n=0, 1, add(b(n-i, k)/i!^k, i=1..n)) end: A:= (n, k)-> n!^k*b(n, k): seq(seq(A(n, d-n), n=0..d), d=0..12); # second Maple program: A:= proc(n, k) option remember; `if`(n=0, 1, add(binomial(n, j)^k*A(j, k), j=0..n-1)) end: seq(seq(A(n, d-n), n=0..d), d=0..12);
A[n_, k_] := A[n, k] = If[n==0, 1, Sum[Binomial[n, j]^k A[j, k], {j, 0, n-1}]]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Dec 03 2020, after 2nd Maple program *)
Rows begin: [1], [1,1], [5,4,1], [55,45,9,1], [1077,880,180,16,1], [32951,26925,5500,500,25,1], [1451723,1186236,242325,22000,1125,36,1],... and equal the term-by-term product of column 0 with the squared binomial coefficients (A008459): [(1)1^2], [(1)1^2,(1)1^2], [(5)1^2,(1)2^2,(1)1^2], [(55)1^2,(5)3^2,(1)3^2,(1)1^2], [(1077)1^2,(55)4^2,(5)6^2,(1)4^2,(1)1^2],... The matrix inverse is [2*I - A008459]: [1], [ -1,1], [ -1,-4,1], [ -1,-9,-9,1], [ -1,-16,-36,-16,1],...
b:= proc(n) option remember; `if`(n=0, 1, add(b(n-i)*binomial(n, i)/i!, i=1..n)) end: T:= (n, k)-> binomial(n, k)^2*b(n-k)*(n-k)!: seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Sep 10 2019
nmax = 10; M = Inverse[2 IdentityMatrix[nmax+1] - Table[Binomial[n, k]^2, {n, 0, nmax}, {k, 0, nmax}]]; T[n_, k_] := M[[n+1, k+1]]; Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 06 2019 *)
{T(n,k)=(matrix(n+1,n+1,i,j,if(i==j,2,0)-binomial(i-1,j-1)^2)^-1)[n+1,k+1]}
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k]^4 a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 13}] nmax = 13; CoefficientList[Series[1/(1 - Sum[x^k/(k!)^4, {k, 1, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!^4
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k]^5 a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 12}] nmax = 12; CoefficientList[Series[1/(1 - Sum[x^k/(k!)^5, {k, 1, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!^5
b:= proc(n) option remember; `if`(n=0, 1, add(b(n-i)/i^2, i=1..n)) end: a:= n-> n!^2*b(n): seq(a(n), n=0..16); # Alois P. Heinz, Jan 04 2024
a[0] = 1; a[n_] := a[n] = (n!)^2 Sum[a[k]/(k! (n - k))^2, {k, 0, n - 1}]; Table[a[n], {n, 0, 16}] nmax = 16; CoefficientList[Series[1/(1 - PolyLog[2, x]), {x, 0, nmax}], x] Range[0, nmax]!^2
The triangle begins n/k|..1.....2.......3........4........5........6 ================================================ .1.|..1 .2.|..1.....4 .3.|..1....18.....36 .4.|..1....68.....432......576 .5.|..1...250....3900....14400....14400 .6.|..1...922...32400...252000...648000...518400 ... T(4,2) = 68: There are 3 compositions of 4 into 2 parts, namely, 4 = 2 + 2 = 1 + 3 = 3 + 1; hence T(4,2) = (4!/(2!*2!))^2 + (4!/(1!*3!))^2 + (4!/(3!*1!))^2 = 36 + 16 + 16 = 68. Matrix identity: A192721 * Pascal's triangle = row reverse of A192722: /...1................\ /..1..............\ |...3.....1...........||..1....1..........| |..19....16.....5.....||..1....2....1.....| |.211...299....65....1||..1....3....3....1| |.....................||..................| = /...1...................\ |...4......1.............| |..36.....18......1......| |.576....432.....68.....1| |........................|
J := unapply(BesselJ(0, 2*sqrt(-1)*sqrt(z)), z): G := 1/(1-x*(J(z)-1)): Gser := simplify(series(G, z = 0, 15)): for n from 1 to 14 do P[n] := n!^2*sort(coeff(Gser, z, n)) od: for n from 1 to 14 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)/i!^2, i=1..n))) end: T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n)*n!^2): seq(T(n), n=1..14); # Alois P. Heinz, Sep 10 2019
b[n_] := b[n] = Expand[If[n == 0, 1, Sum[x b[n-i]/i!^2, {i, 1, n}]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n] n!^2]; Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, Dec 07 2019, after Alois P. Heinz *)
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k]^2 a[k - 1], {k, 1, n}]; Table[a[n], {n, 0, 20}]
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k]^2 2^(n - k - 1) a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 16}] nmax = 16; CoefficientList[Series[2/(3 - BesselI[0, 2 Sqrt[2 x]]), {x, 0, nmax}], x] Range[0, nmax]!^2
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