A023548 -id:A023548 - cp's OEIS frontend

A001924 Apply partial sum operator twice to Fibonacci numbers.

0, 1, 3, 7, 14, 26, 46, 79, 133, 221, 364, 596, 972, 1581, 2567, 4163, 6746, 10926, 17690, 28635, 46345, 75001, 121368, 196392, 317784, 514201, 832011, 1346239, 2178278, 3524546, 5702854, 9227431, 14930317, 24157781, 39088132, 63245948, 102334116, 165580101

Offset: 0

N. J. A. Sloane

Leading coefficients in certain rook polynomials (for n>=2; see p. 18 of the Riordan paper). - Emeric Deutsch, Mar 08 2004
(1, 3, 7, 14, ...) = row sums of triangle A141289. - Gary W. Adamson, Jun 22 2008
a(n) is the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is at most 2. See example below. Generally, the o.g.f. for the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is <= k is: x/((1-x)*(1-2*x+x^(k+1))). Cf. A000217 the case for k=1, A001477 the case for k=0 (counts singleton subsets). - Geoffrey Critzer, Feb 17 2012
-Fibonacci(n-2) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Dec 31 2012
a(n) is the number of bit strings of length n+1 with the pattern 00 and without the pattern 011, see example. - John M. Campbell, Feb 10 2013
From Jianing Song, Apr 28 2025: (Start)
For n >= 2, a(n-2) is the number of subsets of {1,2,...,n} with 2 or more elements that contain no consecutive elements (i.e., such that the difference of successive elements is at least 2). Note that the number of such subsets with k elements is binomial(n+1-k,k), and Sum_{k=2..floor((n+1)/2)} binomial(n+1-k,k) = F(n+2) - binomial(n+1,0) - binomial(n,1) = F(n+2) - (n+1).
If subsets of {1,2,...,n} are required to contain no consecutive elements module n, then the result is A023548(n-3). (End)

			a(5) = 26 because there are 31 nonempty subsets of {1,2,3,4,5} but 5 of these have successive elements that differ by 3 or more: {1,4}, {1,5}, {2,5}, {1,2,5}, {1,4,5}. - _Geoffrey Critzer_, Feb 17 2012
From _John M. Campbell_, Feb 10 2013: (Start)
There are a(5) = 26 bit strings with the pattern 00 and without the pattern 011 of length 5+1:
   000000, 000001, 000010, 000100, 000101, 001000,
   001001, 001010, 010000, 010001, 010010, 010100,
   100000, 100001, 100010, 100100, 100101, 101000, 101001,
   110000, 110001, 110010, 110100, 111000, 111001, 111100.
(End)

J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23.
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).

T. D. Noe, Table of n, a(n) for n = 0..500
Bader AlBdaiwi, On the Number of Cycles in a Graph, arXiv preprint arXiv:1603.01807 [cs.DM], 2016.
Jean-Luc Baril and Jean-Marcel Pallo, Motzkin subposet and Motzkin geodesics in Tamari lattices, 2013.
Ning-Ning Cao and Feng-Zhen Zhao, Some Properties of Hyperfibonacci and Hyperlucas Numbers, J. Int. Seq. 13 (2010) # 10.8.8.
Hung Viet Chu, Various Sequences from Counting Subsets, arXiv:2005.10081 [math.CO], 2020.
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Ligia Loretta Cristea, Ivica Martinjak, and Igor Urbiha, Hyperfibonacci Sequences and Polytopic Numbers, arXiv:1606.06228 [math.CO], 2016.
Emrah Kiliç and Pantelimon Stănică, Generating matrices for weighted sums of second order linear recurrences, JIS 12 (2009) 09.2.7.
Wolfdieter Lang, Problem B-858, Fibonacci Quarterly, 36 (1998), 373-374, Solution, ibid. 37 (1999) 183-184.
Candice A. Marshall, Construction of Pseudo-Involutions in the Riordan Group, Dissertation, Morgan State University, 2017.
Igor Pak, Boris Shapiro, Ilya Smirnov, and Ken-ichi Yoshida, Hilbert-Kunz multiplicity of quadrics via the Ehrhart theory, Stockholm Univ. (Sweden, 2025). See p. 6.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23. [Annotated scanned copy]
Stacey Wagner, Enumerating Alternating Permutations with One Alternating Descent, DePaul Discoveries: Vol. 2: Iss. 1, Article 2.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A000045, A023548, A001891, A133640, A141289.
Right-hand column 4 of triangle A011794.
Cf. A014162, A039834, A077880, A122595, A129696.
Cf. A065220.

