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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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
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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

A005349 Niven (or Harshad, or harshad) numbers: numbers that are divisible by the sum of their digits.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90, 100, 102, 108, 110, 111, 112, 114, 117, 120, 126, 132, 133, 135, 140, 144, 150, 152, 153, 156, 162, 171, 180, 190, 192, 195, 198, 200, 201, 204
Offset: 1

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Author

Keywords

Comments

Both spellings, "Harshad" or "harshad", are in use. It is a Sanskrit word, and in Sanskrit there is no distinction between upper- and lower-case letters. - N. J. A. Sloane, Jan 04 2022
z-Niven numbers are numbers n which are divisible by (A*s(n) + B) where A, B are integers and s(n) is sum of digits of n. Niven numbers have A = 1, B = 0. - Ctibor O. Zizka, Feb 23 2008
A070635(a(n)) = 0. A038186 is a subsequence. - Reinhard Zumkeller, Mar 10 2008
A049445 is a subsequence of this sequence. - Ctibor O. Zizka, Sep 06 2010
Complement of A065877; A188641(a(n)) = 1; A070635(a(n)) = 0. - Reinhard Zumkeller, Apr 07 2011
A001101, the Moran numbers, are a subsequence. - Reinhard Zumkeller, Jun 16 2011
A140866 gives the number of terms <= 10^k. - Robert G. Wilson v, Oct 16 2012
The asymptotic density of this sequence is 0 (Cooper and Kennedy, 1984). - Amiram Eldar, Jul 10 2020
From Amiram Eldar, Oct 02 2023: (Start)
Named "Harshad numbers" by the Indian recreational mathematician Dattatreya Ramchandra Kaprekar (1905-1986) in 1955. The meaning of the word is "giving joy" in Sanskrit.
Named "Niven numbers" by Kennedy et al. (1980) after the Canadian-American mathematician Ivan Morton Niven (1915-1999). During a lecture given at the 5th Annual Miami University Conference on Number Theory in 1977, Niven mentioned a question of finding a number that equals twice the sum of its digits, which appeared in the children's pages of a newspaper. (End)

Examples

			195 is a term of the sequence because it is divisible by 15 (= 1 + 9 + 5).
		

References

  • Paul Dahlenberg and T. Edgar, Consecutive factorial base Niven numbers, Fib. Q., 56:2 (2018), 163-166.
  • D. R. Kaprekar, Multidigital Numbers, Scripta Math., Vol. 21 (1955), p. 27.
  • Robert E. Kennedy and Curtis N. Cooper, On the natural density of the Niven numbers, Abstract 816-11-219, Abstracts Amer. Math. Soc., 6 (1985), 17.
  • Robert E. Kennedy, Terry A. Goodman, and Clarence H. Best, Mathematical Discovery and Niven Numbers, The MATYC Journal, Vol. 14, No. 1 (1980), pp. 21-25.
  • József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 381.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 171.

Crossrefs

Cf. A001102 (a subsequence).
Cf. A118363 (for factorial-base analog).
Cf. A330927, A154701, A141769, A330928, A330929, A330930 (start of runs of 2, 3, ..., 7 consecutive Niven numbers).

Programs

  • GAP
    Filtered([1..230],n-> n mod List(List([1..n],ListOfDigits),Sum)[n]=0); # Muniru A Asiru
  • Haskell
    a005349 n = a005349_list !! (n-1)
    a005349_list = filter ((== 0) . a070635) [1..]
    -- Reinhard Zumkeller, Aug 17 2011, Apr 07 2011
    
  • Magma
    [n: n in [1..250] | n mod &+Intseq(n) eq 0];  // Bruno Berselli, May 28 2011
    
  • Magma
    [n: n in [1..250] | IsIntegral(n/&+Intseq(n))];  // Bruno Berselli, Feb 09 2016
    
  • Maple
    s:=proc(n) local N:N:=convert(n,base,10):sum(N[j],j=1..nops(N)) end:p:=proc(n) if floor(n/s(n))=n/s(n) then n else fi end: seq(p(n),n=1..210); # Emeric Deutsch
  • Mathematica
    harshadQ[n_] := Mod[n, Plus @@ IntegerDigits@ n] == 0; Select[ Range[1000], harshadQ] (* Alonso del Arte, Aug 04 2004 and modified by Robert G. Wilson v, Oct 16 2012 *)
    Select[Range[300],Divisible[#,Total[IntegerDigits[#]]]&] (* Harvey P. Dale, Sep 07 2015 *)
  • PARI
    is(n)=n%sumdigits(n)==0 \\ Charles R Greathouse IV, Oct 16 2012
    
  • Python
    A005349 = [n for n in range(1,10**6) if not n % sum([int(d) for d in str(n)])] # Chai Wah Wu, Aug 22 2014
    
  • Sage
    [n for n in (1..10^4) if sum(n.digits(base=10)).divides(n)] # Freddy Barrera, Jul 27 2018
    

A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

Original entry on oeis.org

1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, 212941, 616227, 1787607, 5196627, 15134931, 44152809, 128996853, 377379369, 1105350729, 3241135527, 9513228123, 27948336381, 82176836301, 241813226151, 712070156203, 2098240353907, 6186675630819
Offset: 0

