A048736 -id:A048736 - cp's OEIS frontend

A276532 a(n) = (a(n-1) * a(n-6) + a(n-2) * a(n-3) * a(n-4) * a(n-5)) / a(n-7), with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1.

1, 1, 1, 1, 1, 1, 1, 2, 3, 5, 11, 41, 371, 7507, 429563, 419408854, 9811194604889, 45615501062085527113, 323645006689468299915979814409, 217332607887523478570092794860281557159140687, 8092345737591989154121803868154457767563221634145658745306515944569

Offset: 0

Seiichi Manyama, Nov 16 2016

nonn

This sequence is one generalization of Dana Scott's sequence (A048736).
a(n) is an integer for all n.
The recursion exhibits the Laurent phenomenon. See A278706 for the exponents of the denominator of the Laurent polynomial. - Michael Somos, Nov 26 2016

Seiichi Manyama, Table of n, a(n) for n = 0..26

Cf. A048736, A006721, A276531, A278706.

def A(k, n)
  a = Array.new(k, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[-1] * a[1] + a[2..-2].inject(:*)
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276532(n)
  A(7, n)
end

a(n) * a(n-7) = a(n-1) * a(n-6) + a(n-2) * a(n-3) * a(n-4) * a(n-5).

a(6-n) = a(n) for all n in Z.

A208219 a(n)=(a(n-1)^3*a(n-3)+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 9, 731, 781235791, 2145650135491172007486084385, 802327342392981520933850619811649523436811893002103478524225246677189521545661182074

Offset: 0

Matthew C. Russell, Apr 25 2012

nonn

This is the case a=1, b=1, c=3, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).

The next term (a(10)) has 258 digits. - Harvey P. Dale, Sep 21 2016

Seiichi Manyama, Table of n, a(n) for n = 0..11
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A048736, A208218, A208222.

y:=proc(n) if n<4 then return 1: fi: return (y(n-1)^3*y(n-3)+y(n-2))/y(n-4): end:
seq(y(n),n=0..9);

RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]^3 a[n-3]+ a[n-2])/ a[n-4]},a,{n,10}] (* Harvey P. Dale, Sep 21 2016 *)

A208220 a(n)=(a(n-1)*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 5, 23, 106, 891, 94289, 46062265, 344980727309, 3442224480935856594, 77458438596193694601268422031, 200130424073190804359006946314196714242380417, 6873796333354760314538446350412794888765818679762438117097006307173727

Offset: 0

Matthew C. Russell, Apr 25 2012

This is the case a=2, b=1, c=1, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).

Seiichi Manyama, Table of n, a(n) for n = 0..21
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A048736.

[n le 4 select 1 else (Self(n-1)*Self(n-3)^2+Self(n-2))/Self(n-4): n in [1..17]]; // Bruno Berselli, Apr 26 2012

y:=proc(n) if n<4 then return 1: fi: return (y(n-1)*y(n-3)^2+y(n-2))/y(n-4): end:
seq(y(n),n=0..16);

a[n_] := a[n] = (a[n - 1]*a[n - 3]^2 + a[n - 2])/a[n - 4];
a[0] = a[1] = a[2] = a[3] = 1;
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Apr 01 2018 *)

A208226 a(n)=(a(n-1)*a(n-3)^4+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 5, 83, 3364, 700861, 6652337263549, 10264082055393717193904815, 736193034562641516492404723890409674438627151, 2057106833431631102316572923185391939849261245309254135929044995902093016346478213863681606

Offset: 0

Matthew C. Russell, Apr 25 2012

nonn

This is the case a=4, b=1, c=1, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).

Seiichi Manyama, Table of n, a(n) for n = 0..16
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A048736, A208223, A208227.

y:=proc(n) if n<4 then return 1: fi: return (y(n-1)*y(n-3)^4+y(n-2))/y(n-4): end:
seq(y(n),n=0..13);

a[n_]:=If[n<4,1, (a[n - 1] *a[n- 3]^4 + a[n - 2])/a[n - 4]]; Table[a[n], {n, 0, 12}] (* Indranil Ghosh, Mar 19 2017 *)
RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]a[n-3]^4+ a[n-2])/ a[n-4]},a,{n,14}] (* Harvey P. Dale, Dec 29 2018 *)

