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.

Showing 1-10 of 13 results. Next

A049347 Period 3: repeat [1, -1, 0].

Original entry on oeis.org

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

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Author

Keywords

Comments

G.f. 1/cyclotomic(3, x) (the third cyclotomic polynomial).
Self-convolution yields (-1)^n*A099254(n). - R. J. Mathar, Apr 06 2008
Hankel transform of A099324. - Paul Barry, Aug 10 2009
A057083(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0..n. - Michael Somos, Apr 29 2012
a(n) appears, together with b(n) = A099837(n+3) in the formula 2*exp(2*Pi*n*I/3) = b(n) + a(n)*sqrt(3)*I, n >= 0, with I = sqrt(-1). See A164116 for the case N=5. - Wolfdieter Lang, Feb 27 2014
The binomial transform is 1, 0, -1, -1, 0, 1, 1, 0, -1, -1.. (see A010891). The inverse binom. transform is 1, -2, 3, -3, 0, 9, -27, 54, -81.. (see A057682). - R. J. Mathar, Feb 25 2023

Examples

			G.f. = 1 - x + x^3 - x^4 + x^6 - x^7 + x^9 - x^10 + x^12 - x^13 + x^15 + ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 175.

Crossrefs

Alternating row sums of A049310 (Chebyshev-S). [Wolfdieter Lang, Nov 04 2011]

Programs

Formula

G.f.: 1/(1+x+x^2).
a(n) = +1 if n mod 3 = 0, a(n) = -1 if n mod 3 = 1, else 0.
a(n) = S(n, -1) = U(n, -1/2) (Chebyshev's U(n, x) polynomials.)
a(n) = 2*sqrt(3)*cos(2*Pi*n/3 + Pi/6)/3. - Paul Barry, Mar 15 2004
a(n) = Sum_{k >= 0} (-1)^(n-k)*C(n-k, k).
Given g.f. A(x), then B(x) = x * A(x) satisfies 0 = f(B(x), B(x^2)) where f(u, v) = u^2 - v + 2*u*v. - Michael Somos, Oct 03 2006
Euler transform of length 3 sequence [-1, 0, 1]. - Michael Somos, Oct 03 2006
a(n) = b(n+1) where b(n) is multiplicative with b(3^e) = 0^e, b(p^e) = 1 if p == 1 (mod 3), b(p^e) = (-1)^e if p == 2 (mod 3). - Michael Somos, Oct 03 2006
From Michael Somos, Oct 03 2006: (Start)
G.f.: (1 - x) /(1 - x^3).
a(n) = -a(1-n) = -a(n-1) - a(n-2) = a(n-3). (End)
From Michael Somos, Apr 29 2012: (Start)
G.f.: 1 / (1 + x / ( 1 - x / (1 + x))).
a(n) = (-1)^n * A010892(n).
a(n) * n! = A194770(n+1).
Revert transform of A001006. Convolution inverse of A130716. MOBIUS transform of A002324. EULER transform is A111317. BIN1 transform of itself. STIRLING transform is A143818(n+2). (End)
a(-n) = A057078(n). a(n) = A106510(n+1) unless n=0. - Michael Somos, Oct 15 2008
G.f. A(x) = 1/(1+x+x^2) = Q(0); Q(k) = 1- x/(1 - x^2/(x^2 - 1 + x/(x - 1 + x^2/(x^2 - 1/Q(k+1))))); (continued fraction 3 kind, 5-step ). - Sergei N. Gladkovskii, Jun 19 2012
a(n) = -1 + floor(67/333*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = -1 + floor(19/26*3^(n+1)) mod 3. - Hieronymus Fischer, Jan 03 2013
a(n) = ceiling((n-1)/3) - ceiling(n/3) + floor(n/3) - floor((n-1)/3). - Wesley Ivan Hurt, Dec 06 2013
a(n) = n + 1 - 3*floor((n+2)/3). - Mircea Merca, Feb 04 2014
a(n) = A102283(n+1) for all n in Z. - Michael Somos, Sep 24 2019
E.g.f.: exp(-x/2)*(3*cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2))/3. - Stefano Spezia, Oct 26 2022

Extensions

Edited by Charles R Greathouse IV, Mar 23 2010

A005169 Number of fountains of n coins.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 9, 15, 26, 45, 78, 135, 234, 406, 704, 1222, 2120, 3679, 6385, 11081, 19232, 33379, 57933, 100550, 174519, 302903, 525734, 912493, 1583775, 2748893, 4771144, 8281088, 14373165, 24946955, 43299485, 75153286, 130440740, 226401112, 392955956, 682038999, 1183789679, 2054659669, 3566196321, 6189714276
Offset: 0

