cp's OEIS Frontend

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.

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A000244 Powers of 3: a(n) = 3^n.

Original entry on oeis.org

1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequences E(1, 3), L(1, 3), P(1, 3), T(1, 3). Essentially same as Pisot sequences E(3, 9), L(3, 9), P(3, 9), T(3, 9). See A008776 for definitions of Pisot sequences.
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2n + 2, s(0) = 1, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004
a(1) = 1, a(n+1) is the least number such that there are a(n) even numbers between a(n) and a(n+1). Generalization for the sequence of powers of k: 1, k, k^2, k^3, k^4, ... There are a(n) multiples of k-1 between a(n) and a(n+1). - Amarnath Murthy, Nov 28 2004
a(n) = sum of (n+1)-th row in Triangle A105728. - Reinhard Zumkeller, Apr 18 2005
With p(n) being the number of integer partitions of n, p(i) being the number of parts of the i-th partition of n, d(i) being the number of different parts of the i-th partition of n, m(i, j) being the multiplicity of the j-th part of the i-th partition of n, Sum_{i = 1..p(n)} being the sum over i and Product_{j = 1..d(i)} being the product over j, one has: a(n) = Sum_{i = 1..p(n)} (p(i)!/(Product_{j = 1..d(i)} m(i, j)!))*2^(p(i) - 1). - Thomas Wieder, May 18 2005
For any k > 1 in the sequence, k is the first prime power appearing in the prime decomposition of repunit R_k, i.e., of A002275(k). - Lekraj Beedassy, Apr 24 2006
a(n-1) is the number of compositions of compositions. In general, (k+1)^(n-1) is the number of k-levels nested compositions (e.g., 4^(n-1) is the number of compositions of compositions of compositions, etc.). Each of the n - 1 spaces between elements can be a break for one of the k levels, or not a break at all. - Franklin T. Adams-Watters, Dec 06 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then a(n) = |S|. - Ross La Haye, Dec 22 2006
From Manfred Boergens, Mar 28 2023: (Start)
With regard to the comment by Ross La Haye:
Cf. A001047 if either nonempty subsets are considered or x is a proper subset of y.
Cf. a(n+1) in A028243 if nonempty subsets are considered and x is a proper subset of y. (End)
If X_1, X_2, ..., X_n is a partition of the set {1, 2, ..., 2*n} into blocks of size 2 then, for n >= 1, a(n) is equal to the number of functions f : {1, 2, ..., 2*n} -> {1, 2} such that for fixed y_1, y_2, ..., y_n in {1, 2} we have f(X_i) <> {y_i}, (i = 1, 2, ..., n). - Milan Janjic, May 24 2007
This is a general comment on all sequences of the form a(n) = [(2^k)-1]^n for all positive integers k. Example 1.1.16 of Stanley's "Enumerative Combinatorics" offers a slightly different version. a(n) in the number of functions f:[n] into P([k]) - {}. a(n) is also the number of functions f:[k] into P([n]) such that the generalized intersection of f(i) for all i in [k] is the empty set. Where [n] = {1, 2, ..., n}, P([n]) is the power set of [n] and {} is the empty set. - Geoffrey Critzer, Feb 28 2009
a(n) = A064614(A000079(n)) and A064614(m)A000079(n). - Reinhard Zumkeller, Feb 08 2010
3^(n+1) = (1, 2, 2, 2, ...) dot (1, 1, 3, 9, ..., 3^n); e.g., 3^3 = 27 = (1, 2, 2, 2) dot (1, 1, 3, 9) = (1 + 2 + 6 + 18). - Gary W. Adamson, May 17 2010
a(n) is the number of generalized compositions of n when there are 3*2^i different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
For n >= 1, a(n-1) is the number of generalized compositions of n when there are 2^(i-1) different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
The sequence in question ("Powers of 3") also describes the number of moves of the k-th disk solving the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] pre-colored Magnetic Tower of Hanoi puzzle (cf. A183111 - A183125).
a(n) is the number of Stern polynomials of degree n. See A057526. - T. D. Noe, Mar 01 2011
Positions of records in the number of odd prime factors, A087436. - Juri-Stepan Gerasimov, Mar 17 2011
Sum of coefficients of the expansion of (1+x+x^2)^n. - Adi Dani, Jun 21 2011
a(n) is the number of compositions of n elements among {0, 1, 2}; e.g., a(2) = 9 since there are the 9 compositions 0 + 0, 0 + 1, 1 + 0, 0 + 2, 1 + 1, 2 + 0, 1 + 2, 2 + 1, and 2 + 2. [From Adi Dani, Jun 21 2011; modified by editors.]
Except the first two terms, these are odd numbers n such that no x with 2 <= x <= n - 2 satisfy x^(n-1) == 1 (mod n). - Arkadiusz Wesolowski, Jul 03 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 3-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Explanation from David Applegate, Feb 20 2017: (Start)
Since the preceding comment appears in a large number of sequences, it might be worth adding a proof.
The number of compositions of n into exactly k parts is binomial(n-1,k-1).
For a p-colored composition of n such that no adjacent parts have the same color, there are exactly p choices for the color of the first part, and p-1 choices for the color of each additional part (any color other than the color of the previous one). So, for a partition into k parts, there are p (p-1)^(k-1) valid colorings.
Thus the number of p-colored compositions of n into exactly k parts such that no adjacent parts have the same color is binomial(n-1,k-1) p (p-1)^(k-1).
The total number of p-colored compositions of n such that no adjacent parts have the same color is then
Sum_{k=1..n} binomial(n-1,k-1) * p * (p-1)^(k-1) = p^n.
To see this, note that the binomial expansion of ((p - 1) + 1)^(n - 1) = Sum_{k = 0..n - 1} binomial(n - 1, k) (p - 1)^k 1^(n - 1 - k) = Sum_{k = 1..n} binomial(n - 1, k - 1) (p - 1)^(k - 1).
(End)
Also, first and least element of the matrix [1, sqrt(2); sqrt(2), 2]^(n+1). - M. F. Hasler, Nov 25 2011
One-half of the row sums of the triangular version of A035002. - J. M. Bergot, Jun 10 2013