Views

Author

Keywords

Comments

Number of ordered trees with n + 1 edges, having root of odd degree and nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 02 2002
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), running from (0,0) to (n,0) (i.e., grand Motzkin paths of length n). For example, a(3) = 7 because we have HHH, HUD, HDU, UDH, DUH, UHD and DHU. - Emeric Deutsch, May 31 2003
Number of lattice paths from (0,0) to (n,n) using steps (2,0), (0,2), (1,1). It appears that 1/sqrt((1 - x)^2 - 4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1). - Joerg Arndt, Jul 01 2011
Number of lattice paths from (0,0) to (n,n) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Binomial transform of A000984, with interpolated zeros. - Paul Barry, Jul 01 2003
Number of leaves in all 0-1-2 trees with n edges, n > 0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) - Emeric Deutsch, Nov 30 2003
a(n) is the number of UDU-free paths of n + 1 upsteps (U) and n downsteps (D) that start U. For example, a(2) = 3 counts UUUDD, UUDDU, UDDUU. - David Callan, Aug 18 2004
Diagonal sums of triangle A063007. - Paul Barry, Aug 31 2004
Number of ordered ballots from n voters that result in an equal number of votes for candidates A and B in a three candidate election. Ties are counted even when candidates A and B lose the election. For example, a(3) = 7 because ballots of the form (voter-1 choice, voter-2 choice, voter-3 choice) that result in equal votes for candidates A and B are the following: (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A) and (C,C,C). - Dennis P. Walsh, Oct 08 2004
a(n) is the number of weakly increasing sequences (a_1,a_2,...,a_n) with each a_i in [n]={1,2,...,n} and no element of [n] occurring more than twice. For n = 3, the sequences are 112, 113, 122, 123, 133, 223, 233. - David Callan, Oct 24 2004
Note that n divides a(n+1) - a(n). In fact, (a(n+1) - a(n))/n = A007971(n+1). - T. D. Noe, Mar 16 2005
Row sums of triangle A105868. - Paul Barry, Apr 23 2005
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having no H steps on the x-axis. Example: a(3) = 7 because we have UDU, UHD, UHH, UHU, UUD, UUH and UUU. - Emeric Deutsch, Oct 07 2007
Equals right border of triangle A152227; starting with offset 1, the row sums of triangle A152227. - Gary W. Adamson, Nov 29 2008
Starting with offset 1 = iterates of M * [1,1,1,...] where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
a(n) is prime for n = 2, 3 and 4, with no others for n <= 10^5 (E. W. Weisstein, Mar 14 2005). It has apparently not been proved that no [other] prime central trinomials exist. - Jonathan Vos Post, Mar 19 2010
a(n) is not divisible by 3 for n whose base-3 representation contains no 2 (A005836).
a(n) = number of (n-1)-lettered words in the alphabet {1,2,3} with as many occurrences of the substring (consecutive subword) [1,2] as those of [2,1]. See the papers by Ekhad-Zeilberger and Zeilberger. - N. J. A. Sloane, Jul 05 2012
a(n) = coefficient of x^n in (1 + x + x^2)^n. - L. Edson Jeffery, Mar 23 2013
a(n) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that (i.) A and B are disjoint and (ii.) A and B contain the same number of elements. For example, a(2) = 3 because we have: ({},{}) ; ({1},{2}) ; ({2},{1}). - Geoffrey Critzer, Sep 04 2013
Also central terms of A082601. - Reinhard Zumkeller, Apr 13 2014
a(n) is the number of n-tuples with entries 0, 1, or 2 and with the sum of entries equal to n. For n=3, the seven 3-tuples are (1,1,1), (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), and (2,1,0). - Dennis P. Walsh, May 08 2015
The series 2*a(n) + 3*a(n+1) + a(n+2) = 2*A245455(n+3) has Hankel transform of L(2n+1)*2^n, offset n = 1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
The series (2*a(n) + 3*a(n+1) + a(n+2))/2 = A245455(n+3) has Hankel transform of L(2n+1), offset n=1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
Conjecture: An integer n > 3 is prime if and only if a(n) == 1 (mod n^2). We have verified this for n up to 8*10^5, and proved that a(p) == 1 (mod p^2) for any prime p > 3 (cf. A277640). - Zhi-Wei Sun, Nov 30 2016
This is the analog for Coxeter type B of Motzkin numbers (A001006) for Coxeter type A. - F. Chapoton, Jul 19 2017
a(n) is also the number of solutions to the equation x(1) + x(2) + ... + x(n) = 0, where x(1), ..., x(n) are in the set {-1,0,1}. Indeed, the terms in (1 + x + x^2)^n that produce x^n are of the form x^i(1)*x^i(2)*...*x^i(n) where i(1), i(2), ..., i(n) are in {0,1,2} and i(1) + i(2) + ... + i(n) = n. By setting j(t) = i(t) - 1 we obtain that j(1), ..., j(n) satisfy j(1) + ... + j(n) =0 and j(t) in {-1,0,1} for all t = 1..n. - Lucien Haddad, Mar 10 2018
If n is a prime greater than 3 then a(n)-1 is divisible by n^2. - Ira M. Gessel, Aug 08 2021
Let f(m) = ceiling((q+log(q))/log(9)), where q = -log(log(27)/(2*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024
Diagonal of the rational function 1 / (1 - x^2 - y^2 - x*y). - Ilya Gutkovskiy, Apr 23 2025

Examples

			For n = 2, (x^2 + x + 1)^2 = x^4 + 2*x^3 + 3*x^2 + 2*x + 1, so a(2) = 3. - _Michael B. Porter_, Sep 06 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 78 and 163, #19.
  • L. Euler, Exemplum Memorabile Inductionis Fallacis, Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 15, p. 59.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 575.
  • P. Henrici, Applied and Computational Complex Analysis. Wiley, NY, 3 vols., 1974-1986. (Vol. 1, p. 42.)
  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 579.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 74.
  • 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. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 6.3.8.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 22.
  • Lin Yang and S.-L. Yang, The parametric Pascal rhombus. Fib. Q., 57:4 (2019), 337-346. See p. 341.

Crossrefs

INVERT transform is A007971. Partial sums are A097893. Squares are A168597.
Main column of A027907. Column k=2 of A305161. Column k=0 of A328347. Column 1 of A201552(?).
Cf. A001006, A002878, A005043, A005717, A082758 (bisection), A273055 (bisection), A102445, A113302, A113303, A113304, A113305 (divisibility of central trinomial coefficients), A152227, A277640.

Programs

  • Haskell
    a002426 n = a027907 n n  -- Reinhard Zumkeller, Jan 22 2013
    
  • Magma
    P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2)^n)): n in [0..26]]; // Bruno Berselli, Jul 05 2011
    
  • Maple
    A002426 := proc(n) local k;
        sum(binomial(n, k)*binomial(n-k, k), k=0..floor(n/2));
    end proc: # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    # Alternatively:
    a := n -> simplify(GegenbauerC(n,-n,-1/2)):
    seq(a(n), n=0..29); # Peter Luschny, May 07 2016
  • Mathematica
    Table[ CoefficientList[ Series[(1 + x + x^2)^n, {x, 0, n}], x][[ -1]], {n, 0, 27}] (* Robert G. Wilson v *)
    a=b=1; Join[{a,b}, Table[c=((2n-1)b + 3(n-1)a)/n; a=b; b=c; c, {n,2,100}]]; Table[Sqrt[-3]^n LegendreP[n,1/Sqrt[-3]],{n,0,26}] (* Wouter Meeussen, Feb 16 2013 *)
    a[ n_] := If[ n < 0, 0, 3^n Hypergeometric2F1[ 1/2, -n, 1, 4/3]]; (* Michael Somos, Jul 08 2014 *)
    Table[4^n *JacobiP[n,-n-1/2,-n-1/2,-1/2], {n,0,29}] (* Peter Luschny, May 13 2016 *)
    a[n_] := a[n] = Sum[n!/((n - 2*i)!*(i!)^2), {i, 0, n/2}]; Table[a[n], {n, 0, 29}] (* Shara Lalo and Zagros Lalo, Oct 03 2018 *)
  • Maxima
    trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
    makelist(trinomial(n,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */
    
  • Maxima
    makelist(ultraspherical(n,-n,-1/2),n,0,12); /* Emanuele Munarini, Dec 20 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, n))};
    
  • PARI
    /* as lattice paths: same as in A092566 but use */
    steps=[[2, 0], [0, 2], [1, 1]];
    /* Joerg Arndt, Jul 01 2011 */
    
  • PARI
    a(n)=polcoeff(sum(m=0, n, (2*m)!/m!^2 * x^(2*m) / (1-x+x*O(x^n))^(2*m+1)), n) \\ Paul D. Hanna, Sep 21 2013
    
  • Python
    from math import comb
    def A002426(n): return sum(comb(n,k)*comb(k,n-k) for k in range(n+1)) # Chai Wah Wu, Nov 15 2022
  • Sage
    A002426 = lambda n: hypergeometric([-n/2, (1-n)/2], [1], 4)
    [simplify(A002426(n)) for n in (0..29)]
    # Peter Luschny, Sep 17 2014
    
  • Sage
    def A():
        a, b, n = 1, 1, 1
        yield a
        while True:
            yield b
            n += 1
            a, b = b, ((3 * (n - 1)) * a + (2 * n - 1) * b) // n
    A002426 = A()
    print([next(A002426) for  in range(30)])  # _Peter Luschny, May 16 2016
    