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Author

Keywords

Comments

A fountain is formed by starting with a row of coins, then stacking additional coins on top so that each new coin touches two in the previous row.
Also the number of Dyck paths for which the sum of the heights of the vertices that terminate an upstep (i.e., peaks and doublerises) is n. Example: a(4)=3 because we have UDUUDD, UUDDUD and UDUDUDUD. - Emeric Deutsch, Mar 22 2008
Also the number of ordered trees with path length n (follows from previous comment via a standard bijection). - Emeric Deutsch, Mar 22 2008
Probably first studied by Jim Propp (unpublished).
Number of compositions of n with c(1) = 1 and c(i+1) <= c(i) + 1. (Slide each row right 1/2 step relative to the row below, and count the columns.) - Franklin T. Adams-Watters, Nov 24 2009
With the additional requirement for weak unimodality one obtains A001524. - Joerg Arndt, Dec 09 2012

Examples

			An example of a fountain with 19 coins:
... O . O O
.. O O O O O O . O
. O O O O O O O O O
From _Peter Bala_, Dec 26 2012: (Start)
F(1/10) = Sum_{n >= 0} a(n)/10^n has the simple continued fraction expansion 1 + 1/(8 + 1/(1 + 1/(8 + 1/(1 + 1/(98 + 1/(1 + 1/(98 + 1/(1 + 1/(998 + 1/(1 + 1/(998 + 1/(1 + ...)))))))))))).
F(-1/10) = Sum_{n >= 0} (-1)^n*a(n)/10^n has the simple continued fraction expansion 1/(1 + 1/(9 + 1/(1 + 1/(9 + 1/(99 + 1/(1 + 1/(99 + 1/(999 + 1/(1 + 1/(999 + 1/(9999 + 1/(1 + ...)))))))))))).
(End)
		

References

  • S. R. Finch, Mathematical Constants, Cambridge, 2003, p. 381.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A001524, A192728, A192729, A192730, A111317, A143951, A285903, A226999 (inverse Euler transform), A291148 (convolution inverse).
First column of A168396. - Franklin T. Adams-Watters, Nov 24 2009
Diagonal of A185646.
Row sums of A047998. Column sums of A138158. - Emeric Deutsch, Mar 22 2008

Programs

  • Haskell
    a005169 0 = 1
    a005169 n = a168396 n 1  -- Reinhard Zumkeller, Sep 13 2013; corrected by R. J. Mathar, Sep 16 2013
  • Maple
    P[0]:=1: for n to 40 do P[n]:=sort(expand(t*(sum(P[j]*P[n-j-1]*t^(n-j-1),j= 0..n-1)))) end do: F:=sort(sum(P[k],k=0..40)): seq(coeff(F,t,j),j=0..36); # Emeric Deutsch, Mar 22 2008
    # second Maple program:
    A005169_G:= proc(x,NK); Digits:=250; Q2:=1;
            for k from NK by -1 to 0 do  Q1:=1-x^k/Q2; Q2:=Q1; od;
            Q3:=Q2; S:=1-Q3;
    end:
    series(A005169_G(x, 20), x, 21); # Sergei N. Gladkovskii, Dec 18 2011
  • Mathematica
    m = 36; p[0] = 1; p[n_] := p[n] = Expand[t*Sum[p[j]*p[n-j-1]*t^(n-j-1), {j, 0, n-1}]]; f[t_] = Sum[p[k], {k, 0, m}]; CoefficientList[Series[f[t], {t, 0, m}], t] (* Jean-François Alcover, Jun 21 2011, after Emeric Deutsch *)
    max = 43; Series[1-Fold[Function[1-x^#2/#1], 1, Range[max, 0, -1]], {x, 0, max}] // CoefficientList[#, x]& (* Jean-François Alcover, Sep 16 2014 *)
    b[n_, i_] := b[n, i] = If[n==0, 1, Sum[b[n-j, j], {j, 1, Min[i+1, n]}]];
    c[n_] :=  b[n, 0] - b[n-1, 0];
    c /@ Range[0, 50] // Accumulate  (* Jean-François Alcover, Nov 14 2020, after Alois P. Heinz in A289080 *)
  • PARI
    /* using the g.f. from p. L1278 of the Glasser, Privman, Svrakic paper */
    N=30;  x='x+O('x^N);
    P(k)=sum(n=0,N, (-1)^n*x^(n*(n+1+k))/prod(j=1,n,1-x^j));
    G=1+x*P(1)/( (1-x)*P(1)-x^2*P(2) );
    Vec(G) /* Joerg Arndt, Feb 10 2011 */
    