List([0..40], n-> Fibonacci(n+4) -n-3); # G. C. Greubel, Jul 08 2019

a001924 n = a001924_list !! n
a001924_list = drop 3 $ zipWith (-) (tail a000045_list) [0..]
-- Reinhard Zumkeller, Nov 17 2013

[Fibonacci(n+4)-(n+3): n in [0..40]]; // Vincenzo Librandi, Jun 23 2016

A001924:=-1/(z**2+z-1)/(z-1)**2; # Conjectured by Simon Plouffe in his 1992 dissertation.
##
a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|-1|-2|3>>^n.
         <<0, 1, 3, 7>>)[1, 1]:
seq(a(n), n=0..40);  # Alois P. Heinz, Oct 05 2012

a[n_]:= Fibonacci[n+4] -3-n; Array[a, 40, 0]  (* Robert G. Wilson v *)
LinearRecurrence[{3,-2,-1,1},{0,1,3,7},40] (* Harvey P. Dale, Jan 24 2015 *)
Nest[Accumulate,Fibonacci[Range[0,40]],2] (* Harvey P. Dale, Jun 15 2016 *)

a(n)=fibonacci(n+4)-n-3 \\ Charles R Greathouse IV, Feb 24 2011

[fibonacci(n+4) -n-3 for n in (0..40)] # G. C. Greubel, Jul 08 2019

From Wolfdieter Lang: (Start)
G.f.: x/((1-x-x^2)*(1-x)^2).
Convolution of natural numbers n >= 1 with Fibonacci numbers F(k).
a(n) = Fibonacci(n+4) - (3+n). (End)
From Henry Bottomley, Jan 03 2003: (Start)
a(n) = a(n-1) + a(n-2) + n = a(n-1) + A000071(n+2).
a(n) = A001891(n) - a(n-1) = n + A001891(n-1).
a(n) = A065220(n+4) + 1 = A000126(n+1) - 1. (End)
a(n) = Sum_{k=0..n} Sum_{i=0..k} Fibonacci(i). - Benoit Cloitre, Jan 26 2003
a(n) = (sqrt(5)/2 + 1/2)^n*(7*sqrt(5)/10 + 3/2) + (3/2 - 7*sqrt(5)/10)*(sqrt(5)/2 - 1/2)^n*(-1)^n - n - 3. - Paul Barry, Mar 26 2003
a(n) = Sum_{k=0..n} Fibonacci(k)*(n-k). - Benoit Cloitre, Jun 07 2004
A107909(a(n)) = A000225(n) = 2^n - 1. - Reinhard Zumkeller, May 28 2005
a(n) - a(n-1) = A101220(1,1,n). - Ross La Haye, May 31 2006
F(n) + a(n-3) = A133640(n). - Gary W. Adamson, Sep 19 2007
a(n) = A077880(-3-n) = 2*a(n-1) - a(n-3) + 1. - Michael Somos, Dec 31 2012
INVERT transform is A122595. PSUM transform is A014162. PSUMSIGN transform is A129696. BINOMIAL transform of A039834 with 0,1 prepended is this sequence. - Michael Somos, Dec 31 2012
a(n) = A228074(n+1,3) for n > 1. - Reinhard Zumkeller, Aug 15 2013
a(n) = Sum_{k=0..n} Sum_{i=0..n} i * C(n-k,k-i). - Wesley Ivan Hurt, Sep 21 2017
E.g.f.: exp(x/2)*(15*cosh(sqrt(5)*x/2) + 7*sqrt(5)*sinh(sqrt(5)*x/2))/5 - exp(x)*(3 + x). - Stefano Spezia, Jun 25 2022

Description improved by N. J. A. Sloane, Jan 01 1997

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

A052940 a(0) = 1; a(n) = 3*2^n - 1, for n > 0.

