A029907 -id:A029907 - cp's OEIS frontend

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799

Offset: 0

N. J. A. Sloane

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) =  x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
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).
Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone and Diego Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
Nathan D. Cahill, John R. D'Errico and John P. Spence, Complex Factorizations of the Fibonacci and Lucas Numbers, Fibonacci Quarterly, 1(41):13-19, 2003.
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Paul Curtz, A269500, SeqFan list, March 2, 2016.
Murray Elder and Arkadius Kalka, Logspace computations for rigid Garside groups, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
Alex Fink, Richard K. Guy and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol. 3, No. 2 (2008), pp. 76-114. See Section 13.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Tian-Xiao He and Louis W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Seong Ju Kim, Ryan Stees and Laura Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
Tanya Khovanova, Recursive Sequences.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
Serge Perrine, Some properties of the equation x^2=5y^2-4, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
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.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019

a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

A002878 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]);
    else
        3*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)

a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011

for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015

a002878 = [1, 4]
for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
print(a002878) # Gennady Eremin, Feb 05 2022

[lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A023610 Convolution of Fibonacci numbers and {F(2), F(3), F(4), ...}.

Original entry on oeis.org

1, 3, 7, 15, 30, 58, 109, 201, 365, 655, 1164, 2052, 3593, 6255, 10835, 18687, 32106, 54974, 93845, 159765, 271321, 459743, 777432, 1312200, 2211025, 3719643, 6248479, 10482351, 17562870, 29391490, 49132669, 82048737, 136884293, 228160495, 379975140, 632293452

Offset: 0

Clark Kimberling

a(n-2) + 1 is the number of (3412,1243)-, (3412,2134)- and (3412,1324)-avoiding involutions in S_n, n>1. - Ralf Stephan, Jul 06 2003
The number of terms in all ordered partitions of (n+1) using only ones and twos. For example, a(3)=15 because there are 15 terms in 1+1+1+1;2+1+1;1+2+1;1+1+2;2+2 - Geoffrey Critzer, Apr 07 2008
a(n) is the number of n-matchings in the graph obtained by a zig-zag triangulation of a convex (2n+1)-gon. Example: a(2)=7 because in the triangulation of the convex pentagon ABCDEA with diagonals AD and AC we have 7 2-matchings: {AB,CD},{AB,DE},{BC,AD},{BC,DE},{BC,EA},{CD,EA} and {DE,AC}. - Emeric Deutsch, Dec 25 2004
Partial sums of A029907. First differences of A002940. - Peter Bala, Oct 24 2007
Equals row sums of triangle A144154. - Gary W. Adamson, Sep 12 2008
Equals the number of 1's in Fibonacci Maximal notation for subsets of
(1, 2, 3, 5, 8, 13, ...) terms. For example (cf. A181630): 4, 5, and 6 are the 3 terms 101, 110, and 111 in Fibonacci Maximal. Total number of 1's for those terms = 7 = a(2). - Gary W. Adamson, Nov 02 2010
a(n) is half the number of strokes needed to draw all the domino tilings of a 2 X (n+2) rectangle. - Roberto Tauraso, Mar 15 2014
a(n) is the total number of 1's in all (n+1)-bit dual Zeckendorf representations of integers (A104326). For example, a(2) = 7 counts the 1's in 101, 110, 111. - Shenghui Yang, Feb 09 2025

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Marcella Anselmo, Giuseppa Castiglione, Manuela Flores, Dora Giammarresi, Maria Madonia, and Sabrina Mantaci, Hypercubes and Isometric Words based on Swap and Mismatch Distance, arXiv:2303.09898 [math.CO], 2023.
Kassie Archer and Noel Bourne, Pattern avoidance in compositions and powers of permutations, arXiv:2505.05218 [math.CO], 2025. See p. 6.
Jean-Luc Baril and José Luis Ramírez, Descent distribution on Catalan words avoiding ordered pairs of Relations, arXiv:2302.12741 [math.CO], 2023.
Jean-Luc Baril and José L. Ramírez, Fibonacci and Catalan paths in a wall, 2023.
Giuseppa Castiglione and Marinella Sciortino, Standard Sturmian words and automata minimization algorithms, Theoretical Computer Science, Volume 601, 11 October 2015, Pages 58-66 ("WORDS 2013").
Tomislav Došlic and Luka Podrug, Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers, arXiv:2203.11761 [math.CO], 2022.
Dmitry Efimov, Hafnian of two-parameter matrices, arXiv:2101.09722 [math.CO], 2021.
Eric S. Egge, Restricted 3412-Avoiding Involutions: Continued Fractions, Chebyshev Polynomials and Enumerations, arXiv:math/0307050 [math.CO], 2003. Sec. 8.
Neven Elezović, Asymptotic Expansions of Central Binomial Coefficients and Catalan Numbers, J. Int. Seq. 17 (2014) # 14.2.1.
Martin Griffiths, A Restricted Random Walk defined via a Fibonacci Process, Journal of Integer Sequences, Vol. 14 (2011), #11.5.4.
László Németh and László Szalay, Explicit solution of system of two higher-order recurrences, arXiv:2408.12196 [math.NT], 2024. See p. 10.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045 (Fibonacci numbers).
Column 1 of triangle A063967.
Cf. A001629, A002940, A004798, A010049, A029907, A104326, A144154, A181630.