  • PARI
    /* As a continued fraction: */
    {a(n)=local(A=1+x,CF);CF=1+x;for(k=0,n,CF=1/(1-x^(n-k+1)*CF+x*O(x^n));A=CF);polcoeff(A,n)} /* Paul D. Hanna */
    
  • PARI
    /* By the Rogers-Ramanujan continued fraction identity: */
    {a(n)=local(A=1+x,P,Q);
    P=sum(m=0,sqrtint(n),(-1)^m*x^(m*(m+1))/prod(k=1,m,1-x^k));
    Q=sum(m=0,sqrtint(n),(-1)^m*x^(m^2)/prod(k=1,m,1-x^k));
    A=P/(Q+x*O(x^n));polcoeff(A,n)}  /* Paul D. Hanna */
    

Formula

A005169(n) = f(n, 1), where f(n, p) = 0 if p > n, 1 if p = n, Sum(1 <= q <= p+1; f(n-p, q)) if p < n. f=A168396.
G.f.: F(t) = Sum_{k>=0} P[k], where P[0]=1, P[n] = t*Sum_{j= 0..n-1} P[j]*P[n-j-1]*t^(n-j-1) for n >= 1. - Emeric Deutsch, Mar 22 2008
G.f.: 1/(1-x/(1-x^2/(1-x^3/(1-x^4/(1-x^5/(...)))))) [given on the first page of the Odlyzko/Wilf reference]. - Joerg Arndt, Mar 08 2011
G.f.: 1/G(0), where G(k)= 1 - x^(k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
G.f.: A(x) = P(x)/Q(x) where
P(x) = Sum_{n>=0} (-1)^n* x^(n*(n+1)) / Product_{k=1..n} (1-x^k),
Q(x) = Sum_{n>=0} (-1)^n* x^(n^2) / Product_{k=1..n} (1-x^k),
due to the Rogers-Ramanujan continued fraction identity. - Paul D. Hanna, Jul 08 2011
From Peter Bala, Dec 26 2012: (Start)
Let F(x) denote the o.g.f. of this sequence. For positive integer n >= 3, the real number F(1/n) has the simple continued fraction expansion 1 + 1/(n-2 + 1/(1 + 1/(n-2 + 1/(1 + 1/(n^2-2 + 1/(1 + 1/(n^2-2 + 1/(1 + ...)))))))), while for n >= 2, F(-1/n) has the simple continued fraction expansion 1/(1 + 1/(n-1 + 1/(1 + 1/(n-1 + 1/(n^2-1 + 1/(1 + 1/(n^2-1 + 1/(n^3-1 + 1/(1 + ...))))))))). Examples are given below. Cf. A111317 and A143951.
(End)
a(n) = c * x^(-n) + O((5/3)^n), where c = 0.312363324596741... and x = A347901 = 0.576148769142756... is the lowest root of the equation Q(x) = 0, Q(x) see above (Odlyzko & Wilf 1988). - Vaclav Kotesovec, Jul 18 2013, updated Sep 24 2020
G.f.: G(0), where G(k)= 1 - x^(k+1)/(x^(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 06 2013
G.f.: 1 - 1/x + 1/(x*W(0)), where W(k)= 1 - x^(2*k+2)/(1 - x^(2*k+1)/W(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 16 2013

Extensions

More terms from David W. Wilson, Apr 30 2001

A014551 Jacobsthal-Lucas numbers.

Original entry on oeis.org

2, 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025, 2047, 4097, 8191, 16385, 32767, 65537, 131071, 262145, 524287, 1048577, 2097151, 4194305, 8388607, 16777217, 33554431, 67108865, 134217727, 268435457, 536870911, 1073741825, 2147483647, 4294967297, 8589934591
Offset: 0

