cp's OEIS Frontend

The row length sequence for this array is [1,5,9,13,17,...] = A016813(n-1), n >= 1.

The g.f. for the k-th column, (with leading zeros and k >= 4) is G(k,x) = x^ceiling(k/4)*P(k,x)/Product_{p = 4..k} (1 - fallfac(p,4)*x), with fallfac(n,m) := A008279(n,m) (falling factorials) and P(k,x) := Sum_{m = 0..kmax(k)} A090221(k,m)*x^m, k >= 4, with kmax(k) := A057353(k-4)= floor(3*(k-4)/4). For the recurrence of the G(k,x) see A090221.

Codara et al., show that T(n,k) gives the number of k-colorings of the graph nK_4 (the disjoint union of n copies of the complete graph K_4). - Peter Bala, Aug 15 2013

An alignment of n strings of various lengths is a way of inserting blank characters into the n strings so that the resulting strings all have the same length. We don't allow insertion of a blank character into the same position in each of the n strings.

In this case, let s_1,...,s_n be n strings each of length 3 over an alphabet A. Let - be a gap symbol not in A and let A' = union of A and {-}. An alignment of the n strings is an n-tuple (s_1',...,s_n') of strings each of length >= 3 over the alphabet A' such that

(a) the strings s_i', 1 <= i <= n, have the same length. This common length is called the length of the alignment.

(b) deleting the gap symbols from s_i' yields the string s_i for 1 <= i <= n

(c) there is no value j such that all the strings s_i', 1 <= i <= n have a gap symbol at position j.

By writing the strings s_i' one under another we can consider an alignment of n strings as an n X L matrix, where L, the length of the alignment, ranges from a minimum value of 3 to a maximum value of 3*n. Each row of the matrix has 3 characters from the alphabet A and (L - 3) gap characters.

For example,

s_1' = ABC------

s_2' = ---DEF---

s_3' = ------GHI

is an alignment (of maximum length L = 9) of three strings s_1 = ABC, s_2 = DEF and s_3 = GHI each of length 3.

For the number of alignments of length k of n strings of length 1 (resp. 2) see A131689 (resp. A122193).

Triangle here T_3(n,m) is such that C(n,m)^3 = Sum_{j=0..n} C(n,j)*T_3(j,m).

Equivalently, lower triangular matrix T_3 such that

|| C(n,m)^3 || = A181583 = P * T_3 = A007318 * T_3.

T_3(n,m) = 0 for n < m and for 3*m < n. In fact:

C(x,m)^q and C(x,m), with m nonnegative and q positive integer, are polynomial in x of degree m*q and m respectively, and C(x,m) is a divisor of C(x,m)^q.

Therefore the Newton series will give C(x,m)^q = T_q(m,m)*C(x,m) + T_q(m+1,m)*C(x,m+1) + ... + T_q(q*m,m)*C(x,q*m), where T_q(n,m) is the n-th forward finite difference of C(x,m)^q at x = 0.

Example:

C(x,2)^3 = x^3*(x-1)^3 / 8 = 1*C(x,2) + 24*C(x,3) + 114*C(x,4) + 180*C(x,5) + 90*C(x,6);

C(5,2)^3 = C(5,3)^3 = 1000 = 1*C(5,2) + 24*C(5,3) + 114*C(5,4) + 180*C(5,5) = 1*C(5,3) + 60*C(5,4) + 690*C(5,5).

So we get the expansion of the 3rd power of the binomial coefficient in terms of the binomial coefficients on the same row.

T_1 is the unitary matrix,

T_2 is the transpose of A109983,

T_3 is this sequence,

T_4, T_5 are A262705, A262706.

Triangle here T_4(n,m) is such that C(n,m)^4 = Sum_{j=0..n} C(n,j)*T_4(j,m).

Equivalently, lower triangular matrix T_4 such that

|| C(n,m)^4 || = A202750 = P * T_4 = A007318 * T_4.

T_4(n,m) = 0 for n < m and for 4*m < n.

Refer to comment to A262704.

Example:

C(x,2)^4 = x^4*(x-1)^4 /16 = 1*C(x,2) + 78*C(x,3) + 978*C(x,4) + 4320*C(x,5) + 8460*C(x,6) + 7560*C(x,7) + 2520*C(x,8);

C(5,2)^4 = C(5,3)^4 = 10000 = 1*C(5,2) + 78*C(5,3) + 978*C(5,4) + 4320*C(5,5) = 1*C(5,3) + 252*C(5,4) + 8730*C(5,5).

Triangle here T_5(n,m) is such that C(n,m)^5 = Sum_{j=0..n} C(n,j)*T_5(j,m).

Equivalently, lower triangular matrix T_5 such that

|| C(n,m)^5 || = P * T_5 = A007318 * T_5.

T_5(n,m) = 0 for n < m and for 5*m < n.

Refer to comment to A262704.

Example:

C(x,2)^5 = x^5*(x-1)^5/32 = 1*C(x,2) + 240*C(x,3) + 6810*C(x,4) + 63540*C(x,5) + 271170*C(x,6) + 604800*C(x,7) + 730800*C(x,8) + 453600*C(x,9) + 113400*C(x,10);

C(5,2)^5 = C(5,3)^5 = 100000 = 1*C(5,2) + 240*C(5,3) + 6810*C(5,4) + 63540*C(5,5) = 1*C(5,3) + 1020*C(5,4) + 94890*C(5,5).