A087127 -id:A087127 - cp's OEIS frontend

A299041 Irregular table: T(n,k) equals the number of alignments of length k of n strings each of length 3.

1, 1, 12, 30, 20, 1, 60, 690, 2940, 5670, 5040, 1680, 1, 252, 8730, 103820, 581700, 1767360, 3087000, 3099600, 1663200, 369600, 1, 1020, 94890, 2615340, 32186070, 214628400, 859992000, 2189325600, 3628409400, 3903900000, 2630628000, 1009008000, 168168000, 1, 4092, 979530, 58061420, 1411122300

Offset: 1

Peter Bala, Feb 02 2018

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).

			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.

P. Bala, Notes on A299041
M. Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
J. Engbers and C. Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522

Row sums A062208. Cf. A014606, A019538, A078741, A087127, A122193, A131689, A174266.

Cf. A086020, A086021, A086022.

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 *)

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*binomial(i,3)^n.
T(n,3) = 1; T(n,3*n) = (3*n)!/6^n = A014606(n)
T(n,k) = binomial(k,3)*( T(n-1,k) + 3*T(n-1,k-1) + 3*T(n-1,k-2) + T(n-1,k-3) ) for 3 <= k <= 3*n with boundary conditions T(n,3) = 1 for n >= 1 and T(n,k) = 0 if (k < 3) or (k > 3*n).
Double e.g.f.: exp(-x)*Sum_{n >= 0} exp(binomial(n,3)*y)*x^n/n! = 1 + (x^3/3!)*y + (x^3/3! + 12*x^4/4! + 30*x^5/5! + 20*x^6/6!)*y^2/2! + ....
n-th row polynomial R(n,x) = Sum_{i >= 3} binomial(i,3)^n*x^i/(1 + x)^(i+1) for n >= 1.
1/(1 - x)*R(n,x/(1 - x)) = Sum_{i >= 3} binomial(i,3)^n*x^i for n >= 1.
R(n,x) = x^3 o x^3 o ... o x^3 (n factors), where o is the black diamond product of power series defined in Dukes and White.
R(n,x) = coefficient of (z_1)^3*...*(z_n)^3 in the expansion of the rational function 1/(1 + x - x*(1 + z_1)*...*(1 + z_n)).
The polynomials Sum_{k = 3..3*n} T(n,k)*x^(k-3)*(1 - x)^(3*n-k) are the row polynomials of A174266.
Sum_{i = 3..n-1} binomial(i,3)^m = Sum_{k = 3..3*m} T(m,k)*binomial(n,k+1) for m >= 1. See Examples below.
x^3*R(n,-1 - x) = (-1)^n*(1 + x)^3*R(n,x).
R(n+1,x) = 1/3!*x^3*(d/dx)^3 ((1 + x)^3*R(n,x)) for n >= 1.
The zeros of R(n,x) belong to the interval [-1, 0].
Row sums R(n,1) = A062208(n); alternating row sums R(n,-1) = (-1)^n.
For k a nonzero integer, the power series A(k,x) := exp( Sum_{n >= 1} 1/k^3*R(n,k)*x^n/n ) appear to have integer coefficients. See the Example section.
Sum_{k = 3..3*n} T(n,k)*binomial(x,k) = ( binomial(x,3) )^n. Equivalently, Sum_{k = 3..3*n} (-1)^(n+k)*T(n,k)*binomial(x+k,k) = ( binomial(x+3,3) )^n. Cf. the Worpitzky-type identity Sum_{k = 1..n} A019538(n,k)* binomial(x,k) = x^n.
Sum_{k = 3..3*n} T(n,k)*binomial(x,k-3) = -binomial(x,3)^n + 3*binomial(x+1,3)^n - 3*binomial(x+2,3)^n + binomial(x+3,3)^n. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane.

A210279 (6n)!/6^n.

Original entry on oeis.org

1, 120, 13305600, 29640619008000, 478741050720092160000, 34111736086958726676480000000, 7973107998754741458076119859200000000, 5019026197962676820927435579005599744000000000

Offset: 0

Mohammad K. Azarian, Apr 12 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..80

Cf. A210278, A210277, A000680, A067630, A084939, A084940, A084941, A084942, A084943, A084944, A087127, A001147, A132101

[Factorial(6*n)/6^n: n in [0..10]]; // Vincenzo Librandi, Feb 15 2013

Table[(6 n)!/6^n, {n, 0, 11}] (* Vincenzo Librandi, Feb 15 2013 *)
With[{nn=50},Take[CoefficientList[Series[1/(1-x^6/6),{x,0,nn}],x] Range[0,nn-2]!,{1,-1,6}]] (* Harvey P. Dale, Sep 25 2023 *)

E.g.f.: 1/(1-x^6/6).