Formula

G.f.: 1/sqrt(1 - 2*x - 3*x^2).
E.g.f.: exp(x)*I_0(2x), where I_0 is a Bessel function. - Michael Somos, Sep 09 2002
a(n) = 2*A027914(n) - 3^n. - Benoit Cloitre, Sep 28 2002
a(n) is asymptotic to d*3^n/sqrt(n) with d around 0.5.. - Benoit Cloitre, Nov 02 2002, d = sqrt(3/Pi)/2 = 0.4886025119... - Alec Mihailovs (alec(AT)mihailovs.com), Feb 24 2005
D-finite with recurrence: a(n) = ((2*n - 1)*a(n-1) + 3*(n - 1)*a(n-2))/n; a(0) = a(1) = 1; see paper by Barcucci, Pinzani and Sprugnoli.
Inverse binomial transform of A000984. - Vladeta Jovovic, Apr 28 2003
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, k/2)*(1 + (-1)^k)/2; a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*binomial(2*k, k). - Paul Barry, Jul 01 2003
a(n) = Sum_{k>=0} binomial(n, 2*k)*binomial(2*k, k). - Philippe Deléham, Dec 31 2003
a(n) = Sum_{i+j=n, 0<=j<=i<=n} binomial(n, i)*binomial(i, j). - Benoit Cloitre, Jun 06 2004
a(n) = 3*a(n-1) - 2*A005043(n). - Joost Vermeij (joost_vermeij(AT)hotmail.com), Feb 10 2005
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, n-k). - Paul Barry, Apr 23 2005
a(n) = (-1/4)^n*Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-2*k, n-k)*(-3)^k. - Philippe Deléham, Aug 17 2005
a(n) = A111808(n,n). - Reinhard Zumkeller, Aug 17 2005
a(n) = Sum_{k=0..n} (((1 + (-1)^k)/2)*Sum_{i=0..floor((n-k)/2)} binomial(n, i)*binomial(n-i, i+k)*((k + 1)/(i + k + 1))). - Paul Barry, Sep 23 2005
a(n) = 3^n*Sum_{j=0..n} (-1/3)^j*C(n, j)*C(2*j, j); follows from (a) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/2)^n*Sum_{j=0..n} 3^j*binomial(n, j)*binomial(2*n-2*j, n) = (3/2)^n*Sum_{j=0..n} (1/3)^j*binomial(n, j)*binomial(2*j, n); follows from (c) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/Pi)*Integral_{x=-1..3} x^n/sqrt((3 - x)*(1 + x)) is moment representation. - Paul Barry, Sep 10 2007
G.f.: 1/(1 - x - 2x^2/(1 - x - x^2/(1 - x - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 05 2009
a(n) = sqrt(-1/3)*(-1)^n*hypergeometric([1/2, n+1], [1], 4/3). - Mark van Hoeij, Nov 12 2009
a(n) = (1/Pi)*Integral_{x=-1..1} (1 + 2*x)^n/sqrt(1 - x^2) = (1/Pi)*Integral_{t=0..Pi} (1 + 2*cos(t))^n. - Eli Wolfhagen, Feb 01 2011
In general, g.f.: 1/sqrt(1 - 2*a*x + x^2*(a^2 - 4*b)) = 1/(1 - a*x)*(1 - 2*x^2*b/(G(0)*(a*x - 1) + 2*x^2*b)); G(k) = 1 - a*x - x^2*b/G(k+1); for g.f.: 1/sqrt(1 - 2*x - 3*x^2) = 1/(1 - x)*(1 - 2*x^2/(G(0)*(x - 1) + 2*x^2)); G(k) = 1 - x - x^2/G(k+1), a = 1, b = 1; (continued fraction). - Sergei N. Gladkovskii, Dec 08 2011
a(n) = Sum_{k=0..floor(n/3)} (-1)^k*binomial(2*n-3*k-1, n-3*k)*binomial(n, k). - Gopinath A. R., Feb 10 2012
G.f.: A(x) = x*B'(x)/B(x) where B(x) satisfies B(x) = x*(1 + B(x) + B(x)^2). - Vladimir Kruchinin, Feb 03 2013 (B(x) = x*A001006(x) - Michael Somos, Jul 08 2014)
G.f.: G(0), where G(k) = 1 + x*(2 + 3*x)*(4*k + 1)/(4*k + 2 - x*(2 + 3*x)*(4*k + 2)*(4*k + 3)/(x*(2 + 3*x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
E.g.f.: exp(x) * Sum_{k>=0} (x^k/k!)^2. - Geoffrey Critzer, Sep 04 2013
G.f.: Sum_{n>=0} (2*n)!/n!^2*(x^(2*n)/(1 - x)^(2*n+1)). - Paul D. Hanna, Sep 21 2013
0 = a(n)*(9*a(n+1) + 9*a(n+2) - 6*a(n+3)) + a(n+1)*(3*a(n+1) + 4*a(n+2) - 3*a(n+3)) + a(n+2)*(-a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = hypergeometric([-n/2, (1-n)/2], [1], 4). - Peter Luschny, Sep 17 2014
a(n) = A132885(n,0), that is, a(n) = A132885(A002620(n+1)). - Altug Alkan, Nov 29 2015
a(n) = GegenbauerC(n,-n,-1/2). - Peter Luschny, May 07 2016
a(n) = 4^n*JacobiP[n,-n-1/2,-n-1/2,-1/2]. - Peter Luschny, May 13 2016
From Alexander Burstein, Oct 03 2017: (Start)
G.f.: A(4*x) = B(-x)*B(3*x), where B(x) is the g.f. of A000984.
G.f.: A(2*x)*A(-2*x) = B(x^2)*B(9*x^2).
G.f.: A(x) = 1 + x*M'(x)/M(x), where M(x) is the g.f. of A001006. (End)
a(n) = Sum_{i=0..n/2} n!/((n - 2*i)!*(i!)^2). [Cf. Lalo and Lalo link. It is Luschny's terminating hypergeometric sum.] - Shara Lalo and Zagros Lalo, Oct 03 2018
From Peter Bala, Feb 07 2022: (Start)
a(n)^2 = Sum_{k = 0..n} (-3)^(n-k)*binomial(2*k,k)^2*binomial(n+k,n-k) and has g.f. Sum_{n >= 0} binomial(2*n,n)^2*x^n/(1 + 3*x)^(2*n+1). Compare with the g.f. for a(n) given above by Hanna.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all prime p and positive integers n and k.
Conjecture: The stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all prime p >= 5 and positive integers n and k. (End)
a(n) = A005043(n) + A005717(n) for n >= 1. - Amiram Eldar, May 17 2024
For even n, a(n) = (n-1)!!* 2^{n/2}/ (n/2)!* 2F1(-n/2,-n/2;1/2;1/4). For odd n, a(n) = n!! *2^(n/2-1/2) / (n/2-1/2)! * 2F1(1/2-n/2,1/2-n/2;3/2;1/4). - R. J. Mathar, Mar 19 2025

A001317 Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.

