A156925 -id:A156925 - cp's OEIS frontend

A050488 a(n) = 3(2^n-1) - 2n.

0, 1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821, 25769803707

Offset: 0

James Sellers, Dec 26 1999

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).
Number of subgroups of C_(2^n) X C_(2^n) (see A060724).
Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008
Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010
(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012
The partial sums of A033484. - J. M. Bergot, Oct 03 2012
Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012
Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013
Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015
Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, Flattened Stirling Permutations, arXiv:2306.13034 [math.CO], 2023. See p. 14.
Tamas Lengyel, On p-adic properties of the Stirling numbers of the first kind, Journal of Number Theory, 148 (2015) 73-94.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

A050487(2^m-1).
Equals (1/2) A051667.
Cf. A000079, A000225, A000295, A005408, A008292, A033484, A054852, A060724, A066810, A125165, A126277, A142963, A156919, A156920, A156925.

List([0..30],n->3*(2^n-1)-2*n); # Muniru A Asiru, Oct 26 2018

a050488 n = sum $ zipWith (*) a000079_list (reverse $ take n a005408_list)
-- Reinhard Zumkeller, Jul 24 2015

[3*(2^n-1) - 2*n: n in [0..30]]; // G. C. Greubel, Oct 23 2018

seq(coeff(series(x*(x+1)/((1-x)^2*(1-2*x)),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 26 2018

Table[3(2^n-1)-2n,{n,0,30}] (* or *) LinearRecurrence[{4,-5,2}, {0,1,5}, 40] (* Harvey P. Dale, Apr 09 2018 *)

a(n)=3*(2^n-1)-2*n \\ Charles R Greathouse IV, Sep 24 2015

for n in range(0, 30): print(3*(2**n-1) - 2*n, end=', ') # Stefano Spezia, Oct 27 2018

Row sums of A125165: (1, 5, 15, 37, ...). Binomial transform of [1, 4, 6, 6, 6, ...] = [1, 5, 15, 37, ...]. 4th diagonal from the right of A126777 = (1, 5, 15, ...). - Gary W. Adamson, Dec 23 2006
a(n) = 2*a(n-1) + (2n-1). - Gary W. Adamson, Sep 30 2007
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n+1) = A156920(n+1,1).
a(n+1) = A156919(n+1,1)/2^n.
a(n+1) = A142963(n+2,1)/2.
a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.
G.f.: z*(1+z)/((1-z)^2*(1-2*z)).
(End)
a(n) = 2*n + 2*a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n+1) = Sum_{k=0..n} A000079(k) * A005408(n-k), convolution of the powers of 2 with the odd numbers. - Reinhard Zumkeller, Mar 08 2012
E.g.f.: exp(x)*(3*exp(x) - 2*x - 3). - Stefano Spezia, May 15 2023

A156920 Triangle of the normalized A142963 and A156919 sequences.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 15, 18, 1, 1, 37, 129, 58, 1, 1, 83, 646, 877, 179, 1, 1, 177, 2685, 8030, 5280, 543, 1, 1, 367, 10002, 56285, 82610, 29658, 1636, 1, 1, 749, 34777, 335162, 919615, 756218, 159742, 4916, 1

Offset: 0

Johannes W. Meijer, Feb 20 2009

The originator sequences are A142963 and A156919.
The Flower Triangle seems to be an appropriate name for the triangular array of this sequence. The zero patterns of the Flower Polynomials of the first, see A156921, the second, see A156925, the third, see A156927, and the fourth kind, see A156933, look like flowers.
The first Maple program generates the Flower Triangle sequence.
The second program generates the Right Hand Columns sequences and the third one generates the Left Hand Column sequences. For an explanation of these two algorithms see A142963.

			The first few rows of the triangle are:
  [1]
  [1, 1]
  [1, 5, 1]
  [1, 15, 18, 1]
  [1, 37, 129, 58, 1]
  [1, 83, 646, 877, 179, 1]

Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104v2 [math.CO]. - From N. J. A. Sloane, Aug 21 2012

Originator sequences A142963, A156919.
Related sequences A156921, A156925, A156927, A156933.
Left hand column sequences A050488, A142965, A142966, A142968.
Right hand column sequences A000340, A156922, A156923, A156924.
Row sums A014307(n+1).