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Author

Keywords

Comments

Also gives the number of points of period n in the subshift of finite type corresponding to the square matrix A=[1,2;1,0] (this is then given by trace(A^n)). - Thomas Ward, Mar 07 2001
Sequence is identical to its signed inverse binomial transform (autosequence of the second kind). - Paul Curtz, Jul 11 2008
a(n) can be expressed in terms of values of the Fibonacci polynomials F_n(x), computed at x=1/sqrt(2). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
Pisano period lengths: 1, 1, 2, 2, 4, 2, 6, 2, 6, 4, 10, 2, 12, 6, 4, 2, 8, 6, 18, 4, ... - R. J. Mathar, Aug 10 2012
Let F(x) = Product_{n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number 1 + F(-1/2) = 2.83717 78068 73232 99799 ... = 2 + 1/(1 + 1/(5 + 1/(7 + 1/(17 + ...)))). See A111317. - Peter Bala, Dec 26 2012
With different signs, 2, -1, 5, -7, 17, -31, 65, -127, 257, -511, 1025, -2047, ... is the Lucas V(-1,-2) sequence. - R. J. Mathar, Jan 08 2013
The identity 2 = 2/2 + 2^2/(2*1) - 2^3/(2*1*5) - 2^4/(2*1*5*7) + 2^5/(2*1*5*7*17) + 2^6/(2*1*5*7*17*31) - - + + can be viewed as a generalized Engel-type expansion of the number 2 to the base 2. Compare with A062510. - Peter Bala, Nov 13 2013
For n >= 2, a(n) is the number of ways to tile a 2 X n strip, where the first two columns have an extra cell at the top, with 1 X 2 dominoes and 2 X 2 squares. Shown here is one of the a(7)=127 ways for the n=7 case:
._.
|_|_________.
| | | |_| |
||__|_|_|_|. - Greg Dresden, Sep 26 2021
Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965) and the French mathematician Édouard Lucas (1842-1891). - Amiram Eldar, Oct 02 2023

References

  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 180, 255.
  • Lind and Marcus, An Introduction to Symbolic Dynamics and Coding, Cambridge University Press, 1995. (General material on subshifts of finite type)
  • Kritkhajohn Onphaeng and Prapanpong Pongsriiam. Jacobsthal and Jacobsthal-Lucas Numbers and Sums Introduced by Jacobsthal and Tverberg. Journal of Integer Sequences, Vol. 20 (2017), Article 17.3.6.
  • Abdelmoumène Zekiri, Farid Bencherif, Rachid Boumahdi, Generalization of an Identity of Apostol, J. Int. Seq., Vol. 21 (2018), Article 18.5.1.

Crossrefs

Cf. A001045 (companion "autosequence"), A019322, A066845, A111317.
Cf. A135440 (first differences), A166920 (partial sums).
Cf. A006995.

Programs

Formula

a(n+1) = 2 * a(n) - (-1)^n * 3.
From Len Smiley, Dec 07 2001: (Start)
a(n) = 2^n + (-1)^n.
G.f.: (2-x)/(1-x-2*x^2). (End)
E.g.f.: exp(x) + exp(-2*x) produces a signed version. - Paul Barry, Apr 27 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-1, 2*k)*3^(2*k)/2^(n-2). - Paul Barry, Feb 21 2003
0, 1, 5, 7 ... is 2^n - 2*0^n + (-1)^n, the 2nd inverse binomial transform of (2^n-1)^2 (A060867). - Paul Barry, Sep 05 2003
a(n) = 2*T(n, i/(2*sqrt(2))) * (-i*sqrt(2))^n with i^2=-1. - Paul Barry, Nov 17 2003
a(n) = A078008(n) + A001045(n+1). - Paul Barry, Feb 12 2004
a(n) = 2*A001045(n+1) - A001045(n). - Paul Barry, Mar 22 2004
a(0)=2, a(1)=1, a(n) = a(n-1) + 2*a(n-2) for n > 1. - Philippe Deléham, Nov 07 2006
a(2*n+1) = Product_{d|(2*n+1)} cyclotomic(d,2). a(2^k*(2*n+1)) = Product_{d|(2*n+1)} cyclotomic(2*d,2^(2^k)). - Miklos Kristof, Mar 12 2007
a(n) = 2^{(n-1)/2}F_{n-1}(1/sqrt(2)) + 2^{(n+2)/2}F_{n-2}(1/sqrt(2)). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
E.g.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k*(k+1)/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
G.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
a(n) = sqrt(9*(A001045)^2 + (-1)^n*2^(n+2)). - Vladimir Shevelev, Mar 13 2013
G.f.: 2 + G(0)*x*(1+4*x)/(2-x), where G(k) = 1 + 1/(1 - x*(9*k-1)/( x*(9*k+8) - 2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 13 2013
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 9*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
For n >= 1: a(n) = A006995(2^((n+2)/2)) when n is even, a(n) = A006995(3*2^((n-1)/2) - 1) when n is odd. - Bob Selcoe, Sep 04 2017
a(n) = J(n) + 4*J(n-1), a(0)=2, where J is A001045. - Yuchun Ji, Apr 23 2019
For n >= 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020
For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 06 2020
From Kai Wang, May 30 2020: (Start)
(2 - a(n+1)/a(n))/9 = Sum_{m>=n} (-2)^m/(a(m)*a(m+1)).
a(n) = 2*A001045(n+1) - A001045(n).
a(n)^2 = a(2*n) + 2*(-2)^n.
a(n)^2 = 9*A001045(n)^2 + 4*(-2)^n.
a(2*n) = 9*A001045(n)^2 + 2*(-2)^n.
2*A001045(m+n) = A001045(m)*a(n) + a(m)*A001045(n).
2*(-2)^n*A001045(m-n) = A001045(m)*a(n) - a(m)*A001045(n).
A001045(m+n) + (-2)^n*A001045(m-n) = A001045(m)*a(n).
A001045(m+n) - (-2)^n*A001045(m-n) = a(m)*A001045(n).
2*a(m+n) = 9*A001045(m)*A001045(n) + a(m)*a(n).
2*(-2)^n*a(m-n) = a(m)*a(n) - 9*A001045(m)*A001045(n).
a(m+n) - (-2)^n*a(m-n) = 9*A001045(m)*A001045(n).
a(m+n) + (-2)^n*a(m-n) = a(m)*a(n).
a(m+n)*a(m-n) - a(m)*a(m) = 9*(-2)^(m-n)*A001045(n)^2.
a(m+1)*a(n) - a(m)*a(n+1) = 9*(-2)^n*A001045(m-n). (End)
a(n) = F(n+1) + F(n-1) + Sum_{k=0..(n-2)} a(k)*F(n-1-k) for F(n) the Fibonacci numbers and for n > 1. - Greg Dresden, Jun 03 2020

