cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A011782 Coefficients of expansion of (1-x)/(1-2*x) in powers of x.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592
Offset: 0

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Author

Lee D. Killough (killough(AT)wagner.convex.com)

Keywords

Comments

Apart from initial term, same as A000079 (powers of 2).
Number of compositions (ordered partitions) of n. - Toby Bartels, Aug 27 2003
Number of ways of putting n unlabeled items into (any number of) labeled boxes where every box contains at least one item. Also "unimodal permutations of n items", i.e., those which rise then fall. (E.g., for three items: ABC, ACB, BCA and CBA are unimodal.) - Henry Bottomley, Jan 17 2001
Number of permutations in S_n avoiding the patterns 213 and 312. - Tuwani Albert Tshifhumulo, Apr 20 2001. More generally (see Simion and Schmidt), the number of permutations in S_n avoiding (i) the 123 and 132 patterns; (ii) the 123 and 213 patterns; (iii) the 132 and 213 patterns; (iv) the 132 and 231 patterns; (v) the 132 and 312 patterns; (vi) the 213 and 231 patterns; (vii) the 213 and 312 patterns; (viii) the 231 and 312 patterns; (ix) the 231 and 321 patterns; (x) the 312 and 321 patterns.
a(n+2) is the number of distinct Boolean functions of n variables under action of symmetric group.
Number of unlabeled (1+2)-free posets. - Detlef Pauly, May 25 2003
Image of the central binomial coefficients A000984 under the Riordan array ((1-x), x*(1-x)). - Paul Barry, Mar 18 2005
Binomial transform of (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...); inverse binomial transform of A007051. - Philippe Deléham, Jul 04 2005
Also, number of rationals in [0, 1) whose binary expansions terminate after n bits. - Brad Chalfan, May 29 2006
Equals row sums of triangle A144157. - Gary W. Adamson, Sep 12 2008
Prepend A089067 with a 1, getting (1, 1, 3, 5, 13, 23, 51, ...) as polcoeff A(x); then (1, 1, 2, 4, 8, 16, ...) = A(x)/A(x^2). - Gary W. Adamson, Feb 18 2010
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 2, 8, 32 and 128, lead to this sequence. For the corner squares these vectors lead to the companion sequence A094373. - Johannes W. Meijer, Aug 15 2010
From Paul Curtz, Jul 20 2011: (Start)
Array T(m,n) = 2*T(m,n-1) + T(m-1,n):
1, 1, 2, 4, 8, 16, ... = a(n)
1, 3, 8, 20, 48, 112, ... = A001792,
1, 5, 18, 56, 160, 432, ... = A001793,
1, 7, 32, 120, 400, 1232, ... = A001794,
1, 9, 50, 220, 840, 2912, ... = A006974, followed with A006975, A006976, gives nonzero coefficients of Chebyshev polynomials of first kind A039991 =
1,
1, 0,
2, 0, -1,
4, 0, -3, 0,
8, 0, -8, 0, 1.
T(m,n) third vertical: 2*n^2, n positive (A001105).
Fourth vertical appears in Janet table even rows, last vertical (A168342 array, A138509, rank 3, 13, = A166911)). (End)
A131577(n) and differences are:
0, 1, 2, 4, 8, 16,
1, 1, 2, 4, 8, 16, = a(n),
0, 1, 2, 4, 8, 16,
1, 1, 2, 4, 8, 16.
Number of 2-color necklaces of length 2n equal to their complemented reversal. For length 2n+1, the number is 0. - David W. Wilson, Jan 01 2012
Edges and also central terms of triangle A198069: a(0) = A198069(0,0) and for n > 0: a(n) = A198069(n,0) = A198069(n,2^n) = A198069(n,2^(n-1)). - Reinhard Zumkeller, May 26 2013
These could be called the composition numbers (see the second comment) since the equivalent sequence for partitions is A000041, the partition numbers. - Omar E. Pol, Aug 28 2013
Number of self conjugate integer partitions with exactly n parts for n>=1. - David Christopher, Aug 18 2014
The sequence is the INVERT transform of (1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...). - Gary W. Adamson, Jul 16 2015
Number of threshold graphs on n nodes [Hougardy]. - Falk Hüffner, Dec 03 2015
Number of ternary words of length n in which binary subwords appear in the form 10...0. - Milan Janjic, Jan 25 2017
a(n) is the number of words of length n over an alphabet of two letters, of which one letter appears an even number of times (the empty word of length 0 is included). See the analogous odd number case in A131577, and the Balakrishnan reference in A006516 (the 4-letter odd case), pp. 68-69, problems 2.66, 2.67 and 2.68. - Wolfdieter Lang, Jul 17 2017
Number of D-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are D-equivalent iff the positions of pattern D are identical in these paths. - Sergey Kirgizov, Apr 08 2018
Number of color patterns (set partitions) for an oriented row of length n using two or fewer colors (subsets). Two color patterns are equivalent if we permute the colors. For a(4)=8, the 4 achiral patterns are AAAA, AABB, ABAB, and ABBA; the 4 chiral patterns are the 2 pairs AAAB-ABBB and AABA-ABAA. - Robert A. Russell, Oct 30 2018
The determinant of the symmetric n X n matrix M defined by M(i,j) = (-1)^max(i,j) for 1 <= i,j <= n is equal to a(n) * (-1)^(n*(n+1)/2). - Bernard Schott, Dec 29 2018
For n>=1, a(n) is the number of permutations of length n whose cyclic representations can be written in such a way that when the cycle parentheses are removed what remains is 1 through n in natural order. For example, a(4)=8 since there are exactly 8 permutations of this form, namely, (1 2 3 4), (1)(2 3 4), (1 2)(3 4), (1 2 3)(4), (1)(2)(3 4), (1)(2 3)(4), (1 2)(3)(4), and (1)(2)(3)(4). Our result follows readily by conditioning on k, the number of parentheses pairs of the form ")(" in the cyclic representation. Since there are C(n-1,k) ways to insert these in the cyclic representation and since k runs from 0 to n-1, we obtain a(n) = Sum_{k=0..n-1} C(n-1,k) = 2^(n-1). - Dennis P. Walsh, May 23 2020
Maximum number of preimages that a permutation of length n + 1 can have under the consecutive-231-avoiding stack-sorting map. - Colin Defant, Aug 28 2020
a(n) is the number of occurrences of the empty set {} in the von Neumann ordinals from 0 up to n. Each ordinal k is defined as the set of all smaller ordinals: 0 = {}, 1 = {0}, 2 = {0,1}, etc. Since {} is the foundational element of all ordinals, the total number of times it appears grows as powers of 2. - Kyle Wyonch, Mar 30 2025

Examples

			G.f. = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 32*x^6 + 64*x^7 + 128*x^8 + ...
    ( -1   1  -1)
det (  1   1   1)  = 4
    ( -1  -1  -1)
		

References

  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education Journal, Vol. 31, No. 1, pp. 24-28, Winter 1997.
  • S. Kitaev, Patterns in Permutations and Words, Springer-Verlag, 2011. see p. 399 Table A.7
  • Xavier Merlin, Methodix Algèbre, Ellipses, 1995, p. 153.