A156920 := proc(n,m): if n=m then 1; elif m=0 then 1 ; elif m<0 or m>n then 0; else (m+1)*procname(n-1, m)+(2*n-2*m+1)*procname(n-1, m-1) ; end if; end proc: seq(seq(A156920(n, m), m=0..n), n=0..8);
RHCnr:=5; RHCmax:=10; RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1),p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx,x,k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr);
LHCnr:=5; LHCmax:=10: LHCend:=LHCnr+LHCmax: for k from LHCnr to LHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!)*x^p/(1-4*x)^(p+1),p=0..k)/ (((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); for n from 0 to nmax do d[n]:= coeff(fx,x,n)/2^n end do: LHC[n]:=d[LHCnr-1] end do: a:=n-> LHC[n]: seq(a(n), n=LHCnr..LHCend-1);

T[, 0] = 1; T[n, n_] = 1; T[n_, m_] := T[n, m] = (m + 1)*T[n - 1, m] + (2*n - 2*m + 1)*T[n - 1, m - 1];
Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017 *)

T(n,m) = (m+1)*T(n-1,m) + (2*n-2*m+1)*T(n-1,m-1) with T(n,m=0) = 1 and T(n,n) = 1, n>=0 and 0 <= m <= n.
From Peter Bala, Jul 22 2012: (Start)
T(n,k) = 1/(2^(n-k))*A156919(n,k).
E.g.f.: 1 + t*x + (t+t^2)*x^2/2! + (t+5*t^2+t^3)*x^3/3! + ... = sqrt(E(x,2*t)), where E(x,t) = (1-t)*exp(x*t)/(exp(x*t)-t*exp(x)) = 1 + t*x + (t+t^2)*x^2/2! + (t+4*t^2+t^3)*x^3/3! + ... is the e.g.f. for the Eulerian numbers A008292.
The row polynomials R(n,x) satisfy 1/sqrt(1-2*x)*(x*d/dx)^n(1/sqrt(1-2*x)) = R(n,x)/(1-2*x)^(n+1). (End)

Minor edits by Johannes W. Meijer, Sep 28 2011

A142963 Triangle read by rows, coefficients of the polynomials P(k, x) = (1/2) Sum_{p=0..k-1} Stirling2(k, p+1)x^p(1-4x)^(k-1-p)(2*p+2)!/(p+1)!.

Original entry on oeis.org

1, 1, 2, 1, 10, 4, 1, 30, 72, 8, 1, 74, 516, 464, 16, 1, 166, 2584, 7016, 2864, 32, 1, 354, 10740, 64240, 84480, 17376, 64, 1, 734, 40008, 450280, 1321760, 949056, 104704, 128, 1, 1498, 139108, 2681296, 14713840, 24198976, 10223488, 629248, 256, 1, 3030, 462264, 14341992

Offset: 1

Wolfdieter Lang, Sep 15 2008

Previous name: Table of coefficients of row polynomials of certain o.g.f.s.
The o.g.f.s G(k, x) for the k-family of sequences S(k, n):= Sum_{p=0..n} p^k*binomial(2*p, p)*binomial(2*(n-p), n-p), k=0,1,... (convolution of two sequences involving the central binomial coefficients) are 1/(1-4*x) for k=0 and 2*x*P(k, x)/(1-4*x)^(k+1) for k=1,2,..., with the row polynomials P(k, x) = Sum_{m=0..k-1} a(n,m)*x^m).
The author was led to compute the sums S(k, n) by a question asked by M. Greiter, Jun 27 2008.
In order to keep the index k>=1 of Sigma(k, n) also for the polynomials P(k, x), their degree is then k-1.

			Triangle starts:
[1]
[1,   2]
[1,  10,     4]
[1,  30,    72,      8]
[1,  74,   516,    464,      16]
[1, 166,  2584,   7016,    2864,     32]
[1, 354, 10740,  64240,   84480,  17376,     64]
[1, 734, 40008, 450280, 1321760, 949056, 104704, 128]
...
P(3,x) = 1+10*x+4*x^2.
G(3,x) = 2*x*(1+10*x+4*x^2)/(1-4*x)^4.

Vincenzo Librandi, Table of n, a(n) for n = 1..210
Wolfdieter Lang, First 10 rows and more.
L. Liu, Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207v5 [math.CO], 2005-2006.

Left hand column sequences 2*A142964, 4*A142965, 8*A142966, 16*A142968.
Row sums A142967.
From Johannes W. Meijer, Feb 20 2009: (Start)
A156919 and this sequence can be mapped onto A156920.
Cf. A156921, A156925, A156927, A156933.
Right hand column sequences 2^n*A000340, 2^n*A156922, 2^n*A156923, 2^n*A156924. (End)
Cf. A142961, A142962.

