A071922 -id:A071922 - cp's OEIS frontend

A072176 Unimodal analog of Fibonacci numbers: a(n+1) = Sum_{k=0..floor(n/2)} A071922(n-k,k).

1, 1, 2, 3, 5, 9, 16, 30, 56, 106, 201, 382, 727, 1384, 2636, 5021, 9565, 18222, 34715, 66137, 126001, 240052, 457338, 871304, 1659978, 3162533, 6025150, 11478911, 21869232, 41664520, 79377833, 151227961, 288114394, 548905795

Offset: 1

Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 30 2002

Based on the observation that F_{n+1} = Sum_{k} binomial (n-k,k). In both cases the sum is extended to 0<=2k<=n.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1,1).

Cf. A071922, A005578.

a:=[1,1,2,3,5];; for n in [6..40] do a[n]:=2*a[n-1]+a[n-2] -2*a[n-3]-a[n-4]+a[n-5]; od; a; # G. C. Greubel, Aug 26 2019

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)) )); // G. C. Greubel, Aug 26 2019

seq(coeff(series(x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)), x, n+1), x, n), n = 1..40); # G. C. Greubel, Aug 26 2019

Rest@CoefficientList[ Series[x(1-x-x^2)/((1-x)(1-x-2x^2+x^4)), {x, 0, 40}], x] (* or *) LinearRecurrence[{2,1,-2,-1,1}, {1,1,2,3,5}, 40] (* Harvey P. Dale, Jun 23 2011 *)

my(x='x+O('x^40)); Vec(x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4))) \\ G. C. Greubel, Aug 26 2019

def A072176_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)) ).list()
a=A072176_list(40); a[1:] # G. C. Greubel, Aug 26 2019

G.f.: x*(1-x-x^2)/((1-x)*(1-x-2*x^2+x^4)).

a(1)=1, a(2)=1, a(3)=2, a(4)=3, a(5)=5, a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Jun 23 2011

A005578 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)-1.

Original entry on oeis.org

1, 1, 2, 3, 6, 11, 22, 43, 86, 171, 342, 683, 1366, 2731, 5462, 10923, 21846, 43691, 87382, 174763, 349526, 699051, 1398102, 2796203, 5592406, 11184811, 22369622, 44739243, 89478486, 178956971, 357913942, 715827883, 1431655766, 2863311531, 5726623062, 11453246123

