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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A026150 a(0) = a(1) = 1; a(n+2) = 2*a(n+1) + 2*a(n).

Original entry on oeis.org

1, 1, 4, 10, 28, 76, 208, 568, 1552, 4240, 11584, 31648, 86464, 236224, 645376, 1763200, 4817152, 13160704, 35955712, 98232832, 268377088, 733219840, 2003193856, 5472827392, 14952042496, 40849739776
Offset: 0

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Author

Keywords

Comments

a(n+1)/A002605(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003
a(n+1)/a(n) converges to 1 + sqrt(3) = 2.732050807568877293.... - Philippe Deléham, Jul 03 2005
Binomial transform of expansion of cosh(sqrt(3)x) (A000244 with interpolated zeros); inverse binomial transform of A001075. - Philippe Deléham, Jul 04 2005
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 3 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(3). - Cino Hilliard, Sep 25 2005
Inverse binomial transform of A001075: (1, 2, 7, 26, 97, 362, ...). - Gary W. Adamson, Nov 23 2007
Starting (1, 4, 10, 28, 76, ...), the sequence is the binomial transform of [1, 3, 3, 9, 9, 27, 27, 81, 81, ...], and inverse binomial transform of A001834: (1, 5, 19, 71, 265, ...). - Gary W. Adamson, Nov 30 2007
[1, 3; 1, 1]^n * [1,0] = [a(n), A002605(n)]. - Gary W. Adamson, Mar 21 2008
(1 + sqrt(3))^n = a(n) + A002605(n)*(sqrt(3)). - Gary W. Adamson, Mar 21 2008
Equals right border of triangle A143908. Also, starting (1, 4, 10, 28, ...) = row sums of triangle A143908 and INVERT transform of (1, 3, 3, 3, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 85, 277, 337 and 340, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A002605 (without the leading 0). - Johannes W. Meijer, Aug 15 2010
Pisano period lengths: 1, 1, 1, 1, 24, 1, 48, 1, 3, 24, 10, 1, 12, 48, 24, 1,144, 3,180, 24, ... - R. J. Mathar, Aug 10 2012
(1 + sqrt(3))^n = a(n) + A002605(n)*sqrt(3), for n >= 0; integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
a(n) is also the number of solutions for cyclic three-dimensional stable matching instances with master preference lists of size n (Escamocher and O'Sullivan 2018). - Guillaume Escamocher, Jun 15 2018
Starting from a(1), first differences of A005665. - Ivan N. Ianakiev, Nov 22 2019
Number of 3-permutations of n elements avoiding the patterns 231, 312. See Bonichon and Sun. - Michel Marcus, Aug 19 2022

Examples

			G.f. = 1 + x + 4*x^2 + 10*x^3 + 28*x^4 + 76*x^5 + 208*x^6 + 568*x^7 + ...
		

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

First differences of A002605.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Haskell
    a026150 n = a026150_list !! n
    a026150_list = 1 : 1 : map (* 2) (zipWith (+) a026150_list (tail
    a026150_list))
    -- Reinhard Zumkeller, Oct 15 2011
    
  • Magma
    [n le 2 select 1 else 2*Self(n-1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018
  • Maple
    with(combstruct):ZL0:=S=Prod(Sequence(Prod(a, Sequence(b))), a):ZL1:=Prod(begin_blockP, Z, end_blockP):ZL2:=Prod(begin_blockLR, Z, Sequence(Prod(mu_length, Z), card>=1), end_blockLR): ZL3:=Prod(begin_blockRL, Sequence(Prod(mu_length, Z), card>=1), Z, end_blockRL):Q:=subs([a=Union(ZL2,ZL2,ZL2), b=ZL1], ZL0), begin_blockP=Epsilon, end_blockP=Epsilon, begin_blockLR=Epsilon, end_blockLR=Epsilon, begin_blockRL=Epsilon, end_blockRL=Epsilon, mu_length=Epsilon:temp15:=draw([S, {Q}, unlabelled], size=15):seq(count([S, {Q}, unlabelled], size=n)/3, n=2..27); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    Expand[Table[((1 + Sqrt[3])^n + (1 - Sqrt[3])^n)/(2), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
    LinearRecurrence[{2, 2}, {1, 1}, 30] (* T. D. Noe, Mar 25 2011 *)
    Round@Table[LucasL[n, Sqrt[2]] 2^(n/2 - 1), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
  • Maxima
    a(n) := if n<=1 then 1 else 2*a(n-1)+2*a(n-2);
    makelist(a(n),n,0,20); /* Emanuele Munarini, Apr 14 2017 */
    
  • PARI
    {a(n) = if( n<0, 0, real((1 + quadgen(12))^n))};
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen2; it = recur_gen2(1,1,2,2); [next(it) for i in range(30)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [lucas_number2(n,2,-2)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n) = (1/2)*((1 + sqrt(3))^n + (1 - sqrt(3))^n). - Benoit Cloitre, Oct 28 2002
G.f.: (1 - x)/(1 - 2*x - 2*x^2).
a(n) = a(n-1) + A083337(n-1). A083337(n)/a(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
From Paul Barry, May 15 2003: (Start)
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)*3^k;
E.g.f.: exp(x)*cosh(sqrt(3)x). (End)
a(n) = Sum_{k=0..n} A098158(n,k)*3^(n - k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1, 1; 3, 1]^n. (1 + sqrt(3))^n = a(n) + A083337(n)/(sqrt(3)). - Gary W. Adamson, Mar 12 2008
a(n) = A080040(n)/2. - Philippe Deléham, Nov 19 2008
If p[1] = 1, and p[i] = 3, (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j + 1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = det A. - Milan Janjic, Apr 29 2010
a(n) = 2 * A052945(n-1). - Vladimir Joseph Stephan Orlovsky, Mar 24 2011
a(n) = round((1 + sqrt(3))^n/2) for n > 0. - Bruno Berselli, Feb 04 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k - 1)/(x*(3*k + 2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
a(n) = (-sqrt(2)*i)^n*T(n,sqrt(2)*i/2), with i = sqrt(-1) and the Chebyshev T-polynomials (A053120). - Wolfdieter Lang, Feb 10 2018

A006012 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - 2*a(n-2), n >= 2.