a023610 n = a023610_list !! n
a023610_list = f [1] $ drop 3 a000045_list where
   f us (v:vs) = (sum $ zipWith (*) us $ tail a000045_list) : f (v:us) vs
-- Reinhard Zumkeller, Jan 18 2014

Table[Sum[Binomial[n - i, i]*(n - i), {i, 0, n}], {n, 1, 33}] (* Geoffrey Critzer, May 04 2009 *)

a(n)=(n+2)*fibonacci(n+4)/5+(n-1)*fibonacci(n+2)/5 \\ Charles R Greathouse IV, Jun 11 2015

def A023610():
    a, b, c, d = 1, 3, 7, 15
    while True:
        yield a
        a, b, c, d = b, c, d, 2*(d-b)+c-a
a = A023610(); [next(a) for i in range(33)]  # Peter Luschny, Nov 20 2013

O.g.f.: (x+1)/(1-x-x^2)^2. - Len Smiley, Dec 11 2001
a(n) = (1/5)*((n+2)*F(n+4) + (n-1)*F(n+2)), with F(n)=A000045(n). - Ralf Stephan, Jul 06 2003
a(n) = Sum_{k=0..n+1} (n-k+1)*binomial(n-k+1, k). - Paul Barry, Nov 05 2005
Recurrence: a(n+2) = a(n+1) + a(n) + Fib(n+4), n >= 0. For n >= 2, a(n-2) = (-1)^n*((-2n+3)*Fib(-n) - (-n)*Fib(-n-1))/5 = (-1)^n*A010049(-n), the second-order Fibonacci numbers of negative index, where Fib(-n) = (-1)^(n+1)*Fib(n). - Peter Bala, Oct 24 2007
a(n) = (n+1)*F(n+2) - A001629(n+1) where F(n) is the n-th Fibonacci number. - Geoffrey Critzer, Apr 07 2008
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4), n >= 4. - L. Edson Jeffery, Mar 29 2013
a(n+1) = A004798(n) + A000045(n+2) for n >= 0. - John Molokach, Jul 04 2013
a(n) = A001629(n+1) + A001629(n+2). - Philippe Deléham, Oct 30 2013
E.g.f.: exp(x/2)*(5*(5 + 7*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(11 + 15*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Dec 04 2023

A010049 Second-order Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 3, 5, 10, 18, 33, 59, 105, 185, 324, 564, 977, 1685, 2895, 4957, 8462, 14406, 24465, 41455, 70101, 118321, 199368, 335400, 563425, 945193, 1583643, 2650229, 4430290, 7398330, 12342849, 20573219, 34262337, 57013865, 94800780, 157517532, 261545777

Offset: 0

N. J. A. Sloane

Number of parts in all compositions of n+1 with no 1's. E.g. a(5)=10 because in the compositions of 6 with no part equal to 1, namely 6,4+2,3+3,2+4,2+2+2, the total number of parts is 10. - Emeric Deutsch, Dec 10 2003

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 83.
Cornelius Gerrit Lekkerkerker, Voorstelling van natuurlijke getallen door een som van getallen van Fibonacci, Simon Stevin, Vol. 29 (1952), pp. 190-195.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Carlos A. Rico A. and Ana Paula Chaves, Double-Recurrence Fibonacci Numbers and Generalizations, arXiv:1903.07490 [math.NT], 2019.
T. Amdeberhan and M. B. Can, M. Jensen, Divisors and specializations of Lucas polynomials, arXiv preprint arXiv:1406.0432 [math.CO], 2014.
Kálmán Liptai, László Németh, Tamás Szakács, and László Szalay, On certain Fibonacci representations, arXiv:2403.15053 [math.NT], 2024. See p. 2.
Mengmeng Liu and Andrew Yezhou Wang, The Number of Designated Parts in Compositions with Restricted Parts, J. Int. Seq., Vol. 23 (2020), Article 20.1.8.
Jia Huang, Compositions with restricted parts, arXiv:1812.11010 [math.CO], 2018.
Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See pp. 2, 50, 57.
Loïc Turban, Lattice animals on a staircase and Fibonacci numbers, arXiv:cond-mat/0011038 [cond-mat.stat-mech], 2000; J. Phys. A 33 (2000) 2587-2595.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Partial sums of A006367.
Cf. A000032, A000045, A001629, A007895, A023610.
Cf. A029907, A094440, A134410, A135830.