Original entry on oeis.org

1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, 4294967297, 12884901891, 21474836485, 64424509455, 73014444049, 219043332147, 365072220245, 1095216660735, 1103806595329, 3311419785987
Offset: 0

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Comments

The members are all palindromic in binary, i.e., a subset of A006995. - Ralf Stephan, Sep 28 2004
J. H. Conway writes (in Math Forum): at least the first 31 numbers give odd-sided constructible polygons. See also A047999. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005 [This observation was also made in 1982 by N. L. White (see letter). - N. J. A. Sloane, Jun 15 2015]
Decimal number generated by the binary bits of the n-th generation of the Rule 60 elementary cellular automaton. Thus: 1; 0, 1, 1; 0, 0, 1, 0, 1; 0, 0, 0, 1, 1, 1, 1; 0, 0, 0, 0, 1, 0, 0, 0, 1; ... . - Eric W. Weisstein, Apr 08 2006
Limit_{n->oo} log(a(n))/n = log(2). - Bret Mulvey, May 17 2008
Equals row sums of triangle A166548; e.g., 17 = (2 + 4 + 6 + 4 + 1). - Gary W. Adamson, Oct 16 2009
Equals row sums of triangle A166555. - Gary W. Adamson, Oct 17 2009
For n >= 1, all terms are in A001969. - Vladimir Shevelev, Oct 25 2010
Let n,m >= 0 be such that no carries occur when adding them. Then a(n+m) = a(n)*a(m). - Vladimir Shevelev, Nov 28 2010
Let phi_a(n) be the number of a(k) <= a(n) and respectively prime to a(n) (i.e., totient function over {a(n)}). Then, for n >= 1, phi_a(n) = 2^v(n), where v(n) is the number of 0's in the binary representation of n. - Vladimir Shevelev, Nov 29 2010
Trisection of this sequence gives rows of A008287 mod 2 converted to decimal. See also A177897, A177960. - Vladimir Shevelev, Jan 02 2011
Converting the rows of the powers of the k-nomial (k = 2^e where e >= 1) term-wise to binary and reading the concatenation as binary number gives every (k-1)st term of this sequence. Similarly with powers p^k of any prime. It might be interesting to study how this fails for powers of composites. - Joerg Arndt, Jan 07 2011
This sequence appears in Pascal's triangle mod 2 in another way, too. If we write it as
1111111...
10101010...
11001100...
10001000...
we get (taking the period part in each row):
.(1) (base 2) = 1
.(10) = 2/3
.(1100) = 12/15 = 4/5
.(1000) = 8/15
The k-th row, treated as a binary fraction, seems to be equal to 2^k / a(k). - Katarzyna Matylla, Mar 12 2011
From Daniel Forgues, Jun 16-18 2011: (Start)
Since there are 5 known Fermat primes, there are 32 products of distinct Fermat primes (thus there are 31 constructible odd-sided polygons, since a polygon has at least 3 sides). a(0)=1 (empty product) and a(1) to a(31) are those 31 non-products of distinct Fermat primes.
It can be proved by induction that all terms of this sequence are products of distinct Fermat numbers (A000215):
a(0)=1 (empty product) are products of distinct Fermat numbers in { };
a(2^n+k) = a(k) * (2^(2^n)+1) = a(k) * F_n, n >= 0, 0 <= k <= 2^n - 1.
Thus for n >= 1, 0 <= k <= 2^n - 1, and
a(k) = Product_{i=0..n-1} F_i^(alpha_i), alpha_i in {0, 1},
this implies
a(2^n+k) = Product_{i=0..n-1} F_i^(alpha_i) * F_n, alpha_i in {0, 1}.
(Cf. OEIS Wiki links below.) (End)
The bits in the binary expansion of a(n) give the coefficients of the n-th power of polynomial (X+1) in ring GF(2)[X]. E.g., 3 ("11" in binary) stands for (X+1)^1, 5 ("101" in binary) stands for (X+1)^2 = (X^2 + 1), and so on. - Antti Karttunen, Feb 10 2016

Examples

			Given a(5)=51, a(6)=85 since a(5) XOR 2*a(5) = 51 XOR 102 = 85.
From _Daniel Forgues_, Jun 18 2011: (Start)
  a(0) = 1 (empty product);
  a(1) = 3 = 1 * F_0 = a(2^0+0) = a(0) * F_0;
  a(2) = 5 = 1 * F_1 = a(2^1+0) = a(0) * F_1;
  a(3) = 15 = 3 * 5 = F_0 * F_1 = a(2^1+1) = a(1) * F_1;
  a(4) = 17 = 1 * F_2 = a(2^2+0) = a(0) * F_2;
  a(5) = 51 = 3 * 17 = F_0 * F_2 = a(2^2+1) = a(1) * F_2;
  a(6) = 85 = 5 * 17 = F_1 * F_2 = a(2^2+2) = a(2) * F_2;
  a(7) = 255 = 3 * 5 * 17 = F_0 * F_1 * F_2 = a(2^2+3) = a(3) * F_2;
  ... (End)
		

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic sequences, Cambridge University Press, 2003, p. 113.
  • Henry Wadsworth Gould, Exponential Binomial Coefficient Series, Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sept. 1961.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Crossrefs

Cf. A038183 (odd bisection, 1D Cellular Automata Rule 90).
Iterates of A048724 (starting from 1).
Row 3 of A048723.
Positions of records in A268389.
Positions of ones in A268669 and A268384 (characteristic function).
Not the same as A045544 nor as A053576.
Cf. A045544.

Programs

  • Haskell
    a001317 = foldr (\u v-> 2*v + u) 0 . map toInteger . a047999_row
    -- Reinhard Zumkeller, Nov 24 2012
    (Scheme, with memoization-macro definec, two variants)
    (definec (A001317 n) (if (zero? n) 1 (A048724 (A001317 (- n 1)))))
    (definec (A001317 n) (if (zero? n) 1 (A048720bi 3 (A001317 (- n 1))))) ;; Where A048720bi implements the dyadic function given in A048720.
    ;; Antti Karttunen, Feb 10 2016
    
  • Magma
    [&+[(Binomial(n, i) mod 2)*2^i: i in [0..n]]: n in [0..41]]; // Vincenzo Librandi, Feb 12 2016
    
  • Maple
    A001317 := proc(n) local k; add((binomial(n,k) mod 2)*2^k, k=0..n); end;
  • Mathematica
    a[n_] := Nest[ BitXor[#, BitShiftLeft[#, 1]] &, 1, n]; Array[a, 42, 0] (* Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007 *)
    NestList[BitXor[#,2#]&,1,50] (* Harvey P. Dale, Aug 02 2021 *)
  • PARI
    a(n)=sum(i=0,n,(binomial(n,i)%2)*2^i)
    
  • PARI
    a=1; for(n=0, 66, print1(a,", "); a=bitxor(a,a<<1) ); \\ Joerg Arndt, Mar 27 2013
    
  • PARI
    A001317(n,a=1)={for(k=1,n,a=bitxor(a,a<<1));a} \\ M. F. Hasler, Jun 06 2016
    
  • PARI
    a(n) = subst(lift(Mod(1+'x,2)^n), 'x, 2); \\ Gheorghe Coserea, Nov 09 2017
    
  • Python
    from sympy import binomial
    def a(n): return sum([(binomial(n, i)%2)*2**i for i in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
    
  • Python
    def A001317(n): return int(''.join(str(int(not(~n&k))) for k in range(n+1)),2) # Chai Wah Wu, Feb 04 2022