Original entry on oeis.org

1, 2, 6, 20, 68, 232, 792, 2704, 9232, 31520, 107616, 367424, 1254464, 4283008, 14623104, 49926400, 170459392, 581984768, 1987020288, 6784111616, 23162405888, 79081400320, 270000789504, 921840357376, 3147359850496
Offset: 0

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Author

Keywords

Comments

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 4, s(2n) = 4. - Herbert Kociemba, Jun 12 2004
a(n-1) counts permutations pi on [n] for which the pairs {i, pi(i)} with i < pi(i), considered as closed intervals [i+1,pi(i)], do not overlap; equivalently, for each i in [n] there is at most one j <= i with pi(j) > i. Counting these permutations by the position of n yields the recurrence relation. - David Callan, Sep 02 2003
a(n) is the sum of (n+1)-th row terms of triangle A140070. - Gary W. Adamson, May 04 2008
The binomial transform is in A083878, the Catalan transform in A084868. - R. J. Mathar, Nov 23 2008
Equals row sums of triangle A152252. - Gary W. Adamson, Nov 30 2008
Counts all paths of length (2*n), n >= 0, starting at the initial node on the path graph P_7, see the second Maple program. - Johannes W. Meijer, May 29 2010
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U_1 and U_3 be the unit-primitive matrices (see [Jeffery])
U_1 = U_(8,1) = [(0,1,0,0); (1,0,1,0); (0,1,0,1); (0,0,2,0)] and
U_3 = U_(8,3) = [(0,0,0,1); (0,0,2,0); (0,2,0,1); (2,0,2,0)]. Then a(n) = (1/4) * Trace(U_1^(2*n)) = (1/2^(n+2)) * Trace(U_3^(2*n)). (See also A084130, A001333.) (End)
Pisano period lengths: 1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... - R. J. Mathar, Aug 10 2012
a(n) is the first superdiagonal of array A228405. - Richard R. Forberg, Sep 02 2013
Conjecture: With offset 1, a(n) is the number of permutations on [n] with no subsequence abcd such that (i) bc are adjacent in position and (ii) max(a,c) < min(b,d). For example, the 4 permutations of [4] not counted by a(4) are 1324, 1423, 2314, 2413. - David Callan, Aug 27 2014
The conjecture of David Callan above is correct - with offset 1, a(n) is the number of permutations on [n] with no subsequence abcd such that (i) bc are adjacent in position and (ii) max(a,c) < min(b,d). - Yonah Biers-Ariel, Jun 27 2017
From Gary W. Adamson, Jul 22 2016: (Start)
A production matrix for the sequence is M =
1, 1, 0, 0, 0, 0, ...
1, 0, 3, 0, 0, 0, ...
1, 0, 0, 3, 0, 0, ...
1, 0, 0, 0, 3, 0, ...
1, 0, 0, 0, 0, 3, ...
...
Take powers of M, extracting the upper left terms; getting the sequence starting: (1, 1, 2, 6, 20, 68, ...). (End)
From Gary W. Adamson, Jul 24 2016: (Start)
The sequence is the INVERT transform of the powers of 3 prefaced with a "1": (1, 1, 3, 9, 27, ...) and is N=3 in an infinite of analogous sequences starting:
N=1 (A000079): 1, 2, 4, 8, 16, 32, ...
N=2 (A001519): 1, 2, 5, 13, 34, 89, ...
N=3 (A006012): 1, 2, 6, 20, 68, 232, ...
N=4 (A052961): 1, 2, 7, 29, 124, 533, ...
N=5 (A154626): 1, 2, 8, 40, 208, 1088, ...
N=6: 1, 2, 9, 53, 326, 2017, ...
... (End)
Number of permutations of length n > 0 avoiding the partially ordered pattern (POP) {1>2, 1>3, 4>2, 4>3} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first and fourth elements are larger than the second and third elements. - Sergey Kitaev, Dec 08 2020
a(n-1) is the number of permutations of [n] that can be obtained by placing n points on an X-shape (two crossing lines with slopes 1 and -1), labeling them 1,2,...,n by increasing y-coordinate, and then reading the labels by increasing x-coordinate. - Sergi Elizalde, Sep 27 2021
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of decreasing sizes, followed by a series of reversals of increasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025
Number of permutations of [n] that are correctly sorted after performing one left-to-right pass and one right-to-left pass of the cocktail sort. - Thomas Baruchel, May 16 2025

References

  • D. H. Greene and D. E. Knuth, Mathematics for the Analysis of Algorithms. Birkhäuser, Boston, 3rd edition, 1990, p. 86.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, Sect 5.4.8 Answer to Exer. 8.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a006012 n = a006012_list !! n
    a006012_list = 1 : 2 : zipWith (-) (tail $ map (* 4) a006012_list)
    (map (* 2) a006012_list)
    -- Reinhard Zumkeller, Oct 03 2011
    
  • Magma
    [n le 2 select n else 4*Self(n-1)- 2*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Apr 05 2011
    
  • Maple
    A006012:=-(-1+2*z)/(1-4*z+2*z**2); # Simon Plouffe in his 1992 dissertation
    with(GraphTheory): G:=PathGraph(7): A:= AdjacencyMatrix(G): nmax:=24; n2:=2*nmax: for n from 0 to n2 do B(n):=A^n; a(n):=add(B(n)[1,k],k=1..7); od: seq(a(2*n),n=0..nmax); # Johannes W. Meijer, May 29 2010
  • Mathematica
    LinearRecurrence[{4,-2},{1,2},50] (* or *) With[{c=Sqrt[2]}, Simplify[ Table[((2+c)^n+(3+2c)(2-c)^n)/(2(2+c)),{n,50}]]] (* Harvey P. Dale, Aug 29 2011 *)
  • PARI
    {a(n) = real(((2 + quadgen(8))^n))}; /* Michael Somos, Feb 12 2004 */
    
  • PARI
    {a(n) = if( n<0, 2^n, 1) * polsym(x^2 - 4*x + 2, abs(n))[abs(n)+1] / 2}; /* Michael Somos, Feb 12 2004 */
    
