A005173 -id:A005173 - cp's OEIS frontend

A003063 a(n) = 3^(n-1) - 2^n.

-1, -1, 1, 11, 49, 179, 601, 1931, 6049, 18659, 57001, 173051, 523249, 1577939, 4750201, 14283371, 42915649, 128878019, 386896201, 1161212891, 3484687249, 10456158899, 31372671001, 94126401611, 282395982049, 847221500579, 2541731610601, 7625329049531, 22876255584049

Offset: 1

Henrik Johansson (Henrik.Johansson(AT)Nexus.SE)

Binomial transform of A000918: (-1, 0, 2, 6, 14, 30, ...). - Gary W. Adamson, Mar 23 2012

This sequence demonstrates 2^n as a loose lower bound for g(n) in Waring's problem. Since 3^n > 2(2^n) for all n > 2, the number 2^(n + 1) - 1 requires 2^n n-th powers for its representation since 3^n is not available for use in the sum: the gulf between the relevant powers of 2 and 3 widens considerably as n gets progressively larger. - Alonso del Arte, Feb 01 2013

			a(3) = 1 because 3^2 - 2^3 = 9 - 8 = 1.
a(4) = 11 because 3^3 - 2^4 = 27 - 16 = 11.
a(5) = 49 because 3^4 - 2^5 = 81 - 32 = 49.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
D. Knuth, Letter to N. J. A. Sloane, date unknown
Index entries for linear recurrences with constant coefficients, signature (5,-6).

Cf. A000918, A056182 (first differences), A064686, A083313, A214091, A369490.
Cf. A363024 (prime terms).
From the third term onward the first differences of A005173.
Difference between two leftmost columns of A090888.
A diagonal in A254027.
Right edge of irregular triangle A252750.

[3^(n-1) -2^n: n in [1..30]]; // G. C. Greubel, Nov 03 2022

Table[3^(n-1) - 2^n, {n, 25}] (* Alonso del Arte, Feb 01 2013 *)
LinearRecurrence[{5,-6},{-1,-1},30] (* Harvey P. Dale, Feb 02 2015 *)

a(n)=3^(n-1)-2^n \\ Charles R Greathouse IV, Oct 07 2015

[3^(n-1) -2^n for n in range(1,31)] # G. C. Greubel, Nov 03 2022

Let b(n) = 2*(3/2)^n - 1. Then a(n) = -b(1-n)*3^(n-1) for n > 0. A083313(n) = A064686(n) = b(n)*2^(n-1) for n > 0. - Michael Somos, Aug 06 2006
From Colin Barker, May 27 2013: (Start)
a(n) = 5*a(n-1) - 6*a(n-2).
G.f.: -x*(1-4*x) / ((1-2*x)*(1-3*x)). (End)
E.g.f.: (1/3)*(2 - 3*exp(2*x) + exp(3*x)). - G. C. Greubel, Nov 03 2022

A few more terms from Alonso del Arte, Feb 01 2013

A094262 Triangle read by rows: T(n,k) is the number of rooted trees with k nodes which are disjoint sets of labels with union {1..n}. If a node has an empty set of labels then it must have at least two children.

Original entry on oeis.org

1, 1, 2, 1, 1, 6, 12, 10, 3, 1, 14, 61, 124, 131, 70, 15, 1, 30, 240, 890, 1830, 2226, 1600, 630, 105, 1, 62, 841, 5060, 16990, 35216, 47062, 40796, 22225, 6930, 945, 1, 126, 2772, 25410, 127953, 401436, 836976, 1196532, 1182195, 795718, 349020, 90090, 10395

Offset: 1

André F. Labossière, Jun 01 2004

The original name for this sequence was "Triangle read by rows giving the coefficients of formulas generating each variety of S2(n,k) (Stirling numbers of 2nd kind). The p-th row (p>=1) contains T(i,p) for i=1 to 2*p-1, where T(i,p) satisfies Sum_{i=1..2*p-1} T(i,p) * C(n-p,i-1)".
The terms of the n-th diagonal sequence of the triangle of Stirling numbers of the second kind A008277, i.e., (Stirling2(N + n - 1,N)), N>=1, are given by a polynomial in N of degree 2*n - 2. This polynomial may be expressed as a linear combination of the falling factorial polynomials binomial(N - n,0), binomial(N - n,1), ... , binomial(N - n,2*n - 2). This table gives the coefficients in these expansions.
The formulas obtained are those for Stirling2(N+1,N) (A000217), Stirling2(N+2,N) (A001296), Stirling2(N+3,N) (A001297), Stirling2(N+4,N) (A001298), Stirling2(N+5,N) (A112494), Stirling2(N+6,N) (A144969) and so on.