A234253 a(n) = Sum_{i=1..n} C(7+i,8)^2.

Original entry on oeis.org

1, 82, 2107, 29332, 274357, 1930726, 10948735, 52357960, 217994860, 808970960, 2723733524, 8436372248, 24304813148, 65712993248, 167965846148, 408373664744, 949291256585, 2119095737210, 4559798912835, 9488531918460, 19148848609485, 37571357310510

Offset: 1

Yahia Kahloune, Dec 22 2013

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (18,-153,816,-3060,8568,-18564,31824,-43758,48620,-43758,31824,-18564,8568,-3060,816,-153,18,-1).

Cf. A087127, A086023, A086025, A086027, A086029.

A234253:=n->add(binomial(7+i,8)^2, i=1..n); seq(A234253(n), n=1..30); # Wesley Ivan Hurt, Dec 23 2013

Table[Sum[Binomial[7 + i, 8]^2, {i, n}], {n, 30}] (* Wesley Ivan Hurt, Dec 23 2013 *)
CoefficientList[Series[(x^8 + 64 x^7 + 784 x^6 + 3136 x^5 + 4900 x^4 + 3136 x^3 + 784 x^2 + 64 x + 1)/(x - 1)^18, {x, 0, 40}], x] (* Vincenzo Librandi, May 06 2014 *)

Vec(x*(x^8 +64*x^7 +784*x^6 +3136*x^5 +4900*x^4 +3136*x^3 +784*x^2 +64*x +1)/(x-1)^18 + O(x^100)) \\ Colin Barker, May 02 2014

G.f.: x*(x^8 +64*x^7 +784*x^6 +3136*x^5 +4900*x^4 +3136*x^3 +784*x^2 +64*x +1) / (x-1)^18. - Colin Barker, May 02 2014

One term corrected and more terms from Colin Barker, May 02 2014

A300729 Number of arrangements on a line of n finite closed intervals (possibly of zero length) with k distinct endpoints (n >= 1, 1 <= k <= 2*n).

Original entry on oeis.org

1, 1, 1, 7, 12, 6, 1, 25, 138, 294, 270, 90, 1, 79, 1056, 5298, 12780, 16020, 10080, 2520, 1, 241, 7050, 70350, 334710, 875970, 1335600, 1184400, 567000, 113400, 1, 727, 44472, 817746, 6849900, 31500180, 87348240, 152643960, 169533000, 116235000, 44906400, 7484400

Offset: 1

Peter Bala, Mar 12 2018

A122193(n,k) equals the number of arrangements on a line of n (nondegenerate) finite closed intervals having k distinct endpoints. The entries T(n,k) of the present table satisfy T(n,k) = A122193(n,k) + A122193(n,k+1). Proof. In an arrangement contributing to T(n,k) either the intervals are all nondegenerate, and there are A122193(n,k) arrangements of this type, or at least one of the intervals in the arrangement is degenerate. The following argument to show there are A122193(n,k+1) arrangements of the latter type is taken from the solution to the problem posed in the 'IBM Ponder This' link.
In an arrangement of n nondegenerate finite closed intervals having k+1 distinct endpoints, the rightmost point is the right endpoint of one or more intervals. If we move each of these right endpoints to coincide with their corresponding left endpoint then we obtain an arrangement of n finite closed intervals with k distinct endpoints, where at least one of the intervals has zero length. The reverse mapping is clear: given an arrangement of n finite closed intervals with k distinct endpoints, where at least one of the intervals has zero length, take each interval of zero length and move all the right endpoints of these degenerate intervals to a single new rightmost point. This produces an arrangement of n nondegenerate finite closed intervals having k+1 distinct endpoints. (End proof)
Most of the properties of the present table now follow from the properties of A122193.
Reading the table by antidiagonals produces A059515.

			Table begins
      |k=0   1   2     3     4      5      6      7     8
---------------------------------------------------------
  n=0 |  1
    1 |  0   1   1
    2 |  0   1   7    12     6
    3 |  0   1  25   138   294    270     90
    4 |  0   1  79  1056  5298  12780  16020  10080  2520
   ...
T(2,3) = 12: The 12 arrangements with 3 endpoints of two (possibly degenerate) intervals [a, A] and [b, B] are
     aA-b-B, b-aA-B, b-B-aA, bB-a-A, a-bB-A, a-A-bB,
     ab-A-B, ab-B-A, a-b-AB, b-a-AB, a-bA-B, b-a-AB.
Here, for example, the notation aA-b-B indicates a = A < b < B, so the interval [a, A] is degenerate and lies to the left of the nondegenerate interval [b, B].
Row 2: (1, 7, 12, 6)
(x*(x + 1)/2)^2 = C(x,1) + 7*C(x,2) + 12*C(x,3) + 6*C(x,4).
Row 3: (1, 25, 138, 294, 270, 90)
(x*(x + 1)/2)^3 = C(x,1) + 25*C(x,2) + 138*C(x,3) + 294*C(x,4) + 270*C(x,5) + 90*C(x,6).
Sums of powers of triangular numbers:
Sum_{i = 1..n-1} (i*(i+1)/2)^2 = C(n,2) + 7*C(n,3) + 12*C(n,4) + 6*C(n,5);
Sum_{i = 1..n-1} (i*(i+1)/2)^3 = C(n,2) + 25*C(n,3) + 138*C(n,4) + 294*C(n,5) + 270*C(n,6) + 90*C(n,7).

