A140071 -id:A140071 - cp's OEIS frontend

A007070 a(n) = 4a(n-1) - 2a(n-2) with a(0) = 1, a(1) = 4.

1, 4, 14, 48, 164, 560, 1912, 6528, 22288, 76096, 259808, 887040, 3028544, 10340096, 35303296, 120532992, 411525376, 1405035520, 4797091328, 16378294272, 55918994432, 190919389184, 651839567872, 2225519493120, 7598398836736, 25942556360704, 88573427769344, 302408598355968

Offset: 0

N. J. A. Sloane, Mira Bernstein, Simon Plouffe

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."
It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002
Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004
a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006
The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006
Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008
Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009
For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Binomial transform of A078057. - R. J. Mathar, Mar 28 2011
Pisano period lengths:  1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012
a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013
From Wolfdieter Lang, Oct 01 2013: (Start)
a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.
The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and  B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)
The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015
From Tom Copeland, Dec 04 2015: (Start)
An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) =  1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.
A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)
a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

			a(3) = 48 = 3 * 4 + 4 + 1 + 1 = 3*a(2) + a(1) + a(0) + 1.
Example for the octagon rho(8) powers: rho(8)^4  = 2 + sqrt(2) = -2*1 + 4*rho(8)^2  = A(5)*1 + A(4)*rho(8) + B(5)*rho(8)^2 + B(4)*rho(8)^3, with a(5) = A106731(1) = -2, B(5) = a(1) = 4, A(4) = 0, B(4) = 0. - _Wolfdieter Lang_, Oct 01 2013

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

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
C. Bautista-Ramos and C. Guillen-Galvan, Fibonacci numbers of generalized Zykov sums, J. Integer Seq., 15 (2012), Article 12.7.8.
A. Bernini, F. Disanto, R. Pinzani, and S. Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007) # 07.9.7.
A. Burstein, S. Kitaev, and T. Mansour, Independent sets in certain classes of (almost) regular graphs, arXiv:math/0310379 [math.CO], 2003.
Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607. (See Cor. 3.7(e).)
Tomislav Doslic and I. Zubac, Counting maximal matchings in linear polymers, Ars Mathematica Contemporanea 11 (2016) 255-276.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
L. Escobar, P. Gallardo, J. González-Anaya, J. L. González, G. Montúfar, and A. H. Morales, Enumeration of max-pooling responses with generalized permutohedra, arXiv:2209.14978 [math.CO], 2022. (See Ex. 5.1)
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
Pamela Fleischmann, Jonas Höfer, Annika Huch, and Dirk Nowotka, alpha-beta-Factorization and the Binary Case of Simon's Congruence, arXiv:2306.14192 [math.CO], 2023.
A. S. Fraenkel and C. Kimberling, Generalized Wythoff arrays, shuffles and interspersions, Discr. Math. 126 (1-3) (1994) 137-149.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Sela Fried, Toufik Mansour, and Mark Shattuck, Counting k-ary words by number of adjacency differences of a prescribed size, arXiv:2504.03013 [math.CO], 2025. See p. 7.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 440
László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222.
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222. [Annotated scanned copy]
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integ. Seqs. Vol. 9 (2006), #06.1.1.
Index entries for sequences related to poker
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-2).

Row sums of A059474. - David W. Wilson, Aug 14 2006
Cf. A000129, A000225, A002307, A002315, A006012 (same recurrence), A007052, A075271, A078057, A106731, A179260, A204089, A228405.
Equals 2 * A003480, n>0.
Row sums of A140071.
Cf. A127672, A265185, A133314.

a007070 n = a007070_list !! n
a007070_list = 1 : 4 : (map (* 2) $ zipWith (-)
   (tail $ map (* 2) a007070_list) a007070_list)
-- Reinhard Zumkeller, Jan 16 2012

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-8); S:=[ ((4+r)^(1+n)-(4-r)^(1+n))/((2^(1+n))*r): n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Vincenzo Librandi, Mar 27 2011

[n le 2 select 3*n-2 else 4*Self(n-1)-2*Self(n-2): n in [1..23]];  // Bruno Berselli, Mar 28 2011

A007070 :=proc(n) option remember; if n=0 then 1 elif n=1 then 4 else 4*procname(n-1)-2*procname(n-2); fi; end:
seq(A007070(n), n=0..30); # Wesley Ivan Hurt, Dec 06 2015