a:=List([0..40],n->Sum([0..n-1],k->(k+1)*Binomial(n-k-1,k)));; Print(a); # Muniru A Asiru, Dec 31 2018

a010049 n = a010049_list !! n
a010049_list = uncurry c $ splitAt 1 a000045_list where
   c us (v:vs) = (sum $ zipWith (*) us (1 : reverse us)) : c (v:us) vs
-- Reinhard Zumkeller, Nov 01 2013

[((2*n+3)*Fibonacci(n)-n*Fibonacci(n-1))/5: n in [0..40]]; // Vincenzo Librandi, Dec 31 2018

with(combinat): A010049 := proc(n) options remember; if n <= 1 then n else A010049(n-1)+A010049(n-2)+fibonacci(n-2); fi; end;

CoefficientList[Series[(z - z^2)/(z^2 + z - 1)^2, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
CoefficientList[Series[x (1 - x) / (1 - x - x^2)^2, {x, 0, 60}], x] (* Vincenzo Librandi, Jun 11 2013 *)
LinearRecurrence[{2, 1, -2, -1}, {0, 1, 1, 3}, 38] (* Amiram Eldar, Jan 11 2020 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-2,1,2]^n*[0;1;1;3])[1,1] \\ Charles R Greathouse IV, Jul 20 2016

def A010049():
    a, b, c, d = 0, 1, 1, 3
    while True:
        yield a
        a, b, c, d = b, c, d, 2*(d-b)+c-a
a = A010049(); [next(a) for i in range(38)]  # Peter Luschny, Nov 20 2013

def A010049(n): return (1/5)*(n*lucas_number2(n-1, 1, -1) + 3*fibonacci(n))
[A010049(n) for n in (0..40)] # G. C. Greubel, Apr 06 2022

First differences of A001629.
From Wolfdieter Lang, May 03 2000: (Start)
a(n) = ((2*n+3)*F(n) - n*F(n-1))/5, F(n)=A000045(n) (Fibonacci numbers) (Turban reference eq.(2.12)).
G.f.: x*(1-x)/(1-x-x^2)^2. (Turban reference eq.(2.10)). (End)
Recurrence: a(0)=0, a(1)=1, a(2)=1, a(n+2) = a(n+1) + a(n) + F(n). - Benoit Cloitre, Sep 02 2002
Set A(n) = a(n+1) + a(n-1), B(n) = a(n+1) - a(n-1). Then A(n+2) = A(n+1) + A(n) + Lucas(n) and B(n+2) = B(n+1) + B(n) + Fibonacci(n). The polynomials F_2(n,-x) = Sum_{k=0..n} C(n,k)*a(n-k)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the polynomials F(n,-x) defined in A094440. For a similar conjecture for polynomials involving the second-order Lucas numbers see A134410. - Peter Bala, Oct 24 2007
a(n) = -A001629(n+2) + 2*A001629(n+1) + A000045(n+1). - R. J. Mathar, Nov 16 2007
Starting (1, 1, 3, 5, 10, ...), = row sums of triangle A135830. - Gary W. Adamson, Nov 30 2007
a(n) = F(n) + Sum_{k=0..n-1} F(k)*F(n-1-k), where F = A000045. - Reinhard Zumkeller, Nov 01 2013
a(n) = Sum_{k=0..n-1} (k+1)*binomial(n-k-1, k). - Peter Luschny, Nov 20 2013
a(n) = Sum_{i=0..n-1} Sum_{j=0..i} F(j-1)*F(i-j-1), where F = A000045. - Carlos A. Rico A., Jul 14 2016
a(n) = Sum_{k = F(n+1)..F(n+2)-1} A007895(k), where F(n) is the n-th Fibonacci number (Lekkerkerker, 1952). - Amiram Eldar, Jan 11 2020
a(n) = (1/5)*(n*A000032(n-1) + 3*A000045(n)). - G. C. Greubel, Apr 06 2022
E.g.f.: 2*exp(x/2)*(5*x*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Dec 04 2023