Peter Bala, Deformations of the Hadamard product of power series
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
IBM Ponder This, Jan 01 2001
Helmut Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.

Cf. A059516 (row sums), A059515, A087127, A122193, A131689.

A300729 := proc (n, k)
add((-1)^(k-i)*binomial(k, i)*((1/2)*i*(i+1))^n, i = 0..k);
end proc:
for n from 0 to 8 do
seq(A300729(n, k), k = 0..2*n)
end do;

T[0, 0] = 1; T[n_, k_] := Sum[(-1)^(k-i)*Binomial[k, i]*((1/2)*i*(i+1))^n, {i, 0, k}]; Table[T[n, k], {n, 1, 6}, {k, 1, 2 n}] // Flatten (* Jean-François Alcover, Mar 16 2018 *)

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * (i*(i+1)/2)^n for 0 <= k <= 2*n.
T(n,k) = A122193(n,k) + A122193(n,k+1).
T(n,k) = k*(k+1)/2*T(n-1,k) + k^2*T(n-1,k-1) + k*(k-1)/2*T(n-1,k-2) for 1 < k <= 2*n with boundary conditions T(0,0) = 1, T(0,n) = 0 for n >= 1; T(n,1) = 1 for n >= 1 and T(n,k) = 0 if k > 2*n.
Double e.g.f.: exp(-x)*Sum_{n>=0} exp( binomial(n+1,2)*y )* x^n/n! = 1 + (x + x^2/2!)*y + (x + 7*x^2/2! + 12*x^3/3! + 6*x^4/4!)*y^2/2! + ....
The n-th row of the table is given by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318 and v_n is the sequence (0, 1, 3^n, 6^n, 10^n, ...) regarded as an infinite column vector, where 1, 3, 6, 10, ... is the sequence of triangular numbers A000217. Cf. A087127 and A122193.
n-th row polynomial R(n,x) = (x + x^2) o ... o (x + x^2) (n factors) where o denotes the black diamond product of power series as defined by Dukes and White.
R(n,x) = Sum_{i >= 1} (i*(i+1)/2)^n*x^i/(1 + x)^(i+1) for n >= 1.
x*R(n,x) = (1 + x)* the n-th row polynomial of A122193 for n >= 1.
(1 + x)*R(n,x) = x * the n-th row polynomial of A087127 for n >= 1.
Sum_{k = 1..2*n} T(n,k)*binomial(x,k) = (binomial(x+1,2))^n for n >= 1.
Sum_{i = 1..n-1} (i*(i+1)/2)^m = Sum_{k = 1..2*m} T(m,k)*binomial(n,k+1) for m >= 1. See Example section below.
R(n,x) = 1/2^n*Sum_{k = 0..n} binomial(n,k)*F(n+k,x), where F(n,x) = Sum_{k = 0..n} k!*Stirling2(n,k)*x^k is the n-th Fubini polynomial, the n-th row polynomial of A131689.
R(n+1,x) = 1/2*(x + x^2) * (d/dx)^2 ( (x + x^2)*R(n,x) ).
R(n,x) = R(n,-1 - x).
The zeros of R(n,x) belong to the interval [-1, 0].
For n >= 1, the alternating sum of the n-th row equals 0.
E.g.f. as a continued fraction: 1/(1 + (x + x^2)*(1 - exp(t))/(1 + (x + x^2)*(1 -exp(2*t))/(1 + (x + x^2)*(1 - exp(3*t))/(1 + ...)))) = 1 + (x + x^2)*t + (x + 7*x^2 + 12*x^3 + 6*x^4)*t^2/2! + ... (use Prodinger, equation 1.1). - Peter Bala, Jun 13 2019

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A210278 (5n)!/5^n.

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A299041 Irregular table: T(n,k) equals the number of alignments of length k of n strings each of length 3.

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A210279 (6n)!/6^n.

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A234253 a(n) = Sum_{i=1..n} C(7+i,8)^2.

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A300729 Number of arrangements on a line of n finite closed intervals (possibly of zero length) with k distinct endpoints (n >= 1, 1 <= k <= 2*n).

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A210280 (7n)!/7^n.

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A210281 (8n)!/8^n.

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