LinearRecurrence[{4,-2}, {1,4}, 30] (* Harvey P. Dale, Sep 16 2014 *)

a(n)=polcoeff(1/(1-4*x+2*x^2)+x*O(x^n),n)

a(n)=if(n<1,1,ceil((2+sqrt(2))*a(n-1)))

[lucas_number1(n,4,2) for n in range(1, 24)]# Zerinvary Lajos, Apr 22 2009

G.f.: 1/(1 - 4*x + 2*x^2).
Preceded by 0, this is the binomial transform of the Pell numbers A000129. Its e.g.f. is then exp(2*x)*sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = ((2+sqrt(2))^(n+1) - (2-sqrt(2))^(n+1))/sqrt(8). - Al Hakanson (hawkuu(AT)gmail.com), Dec 27 2008, corrected Mar 28 2011
a(n) = (2 - sqrt(2))^n*(1/2 - sqrt(2)/2) + (2 + sqrt(2))^n*(1/2 + sqrt(2)/2). - Paul Barry, May 09 2003
a(n) = ceiling((2 + sqrt(2))*a(n-1)). - Benoit Cloitre, Aug 15 2003
a(n) = U(n, sqrt(2))*sqrt(2)^n. - Paul Barry, Nov 19 2003
a(n) = (1/4)*Sum_{r=1..7} sin(r*Pi/8)*sin(r*Pi/2)*(2*cos(r*Pi/8))^(2*n+3). - Herbert Kociemba, Jun 11 2004
a(n) = center term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 1 2 1 / 1 1 1]. M^n * [1 1 1] = [A007052(n) a(n) A007052(n)]. E.g., a(3) = 48 since M^3 * [1 1 1] = [34 48 34], where 34 = A007052(3). - Gary W. Adamson, Dec 18 2004
This is the binomial mean transform of A002307. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(2n) = Sum_{r=0..n} 2^(2n-1-r)*(4*binomial(2n-1,2r) + 3*binomial(2n-1,2r+1)) a(2n-1) = Sum_{r=0..n} 2^(2n-2-r)*(4*binomial(2n-2,2r) + 3*binomial(2n-2,2r+1)). - Jeffrey Liese, Oct 12 2006
a(n) = 3*a(n - 1) + a(n - 2) + a(n - 3) + ... + a(0) + 1. - Gary W. Adamson, Feb 18 2011
G.f.: 1/(1 - 4*x + 2*x^2) = 1/( x*(1 + U(0)) ) - 1/x  where U(k)= 1 - 2^k/(1 - x/(x - 2^k/U(k+1) )); (continued fraction 3rd kind, 3-step). - Sergei N. Gladkovskii, Dec 05 2012
G.f.: A(x) = G(0)/(1-2*x) where G(k) = 1 + 2*x/(1 - 2*x - x*(1-2*x)/(x + (1-2*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 04 2013
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - (1-x)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n-1) = Sum_{k=0..n} binomial(2*n, n+k)*(k|8) where (k|8) is the Kronecker symbol. - Greg Dresden, Oct 11 2022
E.g.f.: exp(2*x)*(cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, May 20 2024

A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 4, 4, 8, 1, 4, 12, 8, 16, 1, 6, 12, 32, 16, 32, 1, 6, 24, 32, 80, 32, 64, 1, 8, 24, 80, 80, 192, 64, 128, 1, 8, 40, 80, 240, 192, 448, 128, 256, 1, 10, 40, 160, 240, 672, 448, 1024, 256, 512, 1, 10, 60, 160, 560, 672, 1792, 1024, 2304, 512, 1024, 1, 12, 60, 280, 560, 1792, 1792, 4608, 2304, 5120, 1024, 2048