Crossrefs

Sequences with g.f.'s of the form ((1-x)/(1-2*x))^k: this sequence (k=1), A045623 (k=2), A058396 (k=3), A062109 (k=4), A169792 (k=5), A169793 (k=6), A169794 (k=7), A169795 (k=8), A169796 (k=9), A169797 (k=10).
Cf. A005418 (unoriented), A122746(n-3) (chiral), A016116 (achiral).
Row sums of triangle A100257.
A row of A160232.
Row 2 of A278984.

Programs

  • Haskell
    a011782 n = a011782_list !! n
    a011782_list = 1 : scanl1 (+) a011782_list
    -- Reinhard Zumkeller, Jul 21 2013
    
  • Magma
    [Floor((1+2^n)/2): n in [0..35]]; // Vincenzo Librandi, Aug 21 2011
    
  • Maple
    A011782:= n-> ceil(2^(n-1)): seq(A011782(n), n=0..50); # Wesley Ivan Hurt, Feb 21 2015
    with(PolynomialTools):  A011782:=seq(coeftayl((1-x)/(1-2*x), x = 0, k),k=0..10^2); # Muniru A Asiru, Sep 26 2017
  • Mathematica
    f[s_] := Append[s, Ceiling[Plus @@ s]]; Nest[f, {1}, 32] (* Robert G. Wilson v, Jul 07 2006 *)
    CoefficientList[ Series[(1-x)/(1-2x), {x, 0, 32}], x] (* Robert G. Wilson v, Jul 07 2006 *)
    Table[Sum[StirlingS2[n, k], {k,0,2}], {n, 0, 30}] (* Robert A. Russell, Apr 25 2018 *)
    Join[{1},NestList[2#&,1,40]] (* Harvey P. Dale, Dec 06 2018 *)
  • PARI
    {a(n) = if( n<1, n==0, 2^(n-1))};
    
  • PARI
    Vec((1-x)/(1-2*x) + O(x^30)) \\ Altug Alkan, Oct 31 2015
    
  • Python
    def A011782(n): return 1 if n == 0 else 2**(n-1) # Chai Wah Wu, May 11 2022
  • Sage
    [sum(stirling_number2(n,j) for j in (0..2)) for n in (0..35)] # G. C. Greubel, Jun 02 2020
    

Formula

a(0) = 1, a(n) = 2^(n-1).
G.f.: (1 - x) / (1 - 2*x) = 1 / (1 - x / (1 - x)). - Michael Somos, Apr 18 2012
E.g.f.: cosh(z)*exp(z) = (exp(2*z) + 1)/2.
a(0) = 1 and for n>0, a(n) = sum of all previous terms.
a(n) = Sum_{k=0..n} binomial(n, 2*k). - Paul Barry, Feb 25 2003
a(n) = Sum_{k=0..n} binomial(n,k)*(1+(-1)^k)/2. - Paul Barry, May 27 2003
a(n) = floor((1+2^n)/2). - Toby Bartels (toby+sloane(AT)math.ucr.edu), Aug 27 2003
G.f.: Sum_{i>=0} x^i/(1-x)^i. - Jon Perry, Jul 10 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(k+1, n-k)*binomial(2*k, k). - Paul Barry, Mar 18 2005
a(n) = Sum_{k=0..floor(n/2)} A055830(n-k, k). - Philippe Deléham, Oct 22 2006
a(n) = Sum_{k=0..n} A098158(n,k). - Philippe Deléham, Dec 04 2006
G.f.: 1/(1 - (x + x^2 + x^3 + ...)). - Geoffrey Critzer, Aug 30 2008
a(n) = A000079(n) - A131577(n).
a(n) = A173921(A000079(n)). - Reinhard Zumkeller, Mar 04 2010
a(n) = Sum_{k=2^n..2^(n+1)-1} A093873(k)/A093875(k), sums of rows of the full tree of Kepler's harmonic fractions. - Reinhard Zumkeller, Oct 17 2010
E.g.f.: (exp(2*x)+1)/2 = (G(0) + 1)/2; G(k) = 1 + 2*x/(2*k+1 - x*(2*k+1)/(x + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
A051049(n) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Apr 18 2012
A008619(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Apr 18 2012
INVERT transform is A122367. MOBIUS transform is A123707. EULER transform of A059966. PSUM transform is A000079. PSUMSIGN transform is A078008. BINOMIAL transform is A007051. REVERT transform is A105523. A002866(n) = a(n)*n!. - Michael Somos, Apr 18 2012
G.f.: U(0), where U(k) = 1 + x*(k+3) - x*(k+2)/U(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 10 2012
a(n) = A000041(n) + A056823(n). - Omar E. Pol, Aug 31 2013
E.g.f.: E(0), where E(k) = 1 + x/( 2*k+1 - x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 25 2013
G.f.: 1 + x/(1 + x)*( 1 + 3*x/(1 + 3*x)*( 1 + 5*x/(1 + 5*x)*( 1 + 7*x/(1 + 7*x)*( 1 + ... )))). - Peter Bala, May 27 2017
a(n) = Sum_{k=0..2} stirling2(n, k).
G.f.: Sum_{j=0..k} A248925(k,j)*x^j / Product_{j=1..k} 1-j*x with k=2. - Robert A. Russell, Apr 25 2018
a(n) = A053120(n, n), n >= 0, (main diagonal of triangle of Chebyshev's T polynomials). - Wolfdieter Lang, Nov 26 2019

Extensions

Additional comments from Emeric Deutsch, May 14 2001
Typo corrected by Philippe Deléham, Oct 25 2008

A001519 a(n) = 3*a(n-1) - a(n-2) for n >= 2, with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437, 1134903170, 2971215073, 7778742049, 20365011074, 53316291173, 139583862445, 365435296162, 956722026041
Offset: 0