Original entry on oeis.org

1, 5, 11, 23, 47, 95, 191, 383, 767, 1535, 3071, 6143, 12287, 24575, 49151, 98303, 196607, 393215, 786431, 1572863, 3145727, 6291455, 12582911, 25165823, 50331647, 100663295, 201326591, 402653183, 805306367, 1610612735, 3221225471, 6442450943, 12884901887

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

A simple regular expression.
Numbers k > 1 such that a(k-1)^2 + a(k) is square, e.g., 5^2 + 11 = 6^2; 11^2 + 23 = 12^2. - Vincenzo Librandi, Aug 06 2010
Numerator of the sum of terms at the n-th level of the Calkin-Wilf tree. - Carl Najafi, Jul 10 2011

Michael De Vlieger, Table of n, a(n) for n = 0..3320
Gennady Eremin, Dyck Numbers, III. Enumeration and bijection with symmetric Dyck paths, arXiv:2302.02765 [math.CO], 2023.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 931
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Apart from initial terms, same as A055010 and A083329.
Subsequence of A036991.
Cf. A000225, A023548, A107909, A134060, A140182.

Concatenation([1], List([1..30], n-> 3*2^n -1)); # G. C. Greubel, Oct 18 2019

[1] cat [3*2^n - 1: n in [1..30]]; // Vincenzo Librandi, Dec 01 2015

spec:= [S,{S=Prod(Sequence(Union(Z,Z)),Union(Sequence(Z),Z,Z))},unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
seq(`if`(n=0,1,3*2^n -1), n=0..30); # G. C. Greubel, Oct 18 2019

Join[{1},Table[3*2^n-1,{n,30}]] (* or *) Join[{1},LinearRecurrence[{3,-2},{5,11},30]] (* Harvey P. Dale, Mar 07 2015 *)

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

Vec((1+2*x-2*x^2)/(-1+2*x)/(-1+x) + O(x^30)) \\ Altug Alkan, Dec 01 2015

print([1] + [(3<Gennady Eremin, Aug 29 2023

[1]+[3*2^n -1 for n in (1..30)] # G. C. Greubel, Oct 18 2019

G.f.: (1+2*x-2*x^2)/((1-x)*(1-2*x)).
a(n) = 3*a(n-1) - 2*a(n-2) for n > 2.
Binomial transform of 3 - 0^n - (-1)^n = (1, 4, 2, 4, 2, 4, 2, ...). - Paul Barry, Jun 30 2003
a(n) = A107909(A023548(n+1)) for n > 1. - Reinhard Zumkeller, May 28 2005
Row sums of triangle A134060. - Gary W. Adamson, Oct 05 2007
Equals row sums of triangle A140182. - Gary W. Adamson, May 11 2008
Equals M*Q, where M is a modified Pascal triangle (1,2) with first term "1" instead of 2; as an infinite lower triangular matrix. Q is the vector (1, 2, 2, 2, ...). - Gary W. Adamson, Nov 30 2015
From Gennady Eremin, Aug 29 2023: (Start)
a(n+1) = 2*a(n) + 1 for n > 0.
a(n) = (A000225(n+1) + A000225(n+2))/2 for n > 0. (End)

More terms from James Sellers, Jun 08 2000

a(30)-a(32) from Vincenzo Librandi, Dec 01 2015

A035612 Horizontal para-Fibonacci sequence: says which column of Wythoff array (starting column count at 1) contains n.

Original entry on oeis.org

1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 6, 1, 2, 3, 1, 4, 1, 2, 7, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 8, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 6, 1, 2, 3, 1, 4, 1, 2, 9, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 6, 1, 2, 3, 1, 4, 1, 2, 7, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 10, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2

Offset: 1

J. H. Conway and N. J. A. Sloane

Ordinal transform of A003603. Removing all 1's from this sequence and decrementing the remaining numbers generates the original sequence. - Franklin T. Adams-Watters, Aug 10 2012
It can be shown that a(n) is the index of the smallest Fibonacci number used in the Zeckendorf representation of n, where f(0)=f(1)=1. - Rachel Chaiser, Aug 18 2017
The asymptotic density of the occurrences of k = 1, 2, ..., is (2-phi)/phi^(k-1), where phi is the golden ratio (A001622). The asymptotic mean of this sequence is 1 + phi (A104457). - Amiram Eldar, Nov 02 2023

			After the first 6 we see "1 2 3 1 4 1 2" then 7.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Paul Curtz, Comments on A035612, Jan 25 2016.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Classic Sequences.

