A076839 -id:A076839 - cp's OEIS frontend

A105734 Duplicate of A076839.

1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1

Offset: 1

dead

Previous name was: For n>2, a(n) > 0 is such that a(n-1)^2+4*a(n-2)*a(n) is a minimal square, with a(1)=1, a(2)=1.

The sequence depends on seed terms a(1) and a(2); if a(1)=1, a(3)=a(2)+1. All(?) sequences end with cycle={1,2,3,2,1} (or {2,4,6,4,2}, which essentially the same cycle) of length=5.

A105822 For n > 2, a(n) > 0 not appeared previously is such that a(n-1)^2+4a(n-2)a(n) = d^2 is a minimal square, a(1)=1, a(2)=2.

Original entry on oeis.org

1, 2, 3, 5, 8, 4, 12, 7, 10, 17, 6, 11, 21, 32, 13, 19, 14, 33, 20, 28, 24, 27, 42, 40, 18, 9, 35, 44, 39, 54, 48, 22, 15, 37, 52, 89, 30, 59, 23, 36, 99, 70, 16, 86, 47, 45, 92, 65, 157, 34, 123, 135, 222, 56, 136, 82, 29, 53, 102, 155, 25, 130, 87, 43, 170, 213, 63, 150, 57

Offset: 1

Zak Seidov, Apr 22 2005

nonn

Is it a permutation of positive integers? Among first 2000 terms, first missing numbers are 233, 349, 394, 443, 449.
The sequence depends on seed terms a(1) and a(2); if a(1) = 1, a(3) = a(2)+1.
Values of d^2 in A105823.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A076839, A105823, A104663 (putative inverse).

N:= 1000: # to get a(1) to a(N)
S:= 'S':
a[1]:= 1: a[2]:= 2:
S[1]:= 1: S[2]:= 1:
for n from 3 to N do
  ds:= map(t -> rhs(op(t)), [msolve(x^2=a[n-1]^2, 4*a[n-2])]);
  xmin:= infinity;
  for d in ds do
    found:= false;
    for y from floor((a[n-1]-d)/(4*a[n-2]))+1 do
      xy:= 4*a[n-2]*y + d;
      cand:= (xy^2 - a[n-1]^2)/(4*a[n-2]);
      if cand >= xmin then found:= false; break fi;
      if not assigned(S[cand]) then found:= true; break fi;
    od:
    if found then xmin:= cand;  fi;
  od:
  a[n]:= xmin;
  S[xmin]:= 1;
od:
seq(a[n],n=1..N); # Robert Israel, May 11 2015

a = {1, 2}; Do[i = 1; While[MemberQ[a, i] || !IntegerQ[Sqrt[a[[-1]]^2 + 4 a[[-2]]*i]], i++]; AppendTo[a, i], {n, 3, 70}]; a (* Ivan Neretin, May 11 2015 *)

A105736 For n>2, a(n) > 0 is such that a(n-1)^2+4a(n-2)a(n) is a minimal square, a(1)=1,a(2)=3.

Original entry on oeis.org

1, 3, 4, 4, 3, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2

Offset: 1

Zak Seidov, Apr 19 2005

nonn

The sequence depends on seed terms a(1) and a(2); if a(1)=1, a(3)=a(2)+1. All(?) sequences end with cycle={1,2,3,2,1} (or {2,4,6,4,2}, which essentially the same cycle) of length=5. This particular sequence merges with A076839, starting with 6th term = 1.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).

Cf. A076839, A105737 - A105746.

More terms from Ray Chandler, May 17 2024

A105746 a(n) = minimal c>0 such that (n+1)^2+4nc = d^2 is a square.

Original entry on oeis.org

3, 2, 4, 6, 8, 3, 12, 14, 16, 6, 20, 4, 24, 9, 7, 30, 32, 12, 36, 5, 8, 15, 44, 11, 48, 18, 52, 9, 56, 6, 60, 62, 15, 24, 20, 10, 72, 27, 16, 21, 80, 7, 84, 14, 11, 33, 92, 20, 96, 36, 23, 15, 104, 39, 12, 8, 24, 42, 116, 19, 120, 45, 39, 126, 35, 13, 132, 20, 31, 17, 140, 9, 144, 54

Offset: 1

Zak Seidov, Apr 19 2005

nonn

There is a family of solutions c = 2*n - 2, d = 3*n-1, which gives maximums at graphic of function c(n).