Form an array with m(0,n) = m(n,0) = 2^n; m(i,j) equals the sum of the terms to the left of m(i,j) and the sum of the terms above m(i,j), which is m(i,j) = Sum_{k=0..j-1} m(i,k) + Sum_{k=0..i-1} m(k,j). The sum of the terms in antidiagonal(n+1) = 4*a(n). - J. M. Bergot, Jul 10 2013
a(n) = A007051(n+1) - A007051(n), and A007051 are the antidiagonal sums of an array defined by m(0,k) = 1 and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to left of m(n, k) plus those above m(n, k). m(1, k) = A000079(k); m(2, k) = A045623(k + 1); m(k + 1, k) = A084771(k). - J. M. Bergot, Jul 16 2013
Define an array to have m(0,k) = 2^k and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to the left of m(n, k) plus those above m(n, k). Row n = 0 of the array comprises A000079, column k = 0 comprises A011782, row n = 1 comprises A001792. Antidiagonal sums of the array are a(n): 1 = 3^0, 1 + 2 = 3^1, 2 + 3 + 4 = 3^2, 4 + 7 + 8 + 8 = 3^3. - J. M. Bergot, Aug 02 2013
The sequence with interspersed zeros and o.g.f. x/(1 - 3*x^2), A(2*k) = 0, A(2*k + 1) = 3^k = a(k), k >= 0, can be called hexagon numbers. This is because the algebraic number rho(6) = 2*cos(Pi/6) = sqrt(3) of degree 2, with minimal polynomial C(6, x) = x^2 - 3 (see A187360, n = 6), is the length ratio of the smaller diagonal and the side in the hexagon. Hence rho(6)^n = A(n-1)*1 + A(n)*rho(6), in the power basis of the quadratic number field Q(rho(6)). One needs also A(-1) = 1. See also a Dec 02 2010 comment and the P. Steinbach reference given in A049310. - Wolfdieter Lang, Oct 02 2013
Numbers k such that sigma(3k) = 3k + sigma(k). - Jahangeer Kholdi, Nov 23 2013
All powers of 3 are perfect totient numbers (A082897), since phi(3^n) = 2 * 3^(n - 1) for n > 0, and thus Sum_{i = 0..n} phi(3^i) = 3^n. - Alonso del Arte, Apr 20 2014
The least number k > 0 such that 3^k ends in n consecutive decreasing digits is a 3-term sequence given by {1, 13, 93}. The consecutive increasing digits are {3, 23, 123}. There are 100 different 3-digit endings for 3^k. There are no k-values such that 3^k ends in '012', '234', '345', '456', '567', '678', or '789'. The k-values for which 3^k ends in '123' are given by 93 mod 100. For k = 93 + 100*x, the digit immediately before the run of '123' is {9, 5, 1, 7, 3, 9, 5, 1, 3, 7, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus we see the digit before '123' will never be a 0. So there are no further terms. - Derek Orr, Jul 03 2014
All elements of A^n where A = (1, 1, 1; 1, 1, 1; 1, 1, 1). - David Neil McGrath, Jul 23 2014
Counts all walks of length n (open or closed) on the vertices of a triangle containing a loop at each vertex starting from any given vertex. - David Neil McGrath, Oct 03 2014
a(n) counts walks (closed) on the graph G(1-vertex;1-loop,1-loop,1-loop). - David Neil McGrath, Dec 11 2014
2*a(n-2) counts all permutations of a solitary closed walk of length (n) from the vertex of a triangle that contains 2 loops on each of the remaining vertices. In addition, C(m,k)=2*(2^m)*B(m+k-2,m) counts permutations of walks that contain (m) loops and (k) arcs. - David Neil McGrath, Dec 11 2014
a(n) is the sum of the coefficients of the n-th layer of Pascal's pyramid (a.k.a., Pascal's tetrahedron - see A046816). - Bob Selcoe, Apr 02 2016
Numbers n such that the trinomial x^(2*n) + x^n + 1 is irreducible over GF(2). Of these only the trinomial for n=1 is primitive. - Joerg Arndt, May 16 2016
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
a(n-1) is also the number of compositions of n if the parts can be runs of any length from 1 to n, and can contain any integers from 1 to n. - Gregory L. Simay, May 26 2017
Also the number of independent vertex sets and vertex covers in the n-ladder rung graph n P_2. - Eric W. Weisstein, Sep 21 2017
Also the number of (not necessarily maximal) cliques in the n-cocktail party graph. - Eric W. Weisstein, Nov 29 2017
a(n-1) is the number of 2-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
a(n) is the number of faces of any dimension (vertices, edges, square faces, etc.) of the n-dimensional hypercube. For example, the 0-dimensional hypercube is a point, and its only face is itself. The 1-dimensional hypercube is a line, which has two vertices and an edge. The 2-dimensional hypercube is a square, which has four vertices, four edges, and a square face. - Kevin Long, Mar 14 2023
Number of pairs (A,B) of subsets of M={1,2,...,n} with union(A,B)=M. For nonempty subsets cf. A058481. - Manfred Boergens, Mar 28 2023
From Jianing Song, Sep 27 2023: (Start)
a(n) is the number of disjunctive clauses of n variables up to equivalence. A disjunctive clause is a propositional formula of the form l_1 OR ... OR l_m, where l_1, ..., l_m are distinct elements in {x_1, ..., x_n, NOT x_1, ..., NOT x_n} for n variables x_1, ... x_n, and no x_i and NOT x_i appear at the same time. For each 1 <= i <= n, we can have neither of x_i or NOT x_i, only x_i or only NOT x_i appearing in a disjunctive clause, so the number of such clauses is 3^n. Viewing the propositional formulas of n variables as functions {0,1}^n -> {0,1}, a disjunctive clause corresponds to a function f such that the inverse image of 0 is of the form A_1 X ... X A_n, where A_i is nonempty for all 1 <= i <= n. Since each A_i has 3 choices ({0}, {1} or {0,1}), we also find that the number of disjunctive clauses of n variables is 3^n.
Equivalently, a(n) is the number of conjunctive clauses of n variables. (End)
The finite subsequence a(2), a(3), a(4), a(5) = 9, 27, 81, 243 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A007283 (see comment there). - Felix Huber, Feb 15 2024

Examples

			G.f. = 1 + 3*x + 9*x^2 + 27*x^3 + 81*x^4 + 243*x^5 + 729*x^6 + 2187*x^7 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A008776 (2*a(n), and first differences).
a(n) = A092477(n, 2) for n > 0.
a(n) = A159991(n) / A009964(n).
Cf. A100772, A035002. Row sums of A125076 and A153279.
a(n) = A217764(0, n).
Cf. A046816, A006521, A014945, A275414 (multisets).
The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).

