A030267 Compose the natural numbers with themselves, A(x) = B(B(x)) where B(x) = x/(1-x)^2 is the generating function for natural numbers.
1, 4, 14, 46, 145, 444, 1331, 3926, 11434, 32960, 94211, 267384, 754309, 2116936, 5914310, 16458034, 45638101, 126159156, 347769719, 956238170, 2623278946, 7181512964, 19622668679, 53522804976, 145753273225, 396323283724, 1076167858046, 2918447861686
Offset: 1
Examples
From _Petros Hadjicostas_, Jun 24 2019: (Start) Recall that with m-color compositions, a part of size m may be colored with one of m colors. We have a(1) = 1 because we only have one colored composition, namely 1_1, that has only 1 part. We have a(2) = 4 because we have the following colored compositions of n = 2: 2_1, 2_2, 1_1 + 1_1; hence, a(2) = 1 + 1 + 2 = 4. We have a(3) = 14 because we have the following colored compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 2_1 + 1_1, 2_2 + 1_1, 1_1 + 1_1 + 1_1; hence, a(3) = 1 + 1 + 1 + 2 + 2 + 2 + 2 + 3 = 14. We have a(14) = 46 because we have the following colored compositions of n = 4: (i) 4_1, 4_2, 4_3, 4_4; with a total of 4 parts. (ii) 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 3_1 + 1_1, 3_2 + 1_1, 3_3 + 1_1, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_1, 2_2 + 2_2; with a total of 2 x 10 = 20 parts. (iii) 1_1 + 1_1 + 2_1, 1_1 + 1_1 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 2_1 + 1_1 + 1_1, 2_2 + 1_1 + 1_1; with a total of 3 x 6 = 18 parts. (iv) 1_1 + 1_1 + 1_1 + 1_1; with a total of 4 parts. Hence, a(4) = 4 + 20 + 18 + 4 = 46. (End)
References
- R. P. Grimaldi, Compositions and the alternate Fibonacci numbers, Congressus Numerantium, 186, 2007, 81-96.
Links
- T. D. Noe, Table of n, a(n) for n = 1..200
- A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (11) (2000), 1421-1427.
- E. Barcucci, A. Del Lungo, S. Fezzi, and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170 (1997), 211-217.
- C. G. Bower, Transforms (2).
- R. X. F. Chen and L. W. Shapiro, On sequences G(n) satisfying G(n)=(d+2)G(n-1)-G(n-2), J. Integer Seq. 10 (2007), Article #07.8.1; see Proposition 17.
- Éva Czabarka, Rigoberto Flórez, and Leandro Junes, Some Enumerations on Non-Decreasing Dyck Paths, Electron. J. Combin., 22(1) (2015), #P1.3.
- Éva Czabarka, Rigoberto Flórez, and Leandro Junes, A Discrete Convolution on the Generalized Hosoya Triangle, J. Integer Seq., 18 (2015), Article #15.1.6.
- Éva Czabarka, Rigoberto Flórez, Leandro Junes, and José L. Ramírez, Enumerations of peaks and valleys on non-decreasing Dyck paths, Discrete Math. 341 (10) (2018), 2789-2807.
- A. Denise and R. Simion, Two combinatorial statistics on Dyck paths, Discrete Math., 137 (1995), 155-176.
- Rigoberto Flórez, Leandro Junes, Luisa M. Montoya, and José L. Ramírez, Counting Subwords in Non-Decreasing Dyck Paths, Journal of Integer Sequences, Vol. 28 (2025), Article 25.1.6. See pp. 6, 14-15, 17, 19.
- Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
- Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
- Milan Janjic, Hessenberg Matrices and Integer Sequences, J. Integer Seq. 13 (2010), Article #10.7.8.
- N. J. A. Sloane, Transforms.
- Index entries for two-way infinite sequences
- Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).
Crossrefs
Programs
-
Maple
with(combinat): L[0]:=2: L[1]:=1: for n from 2 to 60 do L[n]:=L[n-1] +L[n-2] end do: seq(2*fibonacci(2*n)*1/5+(1/5)*n*L[2*n],n=1..30); # Emeric Deutsch, Jul 21 2008
-
Mathematica
Table[Sum[k Binomial[n+k-1,2k-1],{k,n}],{n,30}] (* or *) LinearRecurrence[ {6,-11,6,-1},{1,4,14,46},30] (* Harvey P. Dale, Aug 01 2011 *)
-
PARI
a(n)=(2*n*fibonacci(2*n+1)+(2-n)*fibonacci(2*n))/5
Formula
a(n) = -a(-n) = (2n * F(2n+1) + (2 - n) * F(2n))/5 with F(n) = A000045(n) (Fibonacci numbers).
G.f.: x * (1 - x)^2/(1 - 3*x + x^2)^2.
a(n) = Sum_{k = 1..n} k*C(n + k - 1, 2*k - 1).
a(n) = (2/5)*F(2*n) + (1/5)*n*L(2*n), where F(k) are the Fibonacci numbers (F(0)=0, F(1)=1) and L(k) are the Lucas numbers (L(0) = 2, L(1) = 1). - Emeric Deutsch, Jul 21 2008
a(0) = 1, a(1) = 4, a(2) = 14, a(3) = 46, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - Harvey P. Dale, Aug 01 2011
a(n) = ((3 - sqrt(5))^n*(5*n - 2*sqrt(5)) + (3 + sqrt(5))^n*(5*n + 2*sqrt(5)))/ (25*2^n). - Peter Luschny, Mar 07 2022
E.g.f.: exp(3*x/2)*(15*x*cosh(sqrt(5)*x/2) + sqrt(5)*(4 + 5*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025
Extensions
Name clarified using a comment of the author by Peter Luschny, Aug 03 2019
Comments