Cf. A076839, A105736 - A105745.

A271751 Period 10 zigzag sequence; repeat: [0, 1, 2, 3, 4, 5, 4, 3, 2, 1].

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5

Offset: 0

Wesley Ivan Hurt, Apr 13 2016

Decimal expansion of 11111/900009. - Elmo R. Oliveira, Mar 03 2024

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,-1,1).

Cf. A076839, A117444.

Period k zigzag sequences: A000035 (k=2), A007877 (k=4), A260686 (k=6), A266313 (k=8), this sequence (k=10), A271832 (k=12), A279313 (k=14), A279319 (k=16), A158289 (k=18).

&cat[[0, 1, 2, 3, 4, 5, 4, 3, 2, 1]: n in [0..10]];

a:=n->[0, 1, 2, 3, 4, 5, 4, 3, 2, 1][(n mod 10)+1]: seq(a(n), n=0..100);

CoefficientList[Series[x*(1 + x + x^2 + x^3 + x^4)/(1 - x + x^5 - x^6), {x, 0, 30}], x]

a(n) = abs(n-10*round(n/10)); \\ Altug Alkan, Apr 13 2016

G.f.: x*(1 + x + x^2 + x^3 + x^4)/(1 - x + x^5 - x^6).
a(n) = a(n-1) - a(n-5) + a(n-6) for n>5.
a(n) = abs(n - 10*round(n/10)).
a(n) = Sum_{i=1..n} (-1)^floor((i-1)/5).
a(2n) = 2*abs(A117444(n)).
a(2n+7) = 2*A076839(n)-1 for n>0.
a(n) = a(n-10) for n >= 10. - Wesley Ivan Hurt, Sep 07 2022

A076844 a(1) = a(2) = a(3) = 1; a(n) = (a(n-1) + a(n-2) + 1)/a(n-3) (for n>3).

Original entry on oeis.org

1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1, 1, 1, 3, 5, 9, 5, 3, 1

Offset: 1

N. J. A. Sloane, Nov 21 2002

nonn

Any sequence a(1),a(2),a(3),... defined by the recurrence a(n) = (a(n-1) + a(n-2) + 1)/a(n-3) (for n>3) has period 8. - James Propp, Nov 20 2002. This is the 8-cycle discovered by H. Todd - see Lyness, Note 1847. - N. J. A. Sloane, Jul 19 2020

R. C. Lyness, Note 1581. Cycles, Math. Gazette, 26 (1942), 62.
R. C. Lyness, Note 1847. Cycles, Math. Gaz., 29 (1945), 231-233.
R. C. Lyness, Note 2952. Cycles, Math. Gaz., 45 (1961), 207-209.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 1).

Cf. A076840, A076841, A076839, A076842, A076843.

a := 1; b := 1; c := 1; f := proc(n) option remember; global a,b,c; if n=1 then RETURN(a); fi; if n=2 then RETURN(b); fi; if n=3 then RETURN(c); fi; RETURN((f(n-1)+f(n-2)+1)/f(n-3)); end;

nxt[{a_,b_,c_}]:={b,c,(b+c+1)/a}; Transpose[NestList[nxt,{1,1,1},110]][[1]] (* or *) PadRight[{},110,{1,1,1,3,5,9,5,3}] (* Harvey P. Dale, Jan 13 2015 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 1},{1, 1, 1, 3, 5, 9, 5, 3},105] (* Ray Chandler, Aug 25 2015 *)

A276123 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).

Original entry on oeis.org

1, 1, 1, 4, 10, 55, 154, 868, 2449, 13825, 39025, 220324, 621946, 3511351, 9912106, 55961284, 157971745, 891869185, 2517635809, 14213945668, 40124201194, 226531261495, 639469583290, 3610286238244, 10191389131441, 57538048550401, 162422756519761

Offset: 0

Bruno Langlois, Aug 21 2016

Colin Barker, Table of n, a(n) for n = 0..1000
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences", PhD Dissertation, Mathematics Department, Rutgers University, May 2016.
Index entries for linear recurrences with constant coefficients, signature (0,17,0,-17,0,1).