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Author

Keywords

Comments

This is a bisection of the Fibonacci sequence A000045. a(n) = F(2*n-1), with F(n) = A000045(n) and F(-1) = 1.
Number of ordered trees with n+1 edges and height at most 3 (height=number of edges on a maximal path starting at the root). Number of directed column-convex polyominoes of area n+1. Number of nondecreasing Dyck paths of length 2n+2. - Emeric Deutsch, Jul 11 2001
Terms are the solutions x to: 5x^2-4 is a square, with 5x^2-4 in A081071 and sqrt(5x^2-4) in A002878. - Benoit Cloitre, Apr 07 2002
a(0) = a(1) = 1, a(n+1) is the smallest Fibonacci number greater than the n-th partial sum. - Amarnath Murthy, Oct 21 2002
The fractional part of tau*a(n) decreases monotonically to zero. - Benoit Cloitre, Feb 01 2003
Numbers k such that floor(phi^2*k^2) - floor(phi*k)^2 = 1 where phi=(1+sqrt(5))/2. - Benoit Cloitre, Mar 16 2003
Number of leftist horizontally convex polyominoes with area n+1.
Number of 31-avoiding words of length n on alphabet {1,2,3} which do not end in 3. (E.g., at n=3, we have 111, 112, 121, 122, 132, 211, 212, 221, 222, 232, 321, 322 and 332.) See A028859. - Jon Perry, Aug 04 2003
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=phi=(1+sqrt(5))/2. - Benoit Cloitre, Feb 24 2004
a(1) = 1, a(2) = 2, then the least number such that the square of any term is just less than the geometric mean of its neighbors. a(n+1)*a(n-1) > a(n)^2. - Amarnath Murthy, Apr 06 2004
All positive integer solutions of Pell equation b(n)^2 - 5*a(n+1)^2 = -4 together with b(n)=A002878(n), n >= 0. - Wolfdieter Lang, Aug 31 2004
Essentially same as Pisot sequence E(2,5).
Number of permutations of [n+1] avoiding 321 and 3412. E.g., a(3) = 13 because the permutations of [4] avoiding 321 and 3412 are 1234, 2134, 1324, 1243, 3124, 2314, 2143, 1423, 1342, 4123, 3142, 2413, 2341. - Bridget Tenner, Aug 15 2005
Number of 1324-avoiding circular permutations on [n+1].
A subset of the Markoff numbers (A002559). - Robert G. Wilson v, Oct 05 2005
(x,y) = (a(n), a(n+1)) are the solutions of x/(yz) + y/(xz) + z/(xy) = 3 with z=1. - Floor van Lamoen, Nov 29 2001
Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 1. - Herbert Kociemba, Jun 10 2004
With interpolated zeros, counts closed walks of length n at the start or end node of P_4. a(n) counts closed walks of length 2n at the start or end node of P_4. The sequence 0,1,0,2,0,5,... counts walks of length n between the start and second node of P_4. - Paul Barry, Jan 26 2005
a(n) is the number of ordered trees on n edges containing exactly one non-leaf vertex all of whose children are leaves (every ordered tree must contain at least one such vertex). For example, a(0) = 1 because the root of the tree with no edges is not considered to be a leaf and the condition "all children are leaves" is vacuously satisfied by the root and a(4) = 13 counts all 14 ordered trees on 4 edges (A000108) except (ignore dots)
|..|
.\/.
which has two such vertices. - David Callan, Mar 02 2005
Number of directed column-convex polyominoes of area n. Example: a(2)=2 because we have the 1 X 2 and the 2 X 1 rectangles. - Emeric Deutsch, Jul 31 2006
Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons in extended (1,1)-nanotubes. See Table 1 on page 411 of I. Lukovits and D. Janezic. - Parthasarathy Nambi, Aug 22 2006
Number of free generators of degree n of symmetric polynomials in 3-noncommuting variables. - Mike Zabrocki, Oct 24 2006
Inverse: With phi = (sqrt(5) + 1)/2, log_phi((sqrt(5)*a(n) + sqrt(5*a(n)^2 - 4))/2) = n for n >= 1. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
Consider a teacher who teaches one student, then he finds he can teach two students while the original student learns to teach a student. And so on with every generation an individual can teach one more student then he could before. a(n) starting at a(2) gives the total number of new students/teachers (see program). - Ben Paul Thurston, Apr 11 2007
The Diophantine equation a(n)=m has a solution (for m >= 1) iff ceiling(arcsinh(sqrt(5)*m/2)/log(phi)) != ceiling(arccosh(sqrt(5)*m/2)/log(phi)) where phi is the golden ratio. An equivalent condition is A130255(m)=A130256(m). - Hieronymus Fischer, May 24 2007
a(n+1) = B^(n)(1), n >= 0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 2=`0`, 5=`00`, 13=`000`, ..., in Wythoff code.
Bisection of the Fibonacci sequence into odd-indexed nonzero terms (1, 2, 5, 13, ...) and even-indexed terms (1, 3, 8, 21, ...) may be represented as row sums of companion triangles A140068 and A140069. - Gary W. Adamson, May 04 2008
a(n) is the number of partitions pi of [n] (in standard increasing form) such that Flatten[pi] is a (2-1-3)-avoiding permutation. Example: a(4)=13 counts all 15 partitions of [4] except 13/24 and 13/2/4. Here "standard increasing form" means the entries are increasing in each block and the blocks are arranged in increasing order of their first entries. Also number that avoid 3-1-2. - David Callan, Jul 22 2008
Let P be the partial sum operator, A000012: (1; 1,1; 1,1,1; ...) and A153463 = M, the partial sum & shift operator. It appears that beginning with any randomly taken sequence S(n), iterates of the operations M * S(n), -> M * ANS, -> P * ANS, etc. (or starting with P) will rapidly converge upon a two-sequence limit cycle of (1, 2, 5, 13, 34, ...) and (1, 1, 3, 8, 21, ...). - Gary W. Adamson, Dec 27 2008
Number of musical compositions of Rhythm-music over a time period of n-1 units. Example: a(4)=13; indeed, denoting by R a rest over a time period of 1 unit and by N[j] a note over a period of j units, we have (writing N for N[1]): NNN, NNR, NRN, RNN, NRR, RNR, RRN, RRR, N[2]R, RN[2], NN[2], N[2]N, N[3] (see the J. Groh reference, pp. 43-48). - Juergen K. Groh (juergen.groh(AT)lhsystems.com), Jan 17 2010
Given an infinite lower triangular matrix M with (1, 2, 3, ...) in every column but the leftmost column shifted upwards one row. Then (1, 2, 5, ...) = lim_{n->infinity} M^n. (Cf. A144257.) - Gary W. Adamson, Feb 18 2010
As a fraction: 8/71 = 0.112676 or 98/9701 = 0.010102051334... (fraction 9/71 or 99/9701 for sequence without initial term). 19/71 or 199/9701 for sequence in reverse. - Mark Dols, May 18 2010
For n >= 1, a(n) is the number of compositions (ordered integer partitions) of 2n-1 into an odd number of odd parts. O.g.f.: (x-x^3)/(1-3x^2+x^4) = A(A(x)) where A(x) = 1/(1-x)-1/(1-x^2).
For n > 0, determinant of the n X n tridiagonal matrix with 1's in the super and subdiagonals, (1,3,3,3,...) in the main diagonal, and the rest zeros. - Gary W. Adamson, Jun 27 2011
The Gi3 sums, see A180662, of the triangles A108299 and A065941 equal the terms of this sequence without a(0). - Johannes W. Meijer, Aug 14 2011
The number of permutations for which length equals reflection length. - Bridget Tenner, Feb 22 2012