A142963 := proc(n,m): if n=m+1 then 2^(n-1); elif m=0 then 1 ; elif m<0 or m>n-1 then 0; else (m+1)*procname(n-1, m)+(4*n-4*m-2)*procname(n-1, m-1); end if; end proc: seq(seq(A142963(n,m), m=0..n-1), n=1..9); # Johannes W. Meijer, Sep 28 2011
# Alternatively (assumes offset 0):
p := (n,x) -> (1/2)*add(Stirling2(n+1,k+1)*x^k*(1-4*x)^(n-k)*(2*k+2)!/(k+1)!, k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od;
# Peter Luschny, Jun 18 2017

t[, 0] = 1; t[n, m_] /; m == n-1 := 2^m; t[n_, m_] := (m+1)*t[n-1, m] + (4*n-4*m-2)*t[n-1, m-1]; Table[t[n, m], {n, 1, 10}, {m, 0, n-1}] // Flatten (* Jean-François Alcover, Jun 21 2013, after Johannes W. Meijer *)

G(k, x) = Sum_{p=0..k} S2(k, p)*((2*p)!/p!)*x^p/(1-4*x)^(p+1), k >= 0 (here k >= 1), with the Stirling2 triangle S2(k, p):=A048993(k, p). (Proof from the product of the o.g.f.s of the two convoluted sequences and the normal ordering (x^d_x)^k = Sum_{p=0..k} S2(k, p)*x^p*d_x^p, with the derivative operator d_x.)
a(k,m) = [x^m]P(k, x) = [x^m] ((1-4*x)^(k+1))*G(k,x)/(2*x), k>=1, m=0,1,...,k-1.
For the triangle coefficients the following relation holds: T(n,m) = (m+1)*T(n-1,m) + (4*n-4*m-2)*T(n-1,m-1) with T(n,m=0) = 1 and T(n,m=n-1) = 2^(n-1), n >= 1 and 0 <= m <= n-1. - Johannes W. Meijer, Feb 20 2009
From Peter Bala, Jan 18 2018: (Start)
(x*d/dx)^n (1/(sqrt(1 - 4*x)) = 2*x*P(n,x)/sqrt(1 - 4*x)^(n+1/2) for n >= 1.
x*P(n,x)/(1 - 4*x)^(n+1/2) = (1/2)*Sum_{k >= 1} binomial(2*k,k)* k^n*x^k for n >= 1.
P(n+1,x) = ((4*n - 2)*x + 1)*P(n,x) - x*(4*x - 1)*d/dx(P(n,x)).
Hence the polynomial P(n,x) has all real zeros by Liu et al., Theorem 1.1, Corollary 1.2. (End)

Minor edits by Johannes W. Meijer, Sep 28 2011
A more precise name by Peter Luschny, Jun 18 2017
Name reformulated with offset corrected, edited by Wolfdieter Lang, Aug 23 2019

A156921 FP1 polynomials related to the generating functions of the right hand columns of the A156920 triangle.

Original entry on oeis.org

1, 1, 1, 1, -6, 1, 7, -79, 119, 126, -270, 1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400, 1, 86, -2255, 5784, 300930, -3904584, 20663714, -41517272, -80232259, 657717054

Offset: 0

Johannes W. Meijer, Feb 20 2009

The FP1 polynomials appear in the numerators of the GF1 o.g.f.s. of the right hand columns of A156920. The FP1 can be calculated with the formula for the RHC sequence, see A156920, and the formula for the general structure of the generating function GF1, see below.
An appropriate name for the FP1 polynomials seems to be the flower polynomials of the first kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2, see A156925, and the FP1 resemble each other closely.
A Maple program that generates for a right hand column with a certain RHCnr its GF1 and FP1 can be found below. RHCnr stands for right hand column number and starts from 1.