Programs

Formula

a(n) = 3^n.
a(0) = 1; a(n) = 3*a(n-1).
G.f.: 1/(1-3*x).
E.g.f.: exp(3*x).
a(n) = n!*Sum_{i + j + k = n, i, j, k >= 0} 1/(i!*j!*k!). - Benoit Cloitre, Nov 01 2002
a(n) = Sum_{k = 0..n} 2^k*binomial(n, k), binomial transform of A000079.
a(n) = A090888(n, 2). - Ross La Haye, Sep 21 2004
a(n) = 2^(2n) - A005061(n). - Ross La Haye, Sep 10 2005
a(n) = A112626(n, 0). - Ross La Haye, Jan 11 2006
Hankel transform of A007854. - Philippe Deléham, Nov 26 2006
a(n) = 2*StirlingS2(n+1,3) + StirlingS2(n+2,2) = 2*(StirlingS2(n+1,3) + StirlingS2(n+1,2)) + 1. - Ross La Haye, Jun 26 2008
a(n) = 2*StirlingS2(n+1, 3) + StirlingS2(n+2, 2) = 2*(StirlingS2(n+1, 3) + StirlingS2(n+1, 2)) + 1. - Ross La Haye, Jun 09 2008
Sum_{n >= 0} 1/a(n) = 3/2. - Gary W. Adamson, Aug 29 2008
If p(i) = Fibonacci(2i-2) and if A is the Hessenberg matrix of order n defined by A(i, j) = p(j-i+1), (i <= j), A(i, j) = -1, (i = j+1), and A(i, j) = 0 otherwise, then, for n >= 1, a(n-1) = det A. - Milan Janjic, May 08 2010
G.f. A(x) = M(x)/(1-M(x))^2, M(x) - o.g.f for Motzkin numbers (A001006). - Vladimir Kruchinin, Aug 18 2010
a(n) = A133494(n+1). - Arkadiusz Wesolowski, Jul 27 2011
2/3 + 3/3^2 + 2/3^3 + 3/3^4 + 2/3^5 + ... = 9/8. [Jolley, Summation of Series, Dover, 1961]
a(n) = Sum_{k=0..n} A207543(n,k)*4^(n-k). - Philippe Deléham, Feb 25 2012
a(n) = Sum_{k=0..n} A125185(n,k). - Philippe Deléham, Feb 26 2012
Sum_{n > 0} Mobius(n)/a(n) = 0.181995386702633887827... (see A238271). - Alonso del Arte, Aug 09 2012. See also the sodium 3s orbital energy in table V of J. Chem. Phys. 53 (1970) 348.
a(n) = (tan(Pi/3))^(2*n). - Bernard Schott, May 06 2022
a(n-1) = binomial(2*n-1, n) + Sum_{k >= 1} binomial(2*n, n+3*k)*(-1)^k. - Greg Dresden, Oct 14 2022
G.f.: Sum_{k >= 0} x^k/(1-2*x)^(k+1). - Kevin Long, Mar 14 2023

A307883 Square array read by descending antidiagonals: T(n, k) where column k is the expansion of 1/sqrt(1 - 2*(k+1)*x + ((k-1)*x)^2).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 13, 20, 1, 1, 5, 22, 63, 70, 1, 1, 6, 33, 136, 321, 252, 1, 1, 7, 46, 245, 886, 1683, 924, 1, 1, 8, 61, 396, 1921, 5944, 8989, 3432, 1, 1, 9, 78, 595, 3606, 15525, 40636, 48639, 12870, 1, 1, 10, 97, 848, 6145, 33876, 127905, 281488, 265729, 48620, 1
Offset: 0

Views

Author

Seiichi Manyama, May 02 2019

Keywords

Comments

Column k is the diagonal of the rational function 1 / ((1-x)*(1-y) - k*x*y). - Seiichi Manyama, Jul 11 2020
More generally, column k is the diagonal of the rational function r / ((1-r*x)*(1-r*y) + r-1 - (k+r-1)*r*x*y) for any nonzero real number r. - Seiichi Manyama, Jul 22 2020

Examples

			Square array begins:
  1,   1,    1,     1,      1,      1,      1, ...
  1,   2,    3,     4,      5,      6,      7, ...
  1,   6,   13,    22,     33,     46,     61, ...
  1,  20,   63,   136,    245,    396,    595, ...
  1,  70,  321,   886,   1921,   3606,   6145, ...
  1, 252, 1683,  5944,  15525,  33876,  65527, ...
  1, 924, 8989, 40636, 127905, 324556, 712909, ...
Seen as a triangle T(n, k):
  [0] 1;
  [1] 1, 1;
  [2] 1, 2,  1;
  [3] 1, 3,  6,   1;
  [4] 1, 4, 13,  20,    1;
  [5] 1, 5, 22,  63,   70,     1;
  [6] 1, 6, 33, 136,  321,   252,     1;
  [7] 1, 7, 46, 245,  886,  1683,   924,     1;
  [8] 1, 8, 61, 396, 1921,  5944,  8989,  3432,     1;
  [9] 1, 9, 78, 595, 3606, 15525, 40636, 48639, 12870, 1;
		

Crossrefs

Columns k=0..6 give A000012, A000984, A001850, A069835, A084771, A084772, A098659.
Main diagonal gives A187021.
T(n,n+1) gives A335309.