Offset: 0

Clark Kimberling, Mar 05 2009

Conjecture 1. If n>1 is even then F(n,x) has no real roots.
Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r,
and if n>4 then 0 < -r < n.
Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010]
Cayley (1876) states "We, in fact, find  1 + sin u = 1 + x,  1 - sin 3u = (1 + x)(1 - 2x)^2,  1 + sin 5u = (1 + x)(1 + 2x - 4x^2)^2,  1 - sin 7u = (1 + x)(1 - 4x - 4x^2 + 8x^3)^2, &c.". - Michael Somos, Jun 19 2012
Appears to be the unsigned row reverse of A180870 and A228565. - Peter Bala, Feb 17 2014
From Wolfdieter Lang, Jul 29 2014: (Start)
This triangle is the Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). For Riordan triangles see the W. Lang link 'Sheffer a-and z-sequences' under A006232, also for references. The o.g.f. given by Peter Bala in the formula section refers to the row reversed triangle. The usual information on this triangle, like o.g.f. for the columns, the row sums, the alternating row sums, the recurrences using A- and Z-sequences, etc. follows from this Riordan property. The Riordan proof follows from the given o.g.f. by Peter Bala, call it Grev(x,z), by row reversion: G(x,z) = Grev(1/x,x*z) = (1+z)/(1- 2*x*z - z^2) = G(z)*(1/(1 - x*F(z))) with G(z) = (1+z)/(1-z^2) and F(z) = z*2/(1-z^2). See A244419 for the discussion for a signed version of this triangle.
(End)

			Rows 0 to 8:
1
1 2
1 2 4
1 4 4 8
1 4 12 8 16
1 6 12 32 16 32
1 6 24 32 80 32 64
1 8 24 80 80 192 64 128
1 8 40 80 240 192 448 128 256
(Row 8) = (1, 4*2, 10*4, 10*8, 15*16, 6*32, 7*64, 1*128, 1*256).
First few polynomials:
F(0,x)=1, F(1,x)=x+2, F(2,x)=x^2+2*x+4, F(3,x)=x^3+4*x^2+4*x+8.
The row polynomials R(n,x) start: 1, 1 + 2*x = x*F(1,1/x), 1 + 2*x + 4^x^2 = x^2*F(2,1/x), ...  - _Wolfdieter Lang_, Jul 29 2014

A. Cayley, On an Expression for 1 +- sin(2p+1)u in Terms of sin u, Messenger of Mathematics, 5 (1876), pp. 7-8 = Mathematical Papers Vol. 10, n. 630, pp. 1-2.

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A140070, A140071, A180870, A228565, A078057, A123335, A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980.

T[n_, 0]:= 1; T[n_, n_]:= 2^n; T[n_, k_]:= T[n, k] = T[n-1, k] + (1 + (-1)^(n-k))*T[n-1, k-1]; Table[T[n, k], {n, 0, 15}, {k, 0, n}] (* G. C. Greubel, Sep 24 2018 *)

t(n,k) = if(k==0, 1, if(k==n, 2^n, t(n-1,k) + (1+(-1)^(n-k))*t(n-1,k-1)));
for(n=0,15, for(k=0,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Sep 24 2018

Count the top row as row 0 and let C(n,k) denote the usual binomial
coefficient. For row 2n, define p(0)=C(n,0), p(1)=C(n,1), p(2)=C(n+1,2),
p(3)=C(n+1,3), p(4)=C(n+2,4), p(5)=c(n+2,5),..., until reaching two final
1's: p(2n-1)=C(2n-1,2n-1) and p(2n)=C(2n,2n). Then the k-th number in row
2n is p(k)*2^k. For row 2n+1, define q(0)=C(n,0), q(1)=C(n+1,1),
q(2)=C(n+1,2), q(3)=C(n+2,3),..., until reaching q(2n+1)=C(2n+1,2n+1).
Then the k-th number in row 2n+1 is q(k)*2^k.
From Peter Bala, Jan 17 2014: (Start)
Working with an offset of 0, the o.g.f. is (1 + x*z)/(1 - 2*z - x^2*z^2) = 1 + (x + 2)*z + (x^2 + 2*x + 4)*z^2 + ....
Recurrence equation: T(n,k) = 2*T(n-1,k-1) + T(n-2,k-2) with T(n,0) = 1.
The polynomials G(n,x) defined by G(0,x) = 1 and G(n,x) = x*F(n-1,x) for n >= 1 satisfy G(n,x) = (x + 1)*G(n-1,x) - G(n-1,-x). Cf. A140070 and A140071. (End)
From Wolfdieter Lang, Jul 29 2014: (Start)
O.g.f. for the row polynomials (rising powers of x) R(n,x) = x^n*F(n,1/x): (1+z)/(1 - 2*x*z - z^2). Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). See a comment above.
Recurrence for the row polynomials R(n,x) = (1+x)*R(n-1,x) - (-1)^n*x*R(n-1,-x), n >= 1, R(0,x) = 1.
R(n,x) = Ftilde(n,2*x) + Ftilde(n-1,2*x) with the monic Fibonacci polynomials Ftilde(n,x) given in A168561.
Recurrence for the triangle: R(n,m) = R(n-1,m) + (1 + (-1)^(n-m))*R(n-1,m-1), n >= m >= 1, R(n,m) = 0 if n < m, R(n,0) = 1.
O.g.f. column sequences ((1+x)/(1-x^2))*(2*x/(1-x^2))^m, m >= 0. See A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980, ...
Row sums A078057. Alternating row sums A123335.
(End)