More terms from Emeric Deutsch, Dec 10 2003

A038792 Rectangular array defined by T(i,1) = T(1,j) = 1 for i >= 1 and j >= 1; T(i,j) = max(T(i-1,j) + T(i-1,j-1), T(i-1,j-1) + T(i,j-1)) for i >= 2, j >= 2, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 8, 8, 5, 1, 1, 6, 12, 13, 12, 6, 1, 1, 7, 17, 21, 21, 17, 7, 1, 1, 8, 23, 33, 34, 33, 23, 8, 1, 1, 9, 30, 50, 55, 55, 50, 30, 9, 1, 1, 10, 38, 73, 88, 89, 88, 73, 38, 10, 1, 1, 11, 47, 103, 138, 144, 144, 138, 103, 47, 11, 1

Offset: 1

Clark Kimberling, May 02 2000

Antidiagonal sums: A029907.
Main diagonal: A001519 (odd-indexed Fibonacci numbers).
Next diagonal: A001906 (even-indexed Fibonacci numbers).

			From _Clark Kimberling_, Jun 20 2011: (Start)
Northwest corner begins at (i,j) = (1,1):
  1,   1,   1,   1,   1,   1,   1,   1, ...
  1,   2,   3,   4,   5,   6,   7,   8, ...
  1,   3,   5,   8,  12,  17,  23,  30, ...
  1,   4,   8,  13,  21,  33,  50,  73, ...
  1,   5,  12,  21,  34,  55,  88, 138, ...
  1,   6,  17,  33,  55,  89, 144, 232, ...
  1,   7,  23,  50,  88, 144, 233, 377, ...
(End)
Antidiagonal triangle begins as:
  1;
  1, 1;
  1, 2,  1;
  1, 3,  3,  1;
  1, 4,  5,  4,  1;
  1, 5,  8,  8,  5,  1;
  1, 6, 12, 13, 12,  6,  1;
  1, 7, 17, 21, 21, 17,  7,  1;
  1, 8, 23, 33, 34, 33, 23,  8, 1;
  1, 9, 30, 50, 55, 55, 50, 30, 9, 1;

G. C. Greubel, Antidiagonals n = 1..50, flattened
H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
A. Dil and I. Mezo, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Fibonacci numbers.
Piero Filipponi, Incomplete Fibonacci and Lucas numbers, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 1 that contains the incomplete Fibonacci numbers.
Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40 (4) (2002), 328-338; see Example 4 (appears as a triangular array).
A. Pintér and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.

Cf. A001519, A001906, A029907.
Cf. A000045, A038730, A134511, A324242.
Main diagonal gives A001519.

function t(n,k)
  if k eq 0 or n eq 0 then return 1;
  else return Max(t(n-1,k-1) + t(n-1,k), t(n-1,k-1) + t(n,k-1));
  end if; return t;
end function;
T:= func< n,k | t(n-k, k-1) >;
[T(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Apr 05 2022

G := x*y*(1-x*y)/((x*y+x-1)*(x*y+y-1)); G := convert(series(G, x=0, 11),polynom):
for i from 1 to 10 do series(coeff(G,x,i),y=0,11) od; # Mark van Hoeij, Nov 09 2011
# second Maple program:
G:= x*y*(1-x*y)/((x*y+x-1)*(x*y+y-1)):
T:= (i, j)-> coeff(series(coeff(series(G, y, j+1), y, j), x, i+1), x, i):
seq(seq(T(i, 1+d-i), i=1..d), d=1..12); # Alois P. Heinz, Sep 02 2019
# third Maple program:
T:= proc(i,j) option remember; `if`(i=1 or j=1, 1,
      max(T(i-1,j) + T(i-1,j-1), T(i-1,j-1) + T(i,j-1)))
    end:
seq(seq(T(i, 1+d-i), i=1..d), d=1..12); # Alois P. Heinz, Sep 02 2019

f[i_, 0]:= 1; f[0, i_]:= 1
f[i_, j_]:= f[i,j]= Max[f[i-1,j] +f[i-1,j-1], f[i-1,j-1] +f[i,j-1]];
T[i_, j_]:= f[i-j, j-1];
TableForm[Table[f[i, j], {i,0,7}, {j,0,7}]]
Table[T[i, j], {i,10}, {j,i}]//Flatten (* modified by G. C. Greubel, Apr 05 2022 *)

def t(n,k):
    if (k==0 or n==0): return 1
    else: return max(t(n-1,k-1) + t(n-1,k), t(n-1,k-1) + t(n,k-1))
def A038792(n,k): return t(n-k, k-1)
flatten([[A038792(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Apr 05 2022