Formula

a(n+1) = a(n) XOR 2*a(n), where XOR is binary exclusive OR operator. - Paul D. Hanna, Apr 27 2003
a(n) = Product_{e(j, n) = 1} (2^(2^j) + 1), where e(j, n) is the j-th least significant digit in the binary representation of n (Roberts: see Allouche & Shallit). - Benoit Cloitre, Jun 08 2004
a(2*n+1) = 3*a(2*n). Proof: Since a(n) = Product_{k in K} (1 + 2^(2^k)), where K is the set of integers such that n = Sum_{k in K} 2^k, clearly K(2*n+1) = K(2*n) union {0}, hence a(2*n+1) = (1+2^(2^0))*a(2*n) = 3*a(2*n). - Emmanuel Ferrand and Ralf Stephan, Sep 28 2004
a(32*n) = 3 ^ (32 * n * log(2) / log(3)) + 1. - Bret Mulvey, May 17 2008
For n >= 1, A000120(a(n)) = 2^A000120(n). - Vladimir Shevelev, Oct 25 2010
a(2^n) = A000215(n); a(2^n-1) = a(2^n)-2; for n >= 1, m >= 0,
a(2^(n-1)-1)*a(2^n*m + 2^(n-1)) = 3*a(2^(n-1))*a(2^n*m + 2^(n-1)-2). - Vladimir Shevelev, Nov 28 2010
Sum_{k>=0} 1/a(k) = Product_{n>=0} (1 + 1/F_n), where F_n=A000215(n);
Sum_{k>=0} (-1)^(m(k))/a(k) = 1/2, where {m(n)} is Thue-Morse sequence (A010060).
If F_n is defined by F_n(z) = z^(2^n) + 1 and a(n) by (1/2)*Sum_{i>=0}(1-(-1)^{binomial(n,i)})*z^i, then, for z > 1, the latter two identities hold as well with the replacement 1/2 in the right hand side of the 2nd one by 1-1/z. - Vladimir Shevelev, Nov 29 2010
G.f.: Product_{k>=0} ( 1 + z^(2^k) + (2*z)^(2^k) ). - conjectured by Shamil Shakirov, proved by Vladimir Shevelev
a(n) = A000225(n+1) - A219843(n). - Reinhard Zumkeller, Nov 30 2012
From Antti Karttunen, Feb 10 2016: (Start)
a(0) = 1, and for n > 1, a(n) = A048720(3, a(n-1)) = A048724(a(n-1)).
a(n) = A048723(3,n).
a(n) = A193231(A000079(n)).
For all n >= 0: A268389(a(n)) = n.
(End)

A002379 a(n) = floor(3^n / 2^n).

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 11, 17, 25, 38, 57, 86, 129, 194, 291, 437, 656, 985, 1477, 2216, 3325, 4987, 7481, 11222, 16834, 25251, 37876, 56815, 85222, 127834, 191751, 287626, 431439, 647159, 970739, 1456109, 2184164, 3276246, 4914369, 7371554, 11057332
Offset: 0

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It is an important unsolved problem related to Waring's problem to show that a(n) = floor((3^n-1)/(2^n-1)) holds for all n > 1. This has been checked for 10000 terms and is true for all sufficiently large n, by a theorem of Mahler. [Lichiardopol]
a(n) = floor((3^n-1)/(2^n-1)) holds true at least for 2 <= n <= 305000. - Hieronymus Fischer, Dec 31 2008
a(n) is also the curve length (rounded down) of the Sierpiński arrowhead curve after n iterations, let a(0) = 1. - Kival Ngaokrajang, May 21 2014
a(n) is composite infinitely often (Forman and Shapiro). More exactly, a(n) is divisible by at least one of 2, 5, 7 or 11 infinitely often (Dubickas and Novikas). - Tomohiro Yamada, Apr 15 2017

References

  • R. K. Guy, Unsolved Problems in Number Theory, E19.
  • D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 82.
  • S. S. Pillai, On Waring's problem, J. Indian Math. Soc., 2 (1936), 16-44.
  • 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).

Crossrefs

Cf. A046037, A070758, A070759, A067904 (Composites and Primes).
Cf. A064628 (an analog for 4/3).

Programs

Formula

a(n) = b(n) - (-2/3)^n where b(n) is defined by the recursion b(0):=2, b(1):=5/6, b(n+1):=(5/6)*b(n) + b(n-1). - Hieronymus Fischer, Dec 31 2008
a(n) = (1/2)*(b(n) + sqrt(b(n)^2 - (-4)^n)) (with b(n) as defined above). - Hieronymus Fischer, Dec 31 2008
3^n = a(n)*2^n + A002380(n). - R. J. Mathar, Oct 26 2012
a(n) = -(1/2) + (3/2)^n + arctan(cot((3/2)^n Pi)) / Pi. - Fred Daniel Kline, Apr 14 2018
a(n+1) = round( -(1/2) + (3^n-1)/(2^n-1) ). - Fred Daniel Kline, Apr 14 2018

Extensions

More terms from Robert G. Wilson v, May 11 2004

A001602 Fibonacci entry points: a(n) = smallest m > 0 such that the n-th prime divides Fibonacci(m).

Original entry on oeis.org

3, 4, 5, 8, 10, 7, 9, 18, 24, 14, 30, 19, 20, 44, 16, 27, 58, 15, 68, 70, 37, 78, 84, 11, 49, 50, 104, 36, 27, 19, 128, 130, 69, 46, 37, 50, 79, 164, 168, 87, 178, 90, 190, 97, 99, 22, 42, 224, 228, 114, 13, 238, 120, 250, 129, 88, 67, 270, 139, 28, 284, 147, 44, 310
Offset: 1

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"[a(n)] is called by Lucas the rank of apparition of p and we know it is a divisor of, or equal to prime(n)-1 or prime(n)+1" - Vajda, p. 84. (Note that a(3)=5. This is the only exception.) - Chris K. Caldwell, Nov 03 2008
Every number except 1, 2, 6 and 12 eventually occurs in this sequence. See also A086597(n), the number of primitive prime factors of Fibonacci(n). - T. D. Noe, Jun 13 2008
For each prime p we have an infinite sequence of integers, F(i*a(n))/p, i=1,2,... See also A236479. For primes p >= 3 and exponents j >= 2, with k = a(n) and p = p(n), it appears that F(k*i*p^(j-1))/p^j is an integer, for i >= 0. For p = 2, F(k*i*p^(j-1))/p^(j+1) = integer. - Richard R. Forberg, Jan 26-29 2014 [Comments revised by N. J. A. Sloane, Sep 24 2015]
Let p=prime(n). a(n) is also a divisor of (p-1)/2 (if p mod 5 == 1 or 4) or (p+1)/2 (if p mod 5 == 2 or 3) if and only if p mod 4 = 1. - Seiichi Azuma, Jul 29 2014

Examples

			The 5th prime is 11 and 11 first divides Fib(10)=55, so a(5) = 10.
		

References

  • 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).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Crossrefs

Cf. A051694, A001177, A086597, A194363 (entries Lucas).

Programs

  • Haskell
    import Data.List (findIndex)
    import Data.Maybe (fromJust)
    a001602 n = (+ 1) $ fromJust $
                findIndex ((== 0) . (`mod` a000040 n)) $ tail a000045_list
    -- Reinhard Zumkeller, Apr 08 2012
    