Cf. A072881, A076839, A276175.

I:=[1,1,1,4,10,55]; [n le 6 select I[n] else 17*Self(n-2)-17*Self(n-4)+Self(n-6): n in [1..30]]; // Vincenzo Librandi, Aug 27 2016

LinearRecurrence[{0, 17, 0, -17, 0, 1}, {1, 1, 1, 4, 10, 55}, 40] (* Vincenzo Librandi, Aug 27 2016 *)
nxt[{a_,b_,c_}]:={b,c,((c+1)(b+1))/a}; NestList[nxt,{1,1,1},30][[All,1]] (* Harvey P. Dale, Oct 01 2021 *)

Vec((1+x-16*x^2-13*x^3+10*x^4+4*x^5)/((1-x)*(1+x)*(1-16*x^2+x^4)) + O(x^30)) \\ Colin Barker, Aug 21 2016

a(n) = (9-3*(-1)^n)/2*a(n-1) - a(n-2) - 1.
From Colin Barker, Aug 21 2016: (Start)
a(n) = 17*a(n-2) - 17*a(n-4) + a(n-6) for n > 5.
G.f.: (1 + x - 16*x^2 - 13*x^3 + 10*x^4 + 4*x^5) / ((1-x)*(1+x)*(1 - 16*x^2 + x^4)). (End)
a(2n+1) = A073352(n). a(2n) = A048907(n). - R. J. Mathar, Jul 04 2024

More terms from Colin Barker, Aug 21 2016

A276175 a(n) = (a(n-1)+1)(a(n-2)+1)(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 1, 8, 36, 666, 222111, 685187756, 2819713283228248, 644335913093223286486628176, 5604757351123068775966272886689217889936356651, 14861563788248216173988661093334637018340529129342104300621091389266132702213641

Offset: 0

Bruno Langlois, Aug 23 2016

nonn

Conjecture: a(n) is an integer for all n >= 0. It has been checked by computer for n <= 40. A proof was proposed by 'mercio' as an answer to the MSE question, which however lacks details and heavily relies on computation. [Updated by Max Alekseyev, May 07 2023]

Seiichi Manyama, Table of n, a(n) for n = 0..16
Math.StackExchange.com, Is A276175 integer-only? Aug 27 2016.

Cf. A076839, A276123, A362884.

RecurrenceTable[{a[n] == (a[n - 1] + 1) (a[n - 2] + 1) (a[n - 3] + 1)/a[n - 4],
a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 12}] (* Michael De Vlieger, Aug 25 2016 *)
a[ n_] := With[{m = Max[3 - n, n]}, If[ m < 4, 1, (a[m - 1] + 1) (a[m - 2] + 1) (a[m - 3] + 1)/a[m - 4]]]; (* Michael Somos, Jun 02 2019 *)

a(n) = if (n <=3, 1, (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4)); \\ Michel Marcus, Aug 23 2016

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(1){|s, i| s * (i + 1)}
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276175(n)
  A(4, n)
end # Seiichi Manyama, Aug 23 2016

Let b(n) = (b(n-1)*b(n-2)*b(n-3)+1)/b(n-4) with b(0) = 1/2, b(1) = 4, b(2) = b(3) = 1/2, then a(n) = b(n)*b(n+1)*b(n+2). - Seiichi Manyama, Sep 03 2016
The sequence 4*b(n) is given by A362884. Correspondingly, a(n) = A362884(n) * A362884(n+1) * A362884(n+2) / 64. - Max Alekseyev, May 07 2023
a(n) = a(3-n), 0 = a(n)*a(n+4)*(a(n+4)+1) - a(n+5)*a(n+1)*(a(n+1)+1) for all n in Z. - Michael Somos, Feb 23 2019
log(a(n)) ~ c * A289917^n, where c = 0.26774381278698... - Vaclav Kotesovec, Aug 27 2021

A076840 a(1) = a(2) = 1; a(n) = (a(n-1) + 1)/a(n-2) (for n>2, n odd), (a(n-1)^2 + 1)/a(n-2) (for n>2, n even).