Programs

  • Maple
    # Seen as a triangle read by rows:
    T := (n, k) -> simplify(hypergeom([-k, -k], [1], n - k)):
    seq(lprint(seq(T(n, k), k = 0..n)), n = 0..9);  # Peter Luschny, May 13 2024
  • Mathematica
    T[n_, k_] := Sum[If[k == j == 0, 1, k^j] * Binomial[n, j]^2, {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 13 2021 *)
    (* Seen as a triangle read by rows: *)
    T[n_, k_] := HypergeometricPFQ[{-k, -k}, {1}, n - k];
    Flatten[Table[T[n, k], {n, 0, 10}, {k, 0, n}]] (* Detlef Meya, May 13 2024 *)

Formula

T(n,k) is the coefficient of x^n in the expansion of (1 + (k+1)*x + k*x^2)^n.
T(n,k) = Sum_{j=0..n} k^j * binomial(n,j)^2.
T(n,k) = Sum_{j=0..n} (k-1)^(n-j) * binomial(n,j) * binomial(n+j,j).
n * T(n,k) = (k+1) * (2*n-1) * T(n-1,k) - (k-1)^2 * (n-1) * T(n-2,k).
T(n,k) = hypergeom([-k, -k], [1], n - k), (triangular form). - Detlef Meya, May 13 2024

A103488 a(n) = 2^(n^2-1).

Original entry on oeis.org

1, 8, 256, 32768, 16777216, 34359738368, 281474976710656, 9223372036854775808, 1208925819614629174706176, 633825300114114700748351602688, 1329227995784915872903807060280344576, 11150372599265311570767859136324180752990208
Offset: 1

Views

Author

Zerinvary Lajos, Jul 30 2006

Keywords

Comments

From the classic story: number of grains of wheat on last square of n X n chessboard.
Hankel transform of A059304 and of A084771. - Philippe Deléham, Dec 02 2007
The number of compositions of n^2. - Alois P. Heinz, Feb 22 2020

Crossrefs

Programs

  • Magma
    [2^(n^2-1): n in [1..15]]; // Vincenzo Librandi, Feb 25 2014
  • Maple
    [seq(2^(n^2-1),n=1..14)];
  • Mathematica
    Table[2^(n^2 - 1), {n, 1, 20}] (* Vincenzo Librandi, Feb 25 2014 *)
  • PARI
    a(n) = 2^(n^2-1); \\ Joerg Arndt, Feb 23 2014
    

Formula

a(n) = A000079(A005563(n-1)). - Michel Marcus, Feb 25 2014
a(n) = A011782(A000290(n)). - Alois P. Heinz, Feb 22 2020

A085363 a(0)=1, for n>0: a(n) = 4*9^(n-1) - (1/2)*Sum_{i=1..n-1} a(i)*a(n-i).

Original entry on oeis.org

1, 4, 28, 212, 1676, 13604, 112380, 940020, 7936620, 67494980, 577309148, 4961187092, 42801458764, 370478720356, 3215827927228, 27982214082612, 244004165618220, 2131710838837380, 18654504783815580, 163488269572628820
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Jun 25 2003

Keywords

Comments

Apparently, the number of 2-D directed walks of semilength n starting at (0,0) and ending on the X-axis using steps NE, SE, NW and SW avoiding adjacent NW/SE and adjacent NE/SW. - David Scambler, Jun 20 2013
Form an array with m(0,n) = m(n,0) = 2^n; m(i,j) equals the sum of the terms to the left of m(i,j) and the sum of the terms above m(i,j), which is m(i,j) = Sum_{k-0..j-1} m(i,k) + Sum_{k=0..i-1} m(k,j). m(n,n) = a(n). - J. M. Bergot, Jul 10 2013
From G. C. Greubel, May 22 2020: (Start)
This sequence is part of a class of sequences, for m >= 0, with the properties:
a(n) = 2*m*(4*m+1)^(n-1) - (1/2)*Sum_{k=1..n-1} a(k)*a(n-k).
a(n) = Sum_{k=0..n} m^k * binomial(n-1, n-k) * binomial(2*k, k).
a(n) = (2*m) * Hypergeometric2F1(-n+1, 3/2; 2; -4*m), for n > 0.
n*a(n) = 2*((2*m+1)*n - (m+1))*a(n-1) - (4*m+1)*(n-2)*a(n-2).
(4*m + 1)^n = Sum_{k=0..n} Sum_{j=0..k} a(j)*a(k-j).
G.f.: sqrt( (1 - t)/(1 - (4*m+1)*t) ).
This sequence is the case of m=2. (End)
The number of elements in the free group on two generators of length 2n that are zero exponent sum. - Tey Berendschot, Aug 09 2021

Crossrefs

Cf. A001019 (9^n), A084771, A085362, A085364, diagonal of A348595.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( Sqrt((1-x)/(1-9*x)) )); // G. C. Greubel, May 23 2020
    
  • Maple
    seq(coeff(series(sqrt((1-x)/(1-9*x)), x, n+1), x, n), n = 0..30); # G. C. Greubel, May 23 2020
  • Mathematica
    CoefficientList[Series[Sqrt[(1-x)/(1-9x)], {x, 0, 25}], x]
  • PARI
    my(x='x+O('x^66)); Vec(sqrt((1-x)/(1-9*x)) ) \\ Joerg Arndt, May 10 2013
    
  • Sage
    def A085363_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( sqrt((1-x)/(1-9*x)) ).list()
    A085363_list(30) # G. C. Greubel, May 23 2020

Formula

G.f.: sqrt((1-x)/(1-9*x)).
Sum_{i=0..n} Sum_{j=0..i} a(j)*a(i-j) = 9^n.
From Vladeta Jovovic, Oct 10 2003: (Start)
First differences of A084771.
a(n) = Sum_{k=1..n} 2^k * binomial(n-1, k-1) * binomial(2*k, k). (End)
D-finite with recurrence n*a(n) = (10*n-6)*a(n-1) - (9*n-18)*a(n-2). - Vladeta Jovovic, Jul 16 2004
a(n) ~ 2*sqrt(2)*3^(2*n-1)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 14 2012
a(0) = 1; a(n) = (4/n) * Sum_{k=0..n-1} (n+k) * a(k). - Seiichi Manyama, Mar 28 2023
From Seiichi Manyama, Aug 22 2025: (Start)
a(n) = (1/4)^n * Sum_{k=0..n} 9^k * binomial(2*k,k) * binomial(2*(n-k),n-k)/(1-2*(n-k)).
a(n) = Sum_{k=0..n} (-2)^k * 9^(n-k) * binomial(2*k,k)/(1-2*k) * binomial(n-1,n-k). (End)

A246923 Expansion of g.f.: 1 / AGM(1-9*x, sqrt((1-x)*(1-81*x))).