  • PARI
    Vec((1-2*x)/(1-4*x+2*x^2) + O(x^100)) \\ Altug Alkan, Dec 05 2015
    
  • Python
    l = [1, 2]
    for n in range(2, 101): l.append(4 * l[n - 1] - 2 * l[n - 2])
    print(l)  # Indranil Ghosh, Jul 02 2017
    
  • SageMath
    A006012=BinaryRecurrenceSequence(4,-2,1,2)
    print([A006012(n) for n in range(41)]) # G. C. Greubel, Aug 27 2025

Formula

G.f.: (1-2*x)/(1 - 4*x + 2*x^2).
a(n) = 2*A007052(n-1) = A056236(n)/2.
Limit_{n -> oo} a(n)/a(n-1) = 2 + sqrt(2). - Zak Seidov, Oct 12 2002
From Paul Barry, May 08 2003: (Start)
Binomial transform of A001333.
E.g.f.: exp(2*x)*cosh(sqrt(2)*x). (End)
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)*2^(n-k) = Sum_{k=0..n} binomial(n, k)*2^(n-k/2)(1+(-1)^k)/2. - Paul Barry, Nov 22 2003 (typo corrected by Manfred Scheucher, Jan 17 2023)
a(n) = ((2+sqrt(2))^n + (2-sqrt(2))^n)/2.
a(n) = Sum_{k=0..n} 2^k*A098158(n,k). - Philippe Deléham, Dec 04 2006
a(n) = A007070(n) - 2*A007070(n-1). - R. J. Mathar, Nov 16 2007
a(n) = Sum_{k=0..n} A147703(n,k). - Philippe Deléham, Nov 29 2008
a(n) = Sum_{k=0..n} A201730(n,k). - Philippe Deléham, Dec 05 2011
G.f.: G(0) where G(k)= 1 + 2*x/((1-2*x) - 2*x*(1-2*x)/(2*x + (1-2*x)*2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 10 2012
G.f.: G(0)*(1-2*x)/2, where G(k) = 1 + 1/(1 - 2*x*(4*k+2-x)/( 2*x*(4*k+4-x) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 27 2014
a(-n) = a(n) / 2^n for all n in Z. - Michael Somos, Aug 24 2014
a(n) = A265185(n) / 4, connecting this sequence to the simple Lie algebra B_4. - Tom Copeland, Dec 04 2015
From G. C. Greubel, Aug 27 2025: (Start)
a(n) = 2^((n-2)/2)*( (n+1 mod 2)*A002203(n) + 2*sqrt(2)*(n mod 2)*A000129(n) ).
a(n) = 2^(n/2)*ChebyshevT(n, sqrt(2)). (End)

A063727 a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 8, 24, 80, 256, 832, 2688, 8704, 28160, 91136, 294912, 954368, 3088384, 9994240, 32342016, 104660992, 338690048, 1096024064, 3546808320, 11477712896, 37142659072, 120196169728, 388962975744, 1258710630400
Offset: 0

Views

Author

Klaus E. Kastberg (kastberg(AT)hotkey.net.au), Aug 12 2001

Keywords

Comments

Essentially the same as A085449.
Convergents to 2*golden ratio = (1+sqrt(5)).
Number of ways to tile an n-board with two types of colored squares and four types of colored dominoes.
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - Cino Hilliard, Sep 25 2005
a(n) is also the quasi-diagonal element A(i-1,i)=A(1,i-1) of matrix A(i,j) whose elements in first row A(1,k) and first column A(k,1) equal k-th Fibonacci Fib(k) and the generic element is the sum of adjacent (previous) in row and column minus the absolute value of their difference. - Carmine Suriano, May 13 2010
Equals INVERT transform of A006131: (1, 1, 5, 9, 29, 65, 181, ...). - Gary W. Adamson, Aug 12 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 2's along the three central diagonals. - John M. Campbell, Jul 19 2011
The numbers composing the denominators of the fractional limit to A134972. - Seiichi Kirikami, Mar 06 2012
Pisano period lengths: 1, 1, 8, 1, 5, 8, 48, 1, 24, 5, 10, 8, 42, 48, 40, 1, 72, 24, 18, 5, ... - R. J. Mathar, Aug 10 2012

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 235.
  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

Second row of A234357. Row sums of triangle A016095.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • GAP
    List([0..25],n->2^n*Fibonacci(n+1)); # Muniru A Asiru, Nov 24 2018
  • Magma
    [n le 2 select n else 2*Self(n-1) + 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018
    
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+4*a[n-2]od: seq(a[n], n=1..33); # Zerinvary Lajos, Dec 15 2008
  • Mathematica
    a[n_]:=(MatrixPower[{{1,5},{1,1}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    CoefficientList[Series[1/(1 - 2 x - 4 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 31 2014 *)
    LinearRecurrence[{2, 4}, {1, 2}, 50] (* G. C. Greubel, Jan 07 2018 *)
  • PARI
    s(n)=if(n<2,n+1,(s(n-1)+(s(n-2)*2))*2); for(n=0,32,print(s(n)))
    
  • PARI
    { for (n=0, 200, if (n>1, a=2*a1 + 4*a2; a2=a1; a1=a, if (n, a=a1=2, a=a2=1)); write("b063727.txt", n, " ", a) ) } \\ Harry J. Smith, Aug 28 2009
    
  • SageMath
    [lucas_number1(n,2,-4) for n in range(1, 26)] # Zerinvary Lajos, Apr 22 2009
    