A143951 Number of Dyck paths such that the area between the x-axis and the path is n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 4, 6, 9, 14, 21, 31, 47, 71, 107, 161, 243, 367, 553, 834, 1258, 1898, 2863, 4318, 6514, 9827, 14824, 22361, 33732, 50886, 76762, 115796, 174680, 263509, 397508, 599647, 904579, 1364576, 2058489, 3105269, 4684359, 7066449, 10659877, 16080632, 24257950, 36593598, 55202165, 83273553, 125619799, 189499952
Offset: 0

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Author

Emeric Deutsch, Oct 09 2008

Keywords

Comments

Column sums of A129182.

Examples

			a(5)=3 because we have UDUUDD, UUDDUD and UDUDUDUDUD, where U=(1,1) and D=(1,-1).
From _Peter Bala_, Dec 26 2012: (Start)
F(1/10) = sum {n >= 0} a(n)/10^n has the simple continued fraction expansion 1 + 1/(8 + 1/(1 + 1/(98 + 1/(1 + 1/(998 + 1/(1 + ...)))))).
F(-1/10) = sum {n >= 0} (-1)^n*a(n)/10^n has the simple continued fraction expansion 1/(1 + 1/(10 + 1/(100 + 1/(1000 + ...)))).
(End)
		

Crossrefs

Cf. A129182, A291874 (convolution inverse).

Programs

  • Maple
    g:=1/(1-x/(1-x^3/(1-x^5/(1-x^7/(1-x^9/(1-x^11/(1-x^13/(1-x^15)))))))): gser:= series(g,x=0,45): seq(coeff(gser,x,n),n=0..44);
    # second Maple program:
    b:= proc(x, y, k) option remember;
          `if`(y<0 or y>x or k<0 or k>x^2/2-(y-x)^2/4, 0,
          `if`(x=0, 1, b(x-1, y-1, k-y+1/2) +b(x-1, y+1, k-y-1/2)))
        end:
    a:= n-> add(b(2*n-4*t, 0, n), t=0..n/2):
    seq(a(n), n=0..50);  # Alois P. Heinz, Aug 24 2018
  • Mathematica
    terms = 50; CoefficientList[1/(1+ContinuedFractionK[-x^(2i-1), 1, {i, 1, Sqrt[terms]//Ceiling}]) + O[x]^terms, x] (* Jean-François Alcover, Jul 11 2018 *)
  • PARI
    N=66; q = 'q +O('q^N);
    G(k) = if(k>N, 1, 1 - q^(k+1) / G(k+2) );
    gf = 1 / G(0);
    Vec(gf) \\ Joerg Arndt, Jul 06 2013