Original entry on oeis.org

1, 25, 1089, 60025, 3690241, 241025625, 16359689025, 1140463805625, 81081830657025, 5852177325225625, 427465780890020929, 31528177440967935225, 2344153069158724611841, 175473167541934734763225, 13211212029033949825064769, 999630716942846408773325625
Offset: 0

Views

Author

Paul D. Hanna, Sep 07 2014

Keywords

Comments

In general, the g.f. of the squares of coefficients in g.f. 1/sqrt((1-p*x)*(1-q*x)) is given by
1/AGM(1-p*q*x, sqrt((1-p^2*x)*(1-q^2*x))) = Sum_{n>=0} x^n*( Sum_{k=0..n} p^(n-k)*((q-p)/4)^k*C(n,k)*C(2*k,k) )^2,
and consists of integer coefficients when 4|(q-p).
Here AGM(x,y) = AGM((x+y)/2, sqrt(x*y)) is the arithmetic-geometric mean.

Examples

			G.f.: A(x) = 1 + 25*x + 1089*x^2 + 60025*x^3 + 3690241*x^4 + 241025625*x^5 +...
where the square-root of each term yields A084771:
[1, 5, 33, 245, 1921, 15525, 127905, 1067925, ...],
the g.f. of which is 1/sqrt((1-x)*(1-9*x)).
		

Crossrefs

Programs

  • Magma
    [9^n*Evaluate(LegendrePolynomial(n), 5/3)^2 : n in [0..40]]; // G. C. Greubel, May 30 2023
    
  • Mathematica
    a[n_] := Sum[2^k * Binomial[n, k] * Binomial[2k, k], {k, 0, n}]^2; Array[a, 17, 0] (* Amiram Eldar, Dec 11 2018 *)
    Table[9^n*LegendreP[n, 5/3]^2, {n, 0, 40}] (* G. C. Greubel, May 30 2023 *)
  • PARI
    {a(n,p=1,q=9)=polcoeff( 1 / agm(1-p*q*x, sqrt((1-p^2*x)*(1-q^2*x) +x*O(x^n))), n) }
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n,p=1,q=9)=polcoeff( 1 / sqrt((1-p*x)*(1-q*x) +x*O(x^n)), n)^2 }
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n,p=1,q=9)=sum(k=0,n,p^(n-k)*((q-p)/4)^k*binomial(n,k)*binomial(2*k,k))^2 }
    for(n=0, 20, print1(a(n), ", "))
    
  • SageMath
    [9^n*gen_legendre_P(n, 0, 5/3)^2 for n in range(41)] # G. C. Greubel, May 30 2023

Formula

a(n) = A084771(n)^2 = ( Sum_{k=0..n} 2^k*C(n,k)*C(2*k,k) )^2.
G.f.: 1 / AGM((1-x)*(1+9*x), (1+x)*(1-9*x)) = Sum_{n>=0} a(n)*x^(2*n).
a(n) ~ 3^(4*n+2) / (8*Pi*n). - Vaclav Kotesovec, Sep 27 2019
a(n) = 9^n * ( LegendreP(n, 5/3) )^2. - G. C. Greubel, May 30 2023

A248167 Expansion of g.f.: 1 / AGM(1-33*x, sqrt((1-9*x)*(1-121*x))).

Original entry on oeis.org

1, 49, 3249, 261121, 23512801, 2266426449, 228110356881, 23642146057761, 2502698427758529, 269194720423487089, 29319711378381802609, 3225762406810715071041, 357859427246543331576481, 39977637030683399494792849, 4492572407488016429783217489, 507445676088537643607528136801
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 2014

Keywords

Comments

In general, the g.f. of the squares of coefficients in g.f. 1/sqrt((1-p*x)*(1-q*x)) is given by 1/AGM(1-p*q*x, sqrt((1-p^2*x)*(1-q^2*x))) = Sum_{n>=0} x^n*( Sum_{k=0..n} p^(n-k)*((q-p)/4)^k*C(n,k)*C(2*k,k) )^2, and consists of integer coefficients when 4|(q-p). Here AGM(x, y) = AGM((x+y)/2, sqrt(x*y)) is the arithmetic-geometric mean.

Examples

			G.f.: A(x) = 1 + 49*x + 3249*x^2 + 261121*x^3 + 23512801*x^4 +...
where the square-root of the terms yields A248168:
[1, 7, 57, 511, 4849, 47607, 477609, 4862319, 50026977, ...],
the g.f. of which is 1/sqrt((1-3*x)*(1-11*x)).
		

Crossrefs

Programs

  • Magma
    m:=40;
    A248168:= [n le 2 select 7^(n-1) else (7*(2*n-3)*Self(n-1) - 33*(n-2)*Self(n-2))/(n-1) : n in [1..m+2]];
    A248167:= func< n | (A248168[n+1])^2 >;
    [A248167(n): n in [0..m]]; // G. C. Greubel, May 31 2025
    
  • Mathematica
    a[n_] := Sum[3^(n - k) * 2^k * Binomial[n, k] * Binomial[2k, k], {k, 0, n} ]^2; Array[a, 17, 0] (* Amiram Eldar, Dec 11 2018 *)
  • PARI
    {a(n,p=3,q=11)=polcoeff( 1 / agm(1-p*q*x, sqrt((1-p^2*x)*(1-q^2*x) +x*O(x^n))), n) }
    for(n=0, 20, print1(a(n,3,11), ", "))
    