Formula

a(n) = 2 * A087206(n+1).
From Vladeta Jovovic, Aug 16 2001: (Start)
a(n) = sqrt(5)/10*((1+sqrt(5))^(n+1) - (1-sqrt(5))^(n+1)).
G.f.: 1/(1-2*x-4*x^2). (End)
From Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003: (Start)
a(2*n) = 4*a(n-1)^2 + a(n)^2.
A084057(n+1)/a(n) converges to sqrt(5). (End)
E.g.f.: exp(x)*(cosh(sqrt(5)*x)+sinh(sqrt(5)*x)/sqrt(5)). - Paul Barry, Sep 20 2003
a(n) = 2^n*Fibonacci(n+1). - Vladeta Jovovic, Oct 25 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k+1)*5^k. - Paul Barry, Nov 15 2003
a(n) = U(n, i/2)*(-i*2)^n, i^2=-1. - Paul Barry, Nov 17 2003
Simplified formula: ((1+sqrt(5))^n-(1-sqrt(5))^n)/sqrt(20). Offset 1. a(3)=8. - Al Hakanson (hawkuu(AT)gmail.com), Jan 03 2009
First binomial transform of 1,1,5,5,25,25. - Al Hakanson (hawkuu(AT)gmail.com), Jul 20 2009
a(n) = A(n-1,n) = A(n,n-1); A(i,j) = A(i-1,j) + A(i,j-1) - abs(A(i-1,j) - A(i,j-1)). - Carmine Suriano, May 13 2010
G.f.: G(0) where G(k) = 1 + 2*x*(1+2*x)/(1 - 2*x*(1+2*x)/(2*x*(1+2*x) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 31 2013
G.f.: G(0)/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
G.f.: Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k+2 + 4*x )/( x*(4*k+4 + 4*x ) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Sep 21 2013
Sum_{n>=0} 1/a(n) = A269991. - Amiram Eldar, Feb 01 2021

Extensions

Better description from Jason Earls and Vladeta Jovovic, Aug 16 2001
Incorrect comment removed by Greg Dresden, Jun 02 2020

A140480 RMS numbers: numbers n such that root mean square of divisors of n is an integer.

Original entry on oeis.org

1, 7, 41, 239, 287, 1673, 3055, 6665, 9545, 9799, 9855, 21385, 26095, 34697, 46655, 66815, 68593, 68985, 125255, 155287, 182665, 242879, 273265, 380511, 391345, 404055, 421655, 627215, 730145, 814463, 823537, 876785, 1069895, 1087009, 1166399, 1204281, 1256489
Offset: 1

Views

Author

Ctibor O. Zizka, Jun 29 2008, Jul 11 2008

Keywords

Comments

For any numbers, A and B, both appearing in the sequence, if gcd(A,B)=1, then A*B is also in the sequence. - Andrew Weimholt, Jul 01 2008
The primes in this sequence are the NSW primes (A088165). For the terms less than 2^31, the only powers greater than 1 appearing in the prime factorization of numbers are 3^3 and 13^2. It appears that all terms are +-1 (mod 8). See A224988 for even numbers. - T. D. Noe, Jul 06 2008, Apr 25 2013
A basis for this sequence is given by A002315. This can be considered as the convergents of quasiregular continued fractions or a special 6-ary numeration system (see A. S. Fraenkel) which gives the characterization of positions of some heap or Wythoff game. What is the Sprague-Grundy function of this game?
Sequence generalized: sigma_r-numbers are numbers n for which sigma_r(n)/sigma_0(n) = c^r. Sigma_r(n) denotes sum of r-th powers of divisors of n; c,r positive integers. This sequence are sigma_2-numbers, A003601 are sigma_1-numbers. In a weaker form we have sigma_r(n)/sigma_0(n) = c^t; t is an integer from <1,r>. - Ctibor O. Zizka, Jul 14 2008
The primes in this sequence are prime numerators with an odd index in A001333. The RMS values (A141812) of prime RMS numbers (this sequence) are prime Pell numbers (A000129) with an odd index. - Ctibor O. Zizka, Aug 13 2008
From Ctibor O. Zizka, Aug 30 2008: (Start)
The set of RMS numbers n could be split into subsets according to the number and form of divisors of n. By definition, RMS(n) = sqrt(sigma_2(n) / sigma_0(n)) should be an integer. Now consider some examples. For n prime number, n has 2 divisors [1,n] and we have to solve Pell's equation n^2 = 2*C^2 - 1; C positive integer. The solution is a prime n of the form u(i) = 6*u(i-1) - u(i-2), i >= 2, u(0)=1, u(1)=7, known as an NSW prime (A088165). For n = p_1*p_2, p_1 and p_2 primes, n has 4 divisors {1; p_1; p_2; p_1*p_2}. There are 2 possible cases. Firstly p^2 = (2*C)^2 - 1 which does not hold for any prime p; secondly p_1^2 = 2*C_1^2 - 1 and p_2^2 = 2*C_2^2 - 1; C_1 and C_2 positive integers.
The solution is that p_1 and p_2 are different NSW primes. If n = p^3, divisors of n are {1; p; p^2; p^3} and we have to solve the Diophantine equation (p^8 - 1)/(p - 1) = (2*C)^2. This equation has no solution for any prime p. RMS numbers n with 4 divisors are only of the form n = p_1*p_2, with p_1 and p_2 NSW primes. The general case is n = p_1*...*p_t, n has 2^t divisors, and for t >= 3, NSW primes are not the only solution. If some of the prime divisors are equals p_i = p_j = ... = p_k, the general case n = p_1*...*p_t is "degenerate" because of the multiplicity of prime factors and therefore n has fewer than 2^t divisors. (End)
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878, whose prime terms give A121534. a(1)=5 gives A001834, whose prime terms give A086386. a(1)=6 gives A030221, whose prime terms {29, 139, 3191, ...} are not a sequence on the OEIS. a(1)=7 gives A002315, whose prime terms give A088165. a(1)=8 gives A033890; the OEIS does not have its prime terms as a sequence (do there exist any prime terms?). a(1)=9 gives A057080, whose prime terms {71, 34649, 16908641, ...} are not a sequence in the OEIS. a(1)=10 gives A057081, whose prime terms {389806471, 192097408520951, ...} are not a sequence in the OEIS. - Ctibor O. Zizka, Sep 02 2008
16 of the first 1660 terms are even (the smallest is 2217231104). The first 16 even terms are all divisible by 30976. - Donovan Johnson, Apr 16 2013
All the 83 even terms up to 10^13 (see A224988) are divisible by 30976. - Giovanni Resta, Oct 29 2019

