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.
A000629 := proc(n) local k ; sum( k^n/2^k,k=0..infinity) ; end: A062208 := proc(n) option remember ; local a,stir,ni,n1,n2,n3,stir2,i,j,tmp ; a := 0 ; if n = 0 then RETURN(1) ; fi ; stir := combinat[partition](n) ; stir2 := {} ; for i in stir do if nops(i) <= 3 then tmp := i ; while nops(tmp) < 3 do tmp := [op(tmp),0] ; od: tmp := combinat[permute](tmp) ; for j in tmp do stir2 := stir2 union { j } ; od: fi ; od: for ni in stir2 do n1 := op(1,ni) ; n2 := op(2,ni) ; n3 := op(3,ni) ; a := a+combinat[multinomial](n,n1,n2,n3)*(A000629(3*n1+2*n2+n3)-1/2-2^(3*n1+2*n2+n3)/4)*(-3)^n2*2^n3 ; od: a/(2*6^n) ; end: A136246 := proc(n) local k ; add((-1)^(n-k)*combinat[stirling1](n,k)*A062208(k),k=0..n)/n! ; end: seq(A136246(n),n=0..14) ; # R. J. Mathar, Apr 01 2008
a[n_] := Sum[Binomial[Binomial[j, 3] + n - 1, n] * Sum[(-1)^(i - j)* Binomial[i, j], {i, j, 3n}], {j, 0, 3n}];
a /@ Range[0, 14] (* Jean-François Alcover, Feb 13 2020, after Andrew Howroyd *)
a(n) = {sum(j=0, 3*n, binomial(binomial(j,3)+n-1, n) * sum(i=j, 3*n, (-1)^(i-j)*binomial(i,j)))} \\ Andrew Howroyd, Feb 09 2020
A(2,2) = 13: [(2,2),(1,2),(0,2),(0,1),(0,0)], [(2,2),(1,2),(0,1),(0,0)], [(2,2),(1,2),(1,1),(0,1),(0,0)], [(2,2),(1,2),(1,1),(0,0)], [(2,2),(1,2),(1,1),(1,0),(0,0)], [(2,2),(2,1),(1,1),(0,1),(0,0)], [(2,2),(2,1),(1,1),(0,0)], [(2,2),(2,1),(1,1),(1,0),(0,0)], [(2,2),(2,1),(2,0),(0,1),(0,0)], [(2,2),(2,1),(1,0),(0,0)], [(2,2),(1,1),(0,1),(0,0)], [(2,2),(1,1),(0,0)], [(2,2),(1,1),(1,0),(0,0)]. Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, ... 1, 1, 3, 13, 75, 541, ... 1, 1, 13, 409, 23917, 2244361, ... 1, 1, 63, 16081, 10681263, 14638956721, ... 1, 1, 321, 699121, 5552351121, 117029959485121, ... 1, 1, 1683, 32193253, 3147728203035, 1050740615666453461, ...
A:= (n, k)-> add(add((-1)^i*binomial(j, i)*
binomial(j-i, n)^k, i=0..j), j=0..k*n):
seq(seq(A(n, d-n), n=0..d), d=0..10);
A[, 0] = 1; A[n, k_] := Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, n]^k, {i, 0, j}], {j, 0, k*n}]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Jul 22 2016, after Alois P. Heinz *)
T(n,k) = {my(m=n*k); sum(j=0, m, binomial(j,n)^k*sum(i=j, m, (-1)^(i-j)*binomial(i, j)))} \\ Andrew Howroyd, Jan 23 2020
In case n = 2 this is the Delannoy number a(2) = D(2,2) = 13. a(2) = 13 because if you have two intervals [a1,a2] and [b1,b2], using a for a1 or a2 and b for b1 or b2 and writing c if an a is at the same place as a b, we get the following possibilities: aabb, acb, abab, cab, abc, baab, abba, cc, bac, cba, baba, bca, bbaa.