G.f.: x*y*(1-x*y)/((x*y+x-1)*(x*y+y-1)). - Mark van Hoeij, Nov 09 2011
From Petros Hadjicostas, Sep 02 2019: (Start)
Following Dil and Mezo (2008), define the incomplete Fibonacci numbers by F(n,k) = Sum_{s = 0..k} binomial(n-1-s, s) for n >= 1 and 0 <= k <= floor((n-1)/2).
Then T(i, j) = F(i+j-1, min(i-1, j-1)) for i,j >= 1.
(End)

New description from Benoit Cloitre, Aug 05 2003

Updated from pre-2003 triangular format to present rectangular, from Clark Kimberling, Jun 20 2011

A267788 T(n,k)=Number of nXk 0..1 arrays with every repeated value in every row greater than or equal to, and in every column greater than, the previous repeated value.

Original entry on oeis.org

2, 4, 4, 8, 16, 6, 15, 64, 36, 9, 28, 225, 216, 81, 12, 51, 784, 1056, 729, 144, 16, 92, 2601, 5004, 5081, 1728, 256, 20, 164, 8464, 22110, 34173, 14956, 4096, 400, 25, 290, 26896, 94554, 211555, 122770, 44742, 8000, 625, 30, 509, 84100, 391314, 1262760, 912667

Offset: 1

R. H. Hardin, Jan 20 2016

Table starts
..2....4.....8.....15.......28........51.........92.........164..........290
..4...16....64....225......784......2601.......8464.......26896........84100
..6...36...216...1056.....5004.....22110......94554......391314......1582824
..9...81...729...5081....34173....211555....1262760.....7263481.....40755550
.12..144..1728..14956...122770....912667....6484282....44116906....291598056
.16..256..4096..44742...460598...4245574...37282358...312449872...2540944329
.20..400..8000.102954..1234716..13126812..132388406..1271314080..11831791048
.25..625.15625.238813..3380133..42012357..494152778..5520112546..59723941668
.30..900.27000.472174..7591852.106570618.1413416776.17806098826.217360101006
.36.1296.46656.935890.17155354.272337497.4075463059.57801662876.793861159136

			Some solutions for n=4 k=4
..1..0..1..0....1..0..0..0....0..1..0..0....1..1..1..0....1..1..0..1
..0..1..0..1....1..1..0..1....0..0..0..0....0..0..0..0....0..0..0..0
..0..0..1..0....0..0..1..1....1..1..1..1....0..1..0..1....1..1..1..1
..1..1..1..1....1..0..0..0....1..0..1..0....1..1..1..1....1..0..0..0

R. H. Hardin, Table of n, a(n) for n = 1..200

Column 1 is A002620(n+2).
Column 2 is A030179(n+2).
Row 1 is A029907(n+1).
Row 2 is A267729.

Empirical for column k:
k=1: a(n) = 2*a(n-1) -2*a(n-3) +a(n-4)
k=2: a(n) = 2*a(n-1) +2*a(n-2) -6*a(n-3) +6*a(n-5) -2*a(n-6) -2*a(n-7) +a(n-8)
k=3: [order 12]
k=4: [order 16] for n>18
k=5: [order 20] for n>22
k=6: [order 24] for n>27
Empirical for row n:
n=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) -a(n-4)
n=2: [order 9]
n=3: [order 12]
n=4: [order 93]

A269619 T(n,k)=Number of length-n 0..k arrays with no repeated value differing from the previous repeated value by other than plus two, zero or minus 1.

Original entry on oeis.org

2, 3, 4, 4, 9, 8, 5, 16, 27, 15, 6, 25, 64, 78, 28, 7, 36, 125, 249, 222, 51, 8, 49, 216, 612, 954, 624, 92, 9, 64, 343, 1275, 2956, 3611, 1740, 164, 10, 81, 512, 2370, 7440, 14125, 13544, 4824, 290, 11, 100, 729, 4053, 16218, 43013, 66925, 50442, 13320, 509, 12, 121