Original entry on oeis.org

1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2

Offset: 1

N. J. A. Sloane, Nov 21 2002

Any sequence a(1),a(2),a(3),... defined by the recurrence a(n) = (a(n-1)+1)/a(n-2) (for n>2, n odd), (a(n-1)^2+1)/a(n-2) (for n>2, n even) has period 6. The theory of cluster algebras currently being developed by Fomin and Zelevinsky gives a context for these facts, but it doesn't really explain them in an elementary way. - James Propp, Nov 20 2002

Sergey Fomin and Andrei Zelevinsky, Cluster algebras II: Finite type classification, arXiv:math/0208229 [math.RA], 2002-2003.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. A076839, A076841, A076844.

a := 1; b := 1; f := proc(n) option remember; global a,b; if n=1 then RETURN(a); fi; if n=2 then RETURN(b); fi; if n mod 2 = 1 then RETURN((f(n-1)+1)/f(n-2)); fi; RETURN((f(n-1)^2+1)/f(n-2)); end;

LinearRecurrence[{0, 0, 0, 0, 0, 1}, {1, 1, 2, 5, 3, 2}, 105] (* Jean-François Alcover, Nov 22 2017 *)

A076841 a(1) = a(2) = 1; a(n) = (a(n-1)+1)/a(n-2) (for n>2, n odd), (a(n-1)^3+1)/a(n-2) (for n>2, n even).

Original entry on oeis.org

1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2, 9, 5, 14, 3, 2, 1, 1, 2

Offset: 1

N. J. A. Sloane, Nov 21 2002

nonn

Any sequence a(1),a(2),a(3),... defined by the recurrence a(n) = (a(n-1)+1)/a(n-2) (for n>2, n odd), (a(n-1)^3+1)/a(n-2) (for n>2, n even) has period 8. The theory of cluster algebras currently being developed by Fomin and Zelevinsky gives a context for these facts, but it doesn't really explain them in an elementary way. - James Propp, Nov 20 2002

Sergey Fomin and Andrei Zelevinsky, Cluster algebras II: Finite type classification
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 1).

Cf. A076839, A076840, A076844.

a := 1; b := 1; f := proc(n) option remember; global a,b; if n=1 then RETURN(a); fi; if n=2 then RETURN(b); fi; if n mod 2 = 1 then RETURN((f(n-1)+1)/f(n-2)); fi; RETURN((f(n-1)^3+1)/f(n-2)); end;

LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 1},{1, 1, 2, 9, 5, 14, 3, 2},99] (* Ray Chandler, Aug 25 2015 *)

cp's OEIS Frontend

A105734 Duplicate of A076839.

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A105822 For n > 2, a(n) > 0 not appeared previously is such that a(n-1)^2+4*a(n-2)*a(n) = d^2 is a minimal square, a(1)=1, a(2)=2.

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A105736 For n>2, a(n) > 0 is such that a(n-1)^2+4*a(n-2)*a(n) is a minimal square, a(1)=1,a(2)=3.

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A105746 a(n) = minimal c>0 such that (n+1)^2+4*n*c = d^2 is a square.

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A271751 Period 10 zigzag sequence; repeat: [0, 1, 2, 3, 4, 5, 4, 3, 2, 1].

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Magma

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A076844 a(1) = a(2) = a(3) = 1; a(n) = (a(n-1) + a(n-2) + 1)/a(n-3) (for n>3).

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A276123 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).

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A276175 a(n) = (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

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A076840 a(1) = a(2) = 1; a(n) = (a(n-1) + 1)/a(n-2) (for n>2, n odd), (a(n-1)^2 + 1)/a(n-2) (for n>2, n even).

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A105822 For n > 2, a(n) > 0 not appeared previously is such that a(n-1)^2+4a(n-2)a(n) = d^2 is a minimal square, a(1)=1, a(2)=2.

A105736 For n>2, a(n) > 0 is such that a(n-1)^2+4a(n-2)a(n) is a minimal square, a(1)=1,a(2)=3.

A105746 a(n) = minimal c>0 such that (n+1)^2+4nc = d^2 is a square.

A276175 a(n) = (a(n-1)+1)(a(n-2)+1)(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.