  • PARI
    {a(n,p=3,q=11)=polcoeff( 1 / sqrt((1-p*x)*(1-q*x) +x*O(x^n)), n)^2 }
    for(n=0, 20, print1(a(n,3,11), ", "))
    
  • PARI
    {a(n,p=3,q=11)=sum(k=0,n,p^(n-k)*((q-p)/4)^k*binomial(n,k)*binomial(2*k,k))^2 }
    for(n=0, 20, print1(a(n,3,11), ", "))
    
  • SageMath
    @CachedFunction
    def b(n): # b = A248168
         if (n<2): return 7^n
         else: return (7*(2*n-1)*b(n-1) - 33*(n-1)*b(n-2))//n
    def A248167(n): return (b(n))^2
    print([A248167(n) for n in range(41)]) # G. C. Greubel, May 31 2025

Formula

a(n) = A248168(n)^2 = ( Sum_{k=0..n} 3^(n-k)*2^k * C(n,k) * C(2*k,k) )^2.
G.f.: 1 / AGM((1-3*x)*(1+11*x), (1+3*x)*(1-11*x)) = Sum_{n>=0} a(n)*x^(2*n).
a(n) ~ 11^(2*n + 1) / (8*Pi*n). - Vaclav Kotesovec, Sep 27 2019

A331516 Expansion of 1/(1 - 10*x + 9*x^2)^(3/2).

Original entry on oeis.org

1, 15, 174, 1850, 18855, 187425, 1832460, 17705700, 169569405, 1612842275, 15256106778, 143660483070, 1347716324227, 12603114069525, 117536416879320, 1093553079352200, 10153324144411065, 94098595671581175, 870667876141568070, 8044341506669534850
Offset: 0

Views

Author

Seiichi Manyama, Jan 19 2020

Keywords

Crossrefs

Column 5 of A331514.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 20); Coefficients(R!( 1/(1 - 10*x + 9*x^2)^(3/2))); // Marius A. Burtea, Jan 20 2020
    
  • Magma
    [1/2*&+[3^(n-k+1)*k*Binomial(n+1, k)*Binomial(n+k+1,k):k in [1..n+1]]:n in [0..20]]; // Marius A. Burtea, Jan 20 2020
  • Mathematica
    a[n_] := 1/2 * Sum[3^( n + 1 - k) * k * Binomial[n + 1, k] * Binomial[n + 1 + k, k], {k, 1, n+1}]; Array[a, 20, 0] (* Amiram Eldar, Jan 20 2020 *)
    CoefficientList[Series[1/(1-10x+9x^2)^(3/2),{x,0,20}],x] (* Harvey P. Dale, Nov 04 2021 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(1/(1-10*x+9*x^2)^(3/2))
    
  • PARI
    a(n) = sum(k=1, n+1, 3^(n+1-k)*k*binomial(n+1, k)*binomial(n+1+k, k))/2;
    

Formula

a(n) = 1/2 * Sum_{k=1..n+1} 3^(n+1-k) * k * binomial(n+1,k) * binomial(n+1+k,k).
n * a(n) = 5 * (2*n+1) * a(n-1) - 9 * (n+1) * a(n-2) for n > 1.
a(n) = ((n+2)/2) * Sum_{k=0..n} 4^k * binomial(n+1,k) * binomial(n+1,k+1).
a(n) ~ sqrt(n) * 3^(2*n + 3) / (2^(7/2) * sqrt(Pi)). - Vaclav Kotesovec, Jan 26 2020
From Seiichi Manyama, Aug 20 2025: (Start)
a(n) = (1/4)^n * Sum_{k=0..n} 9^k * (2*k+1) * (2*(n-k)+1) * binomial(2*k,k) * binomial(2*(n-k),n-k).
a(n) = Sum_{k=0..n} 2^k * (2*k+1) * binomial(2*k,k) * binomial(n+2,n-k).
a(n) = Sum_{k=0..n} (-2)^k * 9^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(n+2,n-k).
a(n) = binomial(n+2,2) * A059231(n+1).
a(n) = ((n+2)/2) * Sum_{k=0..floor(n/2)} 4^k * 5^(n-2*k) * binomial(n+1,n-2*k) * binomial(2*k+1,k).
a(n) = Sum_{k=0..n} (5/2)^k * (-9/10)^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(k,n-k). (End)

A051708 Number of ways to move a chess rook from the lower left corner to square (n,n), with the rook moving only up or right.

Original entry on oeis.org

1, 2, 14, 106, 838, 6802, 56190, 470010, 3968310, 33747490, 288654574, 2480593546, 21400729382, 185239360178, 1607913963614, 13991107041306, 122002082809110, 1065855419418690, 9327252391907790, 81744134786314410, 717367363052796678, 6303080714967178962
Offset: 1

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

This sequence arises in connection with mean lengths of ascents and descents in Dyck paths as follows. Let u(n,k) denote the mean length of the k-th ascent taken over all Dyck n-paths (A000108) where it is understood that if a Dyck path has fewer than k ascents, then the length of the k-th ascent is 0. For example, the second ascent in UUDUUUDDDDUD has length 3 and its fourth has length 0. Similarly, let v(n,k) denote the mean length of the k-th descent. Then u(k) := lim_{n->infinity} u(n,k) and v(k) := lim_{n->infinity} v(n,k) both exist. The sequence (u(k)){k>=1} begins 3, 8/3, ... and decreases steadily toward a limit of 2. Analogously, v(k) increases steadily from 4/3 toward the same limit of 2. For all k >= 1, u(k+1) exceeds 2 by the same amount that v(k) falls below 2. The common difference u(k+1) - 2 = 2 - v(k) is a(k+1)/3^(2k-1). Thus the common difference sequence begins 2/3, 14/27, 106/243, ..., for k=1,2,3,... . - _David Callan, Jul 14 2006
Number of ways to partition the 1 X (n-1) grid into triangles, with all vertices on grid points. - Peter Kagey, Nov 30 2018

Examples

			G.f. = x + 2*x^2 + 14*x^3 + 106*x^4 + 838*x^5 + 6802*x^6 + 56190*x^7 + ...
		