Offset: 1

R. H. Hardin, Mar 01 2016

Table starts
...2.....3......4.......5........6.........7.........8..........9.........10
...4.....9.....16......25.......36........49........64.........81........100
...8....27.....64.....125......216.......343.......512........729.......1000
..15....78....249.....612.....1275......2370......4053.......6504.......9927
..28...222....954....2956.....7440.....16218.....31822......57624......97956
..51...624...3611...14125....43013....110099....248143.....507521.....961625
..92..1740..13544...66925...246798....742487...1923796....4447329....9398090
.164..4824..50442..314935..1407232...4979260..14840928...38800210...91490344
.290.13320.186822.1473779..7982022..33232924.113998742..337209090..887591878
.509.36672.688899.6865098.45074673.220896016.872397577.2920747321.8584628259

			Some solutions for n=6 k=4
..1. .2. .0. .2. .1. .4. .3. .4. .2. .2. .0. .2. .2. .2. .2. .0
..0. .3. .3. .1. .4. .0. .0. .0. .2. .1. .0. .1. .0. .2. .4. .3
..4. .3. .2. .1. .1. .3. .0. .4. .1. .0. .4. .3. .1. .2. .4. .1
..3. .3. .2. .2. .4. .2. .4. .0. .3. .3. .0. .3. .4. .3. .3. .1
..0. .3. .1. .3. .0. .1. .3. .4. .1. .1. .2. .1. .4. .1. .4. .1
..0. .3. .0. .4. .4. .1. .4. .2. .1. .0. .1. .0. .3. .4. .2. .2

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A029907(n+1).
Row 1 is A000027(n+1).
Row 2 is A000290(n+1).
Row 3 is A000578(n+1).

Empirical for column k:
k=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) -a(n-4)
k=2: a(n) = 4*a(n-1) -2*a(n-2) -4*a(n-3)
k=3: a(n) = 12*a(n-1) -51*a(n-2) +81*a(n-3) -3*a(n-4) -63*a(n-5) -24*a(n-6) -9*a(n-7)
k=4: [order 7]
k=5: [order 13]
k=6: [order 15]
k=7: [order 17]
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = n^2 + 2*n + 1
n=3: a(n) = n^3 + 3*n^2 + 3*n + 1
n=4: a(n) = n^4 + 4*n^3 + 5*n^2 + 5*n
n=5: a(n) = n^5 + 5*n^4 + 7*n^3 + 12*n^2 + 3*n
n=6: a(n) = n^6 + 6*n^5 + 9*n^4 + 22*n^3 + 9*n^2 + 9*n - 7 for n>2
n=7: a(n) = n^7 + 7*n^6 + 11*n^5 + 35*n^4 + 18*n^3 + 36*n^2 - 19*n - 7 for n>2

A054450 Triangle of partial row sums of unsigned triangle A049310(n,m), n >= m >= 0 (Chebyshev S-polynomials).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 4, 4, 1, 1, 8, 8, 5, 5, 1, 1, 13, 12, 12, 6, 6, 1, 1, 21, 21, 17, 17, 7, 7, 1, 1, 34, 33, 33, 23, 23, 8, 8, 1, 1, 55, 55, 50, 50, 30, 30, 9, 9, 1, 1, 89, 88, 88, 73, 73, 38, 38, 10, 10, 1, 1, 144, 144, 138, 138, 103, 103, 47, 47, 11, 11, 1, 1

Offset: 0

Wolfdieter Lang, Apr 27 2000 and May 08 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is Fib(z)/(1-x*z/(1-z^2)) where Fib(x)=1/(1-x-x^2) = g.f. for A000045(n+1) (Fibonacci numbers without 0).

This is the first member of the family of Riordan-type matrices obtained from the unsigned convolution matrix A049310 by repeated application of the partial row sums procedure.

			Triangle begins as:
   1;
   1,  1;
   2,  1,  1;
   3,  3,  1,  1;
   5,  4,  4,  1,  1;
   8,  8,  5,  5,  1,  1;
  13, 12, 12,  6,  6,  1,  1;
  21, 21, 17, 17,  7,  7,  1,  1;
  34, 33, 33, 23, 23,  8,  8,  1,  1;
  55, 55, 50, 50, 30, 30,  9,  9,  1, 1;
  89, 88, 88, 73, 73, 38, 38, 10, 10, 1, 1;
  ...
Fourth row polynomial (n=3): p(3,x) = 3 + 3*x + x^2 + x^3.

Cf. A000027, A000045, A029907 (row sums), A038793, A038794, A038795, A038796.

Cf. A049310, A052952, A053121, A054451, A054453, A099571, A099572, A152948.