Formula

G.f.: 1/(1 - x/(1 - x^3/(1 - x^5/(1 - x^7/(1 - x^9/(1 - ...
Derivation: the g.f. G(x,z) of Dyck paths, where x marks area and z marks semilength, satisfies G(x,z)=1+x*z*G(x,z)*G(x,x^2*z). Set z=1.
From Peter Bala, Dec 26 2012: (Start)
Let F(x) denote the o.g.f. of this sequence. For positive integer n >= 3, the real number F(1/n) has the simple continued fraction expansion 1 + 1/(n-2 + 1/(1 + 1/(n^2-2 + 1/(1 + 1/(n^3-2 + 1/(1 + ...)))))).
For n >= 1, F(-1/n) has the simple continued fraction expansion
1/(1 + 1/(n + 1/(n^2 + 1/(n^3 + ...)))). Examples are given below. Cf. A005169 and A111317.
(End)
G.f.: A(x) = 1/(1 - x/(1-x + x/(1+x^2 + x^4/(1-x^3 - x^2/(1+x^4 - x^7/(1-x^5 + x^3/(1+x^6 + x^10/(1-x^7 - x^4/(1+x^8 - x^13/(1-x^9 + x^5/(1+x^10 + x^16/(1 + ...)))))))))))), a continued fraction. - Paul D. Hanna, Aug 08 2016
a(n) ~ c / r^n, where r = 0.66290148514884371255690407749133031115536799774051... and c = 0.337761150388539773466092171229604432776662930886727976914... . - Vaclav Kotesovec, Feb 17 2017, corrected Nov 04 2021
From Peter Bala, Jul 04 2019: (Start)
O.g.f. as a ratio of q-series: N(q)/D(q), where N(q) = Sum_{n >= 0} (-1)^n*q^(2*n^2+n)/( (1-q^2)*(1-q^4)*...*(1-q^(2*n)) ) and D(q) = Sum_{n >= 0} (-1)^n*q^(2*n^2-n)/( (1-q^2)*(1-q^4)*...*(1-q^(2*n)) ). Cf. A224704.
D(q) has its least positive (and simple) real zero at x = 0.66290 14851 48843 71255 69040 ....
a(n) ~ c*d^n, where d = 1/x = 1.5085197761707628638804960 ... and c = - N(x)/(x*D'(x)) = 0.3377611503885397734660921 ... (the prime indicates differentiation w.r.t. q). (End)

Extensions

b-file corrected and extended by Alois P. Heinz, Aug 24 2018

A080924 Jacobsthal gap sequence.

Original entry on oeis.org

0, 1, 3, 1, 15, 1, 63, 1, 255, 1, 1023, 1, 4095, 1, 16383, 1, 65535, 1, 262143, 1, 1048575, 1, 4194303, 1, 16777215, 1, 67108863, 1, 268435455, 1, 1073741823, 1, 4294967295, 1, 17179869183, 1, 68719476735, 1, 274877906943, 1, 1099511627775, 1
Offset: 0

Views

Author

Paul Barry, Feb 26 2003

Keywords

Comments

Inverse binomial transform of A080925
From Peter Bala, Dec 26 2012: (Start)
Let F(x) = product {n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number F(1/4) = 0.79761 68651 30459 16010 ... = 1/(1 + 1/(3 + 1/(1 + 1/(15 + 1/(1 + 1/(63 + 1/(1 + 1/(255 + ...)))))))). See A111317. (End)
Also, the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 3", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Apr 19 2017

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[x (1 + 4 x) / ((1 + x) (1 + 2 x) (1 - 2 x)), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 05 2013 *)
    LinearRecurrence[{-1, 4, 4}, {0, 1, 3}, 42] (* Jean-François Alcover, Sep 21 2017 *)

Formula

a(2n) = 3*A001045(2n) = 3*A002450(n) = 4^n-1, a(2n+1)=1.
a(n) = (2^n-2*(-1)^n+(-2)^n)/2.
G.f.: x*(1+4*x)/((1+x)*(1+2*x)*(1-2*x)).
E.g.f.: (exp(2*x)-2*exp(-x)+exp(-2*x))/2.

A094024 Alternating 1 with one less than the powers of 2.