Offset: 0

N. J. A. Sloane

Might be called the "Arima sequence" after Yoriyuki Arima who in 1769 constructed this sequence as the number of moves of the outer ring in the optimal solution for the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1) = u(k) + v(k), v(k+1) = u(k) + w(k), w(k+1) = v(k) + w(k); let M(k) = Max(u(k), v(k), w(k)); then a(n) = M(n). - Benoit Cloitre, Mar 25 2002
Unimodal analog of Fibonacci numbers: a(n+1) = Sum_{k=0..n/2} A071922(n-k, n-2*k). Based on the observation that F_{n+1} = Sum_{k} binomial (n-k, k). - Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 30 2002
Numbers n at which the length of the symmetric signed digit expansion of n with q=2 (i.e., the length of the representation of n in the (-1,0,1)2 number system) increases. - _Ralf Stephan, Jun 30 2003
Row sums of Riordan array (1/(1-x), x/(1-2*x^2)). - Paul Barry, Apr 24 2005
For n > 0, record-values of A107910: a(n) = A107910(A023548(n)). - Reinhard Zumkeller, May 28 2005
2^(n+1) = 2*a(n) + 2*A001045(n) + A000975(n-1); e.g., 2^6 = 64 = 2*a(5) + 2*A001045(5) + 2*A000975(4) = 2*11 + 2*11 + 2*10. Let a(n), A001045(n) and A000975(n-1) = the legs of a triangle (a, b, c). Then a(n-1), A001045(n-1) and A000975(n-2) = (S-c), (S-b), (S-a), where S = the triangle semiperimeter. Example: a(5), A001045(5) and A000975(4) = triangle (a, b, c) = (11, 11, 10). Then a(4), A001045(4), A000975(3) = (S-c), (S-b), (S-a) = (6, 5, 5). - Gary W. Adamson, Dec 24 2007
a(n) is the number of length-n binary representations of a nonnegative integer that is divisible by 3. The initial digits are allowed to be 0's. a(4) = 6 because we have 0000, 0011, 0110, 1001, 1100, 1111. - Geoffrey Critzer, Jan 13 2014
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 0, 1; 0, 1, 1; 1, 1, 0] or of the 3 X 3 matrix [1, 1, 0; 1, 0, 1; 0, 1, 1]. - R. J. Mathar, Feb 04 2014
With 0 prefixed, this sequence is an autosequence of the first kind because the sequence of first differences A001045 is. Its companion is A052950. - Paul Curtz, Dec 18 2018, edited by M. F. Hasler, Dec 21 2018
Apparently, the sequence gives the distinct values taken by A129761, the first differences of fibbinary numbers. - Rémy Sigrist, Oct 26 2019
The sequence with offset 1 can be generated in three steps starting with A158780. First, put in alternate signs (1, -1, 1, -2, 2, -4, ...) and take the inverse; getting (1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, ...). Take the invert transform of the latter, resulting in the sequence. It follows from the inverti transform being 1, 1, 0, 1, 1, 2, 3, ... that (for example), a(9) = 171 = (1, 1, 0, 1, 1, 2, 3, 5, 8) dot (86, 43, 0, 11, 6, 6, 6, 5, 8) = (86 + 43 + 0 + 11 + 6 + 6 + 6 + 5 + 8). A similar procedure is shown in the Aug 08 2019 comment of A006356. - Gary W. Adamson, Feb 04 2022

R. K. Guy, Graphs and the strong law of small numbers. Graph theory, combinatorics and applications. Vol. 2 (Kalamazoo, MI, 1988), 597-614, Wiley-Intersci. Publ., Wiley, New York, 1991.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Sujith Uthsara Kalansuriya Arachchi, Hung Viet Chu, Jiasen Liu, Qitong Luan, Rukshan Marasinghe, and Steven J. Miller, On a Pair of Diophantine Equations, arXiv:2309.04488 [math.NT], 2023.
Joerg Arndt, Matters Computational (The Fxtbook), p. 315.
Ji Young Choi, Ternary Modified Collatz Sequences And Jacobsthal Numbers, Journal of Integer Sequences, Vol. 19 (2016), #16.7.5.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Fan Chung and Shlomo Sternberg, Mathematics and the Buckyball
Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015.
Johann Cigler, Recurrences for certain sequences of binomial sums in terms of (generalized) Fibonacci and Lucas polynomials, arXiv:2212.02118 [math.NT], 2022.
Madeleine Goertz and Aaron Williams, The Quaternary Gray Code and How It Can Be Used to Solve Ziggurat and Other Ziggu Puzzles, arXiv:2411.19291 [math.CO], 2024. See pp. 6, 17.
R. K. Guy, Letter to N. J. A. Sloane, Apr 08 1988 (annotated scanned copy, included with permission)
Clemens Heuberger and Helmut Prodinger, On minimal expansions in redundant number systems: Algorithms and quantitative analysis, Computing 66(2001), 377-393.
Andreas M. Hinz, The Lichtenberg sequence, Fib. Quart., 55 (2017), 2-12.
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Gurmeet Singh Manku and Joe Sawada, A loopless Gray code for minimal signed-binary representations, 13th Annual European Symposium on Algorithms (ESA), LNCS 3669 (2005), 438-447.
Thor Martinsen, Non-Fisherian generalized Fibonacci numbers, Notes Num. Theor. Disc. Math. (2025) Vol. 31, No. 2, 370-389. See pp. 385, 387.
John P. McSorley, Counting structures in the Moebius ladder, Discrete Math., 184 (1998), 137-164.
Hans Olofsen, Blending functions based on trigonometric and polynomial approximations of the Fabius function, The Arctic University of Norway (Narvik, 2019).
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Chloe E. Shiff and Noah A. Rosenberg, Enumeration of rooted binary perfect phylogenies, arXiv:2410.14915 [q-bio.PE], 2024. See p. 16.
Andrew V. Sills and Hua Wang, On the maximal Wiener index and related questions, Discrete Applied Mathematics, Volume 160, Issues 10-11, July 2012, Pages 1615-1623.
Eric Weisstein's World of Mathematics, Walsh Function
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A071922, A072176, A135229.
Bisections: A007583 and A047849.
Cf. also A000975, A001045 (first differences), A129761.
Cf. A006356.