			The first few rows of the "triangle" of the coefficients of the FP1 polynomials.
In the columns the coefficients of the powers of z^m, m=0,1,2,... , appear.
  [1]
  [1]
  [1, 1, -6]
  [1, 7, -79, 119, 126, -270]
  [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
Matrix of the coefficients of the FP1 polynomials. The coefficients in the columns of this matrix are the powers of z^m, m=0,1,2,.. .
  [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
  [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
  [1, 1, -6, 0 ,0, 0, 0, 0, 0, 0]
  [1, 7, -79, 119, 126, -270, 0, 0, 0, 0]
  [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
The first few FP1 polynomials are:
  FP1(z; RHCnr=1) = 1
  FP1(z; RHCnr=2) = 1
  FP1(z; RHCnr =3) = 1+z-6*z^2
Some GF1(z;RHCnr) are:
  GF1(z;RHCnr= 3) = (1+z-6*z^2)/((1-5*z)*(1-3*z)^2*(1-z)^3)
  GF1(z;RHCnr= 4) = (1+7*z-79*z^2+119*z^3+126*z^4-270*z^5)/((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)

Cf. A156920, A156925, A156927, A156933.
For the first few GF1's see A000340, A156922, A156923, A156924.
The number of FP1 terms follow the triangular numbers A000217, with quite surprisingly one exception here a(0)=1.
Abs(Row sums (n)) = A098695(n).
For the polynomials in the denominators of the GF1(z;RHCnr) see A157702.

RHCnr:=4: if RHCnr=1 then RHCmax :=1; else RHCmax:=(RHCnr-1)*(RHCnr)/2 end if: RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1),p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx,x,k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr+1); for nx from 0 to RHCmax do num:=sort(sum(A[t]*z^t, t=0..RHCmax)); nom:=Product((1-(2*u-1)*z)^(RHCnr-u+1),u=1..RHCnr): RHCa:= series(num/nom,z,nx+1); y:=coeff(RHCa,z,nx)-A[nx]; x:=RHC[nx+1]; A[nx]:=x-y; end do: FP1[RHCnr]:=sort(num,z, ascending); GenFun[RHCnr] :=FP1[RHCnr]/product((1-(2*m-1)*z)^(RHCnr-m+1),m=1..RHCnr);

G.f.: GF1(z;RHCnr) := FP1(z;RHCnr)/product((1-(2*m-1)*z)^(RHCnr+1-m),m=1..RHCnr)

Row sums (n) = (-1)^(1+(n+1)*(n+2)/2)*A098695(n).

A156927 FP3 polynomials related to the generating functions of the columns of the A156921 matrix.

Original entry on oeis.org

1, 1, 1, -6, 29, 31, -283, 245, 298, -286, -108, 119, -3106, 29469, -104585, -220481, 3601363, -15487305, 34949165, -39821950, 4356011, 46881744, -51274736, 9005908, 14663472, -5205168, -1456704, -20736

Offset: 0

Johannes W. Meijer, Feb 20 2009

For the matrix of the FP1 polynomials see A156921. The coefficients in the columns of this matrix are the powers of z^m, m=0, 1, 2, ... . The columns are numbered 1, 2, 3... .
The GF3(z;m) generate the sequences of the z^m coefficients. The general structure of the GF3(z;m) is given below.
The FP3(z,m) in the numerator of the GF3(z;m) is a polynomial of a certain degree, let's say k3. The (k3+1) coefficients of this polynomial can be determined one by one by comparing the series expansion of the FP3(z,m) with the values of the powers of z^m in column (m+1). These values can be generated with the GF1 formulas, see A156921.
An appropriate name for the polynomials FP3(z;m) in the numerators of the GF(3;m) seems to be the flower polynomials of the third kind, the FP3, because the zero patterns of these polynomials look like flowers. The zero patterns of the FP3 and the FP4, see A156933, resemble each other closely and look like the zero patterns of the FP1 and FP2.
The sequence of the (k3+1) number of terms of the FP3(z;m) polynomials for m from 0 to 11 is 1, 2, 8, 17, 29, 45, 63, 84, 109, 137, 167, 200.

			The first few rows of the "triangle" of the FP3(z,m) coefficients are:
  [1]
  [1, 1]
  [-6, 29, 31, -283, 245, 298, -286, -108]
The first few FP3 polynomials are:
  FP3(z; m=0) = 1
  FP3(z; m=1) = (1+z)
  FP3(z; m=2) = (-6+29*z+31*z^2-283*z^3+245*z^4+298*z^5-286*z^6-108*z^7)
Some GF3(z;m) are:
  GF3(z;m=1) = z^2*(1+z)/((1-z)^4*(1-2*z))
  GF3(z;m=2) = z^2*(-6+29*z+31*z^2-283*z^3+245*z^4+298*z^5-286*z^6-108*z^7)/((1-z)^7*(1-2*z)^4*(1-3*z))

Cf. A156920, A156921, A156925, A156933.
For the first few GF3's see A156928, A156929, A156930, A156931.
Row sums A156932.
For the polynomials in the denominators of the GF3(z;m) see A157704.