One more term from Harvey P. Dale, Dec 29 2018

A217787 a(n) = (a(n-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, 2, 3, 4, 9, 14, 19, 43, 67, 91, 206, 321, 436, 987, 1538, 2089, 4729, 7369, 10009, 22658, 35307, 47956, 108561, 169166, 229771, 520147, 810523, 1100899, 2492174, 3883449, 5274724, 11940723, 18606722, 25272721, 57211441, 89150161, 121088881

Offset: 0

Michael Somos, Mar 25 2013

This sequence is similar to A005246 whose recursion is a(n) = (a(n-1)*a(n-2) + 1) / a(n-3). - Michael Somos, Feb 10 2017

			G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 9*x^7 + 14*x^8 + 19*x^9 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..4411
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. See (better) also. See Example 6.2.18.
Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-1).

Cf. A003501.

Cf. A005246, A048736, A006720, A072876, A072877, A111459.

[n le 3 select 1 else (Self(n)*Self(n-2)+1)/Self(n-3): n in [0..40]]; // Bruno Berselli, Mar 25 2013

a[ n_] := With[{m = If [n < 0, 3 - n, n]}, SeriesCoefficient[ (1 + x + x^2 - 4 x^3 - 3 x^4 - 2 x^5) / (1 - 5 x^3 + x^6), {x, 0, m}]]; (* Michael Somos, Jan 18 2015 *)
LinearRecurrence[{0,0,5,0,0,-1},{1,1,1,1,2,3},40] (* Harvey P. Dale, Nov 20 2016 *)

{a(n) = if( n<0, n = 3-n); polcoeff( (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6) + x * O(x^n), n)};

G.f.: (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6).
a(n) = a(3-n) for all n in Z.
a(n+3) + a(n-3) = 5*a(n) for all n in Z.
a(n+1) + a(n-1) = a(n) * (2 + [n mod 3 == 0]) for all n in Z.
a(n+3k)+a(n-3k) = A003501(k)*a(n) for n>=3k. - Bruno Berselli, Mar 25 2013

A275174 a(n) = (a(n-4) + a(n-1) * a(n-7)) / a(n-8), a(0) = a(1) = ... = a(7) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 33, 53, 74, 96, 141, 209, 300, 714, 1151, 1611, 2094, 3083, 4578, 6579, 15665, 25257, 35355, 45959, 67673, 100497, 144431, 343906, 554491, 776186, 1008991, 1485711, 2206346, 3170896, 7550257, 12173533, 17040724

Offset: 0

Seiichi Manyama, Jul 19 2016

Inspired by A048736.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,23,0,0,0,0,0,0,-23,0,0,0,0,0,0,1).

Cf. A101879, A048736, A275173.

RecurrenceTable[{a[n] == (a[n - 4] + a[n - 1] a[n - 7])/a[n - 8], a[1] == 1, a[2] == 1, a[3] == 1, a[4] == 1, a[5] == 1, a[6] == 1, a[7] == 1, a[8] == 1}, a, {n, 42}] (* Michael De Vlieger, Jul 19 2016 *)

Vec((1 +x +x^2 +x^3 +x^4 +x^5 +x^6 -22*x^7 -21*x^8 -20*x^9 -19*x^10 -18*x^11 -16*x^12 -13*x^13 +14*x^14 +10*x^15 +7*x^16 +5*x^17 +4*x^18 +3*x^19 +2*x^20) / ((1 -x)*(1 +x +x^2 +x^3 +x^4 +x^5 +x^6)*(1 -22*x^7 +x^14)) + O(x^20)) \\ Colin Barker, Jul 19 2016

def A(k, l, n)
  a = Array.new(k * 2, 1)
  ary = [1]
  while ary.size < n + 1
    break if (a[1] * a[-1] + a[k] * l) % a[0] > 0
    a = *a[1..-1], (a[1] * a[-1] + a[k] * l) / a[0]
    ary << a[0]
  end
  ary
end
def A275174(n)
  A(4, 1, n)
end

G.f.: (1 +x +x^2 +x^3 +x^4 +x^5 +x^6 -22*x^7 -21*x^8 -20*x^9 -19*x^10 -18*x^11 -16*x^12 -13*x^13 +14*x^14 +10*x^15 +7*x^16 +5*x^17 +4*x^18 +3*x^19 +2*x^20) / ((1 -x)*(1 +x +x^2 +x^3 +x^4 +x^5 +x^6)*(1 -22*x^7 +x^14)). - Colin Barker, Jul 19 2016

a(n) = 23*a(n-7) - 23*a(n-14) + a(n-21).