  • Maple
    A001602 := proc(n)
        local i,p;
        p := ithprime(n);
        for i from 1 do
            if modp(combinat[fibonacci](i),p) = 0 then
                return i;
            end if;
        end do:
    end proc: # R. J. Mathar, Oct 31 2015
  • Mathematica
    Table[k=1;While[!Divisible[Fibonacci[k],Prime[n]],k++];k,{n,70}] (* Harvey P. Dale, Feb 15 2012 *)
    (* a fast, but more complicated method *) MatrixPowerMod[mat_, n_, m_Integer] := Mod[Fold[Mod[If[#2 == 1, #1.#1.mat, #1.#1], m] &, mat, Rest[IntegerDigits[n, 2]]], m]; FibMatrix[n_Integer, m_Integer] := MatrixPowerMod[{{0, 1}, {1, 1}}, n, m]; FibEntryPointPrime[p_Integer] := Module[{n, d, k}, If[PrimeQ[p], n = p - JacobiSymbol[p, 5]; d = Divisors[n]; k = 1; While[FibMatrix[d[[k]], p][[1, 2]] > 0, k++]; d[[k]]]]; SetAttributes[FibEntryPointPrime, Listable]; FibEntryPointPrime[Prime[Range[10000]]] (* T. D. Noe, Jan 03 2013 *)
    With[{nn=70,t=Table[{n,Fibonacci[n]},{n,500}]},Transpose[ Flatten[ Table[ Select[t,Divisible[#[[2]],Prime[i]]&,1],{i,nn}],1]][[1]]] (* Harvey P. Dale, May 31 2014 *)
  • PARI
    a(n)=if(n==3,5,my(p=prime(n));fordiv(p^2-1,d,if(fibonacci(d)%p==0, return(d)))) \\ Charles R Greathouse IV, Jul 17 2012
    
  • PARI
    do(p)=my(k=p+[0,-1,1,1,-1][p%5+1],f=factor(k));for(i=1,#f[,1],for(j=1,f[i,2],if((Mod([1,1;1,0],p)^(k/f[i,1]))[1,2], break); k/=f[i,1])); k
    a(n)=do(prime(n))
    apply(do, primes(100)) \\ Charles R Greathouse IV, Jan 03 2013
    
  • Python
    from sympy.ntheory.generate import prime
    def A001602(n):
        a, b, i, p = 0, 1, 1, prime(n)
        while b % p:
            a, b, i = b, (a+b) % p, i+1
        return i # Chai Wah Wu, Nov 03 2015, revised Apr 04 2016.

Formula

a(n) = A001177(prime(n)).
a(n) <= prime(n) + 1. - Charles R Greathouse IV, Jan 02 2013

Extensions

More terms from Jud McCranie

A000127 Maximal number of regions obtained by joining n points around a circle by straight lines. Also number of regions in 4-space formed by n-1 hyperplanes.

Original entry on oeis.org

1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794, 1093, 1471, 1941, 2517, 3214, 4048, 5036, 6196, 7547, 9109, 10903, 12951, 15276, 17902, 20854, 24158, 27841, 31931, 36457, 41449, 46938, 52956, 59536, 66712, 74519, 82993, 92171, 102091, 112792, 124314, 136698
Offset: 1

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a(n) is the sum of the first five terms in the n-th row of Pascal's triangle. - Geoffrey Critzer, Jan 18 2009
{a(k): 1 <= k <= 5} = divisors of 16. - Reinhard Zumkeller, Jun 17 2009
Equals binomial transform of [1, 1, 1, 1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Mar 02 2010
From Bernard Schott, Apr 05 2021: (Start)
As a(n) = 2^(n-1) for n = 1..5, it is misleading to believe that a(n) = 2^(n-1) for n > 5 (see Patrick Popescu-Pampu link); other curiosities: a(6) = 2^5 - 1 and a(10) = 2^8.
The sequence of the first differences is A000125, the sequence of the second differences is A000124, the sequence of the third differences is A000027 and the sequence of the fourth differences is the all 1's sequence A000012 (see J. H. Conway and R. K. Guy reference, p. 80). (End)
a(n) is the number of binary words of length n matching the regular expression 0*1*0*1*0*. A000124 and A000125 count binary words of the form 0*1*0* and 1*0*1*0*, respectively. - Manfred Scheucher, Jun 22 2023

Examples

			a(7)=99 because the first five terms in the 7th row of Pascal's triangle are 1 + 7 + 21 + 35 + 35 = 99. - _Geoffrey Critzer_, Jan 18 2009
G.f. = x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 31*x^6 + 57*x^7 + 99*x^8 + 163*x^9 + ...
		

References

  • R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 28.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, Chap. 3.
  • J. H. Conway and R. K. Guy, Le Livre des Nombres, Eyrolles, 1998, p. 80.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problem 33 pp. 18; 128 Ellipses Paris 2004.
  • A. Deledicq and D. Missenard, A La Recherche des Régions Perdues, Math. & Malices, No. 22 Summer 1995 issue pp. 22-3 ACL-Editions Paris.
  • M. Gardner, Mathematical Circus, pp. 177; 180-1 Alfred A. Knopf NY 1979.
  • M. Gardner, The Colossal Book of Mathematics, 2001, p. 561.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • M. de Guzman, Aventures Mathématiques, Prob. B pp. 115-120 PPUR Lausanne 1990.
  • Ross Honsberger; Mathematical Gems I, Chap. 9.
  • Ross Honsberger; Mathematical Morsels, Chap. 3.
  • Jeux Mathématiques et Logiques, Vol. 3 pp. 12; 51 Prob. 14 FFJM-SERMAP Paris 1988.
  • J. N. Kapur, Reflections of a Mathematician, Chap.36, pp. 337-343, Arya Book Depot, New Delhi 1996.
  • C. D. Miller, V. E. Heeren, J. Hornsby, M. L. Morrow and J. Van Newenhizen, Mathematical Ideas, Tenth Edition, Pearson, Addison-Wesley, Boston, 2003, Cptr 1, 'The Art of Problem Solving, page 6.
  • I. Niven, Mathematics of Choice, pp. 158; 195 Prob. 40 NML 15 MAA 1965.
  • C. S. Ogilvy, Tomorrow's Math, pp. 144-6 OUP 1972.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 252-255.
  • Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, NY, 2007, page 81-87.
  • A. M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 213.
  • 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).

Crossrefs

Programs

  • Haskell
    a000127 = sum . take 5 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(n^4-6*n^3+23*n^2-18*n+24)/24: n in [1..50]]; // Vincenzo Librandi, Feb 16 2015
    
  • Maple
    A000127 := n->(n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24;
    with (combstruct):ZL:=[S, {S=Sequence(U, card=1)}, unlabeled]: seq(count(subs(r=6, ZL), size=m), m=0..41); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    f[n_] := Sum[Binomial[n, i], {i, 0, 4}]; Table[f@n, {n, 0, 40}] (* Robert G. Wilson v, Jun 29 2007 *)
    Total/@Table[Binomial[n-1,k],{n,50},{k,0,4}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,2,4,8,16},50] (* Harvey P. Dale, Aug 24 2011 *)
    Table[(n^4 - 6 n^3 + 23 n^2 - 18 n + 24) / 24, {n, 100}] (* Vincenzo Librandi, Feb 16 2015 *)
    a[ n_] := Binomial[n, 4] + Binomial[n, 2] + 1; (* Michael Somos, Dec 23 2017 *)
  • PARI
    a(n)=(n^4-6*n^3+23*n^2-18*n+24)/24 \\ Charles R Greathouse IV, Mar 22 2016
    
  • PARI
    {a(n) = binomial(n, 4) + binomial(n, 2) + 1}; /* Michael Somos, Dec 23 2017 */
    
  • Python
    def A000127(n): return n*(n*(n*(n - 6) + 23) - 18)//24 + 1 # Chai Wah Wu, Sep 18 2021