Crossrefs

Programs

  • Haskell
    a140480 n = a140480_list !! (n-1)
    a140480_list = filter
        ((== 1) . a010052 . (\x -> a001157 x `div` a000005 x)) a020486_list
    -- Reinhard Zumkeller, Jan 15 2013
  • Mathematica
    rmsQ[n_] := IntegerQ[Sqrt[DivisorSigma[2, n]/DivisorSigma[0, n]]]; m = 160000; sel1 = Select[8*Range[0, m]+1, rmsQ]; sel7 = Select[8*Range[m]-1, rmsQ]; Union[sel1, sel7] (* Jean-François Alcover, Aug 31 2011, after T. D. Noe's comment *)
    Select[Range[1300000],IntegerQ[RootMeanSquare[Divisors[#]]]&] (* Harvey P. Dale, Mar 24 2016 *)

Extensions

More terms from T. D. Noe and Andrew Weimholt, Jul 01 2008

A080764 First differences of A049472, floor(n/sqrt(2)).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0
Offset: 0

Views

Author

Matthew Vandermast, Mar 25 2003

Keywords

Comments

Fixed point of the morphism 0->1, 1->110. - Benoit Cloitre, May 31 2004
As binary constant 0.1101101110110... = 0.85826765646... (A119812), see Fxtbook link. - Joerg Arndt, May 15 2011
Characteristic word with slope 1/sqrt(2) [see J. L. Ramirez et al.]. - R. J. Mathar, Jul 09 2013
From Peter Bala, Nov 22 2013: (Start)
Sturmian word: equals the limit word S(infinity) where S(0) = 0, S(1) = 1 and for n >= 1, S(n+1) = S(n)S(n)S(n-1).
More generally, for k = 0,1,2,..., we can define a sequence of words S_k(n) by S_k(0) = 0, S_k(1) = 0...01 (k 0's) and for n >= 1, S_k(n+1) = S_k(n)S_k(n)S_k(n-1). Then the limit word S_k(infinity) is a Sturmian word whose terms are given by a(n) = floor((n + 2)/(k + sqrt(2))) - floor((n + 1)/(k + sqrt(2))).
This sequence corresponds to the case k = 0. See A159684 (case k = 1) and A171588 (case k = 2). Compare with the Fibonacci words A005614, A221150, A221151 and A221152. See also A230901. (End)
For n > 0: a(A001951(n)) = 1, a(A001952(n)) = 0. - Reinhard Zumkeller, Jul 03 2015
Binary complement of the Pell word A171588. - Michel Dekking, Feb 22 2018

Examples

			From _Peter Bala_, Nov 22 2013: (Start)
The first few Sturmian words S(n) are
S(0) = 0
S(1) = 1
S(2) = 110
S(3) = 110 110 1
S(4) = 1101101 1101101 110
S(5) = 11011011101101110 11011011101101110 1101101
The lengths of the words are [1, 1, 3, 7, 17, 41, ...] = A001333.  (End)
		

Crossrefs

Programs

  • Haskell
    a080764 n = a080764_list !! n
    a080764_list = tail $ zipWith (-) (tail a049472_list) a049472_list
    -- Reinhard Zumkeller, Jul 03 2015
    
  • Maple
    A080764 := proc(n)
        alpha := 1/sqrt(2) ;
        floor((n+2)*alpha)-floor((n+1)*alpha) ;
    end proc: # R. J. Mathar, Jul 09 2013
  • Mathematica
    Nest[ Flatten[ # /. {0 -> 1, 1 -> {1, 1, 0}}] &, {1}, 7] (* Robert G. Wilson v, Apr 16 2005 *)
    NestList[ Flatten[ # /. {0 -> {1}, 1 -> {1, 0, 1}}] &, {1}, 5] // Flatten (* or *)
    t = Table[Floor[n/Sqrt[2]], {n, 111}]; Drop[t, 1] - Drop[t, -1] (* Robert G. Wilson v, Nov 03 2005 *)
    a[ n_] := With[{m = n + 1}, Floor[(m + 1) / Sqrt[2]] - Floor[m / Sqrt[2]]]; (* Michael Somos, Aug 19 2018 *)
  • PARI
    {a(n) = n++; my(k = sqrtint(n*n\2)); n*(n+2) > 2*k*(k+2)}; /* Michael Somos, Aug 19 2018 */
    
  • Python
    from math import isqrt
    def A080764(n): return (isqrt((m:=(n+2)**2)<<1)>>1)-(isqrt(m-(n<<1)-3<<1)>>1) # Chai Wah Wu, May 19 2025

Formula

a(n) = floor((n+2)*sqrt(2)/2) - floor((n+1)*sqrt(2)/2).
a(n) = A188295(n+2) for all n in Z. - Michael Somos, Aug 19 2018

A084057 a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 6, 16, 56, 176, 576, 1856, 6016, 19456, 62976, 203776, 659456, 2134016, 6905856, 22347776, 72318976, 234029056, 757334016, 2450784256, 7930904576, 25664946176, 83053510656, 268766806016, 869747654656, 2814562533376, 9108115685376, 29474481504256
Offset: 0

Views

Author

Paul Barry, May 10 2003

Keywords

Comments

Inverse binomial transform of A001077. Binomial transform of expansion of cosh(sqrt(5)*x) (1,0,5,0,25,...).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - Cino Hilliard, Sep 25 2005
Numerators of fractions in the approximation of the square root of 5 satisfying: a(n) = (a(n-1)+c)/(a(n-1)+1), with c=5 and a(1)=1. For denominators see A063727. - Mark Dols, Jul 24 2009
Equals right border of triangle A143969. (1, 6, 16, 56, ...) = row sums of triangle A143969 and INVERT transform of (1, 5, 5, 5, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 5 types of other natural numbers. - Milan Janjic, Aug 13 2010
From Gary W. Adamson, Jul 30 2016: (Start)
The sequence is case N=1 in an infinite set obtained by taking powers of the 2 X 2 matrix M = [(1,5); (1,N)], then extracting the upper left terms. The infinite set begins:
N=1 (A084057): 1, 6, 16, 56, 176, 576, 1856, ...
N=2 (A108306): 1, 6, 21, 81, 306, 1161, 4401, ...
N=3 (A164549): 1, 6, 26, 116, 516, 2296, 10216, ...
N=4 (A015449): 1, 6, 31, 161, 836, 4341, 22541, ...
N=5 (A000400): 1, 6, 36, 216, 1296, 7776, 46656, ...
N=6 (A049685): 1, 6, 41, 281, 1926, 13201, 90481, ...
N=7 (.......): 1, 6, 46, 356, 2756, 21336, 222712, ...
...
Sequences in the above set can be obtained by taking INVERT transforms of the following:
N=1 INVERT transform of (1, 5, 5, 5, 5, 5, ...
N=2 ..."......"......". (1, 5, 10, 20, 40, 80, ...
N=3 ..."......"......". (1, 5, 15, 45, 135, 405, ...
N=4 ..."......"......". (1, 5, 20, 80, 320, 1280, ...
...
with the pattern (1, 5, N*5, (N^2)*5, (N^3)*5, ...
It appears that the sequence generated from powers (n>0) of the matrix P = [(1,a); (1,b)], (a,b > 0), then extracting the upper left terms, is equal to the INVERT transform of the sequence starting: (1, a, b*a, (b^2)*a, (b^3)*a, ...). (End)