A049310:= func< n,k | ((n+k) mod 2) eq 0 select (-1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;
A054450:= func< n,k | (&+[Abs(A049310(n,j)): j in [k..n]]) >;
[A054450(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

A049310[n_, k_]:= A049310[n, k]= If[n<0, 0, If[k==n, 1, A049310[n-1, k-1] - A049310[n-2, k] ]];
A054450[n_, k_]:= A054450[n, k]= Sum[Abs[A049310[n,j]], {j,k,n}];
Table[A054450[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

@CachedFunction
def A049310(n, k):
    if (n<0): return 0
    elif (k==n): return 1
    else: return A049310(n-1, k-1) - A049310(n-2, k)
def A054450(n,k): return sum( abs(A049310(n,j)) for j in (k..n) )
flatten([[A054450(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

T(n, m) = Sum_{k=m..n} |A049310(n, k)| (sequence of partial row sums in column m).

Column m recursion: T(n, m) = Sum_{j=m..n} T(j-1, m)*|A049310(n-j, 0)| + |A049310(n, m)|, n >= m >= 0, a(n, m) := 0 if n

T(n, 0) = A000045(n+1).

T(n, 1) = A052952(n-1).

T(n, 2) = A054451(n-2).

Sum_{k=0..n} T(n, k) = A029907(n) = A054453(n, 0).

G.f. for column m: Fib(x)*(x/(1-x^2))^m, m >= 0, with Fib(x) = g.f. A000045(n+1).

The corresponding square array has T(n, k) = Sum_{j=0..floor(k/2)} binomial(n+k-j, j). - Paul Barry, Oct 23 2004

From G. C. Greubel, Jul 25 2022: (Start)

T(n, 3) = A099571(n-3).

T(n, 4) = A099572(n-4).

T(n, n) = T(n, n-1) = A000012(n).

T(n, n-2) = A000027(n), n >= 2.

T(n, n-3) = A000027(n), n >= 3.

T(n, n-4) = A152948(n), n >= 4.

T(n, n-5) = A152948(n), n >= 5.

T(n, n-6) = A038793(n), n >= 6.

T(n, n-8) = A038794(n), n >= 8.

T(n, n-10) = A038795(n), n >= 10.

T(n, n-12) = A038796(n), n >= 12. (End)

A269435 T(n,k)=Number of length-n 0..k arrays with no repeated value greater than the previous repeated value.

Original entry on oeis.org

2, 3, 4, 4, 9, 8, 5, 16, 27, 15, 6, 25, 64, 78, 28, 7, 36, 125, 250, 222, 51, 8, 49, 216, 615, 964, 622, 92, 9, 64, 343, 1281, 2995, 3674, 1722, 164, 10, 81, 512, 2380, 7536, 14455, 13868, 4719, 290, 11, 100, 729, 4068, 16408, 44021, 69235, 51917, 12821, 509, 12, 121

Offset: 1

R. H. Hardin, Feb 26 2016

Table starts
...2.....3......4.......5........6.........7.........8..........9.........10
...4.....9.....16......25.......36........49........64.........81........100
...8....27.....64.....125......216.......343.......512........729.......1000
..15....78....250.....615.....1281......2380......4068.......6525.......9955
..28...222....964....2995.....7536.....16408.....32152......58149......98740
..51...622...3674...14455....44021....112476....252932.....516189.....976135
..92..1722..13868...69235...255576....767172...1981512....4566213....9621220
.164..4719..51917..329430..1475871...5209554..15465934...40265487...94574110
.290.12821.192980.1558430..8482276..35236110.120310016..354051015..927338710
.509.34575.712868.7334806.48543777.237479970.933059856.3105016479.9072298237
From Robert Israel, May 30 2019: (Start)
For each of the A000110 partitions pi of the set {1,...,n}, let A_pi(n,k) be the number of length-n 0..k arrays v, such that v(i)=v(j) if and only if i and j are in the same part, and with no repeated value greater than the previous repeated value. There are restrictions on the values in the parts: if two parts a and b each have cardinality >= 2 and a_2 < b_2 (where the parts are indexed in increasing order), then v(b_i) < v(a_i). Thus if there are m partitions with cardinality >= 2, the values on those m parts are decreasing (listing these parts in order of their second entries).  So for a partition with j parts of which m have cardinality >= 2, we have A_pi(n,k) = (k+1)*k*...*(k+2-j)/m!, which is a polynomial in k of degree j.  The partition of largest cardinality is the partition into singletons, which has m=0.  The result is that for each n, T(n,k) is a monic polynomial of degree n.  To verify the "empirical" formula for a row, only n terms in that row need to be computed. (End)