			Row 5 contains 1,30,240,890,1830,2226,1600,630,105, so the formula generating Stirling2(n+4,n) numbers (A001298) will be the following: 1 + 30*(n-5) + 240*C(n-5,2) + 890*C(n-5,3) + 1830*C(n-5,4) + 2226*C(n-5,5) + 1600*C(n-5,6) + 630*C(n-5,7) + 105*C(n-5,8). For example, taking n = 9 gives Stirling2(13,9) = 359502.
Triangle starts:
  1;
  1,  2,   1;
  1,  6,  12,  10,    3;
  1, 14,  61, 124,  131,   70,   15;
  1, 30, 240, 890, 1830, 2226, 1600, 630, 105;
  ...
From _Peter Bala_, Jun 14 2016: (Start)
Connection with row polynomials of A134991:
  R(2,z) = (1 + z)^2*z
  R(3,z) = (1 + z)^2*(z + 3*z^2)
  R(4,z) = (1 + z)^4*(z + 10*z^2 + 15*z^3)
  R(5,z) = (1 + z)^5*(z + 25*z^2 + 105*z^3 + 105*z^4). (End)
From _Andrew Howroyd_, Mar 28 2025: (Start)
The T(3,3) = 12 trees up to relabeling have one of the following 3 forms:
     {}         {1}        {1}
    /  \       /   \        |
  {1} {2,3}   {2}  {3}     {2}
                            |
                           {3}
(End)

Andrew Howroyd, Table of n, a(n) for n = 1..2500 (rows 1..50)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions (with Formulas, Graphs and Mathematical Tables), U.S. Dept. of Commerce, National Bureau of Standards, Applied Math. Series 55, 1964, 1046 pages (9th Printing: November 1970) - Combinatorial Analysis, Table 24.4, Stirling Numbers of the Second Kind (author: Francis L. Miksa), p. 835.
J. Fernando Barbero G., Jesús Salas, Eduardo J. S. Villaseñor, Generalized Stirling permutations and forests: Higher-order Eulerian and Ward numbers, Electronic Journal of Combinatorics 22(3) (2015), #P3.37.
M. Kazarian, KP hierarchy for Hodge integrals, p. 2, arxiv:0809.3263 [math.AG], 18 Sep 2008. [From _Tom Copeland_, Jun 12 2015]
F. R. McMorris and T. Zaslavsky, The number of cladistic characters, Math. Biosciences, 54 (1981), 3-10.
F. R. McMorris and T. Zaslavsky, The number of cladistic characters, Math. Biosciences, 54 (1981), 3-10. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Stirling numbers of the 2nd kind.

Row sums are A005172.
Columns 1..4 are A000012, A000918, A005173, A005174, A005175.
Cf. A008277, A000217, A001296, A001297, A001298, A008517, A134991, A112494, A144969.

row_poly := n -> (1+z)^(n+1)*add(z^k*add((-1)^(m+k)*binomial(n+k,n+m)*Stirling2(n+m,m), m=0..k), k=0..n): T_row := n -> seq(coeff(row_poly(n),z,j),j=1..2*n+1):
seq(T_row(n),n=0..6); # Peter Luschny, Jun 15 2016

Clear[T, q, u]; T[0] = q[1];T[n_] := Sum[m*(u^2*q[m] + 2*u*q[m+1] + q[m+2])*D[T[n-1], q[m]], {m, 1, 2*n+1}]; row[n_] := List @@ Expand[T[n-1]] /. {u -> 1, q[] -> 1}; Table[row[n], {n, 1, 7}] // Flatten (* _Jean-François Alcover, Jun 12 2015 *)

T(n)={my(g=serreverse(log(((1+1/y)*x+1)/exp(x + O(x*x^n))))); [Vecrev(p/y) | p<-Vec(serlaplace(g))]}
{ my(A=T(5)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Mar 28 2025

Apparently, a raising operator for bivariate polynomials P(n,u,z) having these coefficients is R = (u+z)^2 * z * d/dz with P(0,u,z) = z. E.g., R P(1,u,z) = R^2 P(0,u,z) = R^2 z = u^4 z + 6 u^3 z^2 + 12 u^2 z^3 + 10 u z^4 + 3 z^5 = P(2,u,z). See the Kazarian link. - Tom Copeland, Jun 12 2015
Reverse polynomials seem to be generated by 1 + exp[t*(x+1+z)^2*(1+z)d/dz]z evaluated at z = 0. - Tom Copeland, Jun 13 2015
From Peter Bala, Jun 14 2016: (Start)
T(n,k) = k*T(n,k) + 2*(k - 1)*T(n,k-1) + (k - 2)*T(n,k-2).
n-th diagonal of A008277: Stirling2(N + n - 1,N) = Sum_{k = 1..2*n - 1} T(n,k)*binomial(N - n,k - 1) for N = 1,2,3,....
Row polynomials R(n,z) = Sum_{k >= 1} k^(n+k-1)*( z/(1 + z)*exp(-z/(1 + z)) )^k/k!, n = 1,2,..., follows from the formula given in A008277 for the o.g.f.'s of the diagonals of the Stirling numbers of the second kind.
Consequently, R(n+1,z) = (1 + z)^2*z*d/dz(R(n,z)) for n >= 1 as conjectured above by Copeland.
R(n,z) = (1 + z)^n*P(n,z) where P(n,z) are the row polynomials of A134991.
R(n,z) = (1 + z)^(2*n+1)*B(n,z/(1 + z)), where B(n,z) are the row polynomials of the triangle of second-order Eulerian numbers A008517 (see Barbero et al., Section 6, equation 27). (End)
Based on the comment of Bala the row polynomials have the explicit form R(n, z) = (1+z)^(n+1)*Sum_{k=0..n}(z^k*Sum_{m=0..k}((-1)^(m+k)*binomial(n+k, n+m)* Stirling2(n+m,m))). - Peter Luschny, Jun 15 2016
E.g.f. G(x,y) satisfies G(x,y) = y*(exp(x)*exp(G(x,y)) - G(x,y) - 1). - Andrew Howroyd, Mar 28 2025

Edited and Name changed by Peter Bala, Jun 16 2016

Name changed by Andrew Howroyd, Mar 28 2025

cp's OEIS Frontend

A003063 a(n) = 3^(n-1) - 2^n.

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