G.f.: GF3(z;m):= z^p*FP3(z;m)/Product_{k=0..m} (1-(k+1)*z)^(1+3*k).

A142965 One fourth of third column (m=2) of triangle A142963.

Original entry on oeis.org

1, 18, 129, 646, 2685, 10002, 34777, 115566, 372453, 1175290, 3654369, 11245110, 34349005, 104373282, 315969705, 954002878, 2874983541, 8652474378, 26015617585, 78169534470, 234766551261, 704840716978, 2115654610809, 6349329417486, 19052920751365, 57169029907482

Offset: 0

Wolfdieter Lang, Sep 15 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10, -40, 82, -91, 52, -12).

Column m=1: 2*A142964; m=3: 8*A142966.
From Johannes W. Meijer, Feb 20 2009: (Start)
Cf. A156925.
Equals A156920(n+2,2).
Equals A156919(n+2,2)/2^n.
(End)

[35/2+2*n^2+12*n-84*2^n-24*2^n*n+135/2*3^n: n in [0..25]]; // Vincenzo Librandi, Jun 18 2017

LinearRecurrence[{10,-40,82,-91,52,-12}, {1,18,129,646,2685,10002},30] (* or *) CoefficientList[Series[(1+8x-11x^2-6x^3)/((x-1)^3 (2x-1)^2 (3x-1)),{x,0,30}],x] (* Harvey P. Dale, Apr 24 2011 *)

a(n) = A142963(n+3,2)/4.
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n) = 10a(n-1) - 40a(n-2) + 82a(n-3) - 91a(n-4) + 52a(n-5) - 12a(n-6).
a(n) = 35/2 + 2*n^2 + 12*n - 84*2^n - 24*2^n*n + 135/2*3^n
G.f.: (1 + 8*z - 11*z^2 - 6*z^3)/((1-z)^3*(1-2*z)^2*(1-3*z)).
(End)

A142966 Fourth column (m=3) of triangle A142963 divided by 8.

Original entry on oeis.org

1, 58, 877, 8030, 56285, 335162, 1792749, 8904486, 41949645, 190129090, 837258109, 3607669966, 15289404989, 63975698570, 265065915725, 1089837752118, 4454225465325, 18119738464530, 73441531708765, 296814738679390, 1196884383319261, 4817845684107098

Offset: 0

Wolfdieter Lang, Sep 15 2008

Cf. A142963.
Column m=2: 4*A142965.
From Johannes W. Meijer, Feb 20 2009: (Start)
Cf. A156925.
Equals A156920(n+3,3).
Equals A156919(n+3,3)/2^n.
(End)

a(n) = A142963(n+4,3)/8.
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n) = 20a(n-1) - 175*a(n-2) + 882*a(n-3) - 2835*a(n-4) + 6072*a(n-5) - 8777*a(n-6) + 8458*a(n-7) - 5204*a(n-8) + 1848*a(n-9) - 288*a(n-10).
a(n) = -(105/2) - (143/3)*n - 14*n^2 - (4/3)*n^3 + 756*2^n + 48*2^n*n^2 + 384*2^n*n - (3645/2)*3^n - 405*3^n*n + 1120*4^n.
G.f.: (1 + 38*z - 108*z^2 - 242*z^3 + 839*z^4 - 444*z^5 - 180*z^6)/((1-z)^4*(1-2*z)^3*(1-3*z)^2*(1-4*z)).
(End)

A142968 Fifth column (m=4) of triangle A142963 divided by 16=2^4.

Original entry on oeis.org

1, 179, 5280, 82610, 919615, 8284857, 64730022, 457217400, 2999230965, 18608607535, 110625457964, 636103699038, 3562753619915, 19541111960965, 105392471360850, 560747327119908, 2950726075955265, 15387821226034875, 79656442803398680, 409857988825489610