Formula

a(n) = C(n-1, 4) + C(n-1, 3) + ... + C(n-1, 0) = A055795(n) + 1 = C(n, 4) + C(n-1, 2) + n.
a(n) = Sum_{k=0..2} C(n, 2k). - Joel Sanderi (sanderi(AT)itstud.chalmers.se), Sep 08 2004
a(n) = (n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24.
G.f.: (1 - 3*x + 4*x^2 - 2*x^3 + x^4)/(1-x)^5. (for offset 0) - Simon Plouffe in his 1992 dissertation
E.g.f.: (1 + x + x^2/2 + x^3/6 + x^4/24)*exp(x) (for offset 0). [Typos corrected by Juan M. Marquez, Jan 24 2011]
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4. - Harvey P. Dale, Aug 24 2011
a(n) = A000124(A000217(n-1)) - n*A000217(n-2) - A034827(n), n > 1. - Melvin Peralta, Feb 15 2016
a(n) = A223718(-n). - Michael Somos, Dec 23 2017
For n > 2, a(n) = n + 1 + sum_{i=2..(n-2)}sum_{j=1..(n-i)}(1+(i-1)(j-1)). - Alec Jones, Nov 17 2019

Extensions

Formula corrected and additional references from torsten.sillke(AT)lhsystems.com
Additional correction from Jonas Paulson (jonasso(AT)sdf.lonestar.org), Oct 30 2003

A003401 Numbers of edges of regular polygons constructible with ruler (or, more precisely, an unmarked straightedge) and compass.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96, 102, 120, 128, 136, 160, 170, 192, 204, 240, 255, 256, 257, 272, 320, 340, 384, 408, 480, 510, 512, 514, 544, 640, 680, 768, 771, 816, 960, 1020, 1024, 1028, 1088, 1280, 1285
Offset: 1

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Author

Keywords

Comments

The terms 1 and 2 correspond to degenerate polygons.
These are also the numbers for which phi(n) is a power of 2: A209229(A000010(a(n))) = 1. - Olivier Gérard Feb 15 1999
From Stanislav Sykora, May 02 2016: (Start)
The sequence can be also defined as follows: (i) 1 is a member. (ii) Double of any member is also a member. (iii) If a member is not divisible by a Fermat prime F_k then its product with F_k is also a member. In particular, the powers of 2 (A000079) are a subset and so are the Fermat primes (A019434), which are the only odd prime members.
The definition is too restrictive (though correct): The Georg Mohr - Lorenzo Mascheroni theorem shows that constructibility using a straightedge and a compass is equivalent to using compass only. Moreover, Jean Victor Poncelet has shown that it is also equivalent to using straightedge and a fixed ('rusty') compass. With the work of Jakob Steiner, this became part of the Poncelet-Steiner theorem establishing the equivalence to using straightedge and a fixed circle (with a known center). A further extension by Francesco Severi replaced the availability of a circle with that of a fixed arc, no matter how small (but still with a known center).
Constructibility implies that when m is a member of this sequence, the edge length 2*sin(Pi/m) of an m-gon with circumradius 1 can be written as a finite expression involving only integer numbers, the four basic arithmetic operations, and the square root. (End)
If x,y are terms, and gcd(x,y) is a power of 2 then x*y is also a term. - David James Sycamore, Aug 24 2024

Examples

			34 is a term of this sequence because a circle can be divided into exactly 34 parts. 7 is not.
		

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 183.
  • Allan Clark, Elements of Abstract Algebra, Chapter 4, Galois Theory, Dover Publications, NY 1984, page 124.
  • Duane W. DeTemple, "Carlyle circles and the Lemoine simplicity of polygon constructions." The American Mathematical Monthly 98.2 (1991): 97-108. - N. J. A. Sloane, Aug 05 2021
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • B. L. van der Waerden, Modern Algebra. Unger, NY, 2nd ed., Vols. 1-2, 1953, Vol. 1, p. 187.

Crossrefs

Subsequence of A295298. - Antti Karttunen, Nov 27 2017
A004729 and A051916 are subsequences. - Reinhard Zumkeller, Mar 20 2010
Cf. A000079, A004169, A000215, A099884, A019434 (Fermat primes).
Edge lengths of other constructible m-gons: A002194 (m=3), A002193 (4), A182007 (5), A101464 (8), A094214 (10), A101263 (12), A272534 (15), A272535 (16), A228787 (17), A272536 (20).
Positions of zeros in A293516 (apart from two initial -1's), and in A336469, positions of ones in A295660 and in A336477 (characteristic function).
Cf. also A046528.

Programs

  • Haskell
    a003401 n = a003401_list !! (n-1)
    a003401_list = map (+ 1) $ elemIndices 1 $ map a209229 a000010_list
    -- Reinhard Zumkeller, Jul 31 2012
    
  • Mathematica
    Select[ Range[ 1300 ], IntegerQ[ Log[ 2, EulerPhi[ # ] ] ]& ] (* Olivier Gérard Feb 15 1999 *)
    (* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) Take[ Union[ Flatten[ NestList[2# &, Times @@@ Table[ UnrankSubset[n, Join[{1}, Table[2^2^i + 1, {i, 0, 4}]]], {n, 63}], 11]]], 60] (* Robert G. Wilson v, Jun 11 2005 *)
    nn=10; logs=Log[2,{2,3,5,17,257,65537}]; lim2=Floor[nn/logs[[1]]]; Sort[Reap[Do[z={i,j,k,l,m,n}.logs; If[z<=nn, Sow[2^z]], {i,0,lim2}, {j,0,1}, {k,0,1}, {l,0,1}, {m,0,1}, {n,0,1}]][[2,1]]]
    A092506 = {2, 3, 5, 17, 257, 65537}; s = Sort[Times @@@ Subsets@ A092506]; mx = 1300; Union@ Flatten@ Table[(2^n)*s[[i]], {i, 64}, {n, 0, Log2[mx/s[[i]]]}] (* Robert G. Wilson v, Jul 28 2014 *)
  • PARI
    for(n=1,10^4,my(t=eulerphi(n));if(t/2^valuation(t,2)==1,print1(n,", "))); \\ Joerg Arndt, Jul 29 2014
    
  • PARI
    is(n)=n>>=valuation(n,2); if(n<7, return(n>0)); my(k=logint(logint(n,2),2)); if(k>32, my(p=2^2^k+1); if(n%p, return(0)); n/=p; unknown=1; if(n%p==0, return(0)); p=0; if(is(n)==0, 0, "unknown [has large Fermat number in factorization]"), 4294967295%n==0) \\ Charles R Greathouse IV, Jan 09 2022
    
  • PARI
    is(n)=n>>=valuation(n,2); 4294967295%n==0 \\ valid for n <= 2^2^33, conjecturally valid for all n; Charles R Greathouse IV, Jan 09 2022
    
  • Python
    from sympy import totient
    A003401_list = [n for n in range(1,10**4) if format(totient(n),'b').count('1') == 1]
    # Chai Wah Wu, Jan 12 2015

Formula

Terms from 3 onward are computable as numbers such that cototient-of-totient equals the totient-of-totient: Flatten[Position[Table[co[eu[n]]-eu[eu[n]], {n, 1, 10000}], 0]] eu[m]=EulerPhi[m], co[m]=m-eu[m]. - Labos Elemer, Oct 19 2001, clarified by Antti Karttunen, Nov 27 2017
Any product of 2^k and distinct Fermat primes (primes of the form 2^(2^m)+1). - Sergio Pimentel, Apr 30 2004, edited by Franklin T. Adams-Watters, Jun 16 2006
If the well-known conjecture that there are only five prime Fermat numbers F_k=2^{2^k}+1, k=0,1,2,3,4 is true, then we have exactly: Sum_{n>=1} 1/a(n)= 2*Product_{k=0..4} (1+1/F_k) = 4869735552/1431655765 = 3.40147098978.... - Vladimir Shevelev and T. D. Noe, Dec 01 2010
log a(n) >> sqrt(n); if there are finitely many Fermat primes, then log a(n) ~ k log n for some k. - Charles R Greathouse IV, Oct 23 2015

Extensions

Definition clarified by Bill Gosper. - N. J. A. Sloane, Jun 14 2020

A001429 Number of n-node connected unicyclic graphs.