			Some solutions for n=6 k=4
..1. .2. .3. .3. .2. .2. .2. .4. .3. .2. .3. .4. .3. .2. .1. .2
..4. .3. .3. .4. .0. .0. .4. .3. .1. .0. .2. .3. .4. .0. .3. .4
..3. .0. .1. .3. .3. .0. .3. .4. .3. .4. .3. .3. .3. .3. .4. .0
..3. .4. .3. .2. .1. .0. .3. .2. .1. .3. .4. .3. .2. .0. .0. .0
..3. .1. .4. .2. .2. .2. .2. .4. .3. .0. .3. .3. .3. .4. .4. .2
..2. .4. .3. .1. .3. .3. .0. .4. .2. .1. .2. .0. .0. .3. .4. .3

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A029907(n+1).
Column 2 is A268013.
Column 3 is A267975.
Diagonal is A268104.
Row 1 is A000027(n+1).
Row 2 is A000290(n+1).
Row 3 is A000578(n+1).

Empirical for column k:
k=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) -a(n-4)
k=2: a(n) = 6*a(n-1) -9*a(n-2) -4*a(n-3) +9*a(n-4) +6*a(n-5) +a(n-6)
k=3: [order 8]
k=4: [order 10]
k=5: [order 12]
k=6: [order 14]
k=7: [order 16]
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = n^2 + 2*n + 1
n=3: a(n) = n^3 + 3*n^2 + 3*n + 1
n=4: a(n) = n^4 + 4*n^3 + (11/2)*n^2 + (7/2)*n + 1
n=5: a(n) = n^5 + 5*n^4 + (17/2)*n^3 + 8*n^2 + (9/2)*n + 1
n=6: a(n) = n^6 + 6*n^5 + 12*n^4 + (44/3)*n^3 + (23/2)*n^2 + (29/6)*n + 1
n=7: a(n) = n^7 + 7*n^6 + 16*n^5 + (71/3)*n^4 + (139/6)*n^3 + (43/3)*n^2 + (35/6)*n + 1

A267735 T(n,k)=Number of nXk 0..1 arrays with every repeated value in every row and column greater than or equal to the previous repeated value.

Original entry on oeis.org

2, 4, 4, 8, 16, 8, 15, 64, 64, 15, 28, 225, 512, 225, 28, 51, 784, 3375, 3375, 784, 51, 92, 2601, 21952, 39330, 21952, 2601, 92, 164, 8464, 132651, 451278, 451278, 132651, 8464, 164, 290, 26896, 778688, 4705642, 9116396, 4705642, 778688, 26896, 290, 509

Offset: 1

R. H. Hardin, Jan 20 2016

Table starts
...2......4.........8..........15.............28...............51
...4.....16........64.........225............784.............2601
...8.....64.......512........3375..........21952...........132651
..15....225......3375.......39330.........451278..........4705642
..28....784.....21952......451278........9116396........163879101
..51...2601....132651.....4705642......163879101.......4977655636
..92...8464....778688....47293550.....2818160392.....143643342651
.164..26896...4410944...453985015....45839624169....3885150758271
.290..84100..24389000..4225323881...718310821738..100631663710208
.509.259081.131872229.38217977443.10874476927778.2504347486850059

			Some solutions for n=4 k=4
..0..1..1..1....1..1..1..0....0..0..1..0....0..1..1..1....0..0..1..1
..1..0..0..0....0..0..1..1....1..0..1..0....0..0..1..1....0..1..0..0
..1..0..0..1....0..1..1..0....0..0..0..0....1..1..1..1....0..1..0..0
..0..0..1..1....1..1..1..0....0..0..1..1....0..0..1..0....1..1..1..0

R. H. Hardin, Table of n, a(n) for n = 1..180

Column 1 is A029907(n+1).

Empirical for column k:
k=1: a(n) = 2*a(n-1) +a(n-2) -2*a(n-3) -a(n-4)
k=2: [order 9]
k=3: [order 16]
k=4: [order 42]

Showing 1-10 of 31 results. Next

cp's OEIS Frontend

A172050 A008585+A029907.

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A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

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A023610 Convolution of Fibonacci numbers and {F(2), F(3), F(4), ...}.

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A010049 Second-order Fibonacci numbers.

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A038792 Rectangular array defined by T(i,1) = T(1,j) = 1 for i >= 1 and j >= 1; T(i,j) = max(T(i-1,j) + T(i-1,j-1), T(i-1,j-1) + T(i,j-1)) for i >= 2, j >= 2, read by antidiagonals.

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A267788 T(n,k)=Number of nXk 0..1 arrays with every repeated value in every row greater than or equal to, and in every column greater than, the previous repeated value.

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