Offset: 0

Wolfdieter Lang, Sep 15 2008

Column m=3: 8*A142966.
From Johannes W. Meijer, Feb 20 2009: (Start)
Cf. A156925.
Equals A156920(n+4,4).
Equals A156919(n+4,4)/2^n.
(End)

a(n) = A142963(n+5,3)/2^4.
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n) = 35a(n-1) - 560a(n-2) + 5432a(n-3) - 35714a(n-4) + 168542a(n-5) - 589632a(n-6) + 1556776a(n-7) - 3126949a(n-8) + 4777591a(n-9) - 5506936a(n-10) + 4703032a(n-11) - 2881136a(n-12) + 1195632a(n-13) - 300672a(n-14) + 34560a(n-15).
a(n) = (1155/8) + (472/3)*n - 5544*2^n + (120285/4)*3^n - 49280*4^n + (196875/8)*5^n - 64*2^n*n^3 - 864*2^n*n^2 - 3824*2^n*n + (187/3)*n^2 + 1215*3^n*n^2 + 12150*3^n*n - 8960*4^n*n + (32/3)*n^3 + (2/3)*n^4.
G.f.: (1 + 144*z - 425*z^2 - 7382*z^3 + 48451*z^4 - 96764*z^5 - 2559*z^6 + 257002*z^7 - 312444*z^8 + 88344*z^9 + 30240*z^10)/((1-z)^5*(1-2*z)^4*(1-3*z)^3*(1-4*z)^2*(1-5*z)).
(End)

A142964 a(n) = 62^n - 2n - 5.

Original entry on oeis.org

1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821

Offset: 0

Wolfdieter Lang, Sep 15 2008

Previous name was: One half of second column (m=1) of triangle A142963.

Essentially a duplicate of A050488. - Johannes W. Meijer, Feb 20 2009

			a(3) = 6*2^3 - 2*3 - 5 = 37.

Eric Billault, Walter Damin, Robert Ferréol, and Rodolphe Garin, MPSI Classes Prépas - Khôlles de Maths, Exercices corrigés, Ellipses, 2012, exercice 2.22 (1) pp 26, 43-44.

Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A142965 (m=2 column/4).
Equals A050488(n+1).
Equals A156920(n+1,1).
Equals A156919(n+1,1)/2^n.
Cf. A156925, A339771.
Partial sums of A033484.

seq(6*2^n-2*n-5,n=0..40); # Bernard Schott, Dec 16 2020

a[n_]:=6*2^n-2n-5;Array[a,32,0] (* or *) CoefficientList[Series[(1+x)/((1-x)^2*(1-2*x)),{x,0,31}],x] (* or *) LinearRecurrence[{4,-5,2},{1,5,15},32] (* James C. McMahon, Aug 12 2025 *)

Vec((1+z)/((1-z)^2*(1-2*z)) + O(z^50)) \\ Michel Marcus, Jun 18 2017

a(n) = A142693(n+2,1)/2.
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n > 2 with a(0) = 1, a(1) = 5, a(2) = 15.
G.f.: (1+z)/((1-z)^2*(1-2*z)). (End)
a(n) = Sum_{i=0..n} Sum_{j=0..n} 2^min(i,j) (Billault et al) (compare with A339771 that has max instead of min). - Bernard Schott, Dec 16 2020
a(n) = 2*A066524(n+1) - A339771(n). - Kevin Ryde, Dec 17 2020
E.g.f.: 6*exp(2*x) - exp(x)*(5 + 2*x). - Stefano Spezia, Dec 17 2020

New name using a formula of Bernard Schott by Peter Luschny, Dec 17 2020

cp's OEIS Frontend

A050488 a(n) = 3*(2^n-1) - 2*n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Python

Formula

A156920 Triangle of the normalized A142963 and A156919 sequences.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A142963 Triangle read by rows, coefficients of the polynomials P(k, x) = (1/2) Sum_{p=0..k-1} Stirling2(k, p+1)*x^p*(1-4*x)^(k-1-p)*(2*p+2)!/(p+1)!.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A156921 FP1 polynomials related to the generating functions of the right hand columns of the A156920 triangle.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A156927 FP3 polynomials related to the generating functions of the columns of the A156921 matrix.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A142965 One fourth of third column (m=2) of triangle A142963.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A142966 Fourth column (m=3) of triangle A142963 divided by 8.

Views

Author

Keywords

Crossrefs

Formula

A142968 Fifth column (m=4) of triangle A142963 divided by 16=2^4.

Views

Author

A050488 a(n) = 3(2^n-1) - 2n.

A142963 Triangle read by rows, coefficients of the polynomials P(k, x) = (1/2) Sum_{p=0..k-1} Stirling2(k, p+1)x^p(1-4x)^(k-1-p)(2*p+2)!/(p+1)!.

A142964 a(n) = 62^n - 2n - 5.