List([0..40],n->(2^(n+1)+3+(-1)^n)/6); # Muniru A Asiru, Dec 22 2018

[(2^(n+1)+3+(-1)^n)/6: n in [0..40]]; // Vincenzo Librandi, Aug 14 2011

A005578:=-(-1+z+z^2)/((z-1)*(2*z-1)*(z+1)); # Simon Plouffe in his 1992 dissertation
with(combstruct):ZL0:=S=Prod(Sequence(Prod(a, Sequence(b))), a):ZL1:=Prod(begin_blockP, Z, end_blockP):ZL2:=Prod(begin_blockLR, Z, Sequence(Prod(mu_length, Z), card>=1), end_blockLR): ZL3:=Prod(begin_blockRL, Sequence(Prod(mu_length, Z), card>=1), Z, end_blockRL):Q:=subs([a=Union(ZL3), b=ZL3], ZL0), begin_blockP=Epsilon, end_blockP=Epsilon, begin_blockLR=Epsilon, end_blockLR=Epsilon, begin_blockRL=Epsilon, end_blockRL=Epsilon, mu_length=Epsilon:temp15:=draw([S, {Q}, unlabelled], size=15):seq(count([S, {Q}, unlabelled], size=n), n=2..34); # Zerinvary Lajos, Mar 08 2008

a=0; Table[a=2^n-a;(a/2+1)/2,{n,5!}] (* Vladimir Joseph Stephan Orlovsky, Nov 22 2009 *)
LinearRecurrence[{2,1,-2}, {1,1,2}, 40] (* G. C. Greubel, Aug 26 2019 *)

a(n)=(2^(n+1)+3+(-1)^n)/6 \\ Charles R Greathouse IV, Mar 22 2016

print([1+2**n//3 for n in range(40)])  # Gennady Eremin, Feb 05 2022

[(2^(n+1)+3+(-1)^n)/6 for n in (0..40)] # G. C. Greubel, Aug 26 2019

a(n) = ceiling(2^n/3).
a(n) = 1 + floor((2^n)/3) (proof by mathematical induction).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
From Paul Barry, Jul 20 2003: (Start)
a(n) = A001045(n) + A000035(n+1), where A000035 = (0, 1, 0, 1, ...).
G.f.: (1 - x - x^2)/((1-x^2)*(1-2*x)). [Guy, 1988];
E.g.f.: (exp(2*x) - exp(-x))/3 + cosh(x) = (2*exp(2*x) + 3*exp(x) + exp(-x))/6. (End)
The 30 listed terms are given by a(0)=1, a(1)=1 and, for n > 1, by a(n) = a(n-1) + a(n-2) + Sum_{i=0..n-4} Fibonacci(i)*a(n-4-i). - John W. Layman, Jan 07 2000
a(n) = (2^(n+1) + 3 + (-1)^n)/6. - Vladeta Jovovic, Jul 02 2002
Binomial transform of A001045(n-1)(-1)^n + 0^n/2. - Paul Barry, Apr 28 2004
a(n) = (1 + A001045(n+1))/2. - Paul Barry, Apr 28 2004
a(n) = Sum_{k=0..n} (-1)^k*Sum_{j=0..n-k} (if((j-k) mod 2)=0, binomial(n-k, j), 0). - Paul Barry, Jan 25 2005
Let M = the 6 X 6 adjacency matrix of a benzene ring, (reference): [0,1,0,0,0,1; 1,0,1,0,0,0; 0,1,0,1,0,0; 0,0,1,0,1,0; 0,0,0,1,0,1; 1,0,0,0,1,0]. Then a(n) = leftmost nonzero term of M^n * [1,0,0,0,0,0]. E.g.: a(6) = 22 since M^6 * [1,0,0,0,0,0] = [22,0,21,0,21,0]. - Gary W. Adamson, Jun 14 2006
Starting (1, 2, 3, 6, 11, 22, ...), = row sums of triangle A135229. - Gary W. Adamson, Nov 23 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), A000975(n-1)]. - Gary W. Adamson, Dec 24 2007
a(n) = 1 + 2^(n-1) - a(n-1) = a(n-1) + 2*a(n-2) - 1 = a(n-2) + 2^(n-2). - Paul Curtz, Jan 31 2009
a(n) = A023105(n+1) - 1. - Carl Joshua Quines, Jul 17 2019