lambda := proc(p,n) option remember; if n = 1 then if p = 2 then RETURN(1) else RETURN(0) fi; else RETURN((p*(p-1)/2)*(lambda(p,n-1)+2*lambda(p-1,n-1)+lambda(p-2,n-1))) fi; end; A055203 := n->add(lambda(i,n),i=2..2*n); A055203 := proc(n) local k; add(A078739(n,k)*k!,k=0..2*n)/2^n end: seq(A055203(n),n=0..15); # Peter Luschny, Mar 25 2011 # second Maple program: b:= proc(n) option remember; `if`(n=0, 1, add(b(n-j)*binomial(n, j), j=1..n)) end: a:= n-> ceil(add(b(n+k)*binomial(n, k), k=0..n)/2^(n+1)): seq(a(n), n=0..20); # Alois P. Heinz, Jul 10 2018
a[n_] := Sum[((m-1)*m)^n / 2^(m+n+1), {m, 0, Infinity}]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Oct 10 2011, after Vladeta Jovovic *)
With[{r = 2}, Flatten[{1, Table[Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, r]^k, {i, 0, j}], {j, 0, k*r}], {k, 1, 15}]}]] (* Vaclav Kotesovec, Mar 22 2016 *)
Table begins
n\k| 3 4 5 6 7 8 9 10
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
1| 1
2| 1 12 30 20
3| 1 60 690 2940 5670 5040 1680
4| 1 252 8730 103820 581700 1767360 3087000 3099600 ...
...
T(2,5) = 30: An alignment of length 5 will have two gap symbols on each line. There are C(5,2) = 10 ways of choosing the 2 positions to insert the gap symbols in the first string. The second string in the alignment must then have nongap symbols at these two positions leaving three positions in which to insert the remaining 1 nongap symbol, giving in total 10 x 3 = 30 possible alignments of 2 strings of 3 characters. Some examples are
ABC-- ABC-- ABC--
D--EF -D-EF --DEF
Row 2: Sum_{i = 3..n-1} C(i,3)^2 = C(n,4) + 12*C(n,5) + 30*C(n,6) + 20*C(n,7).
Row 3: Sum_{i = 3..n-1} C(i,3)^3 = C(n,4) + 60*C(n,5) + 690*C(n,6) + 2940*C(n,7) + 5670*C(n,8)+ 5040*C(n,9)+ 1680*C(n,10).
exp( Sum_{n >= 1} R(n,2)*x^n/n ) = (1 + x + 153*x^2 + 128793*x^3 + 319155321*x^4 + 1744213657689*x^5 + ....)^8
exp( Sum_{n >= 1} R(n,3)*x^n/n ) = (1 + x + 424*x^2 + 998584*x^3 + 6925040260*x^4 + 105920615923684*x^5 + ....)^27.
seq(seq(add( (-1)^(k-i) *binomial(k,i)*binomial(i,3)^n, i = 0..k ), k = 3..3*n), n = 1..6);
nmax = 6; T[n_, k_] := Sum[(-1)^(k-i) Binomial[k, i] Binomial[i, 3]^n, {i, 0, k}]; Table[T[n, k], {n, 1, nmax}, {k, 3, 3n}] // Flatten (* Jean-François Alcover, Feb 20 2018 *)
A(2, 7) = 48639 since this represents the number of distinct alignments of 2 strings of length 7. All values in A(2,X) can be cross-validated against the Delannoy sequence D(X,X) A001850.
With[{r = 7}, Flatten[{1, Table[Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, r]^k, {i, 0, j}], {j, 0, k*r}], {k, 1, 10}]}]] (* Vaclav Kotesovec, Mar 22 2016 *)
a262809(n, k) = sum(i=0, k*n, sum(j=0, i, (-1)^j*binomial(i, j)*binomial(i-j, n)^k));
my(N=20, x='x+O('x^N)); Vec(exp(sum(k=1, N, a262809(3, k)*x^k/k)))
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