cp's OEIS Frontend

For the symmetric functions a_k and the definition of the triangles S_j(n,k) see a comment in A196841. Here x[1]=1, x[2]=2, and x[j]=j+1 for j=3,...,n. This is the triangle S_3(n,k), n>=0, k=0..n. The first three rows coincide with those of triangle A094638.

a(n) is permanent of Toeplitz matrix

n n-1 n-2 ... 3 2 1

n+1 n n-1 ... 4 3 2

n+2 n+1 n ... 5 4 3

.......

2n-1 2n-2 2n-3 ... n+2 n+1 n. - Vladimir Shevelev, Dec 01 2013

a(n) is also the largest entry in the cycle containing 1, summed over all permutations of {1,2,...,n}. Example: a(3) = 13 because the permutations (1)(2)(3), (1)(23), (12)(3), (13)(2), (123), (132), written in cycle notation, yield 1+1+2+3+3+3=13. - Emeric Deutsch, Nov 10 2008

Row sums of A134433 starting from k=3.

a(n) = sum( (-1)^k*k*A008276(n,k), k=1..n-1).

a(n) = sum( (-1)^k*k*A054654(n,k), k=1..n-2).

For n >= 3, a(n) = number whose factorial base representation (A007623) begins with digits {n-1} and {n-2} followed by n-3 zeros. Viewed in that base, this sequence looks like this: 1, 21, 320, 4300, 54000, 650000, 7600000, 87000000, 980000000, A900000000, BA000000000, ... (where "digits" A and B stand for placeholder values 10 and 11 respectively). - Antti Karttunen, May 07 2015.

Incidentally, the sum of signed coefficients for each k-th row is divisible by (2*k-2)!. Moreover, another variant (but an incomplete one, and sorted differently) of the above sequence is presented in A101751.

Incidentally, the sum of signed coefficients for each k-th row is divisible by (2*k-2)!.

To motivate the definition, consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5, ...), T = A094638 and the list of integers.

Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...) from which factorials may be formed. It's true that

T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4, ...), giving n! for n > 0. Call this sequence the right 1-step factorial (n,+1)!.

Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And,

T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2, ...) = (1,0,0,0,...). Call this the left 1-step factorial (n,-1)!.

Sampling every other integer to the right, we obtain (1,3,5,7,9,...).

T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n > 0, the right 2-step factorial, (n,+2)!.

Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7,...).

T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5, ...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left 2-step factorial, (n,-2)!.

Sampling every 3 steps to the right, we obtain (1,4,7,10,...).

T * c(3) = (1, 1*4, 1*4*7, ...) = (1,4,28,280,...), giving A007559 for n > 0, the right 3-step factorial, (n,+3)!.

Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving

T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8, ...) = (1,-2,10,-80,880,...) = signed A008544 = the left 3-step factorial, (n,-3)!.

The list partition transform A133314 of [1,T * c(t)] gives signed [1,T *c(-t)]. For example:

LPT[1,T*c(1)] = LPT[1,(n,+1)! ] = LPT[A000142] = (1,-1,0,0,0,...) = signed [1,(n,-1)! ]

LPT[1,T*c(2)] = LPT[1,(n,+2)! ] = LPT[A001147] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147) = signed [1,(n,-2)! ]

LPT[1,T*c(3)] = LPT[1,(n,+3)! ] = LPT[A007559] = (1,-1,-2,-10,-80,-880,...) = (1,-A008544) = signed [1,(n,-3)! ]

LPT[1,T*c(-3)] = LPT[1,(n,-3)! ] = signed A007559 = signed [1,(n,+3)! ].

And, e.g.f.[1,T * c(m)] = e.g.f.[1,(n,m)! ] = (1-m*x)^(-1/m).

Also with P(n,t) = Sum_{k=0..n-1} T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n-1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1-tx)^(-1/t).

T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] evaluated at t=x=0.

And, (1-tx)^(-1/t) - 1 is the e.g.f. for plane increasing m-ary trees when t = (m-1), discussed by Bergeron et al. in "Varieties of Increasing Trees" and the book Combinatorial Species and Tree-Like Structures, cited in the OEIS.

The above relations reveal the intimate connections, through T or LPT or sampling, between the right and left step factorials, (n,+m)! and (n,-m)!. The pairs have conjugate interpretations as trees, ignoring signs, which Callan and Lang have noted in several of the OEIS entries above. Also note unsigned (n,-2)! is the diagonal of A001498 and (n,+2)!, the first subdiagonal.