Offset corrected to 0. Cf.s added, keyword easy added by Wolfdieter Lang, Jul 29 2014

A140070 Triangle read by rows, iterates of matrix X * [1,0,0,0,...], where X = an infinite lower bidiagonal matrix with [1,3,1,3,1,3,...] in the main diagonal and [1,1,1,...] in the subdiagonal.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 13, 5, 1, 1, 40, 18, 8, 1, 1, 121, 58, 42, 9, 1, 1, 364, 179, 184, 51, 12, 1, 1, 1093, 543, 731, 235, 87, 13, 1, 1, 3280, 1636, 2736, 966, 496, 100, 16, 1, 1, 9841, 4916, 9844, 3702, 2454, 596, 148, 17, 1, 1, 29524, 14757, 34448, 13546, 11064, 3050, 1040, 165, 20, 1

Offset: 1

Gary W. Adamson, May 04 2008

Row sums = A006012: (1, 2, 6, 20, 68, 232, 792, 2704,...).

Companion to triangle A140071.

			First few rows of the triangle are:
  1;
  1,    1;
  1,    4,    1;
  1,   13,    5,    1;
  1,   40,   18,    8,   1;
  1,  121,   58,   42,   9,   1;
  1,  364,  179,  184,  51,  12,   1;
  1, 1093,  543,  731, 235,  87,  13,  1;
  1, 3280, 1636, 2736, 966, 496, 100, 16, 1;
  ...

Cf. A006012, A140071, A157751.

T:= proc(n, k) option remember;
      `if`(k<0 or k>n, 0, `if`(k=0 or k=n, 1,
      4*T(n-1, k) - 3*T(n-2, k) + T(n-2, k-2)))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Feb 18 2020

With[{m = 10}, CoefficientList[CoefficientList[Series[(1 + (y - 3)*x)/(1 - 4*x - (y^2 - 3)*x^2), {x, 0, m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 18 2020 *)

Triangle read by rows, iterates of matrix X * [1,0,0,0,...], where X = an infinite lower bidiagonal matrix with [1,3,1,3,1,3,...] in the main diagonal and [1,1,1,...] in the subdiagonal; with the rest zeros.
From Peter Bala, Jan 17 2014: (Start)
O.g.f.: (1 + (x - 3)*z)/(1 - 4*z - (x^2 - 3)*z^2) = 1 + (x + 1)*z + (x^2 + 4*x + 1)*z^2 + ....
Recurrence equation: T(n,k) = 4*T(n-1,k) - 3*T(n-2,k) + T(n-2,k-2).
Recurrence equation for row polynomials: R(n,x) = 4*R(n-1,x) + (x^2 - 3)*R(n-2,x) with R(0,x) = 1 and R(1,x) = 1 + x.
Another recurrence equation: R(n,x) = (x + 2)*R(n-1,x) - R(n-1,-x) with R(0,x) = 1. Cf. A157751. (End)

cp's OEIS Frontend

A007070 a(n) = 4*a(n-1) - 2*a(n-2) with a(0) = 1, a(1) = 4.

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A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

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A140070 Triangle read by rows, iterates of matrix X * [1,0,0,0,...], where X = an infinite lower bidiagonal matrix with [1,3,1,3,1,3,...] in the main diagonal and [1,1,1,...] in the subdiagonal.

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A007070 a(n) = 4a(n-1) - 2a(n-2) with a(0) = 1, a(1) = 4.