References

  • Posting to newsgroup rec.puzzles, Dec 03 1999 by Nick Wedd (Nick(AT)maproom.co.uk).

Crossrefs

Main diagonal of the square array given in A035002.
First differences of (A084771-1)/2.
Row d=2 of A181731.

Programs

  • GAP
    a:=[1,2];; for n in [3..25] do a[n]:=((10*n-16)*a[n-1]-(9*n-27)*a[n-2])/(n-1); od; a; # Muniru A Asiru, Nov 30 2018
    
  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( ((x*(1-x))/(Sqrt(1-10*x+9*x^2))+x)/2 )); // G. C. Greubel, Dec 01 2018
  • Maple
    a:= proc(n) option remember;
          `if`(n<3, n, ((10*n-16)*a(n-1)-(9*n-27)*a(n-2))/(n-1))
        end:
    seq(a(n), n=1..30);  # Alois P. Heinz, Jul 21 2012
  • Mathematica
    CoefficientList[Series[(9*x^2 - Sqrt[9*x^2-10*x+1]*x-x) / (2*(9*x-1)), {x,0,20}],x] // Rest (* Jean-François Alcover, Mar 30 2011, after g.f. given by Ralf Stephan *)
    RecurrenceTable[{a[1]==1,a[2]==2,a[n]==((10n-16)a[n-1]-(9n-27)a[n-2])/ (n-1)},a,{n,30}] (* Harvey P. Dale, Sep 28 2013 *)
  • Maxima
    a(n):=sum(binomial(n-1,n-i)*sum(binomial(k+i,i)*binomial(n-1,n-k),k,0,n),i,0,n); /* Vladimir Kruchinin, Apr 20 2015 */
    
  • PARI
    {a(n) = if( n<1, 0, n--; polcoeff( 1/2 + (1 - x) / (2 * sqrt( 1 - 10*x + 9*x^2 + x * O(x^n) ) ), n ) )} /* Michael Somos, Jan 08 2011 */
    
  • PARI
    a(n) = n--; sum(i=0,n, binomial(n-1, n-i)*sum(k=0, n, binomial(k+i, i)*binomial(n-1, n-k))); \\ Michel Marcus, Apr 20 2015
    

Formula

G.f.: ((x*(1-x))/(sqrt(1-10*x+9*x^2)) + x)/2. - Ralf Stephan, Mar 23 2004; confirmed by Martin J. Erickson, Oct 05 2007
D-finite with recurrence a(1)=1; a(2)=2; a(n) = ((10*n-16)*a(n-1) - (9*n-27)*a(n-2)) / (n-1), for n >= 3. - Martin J. Erickson (erickson(AT)truman.edu), Nov 12 2007
a(n) is asymptotic to (sqrt(2)/27)*9^n/(sqrt(Pi*n)). - Martin J. Erickson, Nov 09 2007
G.f.: A(x) satisfies 2 * x^3 = (1 - 9*x) * A(x) * (A(x) - x). - Michael Somos, Jan 08 2011
a(n+1) = Sum_{i=0..n} (C(n-1,n-i)*Sum_{k=0..n} (C(k+i,i)*C(n-1,n-k))). - Vladimir Kruchinin, Apr 20 2015
a(n) = Sum_{k=0..n} (k+1)*C(n-2,k-1)*hypergeom([2+k,2-n],[2],-1) for n >= 2. - Peter Luschny, Apr 20 2015
a(n) = ((-1)^(n-1) * 4^(n-1)) / (48*(n-1)*n) * ( -(4*(n-1)^2 + 16*(n-1) + 28)*JacobiP(n-2, -2*(n-1)-1, 2, -1/2) + (n+2)*(n-2)*JacobiP(n-3, -2*(n-1), 3, -1/2) ) for n > 1. - Alexander R. Povolotsky, Apr 26 2025

Extensions

More terms from James Sellers, Dec 08 1999

A289147 Number of (n+1) X (n+1) binary matrices M with at most one 1 in each of the first n rows and each of the first n columns and M[n+1,n+1] = 0.

Original entry on oeis.org

1, 5, 34, 286, 2840, 32344, 414160, 5876336, 91356544, 1542401920, 28075364096, 547643910400, 11389266525184, 251428006132736, 5869482147358720, 144413021660821504, 3733822274973040640, 101181690628832198656, 2867011297057247002624, 84764595415605494743040
Offset: 0

Views

Author

Alois P. Heinz, Jun 26 2017

Keywords

Comments

Number of marriage patterns between a labeled set X of n women and a labeled set Y of n men (all heterosexual): some couples can be formed where one partner is from X and the other from Y, some members of X and Y marry external (unlabeled) partners, and some do not marry.

Examples

			a(1) = 5:
[0 0]  [1 0]  [0 1]  [0 0]  [0 1]
[0 0]  [0 0]  [0 0]  [1 0]  [1 0] .
.
a(2) = 34:
[0 0 0]  [0 0 0]  [0 0 0]  [0 0 0]  [0 0 0]  [0 0 0]  [0 0 0]
[0 0 0]  [0 0 0]  [0 0 0]  [0 0 0]  [0 0 1]  [0 0 1]  [0 0 1]
[0 0 0]  [0 1 0]  [1 0 0]  [1 1 0]  [0 0 0]  [0 1 0]  [1 0 0]
.
[0 0 0]  [0 0 0]  [0 0 0]  [0 0 0]  [0 0 0]  [0 0 1]  [0 0 1]
[0 0 1]  [0 1 0]  [0 1 0]  [1 0 0]  [1 0 0]  [0 0 0]  [0 0 0]
[1 1 0]  [0 0 0]  [1 0 0]  [0 0 0]  [0 1 0]  [0 0 0]  [0 1 0]
.
[0 0 1]  [0 0 1]  [0 0 1]  [0 0 1]  [0 0 1]  [0 0 1]  [0 0 1]
[0 0 0]  [0 0 0]  [0 0 1]  [0 0 1]  [0 0 1]  [0 0 1]  [0 1 0]
[1 0 0]  [1 1 0]  [0 0 0]  [0 1 0]  [1 0 0]  [1 1 0]  [0 0 0]
.
[0 0 1]  [0 0 1]  [0 0 1]  [0 1 0]  [0 1 0]  [0 1 0]  [0 1 0]
[0 1 0]  [1 0 0]  [1 0 0]  [0 0 0]  [0 0 0]  [0 0 1]  [0 0 1]
[1 0 0]  [0 0 0]  [0 1 0]  [0 0 0]  [1 0 0]  [0 0 0]  [1 0 0]
.
[0 1 0]  [1 0 0]  [1 0 0]  [1 0 0]  [1 0 0]  [1 0 0]
[1 0 0]  [0 0 0]  [0 0 0]  [0 0 1]  [0 0 1]  [0 1 0]
[0 0 0]  [0 0 0]  [0 1 0]  [0 0 0]  [0 1 0]  [0 0 0]  .
		