Number of nonisomorphic graded posets with 0 and 1 and uniform Hasse graph of rank n+1, with exactly 2 elements of each rank between 0 and 1. (Uniform used in the sense of Retakh, Serconek and Wilson. Graded used in R. Stanley's sense that all maximal chains have the same length.)
HANKEL transform of sequence and the sequence omitting a(0) is the sequence A019590(n). This is the unique sequence with that property. - Michael Somos, May 03 2012
The number of Dyck paths of length 2n and height at most 3. - Ira M. Gessel, Aug 06 2012
Pisano period lengths: 1, 3, 4, 3, 10, 12, 8, 6, 12, 30, 5, 12, 14, 24, 20, 12, 18, 12, 9, 30, ... - R. J. Mathar, Aug 10 2012
Primes in the sequence are 2, 5, 13, 89, 233, 1597, 28657, ... (apparently A005478 without the 3). - R. J. Mathar, May 09 2013
a(n+1) is the sum of rising diagonal of the Pascal triangle written as a square - cf. comments in A085812. E.g., 13 = 1+5+6+1. - John Molokach, Sep 26 2013
a(n) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 1, 1, 1; 0, 1, 1] or [1, 1, 1; 0, 1, 1; 1, 1, 1] or [1, 1, 0; 1, 1, 1; 1, 1, 1] or [1, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Except for the initial term, positive values of x (or y) satisfying x^2 - 3xy + y^2 + 1 = 0. - Colin Barker, Feb 04 2014
Except for the initial term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 64 = 0. - Colin Barker, Feb 16 2014
Positive values of x such that there is a y satisfying x^2 - xy - y^2 - 1 = 0. - Ralf Stephan, Jun 30 2014
a(n) is also the number of permutations simultaneously avoiding 231, 312 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
(1, a(n), a(n+1)), n >= 0, are Markoff triples (see A002559 and Robert G. Wilson v's Oct 05 2005 comment). In the Markoff tree they give one of the outer branches. Proof: a(n)*a(n+1) - 1 = A001906(2*n)^2 = (a(n+1) - a(n))^2 = a(n)^2 + a(n+1)^2 - 2*a(n)*a(n+1), thus 1^2 + a(n)^2 + a(n+1)^2 = 3*a(n)*a(n+1). - Wolfdieter Lang, Jan 30 2015
For n > 0, a(n) is the smallest positive integer not already in the sequence such that a(1) + a(2) + ... + a(n) is a Fibonacci number. - Derek Orr, Jun 01 2015
Number of vertices of degree n-2 (n >= 3) in all Fibonacci cubes, see Klavzar, Mollard, & Petkovsek. - Emeric Deutsch, Jun 22 2015
Except for the first term, this sequence can be generated by Corollary 1 (ii) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015
Precisely the numbers F(n)^k + F(n+1)^k that are also Fibonacci numbers with k > 1, see Luca & Oyono. - Charles R Greathouse IV, Aug 06 2015
a(n) = MA(n) - 2*(-1)^n where MA(n) is exactly the maximum area of a quadrilateral with lengths of sides in order L(n-2), L(n-2), F(n+1), F(n+1) for n > 1 and L(n)=A000032(n). - J. M. Bergot, Jan 28 2016
a(n) is the number of bargraphs of semiperimeter n+1 having no valleys (i.e., convex bargraphs). Equivalently, number of bargraphs of semiperimeter n+1 having exactly 1 peak. Example: a(5) = 34 because among the 35 (=A082582(6)) bargraphs of semiperimeter 6 only the one corresponding to the composition [2,1,2] has a valley. - Emeric Deutsch, Aug 12 2016
Integers k such that the fractional part of k*phi is less than 1/k. See Byszewski link p. 2. - Michel Marcus, Dec 10 2016
Number of words of length n-1 over {0,1,2,3} in which binary subwords appear in the form 10...0. - Milan Janjic, Jan 25 2017
With a(0) = 0 this is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the Fibonacci sequence A000045. See a Feb 17 2017 comment on A097805. - Wolfdieter Lang, Feb 17 2017
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) < e(k). [Martinez and Savage, 2.12] - Eric M. Schmidt, Jul 17 2017
Number of permutations of [n] that avoid the patterns 321 and 2341. - Colin Defant, May 11 2018
The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j^2 and j divides 1+i^2. In fact, the pairs (a(n), a(n+1)), n > 0, are all the solutions. - Tomohiro Yamada, Dec 23 2018
Number of permutations in S_n whose principal order ideals in the Bruhat order are lattices (equivalently, modular, distributive, Boolean lattices). - Bridget Tenner, Jan 16 2020
From Wolfdieter Lang, Mar 30 2020: (Start)
a(n) is the upper left entry of the n-th power of the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from A322602: a(n) = ((M_2)^n)[1,1].
Proof: (M_2)^2 = 3*M + 1_2 (with the 2 X 2 unit matrix 1_2) from the characteristic polynomial of M_2 (see a comment in A322602) and the Cayley-Hamilton theorem. The recurrence M^n = M*M^(n-1) leads to (M_n)^n = S(n, 3)*1_2 + S(n-a, 3)*(M - 3*1_2), for n >= 0, with S(n, 3) = F(2(n+1)) = A001906(n+1). Hence ((M_2)^n)[1,1] = S(n, 3) - 2*S(n-1, 3) = a(n) = F(2*n-1) = (1/(2*r+1))*r^(2*n-1)*(1 + (1/r^2)^(2*n-1)), with r = rho(5) = A001622 (golden ratio) (see the first Aug 31 2004 formula, using the recurrence of S(n, 3), and the Michael Somos Oct 28 2002 formula). This proves a conjecture of Gary W. Adamson in A322602.
The ratio a(n)/a(n-1) converges to r^2 = rho(5)^2 = A104457 for n -> infinity (see the a(n) formula in terms of r), which is one of the statements by Gary W. Adamson in A322602. (End)
a(n) is the number of ways to stack coins with a bottom row of n coins such that any coin not on the bottom row touches exactly two coins in the row below, and all the coins on any row are contiguous [Wilf, 2.12]. - Greg Dresden, Jun 29 2020
a(n) is the upper left entry of the (2*n)-th power of the 4 X 4 Jacobi matrix L with L(i,j)=1 if |i-j| = 1 and L(i,j)=0 otherwise. - Michael Shmoish, Aug 29 2020
All positive solutions of the indefinite binary quadratic F(1, -3, 1) := x^2 - 3*x*y + y^2, of discriminant 5, representing -1 (special Markov triples (1, y=x, z=y) if y <= z) are [x(n), y(n)] = [abs(F(2*n+1)), abs(F(2*n-1))], for n = -infinity..+infinity. (F(-n) = (-1)^(n+1)*F(n)). There is only this single family of proper solutions, and there are no improper solutions. [See also the Floor van Lamoen Nov 29 2001 comment, which uses this negative n, and my Jan 30 2015 comment.] - Wolfdieter Lang, Sep 23 2020
These are the denominators of the lower convergents to the golden ratio, tau; they are also the numerators of the upper convergents (viz. 1/1 < 3/2 < 8/5 < 21/13 < ... < tau < ... 13/8 < 5/3 < 2/1). - Clark Kimberling, Jan 02 2022
a(n+1) is the number of subgraphs of the path graph on n vertices. - Leen Droogendijk, Jun 17 2023
For n > 4, a(n+2) is the number of ways to tile this 3 x n "double-box" shape with squares and dominos (reflections or rotations are counted as distinct tilings). The double-box shape is made up of two horizontal strips of length n, connected by three vertical columns of length 3, and the center column can be located anywhere not touching the two outside columns.
_ _ _ _
|||_|||_|||_|||_|||
|| _ |_| _ _ ||
|||_|||_|||_|||_|||. - Greg Dresden and Ruishan Wu, Aug 25 2024
a(n+1) is the number of integer sequences a_1, ..., a_n such that for any number 1 <= k <= n, (a_1 + ... + a_k)^2 = a_1^3 + ... + a_k^3. - Yifan Xie, Dec 07 2024