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

a(n) = A087131(n)/2.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 31 2016
  • Mathematica
    f[n_] := Simplify[((1 + Sqrt[5])^n + (1 - Sqrt[5])^n)/2]; Array[f, 28, 0] (* Or *)
    LinearRecurrence[{2, 4}, {1, 1}, 28] (* Robert G. Wilson v, Sep 18 2013 *)
    RecurrenceTable[{a[1] == 1, a[2] == 1, a[n] == 2 a[n-1] + 4 a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Jul 31 2016 *)
    Table[2^(n-1) LucasL[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 19 2016 *)
  • PARI
    lucas(n)=fibonacci(n-1)+fibonacci(n+1)
    a(n)=lucas(n)/2*2^n \\ Charles R Greathouse IV, Sep 18 2013
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,1,2,4, lambda n: 0); [next(it) for i in range(1,26)] # Zerinvary Lajos, Jul 09 2008
    
  • Sage
    [lucas_number2(n,2,-4)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n) = ((1+sqrt(5))^n + (1-sqrt(5))^n)/2.
G.f.: (1-x) / (1-2*x-4*x^2).
E.g.f.: exp(x) * cosh(sqrt(5)*x).
a(2n+1) = 2*a(n)*a(n+1) - (-4)^n. - Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*5^k . - Paul Barry, Jul 25 2004
a(n) = Sum_{k=0..n} A098158(n,k)*5^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = 2^(n-1)*A000032(n). - Mark Dols, Jul 24 2009
If p(1)=1, and p(i)=5 for i>1, and if A is the Hessenberg matrix of order n defined by: A(i,j) = p(j-i+1) for i<=j, A(i,j):=-1, (i=j+1), and A(i,j):=0 otherwise, then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = A063727(n) - A063272(n-1). - R. J. Mathar, Jun 06 2019
a(n) = 1 + 5*A014335(n). - R. J. Mathar, Jun 06 2019
Sum_{n>=1} 1/a(n) = A269992. - Amiram Eldar, Feb 01 2021

A088165 NSW primes: NSW numbers that are also prime.

Original entry on oeis.org

7, 41, 239, 9369319, 63018038201, 489133282872437279, 19175002942688032928599, 123426017006182806728593424683999798008235734137469123231828679
Offset: 1

Views

Author

Christian Schroeder, Sep 21 2003

Keywords

Comments

Next term a(9) is too large (99 digits) to include here. - Ray Chandler, Sep 21 2003
These primes are the prime RMS numbers (A140480): primes p such that (1+p^2)/2 is a square r^2. Then r is a Pell number, A000129. - T. D. Noe, Jul 01 2008
Also prime numerators with an odd index in A001333. - Ctibor O. Zizka, Aug 13 2008
r in the above note of T. D. Noe is a prime Pell number (A000129) with an odd index. - Ctibor O. Zizka, Aug 13 2008
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1)-3, x = (1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in the OEIS: a(1)=4 gives A002878, primes in it A121534. a(1)=5 gives A001834, primes in it A086386. a(1)=6 gives A030221, primes in it not in the OEIS {29, 139, 3191, ...}. a(1)=7 gives A002315, primes in it A088165. a(1)=8 gives A033890, primes in it not in the OEIS (do there exist any ?). a(1)=9 gives A057080, primes in it not in the OEIS {71, 34649, 16908641, ...}. a(1)=10 gives A057081, primes in it not in the OEIS {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008

References

  • Paulo Ribenboim, The New Book of Prime Number Records, 3rd edition, Springer-Verlag, New York, 1995, pp. 367-369.

Crossrefs

Cf. A002315 (NSW numbers), A005850 (indices for NSW primes).

Programs

  • PARI
    w=3+quadgen(32); forprime(p=2,1e3, if(ispseudoprime(t=imag((1+w)*w^p)), print1(t", "))) \\ Charles R Greathouse IV, Apr 29 2015

Formula

a(n) mod A005850(n) = 1. - Altug Alkan, Mar 17 2016

Extensions

More terms from Ray Chandler, Sep 21 2003

A002531 a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1); a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 5, 7, 19, 26, 71, 97, 265, 362, 989, 1351, 3691, 5042, 13775, 18817, 51409, 70226, 191861, 262087, 716035, 978122, 2672279, 3650401, 9973081, 13623482, 37220045, 50843527, 138907099, 189750626, 518408351, 708158977, 1934726305
Offset: 0

Views

Author

Keywords

Comments

Numerators of continued fraction convergents to sqrt(3), for n >= 1.
For the denominators see A002530.
Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the convergents 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to sqrt(3). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003
In the Murthy comment if we take a = 0, b = 1 then the denominator of the reduced fraction is a(n+1). A083336(n)/a(n+1) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 26 2003
If signs are disregarded, all terms of A002316 appear to be elements of this sequence. - Creighton Dement, Jun 11 2007
2^(-floor(n/2))*(1 + sqrt(3))^n = a(n) + A002530(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
Let T(n) = A000034(n), U(n) = A002530(n), V(n) = a(n), x(n) = U(n)/V(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (x(n) + x(m))/(1 + 3*x(n)*x(m)). - Michael Somos, Nov 29 2022

Examples

			1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 19.
Convergents are 1, 2, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571, 1351/780, 3691/2131, ... = A002531/A002530.
G.f. = 1 + x + 2*x^2 + 5*x^3 + 7*x^4 + 19*x^5 + 26*x^6 + 71*x^7 + ... - _Michael Somos_, Mar 22 2022
		

References

  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
  • 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).
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Crossrefs

Bisections are A001075 and A001834.
Cf. A002530 (denominators), A048788.
Cf. A002316.