Original entry on oeis.org

1, 1, 1, 3, 1, 7, 1, 15, 1, 31, 1, 63, 1, 127, 1, 255, 1, 511, 1, 1023, 1, 2047, 1, 4095, 1, 8191, 1, 16383, 1, 32767, 1, 65535, 1, 131071, 1, 262143, 1, 524287, 1, 1048575, 1, 2097151, 1, 4194303, 1, 8388607, 1, 16777215, 1, 33554431, 1, 67108863, 1
Offset: 0

Views

Author

Paul Barry, Apr 22 2004

Keywords

Comments

Inverse binomial transform of A052542. Partial sums are A075427.
Let F(x) = product {n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number F(1/2) = 0.64227 25013 85234 96714 ... = 1/(1 + 1/(1 + 1/(1 + 1/(3 + 1/(1 + 1/(7 + 1/(1 + 1/(15 + ...)))))))). See A111317. - Peter Bala, Dec 26 2012

Crossrefs

Programs

  • Magma
    [Ceiling((-1)^n+((Sqrt(2))^n-(-Sqrt(2))^n)/Sqrt(2)): n in [0..50]]; // Vincenzo Librandi, Aug 17 2011
    
  • Mathematica
    LinearRecurrence[{-1, 2, 2}, {1, 1, 1}, 60] (* Jean-François Alcover, Jul 02 2018 *)
  • PARI
    a(n)=(1-(-1)^n)*2^floor(n/2)+(-1)^n

Formula

G.f.: (1+2*x) / ((1+x) * (1-2*x^2)).
E.g.f.: exp(-x) + 2*sinh(sqrt(2)*x) / sqrt(2).
a(n) = (-1)^n + ((sqrt(2))^n - (-sqrt(2))^n) / sqrt(2).
a(n) = (1-(-1)^n) * 2^floor(n/2) + (-1)^n. - Ralf Stephan, Aug 19 2013
a(n) = -a(n-1) + 2*a(n-2) + 2*a(n-3). - Andrew Howroyd, Feb 21 2018

Extensions

Better name from Ralf Stephan, Aug 19 2013
Even terms for n >= 60 corrected in b-file by Andrew Howroyd, Feb 21 2018

A358369 Euler transform of 2^floor(n/2), (A016116).

Original entry on oeis.org

1, 1, 3, 5, 12, 20, 43, 73, 146, 250, 475, 813, 1499, 2555, 4592, 7800, 13761, 23253, 40421, 67963, 116723, 195291, 332026, 552882, 932023, 1544943, 2585243, 4267081, 7094593, 11662769, 19281018, 31575874, 51937608, 84753396, 138772038, 225693778, 368017636
Offset: 0

Views

Author

Peter Luschny, Nov 17 2022

Keywords

Crossrefs

Sequences that can be represented as a EulerTransform(BinaryRecurrenceSequence()) include A000009, A000041, A000712, A001970, A002513, A010054, A015128, A022567, A034691, A111317, A111335, A117410, A156224, A166861, A200544, A261031, A261329, A358449.

Programs

  • Maple
    BinaryRecurrenceSequence := proc(b, c, u0:=0, u1:=1) local u;
    u := proc(n) option remember; if n < 2 then return [u0, u1][n + 1] fi;
    b*u(n - 1) + c*u(n - 2) end; u end:
    EulerTransform := proc(a) local b;
    b := proc(n) option remember; if n = 0 then return 1 fi; add(add(d * a(d),
    d = NumberTheory:-Divisors(j)) * b(n-j), j = 1..n) / n end; b end:
    a := EulerTransform(BinaryRecurrenceSequence(0, 2, 1)): seq(a(n), n=0..36);
  • Python
    from typing import Callable
    from functools import cache
    from sympy import divisors
    def BinaryRecurrenceSequence(b:int, c:int, u0:int=0, u1:int=1) -> Callable:
        @cache
        def u(n: int) -> int:
            if n < 2:
                return [u0, u1][n]
            return b * u(n - 1) + c * u(n - 2)
        return u
    def EulerTransform(a: Callable) -> Callable:
        @cache
        def b(n: int) -> int:
            if n == 0:
                return 1
            s = sum(sum(d * a(d) for d in divisors(j)) * b(n - j)
                for j in range(1, n + 1))
            return s // n
        return b
    b = BinaryRecurrenceSequence(0, 2, 1)
    a = EulerTransform(b)
    print([a(n) for n in range(37)])
  • Sage
    # uses[EulerTransform from A166861]
    b = BinaryRecurrenceSequence(0, 2, 1)
    a = EulerTransform(b)
    print([a(n) for n in range(37)])
    

A087404 a(n) = 4*a(n-1) + 5*a(n-2) for n > 1, with a(0) = 2 and a(1) = 4.