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, 4*n+1,
          (2*n+3)*a(n-1)-(n-1)^2*a(n-2))
        end:
    seq(a(n), n=0..25);
    # second Maple program:
    a:= n-> n-> n! * add(binomial(n, i)*4^i/i!, i=0..n):
    seq(a(n), n=0..25);
    # third Maple program:
    a:= n-> n!* simplify(LaguerreL(n, -4), 'LaguerreL'):
    seq(a(n), n=0..25);
  • Mathematica
    Table[n! LaguerreL[n, -4], {n, 0, 30}] (* Indranil Ghosh, Jul 06 2017 *)
  • Python
    from mpmath import *
    mp.dps=150
    l=chop(taylor(lambda x:exp(4*x/(1-x))/(1-x), 0, 31))
    print([int(fac(i)*l[i]) for i in range(len(l))]) # Indranil Ghosh, Jul 06 2017
    # or #
    from mpmath import *
    mp.dps=100
    def a(n): return int(fac(n)*laguerre(n, 0, -4))
    print([a(n) for n in range(31)]) # Indranil Ghosh, Jul 06 2017

Formula

E.g.f.: exp(4*x/(1-x))/(1-x).
a(n) = Sum_{i=0..n} i! * (2^(n-i)*binomial(n,i))^2.
a(n) = Sum_{i=0..n} (n-i)! * 4^i * binomial(n,i)^2.
a(n) = n! * Sum_{i=0..n} 4^i/i! * binomial(n,i).
a(n) = (2*n+3)*a(n-1)-(n-1)^2*a(n-2) for n>=2, a(n) = 4*n+1 for n<2.
a(n) = n! * Laguerre(n,-4) = n! * A160611(n)/A160612(n).
a(n) ~ exp(-2 + 4*sqrt(n) - n) * n^(n + 1/4) / 2 * (1 + 163/(96*sqrt(n))). - Vaclav Kotesovec, Nov 13 2017
Sum_{n>=0} a(n) * x^n / (n!)^2 = exp(x) * Sum_{n>=0} 4^n * x^n / (n!)^2. - Ilya Gutkovskiy, Jul 17 2020

A248168 Expansion of g.f. 1/sqrt((1-3*x)*(1-11*x)).

Original entry on oeis.org

1, 7, 57, 511, 4849, 47607, 477609, 4862319, 50026977, 518839783, 5414767897, 56795795679, 598213529809, 6322787125207, 67026654455433, 712352213507151, 7587639773475777, 80977812878889927, 865716569022673401, 9269461606674304959, 99387936492243451569, 1066975862517563301303
Offset: 0

Views

Author

Paul D. Hanna, Oct 03 2014

Keywords

Examples

			G.f.: A(x) = 1 + 7*x + 57*x^2 + 511*x^3 + 4849*x^4 + 47607*x^5 +...
where A(x)^2 = 1/((1-3*x)*(1-11*x)):
A(x)^2 = 1 + 14*x + 163*x^2 + 1820*x^3 + 20101*x^4 + 221354*x^5 +...
		

Crossrefs

Programs

  • Magma
    [n le 2 select 7^(n-1) else (7*(2*n-3)*Self(n-1) - 33*(n-2)*Self(n-2))/(n-1) : n in [1..40]]; // G. C. Greubel, May 31 2025
    
  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-3*x)*(1-11*x)], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 03 2014 *)
  • PARI
    {a(n)=polcoeff( 1 / sqrt((1-3*x)*(1-11*x) +x*O(x^n)), n) }
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    {a(n)=polcoeff( (1 + 7*x + 4*x^2 +x*O(x^n))^n, n) }
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    {a(n)=sum(k=0,n, 3^(n-k)*2^k*binomial(n,k)*binomial(2*k,k))}
    for(n=0, 25, print1(a(n), ", "))
    
  • SageMath
    @CachedFunction
    def A248168(n):
         if (n<2): return 7^n
         else: return (7*(2*n-1)*A248168(n-1) - 33*(n-1)*A248168(n-2))//n
    print([A248168(n) for n in range(41)]) # G. C. Greubel, May 31 2025

Formula

a(n) equals the central coefficient in (1 + 7*x + 4*x^2)^n, n>=0.
a(n) = Sum_{k=0..n} 3^(n-k) * 2^k * C(n,k) * C(2*k,k).
a(n) = Sum_{k=0..n} 11^(n-k) * (-2)^k * C(n,k) * C(2*k,k). - Paul D. Hanna, Apr 20 2019
a(n)^2 = A248167(n), which gives the coefficients in 1 / AGM(1-3*11*x, sqrt((1-3^2*x)*(1-11^2*x))).
Equals the binomial transform of 2^n*A026375(n).
Equals the second binomial transform of A084771.
Equals the third binomial transform of A059304(n) = 2^n*(2*n)!/(n!)^2.
a(n) ~ 11^(n+1/2)/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 03 2014
D-finite with recurrence: n*a(n) +7*(-2*n+1)*a(n-1) +33*(n-1)*a(n-2)=0. [Belbachir]
a(n) = (1/4)^n * Sum_{k=0..n} 3^k * 11^(n-k) * binomial(2*k,k) * binomial(2*(n-k),n-k). - Seiichi Manyama, Aug 18 2025
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