Cf. A019586, A035513, A035614.
Cf. A000045, A003603.
Cf. A003714, A007814, A022340, A022342, A026274.
Cf. A000012, A000027, A001045, A001610, A003622, A023548, A255671, A268034.
Cf. A001622, A104457, A132338.

a035612 = a007814 . a022340
-- Reinhard Zumkeller, Jul 20 2015, Mar 10 2013

f[1] = {1}; f[2] = {1, 2}; f[n_] := f[n] = Join[f[n-1], Most[f[n-2]], {n}]; f[11] (* Jean-François Alcover, Feb 22 2012 *)

The segment between the first M and the first M+1 is given by the segment before the first M-1.
a(A022342(n)) > 1; a(A026274(n) + 1) = 1. - Reinhard Zumkeller, Jul 20 2015
a(n) = v2(A022340(n)), where v2(n) = A007814(n), the dyadic valuation of n. - Ralf Stephan, Jun 20 2004. In other words, a(n) = A007814(A003714(n)) + 1, which is certainly true. - Don Reble, Nov 12 2005
From Rachel Chaiser, Aug 18 2017: (Start)
a(n) = a(p(n))+1 if n = b(p(n)) where p(n) = floor((n+2)/phi)-1 and b(n) = floor((n+1)*phi)-1 where phi=(1+sqrt(5))/2; a(n)=1 otherwise.
a(n) = 3 - n_1 + s_z(n-1) - s_z(n) + s_z(p(n-1)) - s_z(p(n)), where s_z(n) is the Zeckendorf sum of digits of n (A007895), and n_1 is the least significant digit in the Zeckendorf representation of n. (End)

Formula corrected by Tom Edgar, Jul 09 2018

A107909 Numbers having no consecutive zeros or no consecutive ones in binary representation.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 26, 27, 29, 30, 31, 32, 33, 34, 36, 37, 40, 41, 42, 43, 45, 46, 47, 53, 54, 55, 58, 59, 61, 62, 63, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 86, 87, 90, 91, 93, 94, 95, 106, 107, 109, 110, 111

Offset: 0

Reinhard Zumkeller, May 28 2005

Union of A003754 and A003714, complement of A107911;
a(A023548(n+2)) = A052940(n+1) for n>0;
a(A001924(n)) = A000225(n) = 2^n - 1;
a(A000126(n)) = A000079(n) = 2^n for n>0;
A107910(n) = a(n+1) - a(n).

Ivan Neretin, Table of n, a(n) for n = 0..10000
J. M. Dusel, Balanced parabolic quotients and branching rules for Demazure crystals, 2014.
Index entries for 2-automatic sequences.

Cf. A007088, A107907.

foreach $n(1..100){$_=sprintf("%b",$n); print "$n\n" if !m/11/||!m/00/}
# Ivan Neretin, May 01 2016

A213579 Rectangular array: (row n) = bc, where b(h) = F(h), c(h) = n-1+h, where F=A000045 (Fibonacci numbers), n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 3, 2, 7, 5, 3, 14, 11, 7, 4, 26, 21, 15, 9, 5, 46, 38, 28, 19, 11, 6, 79, 66, 50, 35, 23, 13, 7, 133, 112, 86, 62, 42, 27, 15, 8, 221, 187, 145, 106, 74, 49, 31, 17, 9, 364, 309, 241, 178, 126, 86, 56, 35, 19, 10, 596, 507, 397, 295, 211, 146, 98, 63, 39, 21

Offset: 1

Clark Kimberling, Jun 18 2012

Principal diagonal: A213580.
Antidiagonal sums: A053808.
Row 1, (1,1,2,3,5,...)**(1,2,3,4,...): A001924.
Row 2, (1,1,2,3,5,...)**(2,3,4,5,...): A023548.
Row 3, (1,1,2,3,5,...)**(3,4,5,6,...): A023552.
Row 4, (1,1,2,3,5,...)**(4,5,6,7,...): A210730.
Row 5, (1,1,2,3,5,...)**(5,6,7,8,...): A210731.
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....3....7....14...26...46
2....5....11...21...38...66
3....7....15...28...50...86
4....9....19...35...62...106
5....11...23...42...74...126
6....13...27...49...86...146

Clark Kimberling, Antidiagonas n = 1..60, flattened

Cf. A000045, A213500.