Examples

			a(3) = 13: there are 14 ordered trees with 4 edges; all of them, except for the path with 4 edges, have height at most 3.
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 13,15.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 188.
  • N. G. de Bruijn, D. E. Knuth, and S. O. Rice, The average height of planted plane trees, in: Graph Theory and Computing (ed. T. C. Read), Academic Press, New York, 1972, pp. 15-22.
  • GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 92.
  • Jurgen Groh, Computerimprovisation mit Markoffketten und "kognitiven Algorithmen", Studienarbeit, Technische Hochschule Darmstadt, 1987.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 39.
  • 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).
  • R. Stanley, Enumerative combinatorics, Vol. 1. Cambridge University Press, Cambridge, 1997, pp. 96-100.
  • H. S. Wilf, Generatingfunctionology, 3rd ed., A K Peters Ltd., Wellesley, MA, 2006, p. 41.

Crossrefs

Fibonacci A000045 = union of this sequence and A001906.
a(n)= A060920(n, 0).
Row 3 of array A094954.
Equals A001654(n+1) - A001654(n-1), n > 0.
A122367 is another version. Inverse sequences A130255 and A130256. Row sums of A140068, A152251, A153342, A179806, A179745, A213948.

Programs

  • GAP
    a:=[1,1];; for n in [3..10^2] do a[n]:=3*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Sep 27 2017
  • Haskell
    a001519 n = a001519_list !! n
    a001519_list = 1 : zipWith (-) (tail a001906_list) a001906_list
    -- Reinhard Zumkeller, Jan 11 2012
    a001519_list = 1 : f a000045_list where f (_:x:xs) = x : f xs
    -- Reinhard Zumkeller, Aug 09 2013
    
  • Magma
    [1] cat [(Lucas(2*n) - Fibonacci(2*n))/2: n in [1..50]]; // Vincenzo Librandi, Jul 02 2014
    
  • Maple
    A001519:=-(-1+z)/(1-3*z+z**2); # Simon Plouffe in his 1992 dissertation; gives sequence without an initial 1
    A001519 := proc(n) option remember: if n=0 then 1 elif n=1 then 1 elif n>=2 then 3*procname(n-1)-procname(n-2) fi: end: seq(A001519(n), n=0..28); # Johannes W. Meijer, Aug 14 2011
  • Mathematica
    Fibonacci /@ (2Range[29] - 1) (* Robert G. Wilson v, Oct 05 2005 *)
    LinearRecurrence[{3, -1}, {1, 1}, 29] (* Robert G. Wilson v, Jun 28 2012 *)
    a[ n_] := With[{c = Sqrt[5]/2}, ChebyshevT[2 n - 1, c]/c]; (* Michael Somos, Jul 08 2014 *)
    CoefficientList[ Series[(1 - 2x)/(1 - 3x + x^2), {x, 0, 30}], x] (* Robert G. Wilson v, Feb 01 2015 *)
  • Maxima
    a[0]:1$ a[1]:1$ a[n]:=3*a[n-1]-a[n-2]$ makelist(a[n],n,0,30); /* Martin Ettl, Nov 15 2012 */
    
  • PARI
    {a(n) = fibonacci(2*n - 1)}; /* Michael Somos, Jul 19 2003 */
    
  • PARI
    {a(n) = real( quadgen(5) ^ (2*n))}; /* Michael Somos, Jul 19 2003 */
    
  • PARI
    {a(n) = subst( poltchebi(n) + poltchebi(n - 1), x, 3/2) * 2/5}; /* Michael Somos, Jul 19 2003 */
    
  • Sage
    [lucas_number1(n,3,1)-lucas_number1(n-1,3,1) for n in range(30)] # Zerinvary Lajos, Apr 29 2009
    