Edited by N. J. A. Sloane, Jun 20 2015

A210219 Triangle of coefficients of polynomials u(n,x) jointly generated with A210220; see the Formula section.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 7, 1, 5, 16, 22, 11, 1, 6, 25, 50, 46, 16, 1, 7, 36, 95, 130, 86, 22, 1, 8, 49, 161, 295, 296, 148, 29, 1, 9, 64, 252, 581, 791, 610, 239, 37, 1, 10, 81, 372, 1036, 1792, 1897, 1163, 367, 46, 1, 11, 100, 525, 1716, 3612, 4900, 4166, 2083, 541, 56, 1

Offset: 1

Clark Kimberling, Mar 19 2012

First two terms in row n:  n,(n-1)^2; last term: 1.
Period of alternating row sums: (1,1,0).
For a discussion and guide to related arrays, see A208510.
Apparently this is A071920 without the marginal zeros, read by downwards antidiagonals, or T(n,k) = A071922(n,k). - R. J. Mathar, May 17 2014

			First five rows:
  1
  2...1
  3...4....1
  4...9....7....1
  5...16...22...11...1
First three polynomials u(n,x): 1, 2 + x, 3 + 4x + x^2.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Cf. A210220, A208510, A001906 (row sums).

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]     (* A210219 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]     (* A210220 *)
(* alternative program *)
T[n_,k_] := Sum[Binomial[k, 2*j]*Binomial[n-j, k], {j, 0, Floor[k/2]}]; Flatten[Table[T[n, k],{n, 1, 11}, {k, 1, n}]] (* Detlef Meya, Dec 05 2023 *)

T(n,k) = sum(j=0, k\2, binomial(k,2*j)*binomial(n-j,k)) \\ Andrew Howroyd, Jan 01 2024

u(n,x) = x*u(n-1,x) + v(n-1,x) + 1, v(n,x) = x*u(n-1,x) + (x+1)*v(n-1,x) + 1, where u(1,x)=1, v(1,x)=1.

T(n,k) = Sum_{j=0..floor(k/2)} binomial(k,2*j)*binomial(n-j,k). - Detlef Meya, Dec 05 2023

cp's OEIS Frontend

A072176 Unimodal analog of Fibonacci numbers: a(n+1) = Sum_{k=0..floor(n/2)} A071922(n-k,k).

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A005578 a(2*n) = 2*a(2*n-1), a(2*n+1) = 2*a(2*n)-1.

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A210219 Triangle of coefficients of polynomials u(n,x) jointly generated with A210220; see the Formula section.

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A072285 Numerators of inverse unimodal analog of binomial coefficients: binomial(n,m) = Sum_{k=0..n-m} a(2*k+m-1, 2*k).

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A072286 Denominators of inverse unimodal analog of binomial coefficients: binomial(n,m) = Sum_{k=0..n-m} a(2*k+m-1, 2*k).

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A005578 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)-1.

A072285 Numerators of inverse unimodal analog of binomial coefficients: binomial(n,m) = Sum_{k=0..n-m} a(2k+m-1, 2k).

A072286 Denominators of inverse unimodal analog of binomial coefficients: binomial(n,m) = Sum_{k=0..n-m} a(2k+m-1, 2k).