Flat(List([1..12], n-> List([1..n], k-> Fibonacci(k+3) + n*Fibonacci(k+2) -(n+k+2) ))); # G. C. Greubel, Jul 08 2019

[[Fibonacci(k+3) + n*Fibonacci(k+2) -(n+k+2): k in [1..n]]: n in [1..12]]; // G. C. Greubel, Jul 08 2019

(* First program *)
b[n_]:= Fibonacci[n]; c[n_]:= n;
T[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
TableForm[Table[T[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[T[n-k+1, k], {n, 12}, {k, n, 1, -1}]] (* A213579 *)
r[n_]:= Table[T[n, k], {k, 40}]
d = Table[T[n, n], {n, 1, 40}] (* A213580 *)
s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A053808 *)
(* Second program *)
Table[Fibonacci[n-k+4] +k*Fibonacci[n-k+3] -(n+3), {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jul 08 2019 *)

t(n,k) = fibonacci(n-k+4) + k*fibonacci(n-k+3) - (n+3);
for(n=1,12, for(k=1,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Jul 08 2019

[[fibonacci(k+3) + n*fibonacci(k+2) -(n+k+2) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 08 2019

T(n,k) = 3*T(n,k-1) - 2*T(n,k-2) - T(n,k-3) + T(n,k-4).
G.f. for row n: f(x)/g(x), where f(x) = n - (n-1)*x and g(x) = (1-x-x^2) *(1-x)^2.
T(n, k) = Fibonacci(k+3) + n*Fibonacci(k+2) - (n+k+2). - G. C. Greubel, Jul 08 2019

A082775 Convolution of natural numbers >= 2 and the partition numbers (A000041).

Original entry on oeis.org

2, 5, 11, 21, 38, 64, 105, 165, 254, 381, 562, 813, 1162, 1636, 2279, 3139, 4285, 5794, 7776, 10353, 13694, 17992, 23502, 30520, 39433, 50687, 64855, 82607, 104785, 132375, 166608, 208921, 261090, 325196, 403779, 499818, 616928, 759335, 932135

Offset: 2

Alford Arnold, May 22 2003

Contribution from George Beck, Jan 08 2011: (Start)
The number of multiset partitions of the n-multiset M={0,0,...,0,1,2} (with n-2 zeros) is sum_{k=0..(n-2)}( (n-k) * p(k) ) where p(k) is the number of partitions of k.
Proof:
For each k = 0, 1, ..., n-2, partition k zeros and add the remaining n-k-2 zeros to the block {1, 2}, to give p(k) partitions.
For each k, partition k zeros and add the remaining n-k-2 zeros to the two blocks {1} and {2} in all possible 1 + n-k-2 ways, which gives (1 + n-k-2) * p(k) partitions.
Together, the number of partitions of M is sum_{k=0..n-2}( (n-k) * p(k) ). (End)
A082775 is the special case of A126442 with n-k = 2.

			a(7) = 64 because (7,5,3,2,1,1) dot (2,3,4,5,6,7) = 14+15+12+10+6+7= 64.

Cf. A023548, A126442, A346822.

Column k=2 of A346426.

f[n_] := Sum[(n - k) PartitionsP[k], {k, 0, n - 2}]; Array[f, 39, 2]

a(n) = a(n-1) + A000041(n) + A000070(n) for n>1. - Alford Arnold, Dec 10 2007
a(n) = n*A000070(n-2) - A182738(n-2) for n>2. - Vaclav Kotesovec, Jun 23 2015
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (2*Pi^2). - Vaclav Kotesovec, Jun 23 2015

More terms from Ray Chandler, Oct 11 2003

A292030 Table read by ascending antidiagonals: T(n,k) = A000045(k+1)*n + A000045(k).

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 3, 3, 3, 2, 4, 4, 5, 5, 3, 5, 5, 7, 8, 8, 5, 6, 6, 9, 11, 13, 13, 8, 7, 7, 11, 14, 18, 21, 21, 13, 8, 8, 13, 17, 23, 29, 34, 34, 21, 9, 9, 15, 20, 28, 37, 47, 55, 55, 34, 10, 10, 17, 23, 33, 45, 60, 76, 89, 89, 55, 11, 11, 19, 26, 38, 53, 73, 97, 123, 144, 144, 89

Offset: 0

Ely Golden, Sep 07 2017

T(n,k) is entry k in the Fibonacci-like sequence which starts n, n+1, i.e., T(n,0) = n, T(n,1) = n+1, T(n,k) = T(n,k-1) + T(n,k-2).
Therefore, T(0,k), T(1,k), and T(2,k) are equivalent to A000045(k), A000045(k+2), and A000045(k+3) respectively.
Any two rows T(x,.) and T(y,.) contain infinitely many differing terms for any values of x and y, provided that max(x,y) >= 3.
Column k is the list of all positive integers congruent to A000045(k) mod A000045(k+1), with the exception of row 0 which is the same as row 1 with 0 included.