Original entry on oeis.org

1, 2, 5, 13, 33, 89, 240, 657, 1806, 5026, 13999, 39260, 110381, 311465, 880840, 2497405, 7093751, 20187313, 57537552, 164235501, 469406091, 1343268050, 3848223585, 11035981711, 31679671920, 91021354454, 261741776369, 753265624291, 2169441973139, 6252511838796
Offset: 3

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Keywords

Comments

Also unlabeled connected simple graphs with n vertices and n edges. The labeled version is A057500. - Gus Wiseman, Feb 12 2024

Examples

			From _Gus Wiseman_, Feb 12 2024: (Start)
Representatives of the a(3) = 1 through a(6) = 13 simple graphs:
  {12,13,23}  {12,13,14,23}  {12,13,14,15,23}  {12,13,14,15,16,23}
              {12,13,24,34}  {12,13,14,23,25}  {12,13,14,15,23,26}
                             {12,13,14,23,45}  {12,13,14,15,23,46}
                             {12,13,14,25,35}  {12,13,14,15,26,36}
                             {12,13,24,35,45}  {12,13,14,23,25,36}
                                               {12,13,14,23,25,46}
                                               {12,13,14,23,45,46}
                                               {12,13,14,23,45,56}
                                               {12,13,14,25,26,35}
                                               {12,13,14,25,35,46}
                                               {12,13,14,25,35,56}
                                               {12,13,14,25,36,56}
                                               {12,13,24,35,46,56}
(End)
		

References

  • R. C. Read and R. J. Wilson, An Atlas of Graphs, Oxford, 1998.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 150.
  • 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).

Crossrefs

For at most one cycle we have A005703, labeled A129271, complement A140637.
Next-to-main diagonal of A054924. Cf. A000055.
The labeled version is A057500, connected case of A137916.
This is the connected case of A137917 and A236570.
Row k = 1 of A137918.
The version with loops is A368983.
A001349 counts unlabeled connected graphs.
A001434 and A006649 count unlabeled graphs with # vertices = # edges.
A006129 counts covering graphs, unlabeled A002494.

Programs

  • Mathematica
    Needs["Combinatorica`"];
    nn=30;s[n_,k_]:=s[n,k]=a[n+1-k]+If[n<2k,0,s[n-k,k]];a[1]=1;a[n_]:=a[n]=Sum[a[i]s[n-1,i]i,{i,1,n-1}]/(n-1);rt=Table[a[i],{i,1,nn}];Apply[Plus,Table[Take[CoefficientList[CycleIndex[DihedralGroup[n],s]/.Table[s[j]->Table[Sum[rt[[i]]x^(k*i),{i,1,nn}],{k,1,nn}][[j]],{j,1,nn}],x],nn],{n,3,nn}]]  (* Geoffrey Critzer, Oct 12 2012, after code given by Robert A. Russell in A000081 *)
    (* Second program: *)
    TreeGf[nn_] := Module[{A}, A = Table[1, {nn}]; For[n = 1, n <= nn 1, n++, A[[n + 1]] = 1/n * Sum[Sum[ d*A[[d]], {d, Divisors[k]}]*A[[n - k + 1]], {k, 1, n}]]; x A.x^Range[0, nn-1]];
    seq[n_] := Module[{t, g}, If[n < 3, {}, t = TreeGf[n - 2]; g[e_] := Normal[t + O[x]^(Quotient[n, e]+1)] /. x -> x^e  + O[x]^(n+1); Sum[Sum[ EulerPhi[d]*g[d]^(k/d), {d, Divisors[k]}]/k + If[OddQ[k], g[1]* g[2]^Quotient[k, 2], (g[1]^2 + g[2])*g[2]^(k/2-1)/2], {k, 3, n}]]/2 // Drop[CoefficientList[#, x], 3]&];
    seq[32] (* Jean-François Alcover, Oct 05 2019, after Andrew Howroyd's PARI code *)
  • PARI
    \\ TreeGf gives gf of A000081
    TreeGf(N)={my(A=vector(N, j, 1)); for (n=1, N-1, A[n+1] = 1/n * sum(k=1, n, sumdiv(k, d, d*A[d]) * A[n-k+1] ) ); x*Ser(A)}
    seq(n)={if(n<3, [], my(t=TreeGf(n-2)); my(g(e)=subst(t + O(x*x^(n\e)),x,x^e) + O(x*x^n)); Vec(sum(k=3, n, sumdiv(k, d, eulerphi(d)*g(d)^(k/d))/k + if(k%2, g(1)*g(2)^(k\2), (g(1)^2+g(2))*g(2)^(k/2-1)/2))/2))} \\ Andrew Howroyd, May 05 2018

Formula

a(n) = A068051(n) - A027852(n) - A000081(n).

Extensions

More terms from Ronald C. Read
a(27) corrected, more terms, formula from Christian G. Bower, Feb 12 2002
Edited by Charles R Greathouse IV, Oct 05 2009
Terms a(30) and beyond from Andrew Howroyd, May 05 2018

A005183 a(n) = n*2^(n-1) + 1.

Original entry on oeis.org

1, 2, 5, 13, 33, 81, 193, 449, 1025, 2305, 5121, 11265, 24577, 53249, 114689, 245761, 524289, 1114113, 2359297, 4980737, 10485761, 22020097, 46137345, 96468993, 201326593, 419430401, 872415233, 1811939329, 3758096385, 7784628225, 16106127361, 33285996545
Offset: 0

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Comments

a(n-1) is the number of permutations of length n which avoid the patterns 132, 4312. - Lara Pudwell, Jan 21 2006
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) <= e(j) >= e(k) and e(i) != e(k). [Martinez and Savage, 2.11] - Eric M. Schmidt, Jul 17 2017
Indices of records in A066099. Also, indices of "cusps" in the graph of A030303 giving positions of 1's in the binary Champernowne word A030190. - M. F. Hasler, Oct 12 2020

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

Main diagonal of the array defined by T(0, j)=j+1 j>=0, T(i, 0)=i+1 i>=0, T(i, j)=T(i-1, j-1)+T(i-1, j)-1. - Benoit Cloitre, Jun 17 2003
G.f.: (1 -3*x +3*x^2)/((1-x)*(1-2*x)^2). - Lara Pudwell, Jan 21 2006
E.g.f.: exp(x) +x*exp(2*x). - Joerg Arndt, May 22 2013
Binomial transform of A028310. a(n) = 1 + Sum{k=0..n} C(n, k)*k = 1 + A001787(n). - Paul Barry, Jul 21 2003
a(n) = Sum_{k=0..2^n} A000120(k) = A000788(2^n). - Benoit Cloitre, Sep 25 2003
Row sums of triangle A134399. - Gary W. Adamson, Oct 23 2007
a(n) = A000788(A000079(n)). - Reinhard Zumkeller, Mar 04 2010
a(n) = 2*a(n-1) +2^(n-1) -1 (with a(0)=1). - Vincenzo Librandi, Dec 31 2010

Extensions

More terms from Lara Pudwell, Jan 21 2006
Edited by N. J. A. Sloane at the suggestion of Jim Propp, Jul 14 2007
Showing 1-10 of 28 results. Next