Programs

  • GAP
    a:=[1,1,2,5];; for n in [5..40] do a[n]:=4*a[n-2]-a[n-4]; od; a; # G. C. Greubel, Nov 16 2018
  • Magma
    m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1 +x-2*x^2+x^3)/(1-4*x^2+x^4))); // G. C. Greubel, Nov 16 2018
    
  • Maple
    A002531 := proc(n) option remember; if n=0 then 0 elif n=1 then 1 elif n=2 then 1 elif type(n,odd) then A002531(n-1)+A002531(n-2) else 2*A002531(n-1)+A002531(n-2) fi; end; [ seq(A002531(n), n=0..50) ];
    with(numtheory): tp := cfrac (tan(Pi/3),100): seq(nthnumer(tp,i), i=-1..32 ); # Zerinvary Lajos, Feb 07 2007
    A002531:=(1+z-2*z**2+z**3)/(1-4*z**2+z**4); # Simon Plouffe; see his 1992 dissertation
  • Mathematica
    Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[3], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 01 2006 *)
    Join[{1},Numerator[Convergents[Sqrt[3],40]]] (* Harvey P. Dale, Jan 23 2012 *)
    CoefficientList[Series[(1 + x - 2 x^2 + x^3)/(1 - 4 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Nov 01 2014 *)
    LinearRecurrence[{0, 4, 0, -1}, {1, 1, 2, 5}, 35] (* Robert G. Wilson v, Feb 11 2018 *)
    a[ n_] := ChebyshevT[n, Sqrt[-1/2]]*Sqrt[2]^Mod[n,2]/I^n //Simplify; (* Michael Somos, Mar 22 2022 *)
    a[ n_] := If[n<0, (-1)^n*a[-n], SeriesCoefficient[ (1 + x - 2*x^2 + x^3) / (1 - 4*x^2 + x^4), {x, 0, n}]]; (* Michael Somos, Sep 23 2024 *)
  • PARI
    a(n)=contfracpnqn(vector(n,i,1+(i>1)*(i%2)))[1,1]
    
  • PARI
    apply( {A002531(n,w=quadgen(12))=real((2+w)^(n\/2)*if(bittest(n, 0), w-1, 1))}, [0..30]) \\ M. F. Hasler, Nov 04 2019
    
  • PARI
    {a(n) = if(n<0, (-1)^n*a(-n), polcoeff( (1 + x - 2*x^2 + x^3) / (1 - 4*x^2 + x^4) + x*O(x^n), n))}; /* Michael Somos, Sep 23 2024 */
    
  • Sage
    s=((1+x-2*x^2+x^3)/(1-4*x^2+x^4)).series(x,40); s.coefficients(x, sparse=False) # G. C. Greubel, Nov 16 2018
    

Formula

G.f.: (1 + x - 2*x^2 + x^3)/(1 - 4*x^2 + x^4).
a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1), n > 0.
a(2*n) = (1/2)*((2 + sqrt(3))^n+(2 - sqrt(3))^n); a(2*n) = A003500(n)/2; a(2*n+1) = round(1/(1 + sqrt(3))*(2 + sqrt(3))^n). - Benoit Cloitre, Dec 15 2002
a(n) = ((1 + sqrt(3))^n + (1 - sqrt(3))^n)/(2*2^floor(n/2)). - Bruno Berselli, Nov 10 2011
a(n) = A080040(n)/(2*2^floor(n/2)). - Ralf Stephan, Sep 08 2013
a(2*n) = (-1)^n*T(2*n,u) and a(2*n+1) = (-1)^n*1/u*T(2*n+1,u), where u = sqrt(-1/2) and T(n,x) denotes the Chebyshev polynomial of the first kind. - Peter Bala, May 01 2012
a(n) = (-sqrt(2)*i)^n*T(n, sqrt(2)*i/2)*2^(-floor(n/2)) = A026150(n)*2^(-floor(n/2)), n >= 0, with i = sqrt(-1) and the Chebyshev T polynomials (A053120). - Wolfdieter Lang, Feb 10 2018
From Franck Maminirina Ramaharo, Nov 14 2018: (Start)
a(n) = ((1 - sqrt(2))*(-1)^n + 1 + sqrt(2))*(((sqrt(2) - sqrt(6))/2)^n + ((sqrt(6) + sqrt(2))/2)^n)/4.
E.g.f.: cosh(sqrt(3/2)*x)*(sqrt(2)*sinh(x/sqrt(2)) + cosh(x/sqrt(2))). (End)
a(n) = (-1)^n*a(-n) for all n in Z. - Michael Somos, Mar 22 2022
a(n) = 4*a(n-2) - a(n-4). - Boštjan Gec, Sep 21 2023

Extensions

Name edited (as by discussion in A002530) by M. F. Hasler, Nov 04 2019

A002532 a(n) = 2*a(n-1) + 5*a(n-2), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 9, 28, 101, 342, 1189, 4088, 14121, 48682, 167969, 579348, 1998541, 6893822, 23780349, 82029808, 282961361, 976071762, 3366950329, 11614259468, 40063270581, 138197838502, 476712029909, 1644413252328, 5672386654201, 19566839570042, 67495612411089
Offset: 0

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Comments

The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 6 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(6). - Cino Hilliard, Sep 25 2005
For n>=2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and five types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are five types of twins; namely, both F (identical twins), both F (fraternal twins), both M (identical), both M (fraternal), or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=9, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F identical twins; (6) F,F fraternal twins; (7) M,M identical twins; (8) M,M fraternal twins; or (9) M,F twins (emphasizing that birth order is irrelevant here when children are the same gender, when two children are within the same pair of twins and when pairs of twins have both the same gender(s) and identical-vs-fraternal characteristics). - Rick L. Shepherd, Sep 19 2004
Pisano period lengths: 1, 2, 3, 4, 4, 6, 24, 8, 3, 4, 120, 12, 56, 24, 12, 16, 288, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012

Examples

			G.f. = x + 2*x^2 + 9*x^3 + 28*x^4 + 101*x^5 + 342*x^6 + 1189*x^7 + ...
		