Formula

G.f.: (1-2*x)/(1-3*x+x^2).
G.f.: 1 / (1 - x / (1 - x / (1 - x))). - Michael Somos, May 03 2012
a(n) = A001906(n+1) - 2*A001906(n).
a(n) = a(1-n) for all n in Z.
a(n+2) = (a(n+1)^2+1)/a(n) with a(1)=1, a(2)=2. - Benoit Cloitre, Aug 29 2002
a(n) = (phi^(2*n-1) + phi^(1-2*n))/sqrt(5) where phi=(1+sqrt(5))/2. - Michael Somos, Oct 28 2002
a(n) = A007598(n-1) + A007598(n) = A000045(n-1)^2 + A000045(n)^2 = F(n)^2 + F(n+1)^2. - Henry Bottomley, Feb 09 2001
a(n) = Sum_{k=0..n} binomial(n+k, 2*k). - Len Smiley, Dec 09 2001
a(n) ~ (1/5)*sqrt(5)*phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = Sum_{k=0..n} C(n, k)*F(k+1). - Benoit Cloitre, Sep 03 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 1)=a(n) (this comment is essentially the same as that of L. Smiley). - Benoit Cloitre, Nov 10 2002
a(n) = (1/2)*(3*a(n-1) + sqrt(5*a(n-1)^2-4)). - Benoit Cloitre, Apr 12 2003
Main diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). - Benoit Cloitre, Aug 05 2003
Hankel transform of A002212. E.g., Det([1, 1, 3;1, 3, 10;3, 10, 36]) = 5. - Philippe Deléham, Jan 25 2004
Solutions x > 0 to equation floor(x*r*floor(x/r)) = floor(x/r*floor(x*r)) when r=phi. - Benoit Cloitre, Feb 15 2004
a(n) = Sum_{i=0..n} binomial(n+i, n-i). - Jon Perry, Mar 08 2004
a(n) = S(n-1, 3) - S(n-2, 3) = T(2*n-1, sqrt(5)/2)/(sqrt(5)/2) with S(n, x) = U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first kind. See triangle A049310, resp. A053120. - Wolfdieter Lang, Aug 31 2004
a(n) = ((-1)^(n-1))*S(2*(n-1), i), with the imaginary unit i and S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. - Wolfdieter Lang, Aug 31 2004
a(n) = Sum_{0<=i_1<=i_2<=n} binomial(i_2, i_1)*binomial(n, i_1+i_2). - Benoit Cloitre, Oct 14 2004
a(n) = L(n,3), where L is defined as in A108299; see also A002878 for L(n,-3). - Reinhard Zumkeller, Jun 01 2005
a(n) = a(n-1) + Sum_{i=0..n-1} a(i)*a(n) = F(2*n+1)*Sum_{i=0..n-1} a(i) = F(2*n). - Andras Erszegi (erszegi.andras(AT)chello.hu), Jun 28 2005
The i-th term of the sequence is the entry (1, 1) of the i-th power of the 2 X 2 matrix M = ((1, 1), (1, 2)). - Simone Severini, Oct 15 2005
a(n-1) = (1/n)*Sum_{k=0..n} B(2*k)*F(2*n-2*k)*binomial(2*n, 2*k) where B(2*k) is the (2*k)-th Bernoulli number. - Benoit Cloitre, Nov 02 2005
a(n) = A055105(n,1) + A055105(n,2) + A055105(n,3) = A055106(n,1) + A055106(n,2). - Mike Zabrocki, Oct 24 2006
a(n) = (2/sqrt(5))*cosh((2n-1)*psi), where psi=log(phi) and phi=(1+sqrt(5))/2. - Hieronymus Fischer, Apr 24 2007
a(n) = (phi+1)^n - phi*A001906(n) with phi=(1+sqrt(5))/2. - Reinhard Zumkeller, Nov 22 2007
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3); a(n) = ((sqrt(5) + 5)/10)*(3/2 + sqrt(5)/2)^(n-2) + ((-sqrt(5) + 5)/10)*(3/2 - sqrt(5)/2)^(n-2). - Antonio Alberto Olivares, Mar 21 2008
a(n) = A147703(n,0). - Philippe Deléham, Nov 29 2008
Sum_{n>=0} atan(1/a(n)) = (3/4)*Pi. - Jaume Oliver Lafont, Feb 27 2009
With X,Y defined as X = ( F(n) F(n+1) ), Y = ( F(n+2) F(n+3) ), where F(n) is the n-th Fibonacci number (A000045), it follows a(n+2) = X.Y', where Y' is the transpose of Y (n >= 0). - K.V.Iyer, Apr 24 2009
From Gary Detlefs, Nov 22 2010: (Start)
a(n) = Fibonacci(2*n+2) mod Fibonacci(2*n), n > 1.
a(n) = (Fibonacci(n-1)^2 + Fibonacci(n)^2 + Fibonacci(2*n-1))/2. (End)
INVERT transform is A166444. First difference is A001906. Partial sums is A055588. Binomial transform is A093129. Binomial transform of A000045(n-1). - Michael Somos, May 03 2012
a(n) = 2^n*f(n;1/2), where f(n;d), n=0,1,...,d, denote the so-called delta-Fibonacci numbers (see Witula et al. papers and comments in A000045). - Roman Witula, Jul 12 2012
a(n) = (Fibonacci(n+2)^2 + Fibonacci(n-3)^2)/5. - Gary Detlefs, Dec 14 2012
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 - x/(x*k + 1 )/Q(k+1); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 23 2013
G.f.: (1-2*x)*G(0)/(2-3*x), where G(k) = 1 + 1/( 1 - x*(5*k-9)/(x*(5*k-4) - 6/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 19 2013
G.f.: 1 + x*(1-x^2)*Q(0)/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + 2*x - x^2)/( x*(4*k+4 + 2*x - x^2 ) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 11 2013
G.f.: Q(0,u), where u=x/(1-x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
Sum_{n>=2} 1/(a(n) - 1/a(n)) = 1. Compare with A001906, A007805 and A097843. - Peter Bala, Nov 29 2013
Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n). Then for n > 0, a(n) = F(n)*L(n-1) + (-1)^n. - Charlie Marion, Jan 01 2014
a(n) = A238731(n,0). - Philippe Deléham, Mar 05 2014
1 = a(n)*a(n+2) - a(n+1)*a(n+1) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = (L(2*n+4) + L(2*n-6))/25 for L(n)=A000032(n). - J. M. Bergot, Dec 30 2014
a(n) = (L(n-1)^2 + L(n)^2)/5 with L(n)=A000032(n). - J. M. Bergot, Dec 31 2014
a(n) = (L(n-2)^2 + L(n+1)^2)/10 with L(n)=A000032(n). - J. M. Bergot, Oct 23 2015
a(n) = 3*F(n-1)^2 + F(n-3)*F(n) - 2*(-1)^n. - J. M. Bergot, Feb 17 2016
a(n) = (F(n-1)*L(n) + F(n)*L(n-1))/2 = (A081714(n-1) + A128534(n))/2. - J. M. Bergot, Mar 22 2016
E.g.f.: (2*exp(sqrt(5)*x) + 3 + sqrt(5))*exp(-x*(sqrt(5)-3)/2)/(5 + sqrt(5)). - Ilya Gutkovskiy, Jul 04 2016
a(n) = ((M_2)^n)[1,1] = S(n, 3) - 2*S(n-1, 3), with the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from A322602. For a proof see the Mar 30 2020 comment above. - Wolfdieter Lang, Mar 30 2020
Sum_{n>=1} 1/a(n) = A153387. - Amiram Eldar, Oct 05 2020
a(n+1) = Product_{k=1..n} (1 + 4*cos(2*Pi*k/(2*n + 1))^2). Special case of A099390. - Greg Dresden, Oct 16 2021
a(n+1) = 4^(n+1)*Sum_{k >= n} binomial(2*k,2*n)*(1/5)^(k+1). Cf. A102591. - Peter Bala, Nov 29 2021
a(n) = cosh((2*n-1)*arcsinh(1/2))/sqrt(5/4). - Peter Luschny, May 21 2022
From J. M. Bergot, May 27 2022: (Start)
a(n) = F(n-1)*L(n) - (-1)^n where L(n)=A000032(n) and F(n)=A000045(n).
a(n) = (L(n-1)^2 + L(n-1)*L(n+1))/5 + (-1)^n.
a(n) = 2*(area of a triangle with vertices at (L(n-2), L(n-1)), (F(n), F(n-1)), (L(n), L(n+1))) + 5*(-1)^n for n > 2. (End)
a(n) = A059929(n-1)+A059929(n-2), n>1. - R. J. Mathar, Jul 09 2024

Extensions

Entry revised by N. J. A. Sloane, Aug 24 2006, May 13 2008

A106566 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, ... ] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ... ] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 5, 3, 1, 0, 14, 14, 9, 4, 1, 0, 42, 42, 28, 14, 5, 1, 0, 132, 132, 90, 48, 20, 6, 1, 0, 429, 429, 297, 165, 75, 27, 7, 1, 0, 1430, 1430, 1001, 572, 275, 110, 35, 8, 1, 0, 4862, 4862, 3432, 2002, 1001, 429, 154, 44, 9, 1
Offset: 0

Views

Author

Philippe Deléham, May 30 2005

Keywords

Comments

Catalan convolution triangle; g.f. for column k: (x*c(x))^k with c(x) g.f. for A000108 (Catalan numbers).
Riordan array (1, xc(x)), where c(x) the g.f. of A000108; inverse of Riordan array (1, x*(1-x)) (see A109466).
Diagonal sums give A132364. - Philippe Deléham, Nov 11 2007

Examples

			Triangle begins:
  1;
  0,   1;
  0,   1,   1;
  0,   2,   2,  1;
  0,   5,   5,  3,  1;
  0,  14,  14,  9,  4,  1;
  0,  42,  42, 28, 14,  5, 1;
  0, 132, 132, 90, 48, 20, 6, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0, 1,
  0, 1, 1,
  0, 1, 1, 1,
  0, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
		

Crossrefs

The three triangles A059365, A106566 and A099039 are the same except for signs and the leading term.
See also A009766, A033184, A059365 for other versions.
The following are all versions of (essentially) the same Catalan triangle: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.