			T(4,2) = 9 because 9 is element 2 in the Fibonacci-like sequence starting with 4 and 5.
The array begins:
  0, 1,  1,  2,  3,  5,  8, ...
  1, 2,  3,  5,  8, 13, 21, ...
  2, 3,  5,  8, 13, 21, 34, ...
  3, 4,  7, 11, 18, 29, 47, ...
  4, 5,  9, 14, 23, 37, 60, ...
  5, 6, 11, 17, 28, 45, 73, ...
  6, 7, 13, 20, 33, 53, 86, ...

Ely Golden, Table of n, a(n) for n = 0..10000(contains 140 antidiagonals, flattened).
Ely Golden, List of ordered pairs (j,k) such that T(j,k)=n for n = 0..10000

Cf. A000045, A292031, A292032.

A023548 gives the antidiagonal sums.

T[n_,k_]:=SeriesCoefficient[(x+y-x*y)/((1-x)^2(1-y-y^2)),{x,0,n},{y,0,k}]; Table[T[n-k,k],{n,0,11},{k,0,n}]//Flatten (* Stefano Spezia, May 21 2025 *)

T(n, k) = fibonacci(k+1)*n + fibonacci(k);
tabl(nn) = matrix(nn+1,nn+1, i,j, T(i-1,j-1)); \\ Michel Marcus, Sep 27 2017

def nextAntidiagonal(d):
  if(d[0]==0): return [d[1]+1,0]
  return [d[0]-1,d[1]+1]
def fibonacciM(m,n):
  f0,f1=0,1
  if(n<0):
    for _ in range(-n): f1,f0=f0,f1-f0
  else:
    for _ in range(n): f0,f1=f1,f0+f1
  return f1*m+f0
d=[0,0]
for i in range(10001):
  print(str(i)+" "+str(fibonacciM(d[0],d[1])))
  d=nextAntidiagonal(d)

G.f.: (x + y - x*y)/((1 - x)^2*(1 - y - y^2)). - Stefano Spezia, May 21 2025

A033818 Convolution of natural numbers n >= 1 with Lucas numbers L(k) for k >= -2.

Original entry on oeis.org

3, 5, 9, 14, 22, 34, 53, 83, 131, 208, 332, 532, 855, 1377, 2221, 3586, 5794, 9366, 15145, 24495, 39623, 64100, 103704, 167784, 271467, 439229, 710673, 1149878, 1860526, 3010378, 4870877, 7881227, 12752075, 20633272, 33385316, 54018556, 87403839, 141422361, 228826165, 370248490

Offset: 1

Wolfdieter Lang

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A000032, A000045, A023537, A023548, A033811.

List([1..50], n-> Lucas(1,-1,n+1)[2] +n-1) # G. C. Greubel, Jun 01 2019

[Lucas(n+1) +n-1: n in [1..50]]; // G. C. Greubel, Jun 01 2019

LinearRecurrence[{3,-2,-1,1}, {3,5,9,14}, 50] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2011, modified by G. C. Greubel, Jun 01 2019 *)
Table[LucasL[n+1] +n-1, {n,1,50}] (* G. C. Greubel, Jun 01 2019 *)

{a(n) = fibonacci(n+2) + fibonacci(n) + n-1}; \\ G. C. Greubel, Jun 01 2019

[lucas_number2(n+1,1,-1) +n-1 for n in (1..50)] # G. C. Greubel, Jun 01 2019

a(n) = L(1)*(F(n+1) - 1) + L(0)*F(n) - L(-1)*n, F(n): Fibonacci (A000045), L(n): Lucas (A000032) with L(-n) = (-1)^n*L(n).
G.f.: x*(3-4*x)/((1-x-x^2)*(1-x)^2).
a(n) = Lucas(n+1) + n - 1. - G. C. Greubel, Jun 01 2019

Terms a(31) onward added by G. C. Greubel, Jun 01 2019

cp's OEIS Frontend

A001924 Apply partial sum operator twice to Fibonacci numbers.

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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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A052940 a(0) = 1; a(n) = 3*2^n - 1, for n > 0.

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A035612 Horizontal para-Fibonacci sequence: says which column of Wythoff array (starting column count at 1) contains n.

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A107909 Numbers having no consecutive zeros or no consecutive ones in binary representation.

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A213579 Rectangular array: (row n) = b**c, where b(h) = F(h), c(h) = n-1+h, where F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.

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

A213579 Rectangular array: (row n) = bc, where b(h) = F(h), c(h) = n-1+h, where F=A000045 (Fibonacci numbers), n>=1, h>=1, and = convolution.