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
  • 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).
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Crossrefs

Cf. A015581 (similar application, but no distinguishing identical vs. fraternal twins).
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    [Floor(((1+Sqrt(6))^n-(1-Sqrt(6))^n)/(2*Sqrt(6))): n in [0..30]]; // Vincenzo Librandi, Aug 15 2011
    
  • Magma
    [n le 2 select n-1 else 2*Self(n-1) + 5*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
  • Maple
    A002532:=-z/(-1+2*z+5*z**2); # Conjectured by Simon Plouffe in his 1992 dissertation
    # second program
    seq(simplify(2^(n-1) * hypergeom([1 - (1/2)*n, 1/2 - (1/2)*n], [1 - n], -5)), n = 2..25); # Peter Bala, Jul 06 2025
  • Mathematica
    Expand[Table[((1 + Sqrt[6])^n - (1 - Sqrt[6])^n)/(2Sqrt[6]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
    a[n_]:=(MatrixPower[{{1,2},{1,-3}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    LinearRecurrence[{2,5},{0,1},30] (* Harvey P. Dale, Nov 03 2011 *)
  • PARI
    Vec(1/(1-2*x-5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Apr 17 2012
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen2; it = recur_gen2(0,1,2,5); [next(it) for i in range(30)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [lucas_number1(n,2,-5) for n in range(0, 26)] # Zerinvary Lajos, Apr 22 2009
    

Formula

From Mario Catalani (mario.catalani(AT)unito.it), Jun 14 2003: (Start)
a(2*n+1) = 5*a(n)^2 + a(n+1)^2.
6*a(2*n+1) = 5*A002533(n)^2 + A002533(n+1)^2. (End)
From Paul Barry, Sep 20 2003: (Start)
G.f.: x/(1-2*x-5*x^2).
E.g.f.: exp(x)*sinh(sqrt(6)*x)/sqrt(6).
a(n) = ((1+sqrt(6))^n - (1-sqrt(6))^n)/(2*sqrt(6)). (End)
a(n) = Sum_{k=0..n} binomial(n, 2*k+1)*6^k. - Paul Barry, Sep 29 2004
G.f.: G(0)*x/(2*(1-x)), where G(k)= 1 + 1/(1 - x*(6*k-1)/(x*(6*k+5) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
From Peter Bala, Jul 06 2025: (Start)
For n >= 0, a(n+1) = (2^n) * Sum_{k = 0..floor(n/2)} binomial(n-k, k) * (5/4)^k.
For n >= 2, a(n) = 2^(n-1) * hypergeom([1 - (1/2)*n, 1/2 - (1/2)*n], [1 - n], -5).
Sum_{n >= 1} (-5)^n/(a(n)*a(n+1)) = -(sqrt(6) - 1).
Sum_{n >= 1} 5^n/(a(n)*a(n+2)) = 5/4; Sum_{n >= 1} 5^n/(a(n)*a(n+4)) = 755/7056.
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) = x + 2*x^2 + 9*x^3 + 28*x^4 + .... (End)

Extensions

More terms from Rick L. Shepherd, Sep 19 2004

A048654 a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 9, 22, 53, 128, 309, 746, 1801, 4348, 10497, 25342, 61181, 147704, 356589, 860882, 2078353, 5017588, 12113529, 29244646, 70602821, 170450288, 411503397, 993457082, 2398417561, 5790292204
Offset: 0

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Generalized Pellian with second term equal to 4.
The generalized Pellian with second term equal to s has the terms a(n) = A000129(n)*s + A000129(n-1). The generating function is -(1+s*x-2*x)/(-1+2*x+x^2). - R. J. Mathar, Nov 22 2007

Crossrefs

Programs

  • Haskell
    a048654 n = a048654_list !! n
    a048654_list =
       1 : 4 : zipWith (+) a048654_list (map (* 2) $ tail a048654_list)
    -- Reinhard Zumkeller, Aug 01 2011
    
  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!((1+2*x)/(1-2*x-x^2))); // G. C. Greubel, Jul 26 2018
    
  • Mathematica
    LinearRecurrence[{2,1},{1,4},30] (* Harvey P. Dale, Jul 27 2011 *)
  • Maxima
    a[0]:1$
    a[1]:4$
    a[n]:=2*a[n-1]+a[n-2]$
    A048654(n):=a[n]$
    makelist(A048654(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • PARI
    a(n)=(([0, 1; 1,2]^n)*[1,4]~)[1] \\ Charles R Greathouse IV, May 18 2015
    
  • SageMath
    [lucas_number1(n+1,2,-1) +2*lucas_number1(n,2,-1) for n in (0..40)] # G. C. Greubel, Aug 09 2022

Formula

a(n) = ((3+sqrt(2))*(1+sqrt(2))^n - (3-sqrt(2))*(1-sqrt(2))^n)/2*sqrt(2).
a(n) = 2*A000129(n+2) - 3*A000129(n+1). - Creighton Dement, Oct 27 2004
G.f.: (1+2*x)/(1-2*x-x^2). - Philippe Deléham, Nov 03 2008
a(n) = binomial transform of 1, 3, 2, 6, 4, 12, ... . - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
E.g.f.: exp(x)*cosh(sqrt(2)*x) + 3*exp(x)*sinh(sqrt(2)*x)/sqrt(2). - Vaclav Kotesovec, Feb 16 2015
a(n) is the denominator of the continued fraction [4, 2, ..., 2, 4] with n-1 2's in the middle. For the numerators, see A221174. - Greg Dresden and Tongjia Rao, Sep 02 2021
a(n) = A001333(n) + A000129(n). - G. C. Greubel, Aug 09 2022
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