Programs

  • Magma
    A106566:= func< n,k | n eq 0 select 1 else (k/n)*Binomial(2*n-k-1, n-k) >;
    [A106566(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 06 2021
    
  • Maple
    A106566 := proc(n,k)
        if n = 0 then
            1;
        elif k < 0 or k > n then
            0;
        else
            binomial(2*n-k-1,n-k)*k/n ;
        end if;
    end proc: # R. J. Mathar, Mar 01 2015
  • Mathematica
    T[n_, k_] := Binomial[2n-k-1, n-k]*k/n; T[0, 0] = 1; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2017 *)
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1&, #(1-Sqrt[1-4#])/(2#)&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
  • PARI
    {T(n, k) = if( k<=0 || k>n, n==0 && k==0, binomial(2*n - k, n) * k/(2*n - k))}; /* Michael Somos, Oct 01 2022 */
  • Sage
    def A106566(n, k): return 1 if (n==0) else (k/n)*binomial(2*n-k-1, n-k)
    flatten([[A106566(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 06 2021
    

Formula

T(n, k) = binomial(2n-k-1, n-k)*k/n for 0 <= k <= n with n > 0; T(0, 0) = 1; T(0, k) = 0 if k > 0.
T(0, 0) = 1; T(n, 0) = 0 if n > 0; T(0, k) = 0 if k > 0; for k > 0 and n > 0: T(n, k) = Sum_{j>=0} T(n-1, k-1+j).
Sum_{j>=0} T(n+j, 2j) = binomial(2n-1, n), n > 0.
Sum_{j>=0} T(n+j, 2j+1) = binomial(2n-2, n-1), n > 0.
Sum_{k>=0} (-1)^(n+k)*T(n, k) = A064310(n). T(n, k) = (-1)^(n+k)*A099039(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for x = 0,1,2,3,4,5,6,7,8 respectively.
Sum_{k>=0} T(n, k)*x^(n-k) = C(x, n); C(x, n) are the generalized Catalan numbers.
Sum_{j=0..n-k} T(n+k,2*k+j) = A039599(n,k).
Sum_{j>=0} T(n,j)*binomial(j,k) = A039599(n,k).
Sum_{k=0..n} T(n,k)*A000108(k) = A127632(n).
Sum_{k=0..n} T(n,k)*(x+1)^k*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x= 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Aug 25 2007
Sum_{k=0..n} T(n,k)*A000108(k-1) = A121988(n), with A000108(-1)=0. - Philippe Deléham, Aug 27 2007
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Oct 27 2007
T(n,k)*2^(n-k) = A110510(n,k); T(n,k)*3^(n-k) = A110518(n,k). - Philippe Deléham, Nov 11 2007
Sum_{k=0..n} T(n,k)*A000045(k) = A109262(n), A000045: Fibonacci numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A000129(k) = A143464(n), A000129: Pell numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A100335(k) = A002450(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A100334(k) = A001906(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A099322(k) = A015565(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A106233(k) = A003462(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A151821(k+1) = A100320(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A082505(k+1) = A144706(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A000045(2k+2) = A026671(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A122367(k) = A026726(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A008619(k) = A000958(n+1). - Philippe Deléham, Nov 15 2009
Sum_{k=0..n} T(n,k)*A027941(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
G.f.: Sum_{n>=0, k>=0} T(n, k)*x^k*z^n = 1/(1 - x*z*c(z)) where c(z) the g.f. of A000108. - Michael Somos, Oct 01 2022

Extensions

Formula corrected by Philippe Deléham, Oct 31 2008
Corrected by Philippe Deléham, Sep 17 2009
Corrected by Alois P. Heinz, Aug 02 2012

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449
Offset: 0

Views

Author

Bruno Berselli, Feb 25 2014

Keywords

Comments

First bisection of A041611.

Crossrefs

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

Programs

  • Magma
    [n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];
    
  • Magma
    R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020
    
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]
  • PARI
    a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016
  • Sage
    m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
    print(f.coefficients())
    

Formula

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A025169 a(n) = 2*Fibonacci(2*n+2).

Original entry on oeis.org

2, 6, 16, 42, 110, 288, 754, 1974, 5168, 13530, 35422, 92736, 242786, 635622, 1664080, 4356618, 11405774, 29860704, 78176338, 204668310, 535828592, 1402817466, 3672623806, 9615053952, 25172538050, 65902560198, 172535142544
Offset: 0

Views

Author

Keywords

Comments

The pairs (x, y) = (a(n), a(n+1)) satisfy x^2 + y^2 = 3*x*y + 4. - Michel Lagneau, Feb 01 2014

Crossrefs

Programs

  • GAP
    List([0..30], n-> 2*Fibonacci(2*n+2) ); # G. C. Greubel, Jan 16 2020
  • Haskell
    a025169 n = a025169_list !! n
    a025169_list = 2 : 6 : zipWith (-) (map (* 3) $ tail a025169_list) a025169_list
    -- Reinhard Zumkeller, Apr 08 2012
    
  • Magma
    [2*Fibonacci(2*n+2): n in [0..30]]; // Vincenzo Librandi, Jul 11 2011
    
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( 2/(1-3*x + x^2) )); // Marius A. Burtea, Jan 16 2020
    
  • Maple
    seq( 2*fibonacci(2*n+2), n=0..30); # G. C. Greubel, Jan 16 2020
  • Mathematica
    Table[2Fibonacci[2n+2], {n,0,30}] (* or *)
    CoefficientList[Series[2/(1-3x+x^2), {x,0,30}], x] (* Michael De Vlieger, Mar 09 2016 *)
    LinearRecurrence[{3, -1}, {2, 6}, 30] (* Jean-François Alcover, Sep 27 2017 *)
  • PARI
    a(n)=2*fibonacci(2*n+2)
    
  • Sage
    [2*fibonacci(2*n+2) for n in (0..30)] # G. C. Greubel, Jan 16 2020
    

Formula

G.f.: 2/(1 - 3*x + x^2).
a(n) = 3*a(n-1) - a(n-2).
a(n) = 2*A001906(n+1).
a(n) = A111282(n+2). - Reinhard Zumkeller, Apr 08 2012
a(n) = Fibonacci(2*n+1) + Lucas(2*n+1). - Bruno Berselli, Oct 13 2017

Extensions

Better description from Michael Somos

A081068 a(n) = (Lucas(4*n+2) + 2)/5, or Fibonacci(2*n+1)^2, or A081067(n)/5.

Original entry on oeis.org

1, 4, 25, 169, 1156, 7921, 54289, 372100, 2550409, 17480761, 119814916, 821223649, 5628750625, 38580030724, 264431464441, 1812440220361, 12422650078084, 85146110326225, 583600122205489, 4000054745112196, 27416783093579881
Offset: 0

Views

Author

R. K. Guy, Mar 04 2003

Keywords

Comments

Indices of 12-gonal numbers which are also squares (A342709). - Bernard Schott, Mar 19 2021
Values of y in solutions of x^2 = 5*y^2 - 4*y in positive integers. See A360467 for how this relates to a problem regarding the subdivision of a square into four triangles of integer area. - Alexander M. Domashenko, Feb 26 2023
And the corresponding x values of x^2 = 5*y^2 - 4*y are in A033890. - Bernard Schott, Feb 26 2023

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 19.
  • Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Crossrefs

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081067.