Original entry on oeis.org

2, 4, 26, 124, 626, 3124, 15626, 78124, 390626, 1953124, 9765626, 48828124, 244140626, 1220703124, 6103515626, 30517578124, 152587890626, 762939453124, 3814697265626, 19073486328124, 95367431640626, 476837158203124, 2384185791015626, 11920928955078124, 59604644775390626
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Sep 01 2003

Keywords

Comments

Let F(x) = Product_{n>=0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number 1 + F(-1/5) = 2.24761 97788 60361 46849 ... = 2 + 1/(4 + 1/(26 + 1/(124 + 1/(626 + ...)))). See A111317. - Peter Bala, Dec 26 2012

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(2 - 4x)/(1 - 4x - 5x^2), {x, 0, 25}], x]
    LinearRecurrence[{4,5},{2,4},30] (* Harvey P. Dale, May 13 2022 *)
  • Sage
    [lucas_number2(n,4,-5) for n in range(0, 22)] # Zerinvary Lajos, May 14 2009

Formula

G.f.: (2 - 4*x)/(1 - 4*x - 5*x^2).
a(n) = 5^n + (-1)^n.
From Elmo R. Oliveira, Aug 23 2024: (Start)
E.g.f.: exp(-x)*(exp(6*x) + 1).
a(n) = 2*A081340(n). (End)

Extensions

a(22)-a(24) from Elmo R. Oliveira, Aug 23 2024

A084182 a(n) = 3^n + (-1)^n - [1/(n+1)], where [] represents the floor function.

Original entry on oeis.org

1, 2, 10, 26, 82, 242, 730, 2186, 6562, 19682, 59050, 177146, 531442, 1594322, 4782970, 14348906, 43046722, 129140162, 387420490, 1162261466, 3486784402, 10460353202, 31381059610, 94143178826, 282429536482, 847288609442, 2541865828330, 7625597484986
Offset: 0

Views

Author

Paul Barry, May 19 2003

Keywords

Comments

Binomial transform of A084181.
From Peter Bala, Dec 26 2012: (Start)
Let F(x) = product {n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number F(-1/3) = 1.47627 73316 74531 44215 ... = 1 + 1/(2 + 1/(10 + 1/(26 + 1/(82 + ...)))). See A111317.
(End)

Crossrefs

Except for leading term, same as A102345.

Programs

  • Mathematica
    LinearRecurrence[{2,3},{1,2,10},30] (* Harvey P. Dale, Apr 27 2016 *)

Formula

a(n) = 3^n + (-1)^n - 0^n.
G.f.: (1+3*x^2)/((1+x)*(1-3*x)).
E.g.f.: exp(3*x)-exp(0)+exp(-x).
a(n) = 2 * A046717(n) for n >= 1.

A111165 Let qf(a,q) = Product(1-a*q^j,j=0..infinity); g.f. is qf(q,q^3)/qf(q^2,q^3).

Original entry on oeis.org

1, -1, 1, -1, 0, 1, -1, 0, 1, -2, 2, 0, -2, 2, -1, -1, 3, -2, -1, 3, -3, 0, 4, -5, 2, 3, -6, 4, 2, -7, 6, 0, -7, 9, -2, -7, 10, -5, -6, 13, -8, -5, 15, -13, -1, 16, -17, 2, 16, -22, 8, 16, -27, 14, 12, -30, 22, 9, -34, 29, 3, -36, 39, -5, -37, 47, -14, -36, 58, -26, -33, 66, -41, -26, 75, -56, -18, 81, -74, -4, 87, -94, 12, 87, -113, 34
Offset: 0

Views

Author

N. J. A. Sloane, Nov 09 2005

Keywords

Crossrefs

Cf. A111375. Convolution inverse of A111317.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1,
          add(add(d*[0, -1, 1][irem(d, 3)+1],
          d=numtheory[divisors](j))*a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..100);  # Alois P. Heinz, Apr 02 2014
  • Mathematica
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d*{0, -1, 1}[[Mod[d, 3]+1]], {d, Divisors[j]}]*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Apr 28 2014, after Alois P. Heinz *)
  • PARI
    {a(n) = if( n<0, 0, polcoeff( prod(k=0, n\3, (1 - x^(3*k+1)) / (1 - x^(3*k+2)), 1 + x * O(x^n)), n))} /* Michael Somos, Dec 23 2007 */

Formula

Euler transform of period 3 sequence [ -1, 1, 0, ...]. - Michael Somos, Dec 23 2007
G.f.: Product_{k>=0} (1 - x^(3*k+1)) / (1 - x^(3*k+2)).
Showing 1-10 of 13 results. Next