A206282 a(n) = ( a(n-1) * a(n-3) + a(n-2) ) / a(n-4), a(1) = a(2) = 1, a(3) = -1, a(4) = -4.

Original entry on oeis.org

1, 1, -1, -4, -5, 1, 9, 11, -4, -25, -31, 9, 64, 79, -25, -169, -209, 64, 441, 545, -169, -1156, -1429, 441, 3025, 3739, -1156, -7921, -9791, 3025, 20736, 25631, -7921, -54289, -67105, 20736, 142129, 175681, -54289, -372100, -459941, 142129, 974169, 1204139

Offset: 1

Michael Somos, Feb 05 2012

This satisfies the same recurrence as Dana Scott's sequence A048736.

			G.f. = x + x^2 - x^3 - 4*x^4 - 5*x^5 + x^6 + 9*x^7 + 11*x^8 - 4*x^9 - 25*x^10 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,-2,0,0,2,0,0,1).

Cf. A000045, A048736, A110034, A110035, A126116.

a206282 n = a206282_list !! (n-1)
a206282_list = 1 : 1 : -1 : -4 :
   zipWith div
     (zipWith (+)
       (zipWith (*) (drop 3 a206282_list)
                    (drop 1 a206282_list))
       (drop 2 a206282_list))
     a206282_list
-- Same program as in A048736, see comment.
-- Reinhard Zumkeller, Feb 08 2012

I:=[1,1,-1,-4]; [n le 4 select I[n] else (Self(n-1)*Self(n-3) + Self(n-2))/Self(n-4): n in [1..30]]; // G. C. Greubel, Aug 12 2018

CoefficientList[Series[x*(1+x)*(1-x^2)*(1+x^3)/(1-2*x^2-2*x^4-2*x^6+x^8 ), {x,0,50}], x] (* or *) RecurrenceTable[{a[n] == ( a[n-1]*a[n-3] + a[n-2] )/a[n-4], a[1] == a[2] == 1, a[3] == -1, a[4] == -4}, a, {n,1,50}] (* G. C. Greubel, Aug 12 2018 *)

{a(n) = my(k = n\3); (-1)^k * if( n%3 == 0, fibonacci( k )^2, n%3 == 1, fibonacci( k+2 )^2, fibonacci( k ) * fibonacci( k+3 ) + fibonacci( k+1 ) * fibonacci( k+2 ))};

x='x+O('x^30); Vec(x*(1+x)*(1-x^2)*(1+x^3)/(1-2*x^2-2*x^4 -2*x^6 +x^8 )) \\ G. C. Greubel, Aug 12 2018

G.f.: x * (1 + x - x^2 - 2*x^3 - 3*x^4 - x^5 - x^6 - x^7) / (1 + 2*x^3 - 2*x^6 - x^9).
a(n) = a(-5 - n) = a(n+2) * a(n-2) - a(n+1) * a(n-1) for all n in Z.
a(3*n) = (-1)^n * F(n)^2, a(3*n + 1) = (-1)^n * F(n + 2)^2 where F = Fibonacci A000045.
a(6*n - 4) = - A110034(2*n), a(6*n - 1) = - A110035(2*n), a(3*n + 2) = (-1)^n * A126116(2*n + 3).

cp's OEIS Frontend

A276532 a(n) = (a(n-1) * a(n-6) + a(n-2) * a(n-3) * a(n-4) * a(n-5)) / a(n-7), with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1.

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A208219 a(n)=(a(n-1)^3*a(n-3)+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

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A208220 a(n)=(a(n-1)*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

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A208226 a(n)=(a(n-1)*a(n-3)^4+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

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A217787 a(n) = (a(n-1)*a(n-3) + 1) / a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

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A275174 a(n) = (a(n-4) + a(n-1) * a(n-7)) / a(n-8), a(0) = a(1) = ... = a(7) = 1.

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A206282 a(n) = ( a(n-1) * a(n-3) + a(n-2) ) / a(n-4), a(1) = a(2) = 1, a(3) = -1, a(4) = -4.

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