Programs

  • Magma
    I:=[1, 4, 25]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012
    
  • Magma
    [(Lucas(4*n+2) + 2)/5: n in [0..30]]; // G. C. Greubel, Dec 17 2017
    
  • Maple
    luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,(luc(4*n+2)+2)/5) od: # James Sellers, Mar 05 2003
  • Mathematica
    CoefficientList[Series[-(1-4*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,4,25},50] (* Vincenzo Librandi, Jun 26 2012 *)
    Table[(LucasL[4*n+2] + 2)/5, {n,0,30}] (* G. C. Greubel, Dec 17 2017 *)
  • PARI
    main(size)={ return(concat([1],vector(size,n,fibonacci(2*n+1)^2))) } /* Anders Hellström, Jul 11 2015 */
    
  • PARI
    for(n=0,30, print1(fibonacci(2*n+1)^2, ", ")) \\ G. C. Greubel, Dec 17 2017

Formula

a(n) = A001519(n+1)^2 = A122367(n)^2 = A058038(n) + 1.
a(n) = A103433(n+1) - A103433(n).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(2*n)*Fibonacci(2*n+2) +1. - Gary Detlefs, Apr 01 2012
G.f.: (1-4*x+x^2)/((1-x)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012
Sum_{n>=0} 1/(a(n) + 1) = 1/3*sqrt(5). - Peter Bala, Nov 30 2013
Sum_{n>=0} 1/a(n) = sqrt(5) * Sum_{n>=1} (-1)^(n+1)*n/Fibonacci(2*n) (Jennings, 1994). - Amiram Eldar, Oct 30 2020
Product_{n>=1} (1 + 1/a(n)) = phi^2/2 (A239798), where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021

Extensions

More terms from James Sellers, Mar 05 2003

A026726 a(n) = T(2n,n), T given by A026725.

Original entry on oeis.org

1, 2, 7, 27, 108, 440, 1812, 7514, 31307, 130883, 548547, 2303413, 9686617, 40783083, 171868037, 724837891, 3058850316, 12915186640, 54554594416, 230526280814, 974414815782, 4119854160332, 17422801069670, 73695109608352, 311768697325788, 1319136935150530
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • GAP
    List([0..30], n-> Sum([0..n], k-> (2*k+1)*Binomial(2*n,n-k)*
    Fibonacci(k+1)/(n+k+1) )); # G. C. Greubel, Jul 16 2019
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 4*x*(1-Sqrt(1-4*x))/(8*x^2-(1-Sqrt(1-4*x))^3) )); // G. C. Greubel, Jul 16 2019
    
  • Maple
    A026726 := proc(n)
        A026725(2*n,n) ;
    end proc:
    seq(A026726(n),n=0..10) ; # R. J. Mathar, Oct 26 2019
  • Mathematica
    CoefficientList[Series[4*x*(1-Sqrt[1-4*x])/(8*x^2-(1-Sqrt[1-4*x])^3), {x,0,30}], x] (* G. C. Greubel, Jul 16 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec(4*x*(1-sqrt(1-4*x))/(8*x^2-(1-sqrt(1-4*x))^3)) \\ G. C. Greubel, Jul 16 2019
    
  • Sage
    (4*x*(1-sqrt(1-4*x))/(8*x^2-(1-sqrt(1-4*x))^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 16 2019
    

Formula

From Philippe Deléham, Feb 11 2009: (Start)
a(n) = Sum_{k=0..n} A039599(n,k)*A000045(k+1).
a(n) = Sum_{k=0..n} A106566(n,k)*A122367(k). (End)
From Philippe Deléham, Feb 02 2014: (Start)
a(n) = Sum_{k=0..n} A236843(n+k,2*k).
a(n) = Sum_{k=0..n} A236830(n,k).
a(n) = A236830(n+1,1).
a(n) = A165407(3*n).
G.f.: C(x)/(1-x*C(x)^3), C(x) the g.f. of A000108. (End)
n*(5*n-11)*a(n) +2*(-20*n^2+59*n-30)*a(n-1) +15*(5*n^2-19*n+16)*a(n-2) +2*(5*n-6)*(2*n-5)*a(n-3)=0. - R. J. Mathar, Oct 26 2019
n*a(n) +(-7*n+4)*a(n-1) +(7*n-2)*a(n-2) +(19*n-60)*a(n-3) +2*(2*n-7)*a(n-4)=0. - R. J. Mathar, Oct 26 2019

A187617 Array T(m,n) read by antidiagonals: number of domino tilings of the 2m X 2n grid (m>=0, n>=0).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 5, 1, 1, 13, 36, 13, 1, 1, 34, 281, 281, 34, 1, 1, 89, 2245, 6728, 2245, 89, 1, 1, 233, 18061, 167089, 167089, 18061, 233, 1, 1, 610, 145601, 4213133, 12988816, 4213133, 145601, 610, 1
Offset: 0

Views

Author

N. J. A. Sloane, Mar 11 2011

Keywords

Comments

A099390 is the main entry for this problem.
The even-indexed rows and columns of the square array in A187596.
Row (and column) 2 is given by A122367. - Nathaniel Johnston, Mar 22 2011

Examples

			The array begins:
  1,  1,     1,       1,          1,            1, ...
  1,  2,     5,      13,         34,           89, ...
  1,  5,    36,     281,       2245,        18061, ...
  1, 13,   281,    6728,     167089,      4213133, ...
  1, 34,  2245,  167089,   12988816,   1031151241, ...
  1, 89, 18061, 4213133, 1031151241, 258584046368, ...
		

Crossrefs

A187618 is the triangle version.
Main diagonal is A004003. Second and third rows give A001519, A188899.

Programs

  • Maple
    ft:=(m,n)->
    2^(m*n/2)*mul( mul(
    (cos(Pi*i/(n+1))^2+cos(Pi*j/(m+1))^2), j=1..m/2), i=1..n/2);
    T:=(m,n)->round(evalf(ft(m,n),300));
  • Mathematica
    T[m_, n_] := Product[2(2 + Cos[(2j Pi)/(2m+1)] + Cos[(2k Pi)/(2n+1)]), {j, 1, m}, {k, 1, n}];
    Table[T[m-n, n] // Round, {m, 0, 8}, {n, 0, m}] // Flatten (* Jean-François Alcover, Aug 05 2018 *)
  • PARI
    default(realprecision, 120);
    {T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*cos(a*Pi/(2*n+1))^2+4*cos(b*Pi/(2*k+1))^2)))} \\ Seiichi Manyama, Jan 09 2021

Extensions

More terms from Nathaniel Johnston, Mar 22 2011

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

Original entry on oeis.org

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024
Offset: 0

Views

Author

Wolfdieter Lang, Jul 29 2014

Keywords

Comments

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
[1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

Examples

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014
		

Crossrefs

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).

Programs

  • Mathematica
    T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
    Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)
  • Sage
    def T(n, k):
        if k == 0: return (-1)^(n//2)
        if (0 <= n and n < k) or (n == -1 and k == 1): return 0
        return 2*T(n-1, k-1) - T(n-2, k)
    for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

Formula

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A154811 a(n) = Fibonacci(2*n+1) mod 9.

Original entry on oeis.org

1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1
Offset: 0

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Author

Paul Curtz, Jan 15 2009

Keywords

Comments

Periodic with period length 12.

Crossrefs

Programs

Formula

a(n) = A001519(n+1) mod 9 = A122367(n) mod 9 = |A099496(n)| mod 9.
From R. J. Mathar, Apr 10 2009: (Start)
a(n) = a(n-1) - a(n-6) + a(n-7).
G.f.: -(1 + x + 3*x^2 - x^3 + 3*x^4 + x^5 + x^6)/((x - 1)*(x^2 + 1)*(x^4 - x^2 + 1)). (End)

Extensions

Edited by R. J. Mathar, Jan 23 2009
Typo in A-number in first formula corrected by R. J. Mathar, Feb 23 2009
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