Bruno Langlois - cp's OEIS frontend

A354597 a(n) is the smallest number k>0 such that -n is not a quadratic residue modulo k.

3, 4, 5, 3, 4, 4, 3, 5, 4, 3, 7, 5, 3, 4, 7, 3, 4, 4, 3, 11, 4, 3, 5, 9, 3, 4, 5, 3, 4, 4, 3, 5, 4, 3, 8, 7, 3, 4, 7, 3, 4, 4, 3, 7, 4, 3, 5, 5, 3, 4, 7, 3, 4, 4, 3, 11, 4, 3, 8, 7, 3, 4, 5, 3, 4, 4, 3, 5, 4, 3, 7, 5, 3, 4, 8, 3, 4, 4, 3, 11, 4, 3, 5, 9, 3, 4, 5, 3, 4, 4, 3, 5, 4, 3, 7, 9, 3, 4, 7, 3

Offset: 1

Bruno Langlois, Jul 08 2022

nonn

All values are prime powers, and every prime power except 2 appears in the sequence. This can be proved using the Chinese remainder theorem.

Cf. A139401.

a(n) = my(k=2); while (issquare(Mod(-n, k)), k++); k; \\ Michel Marcus, Jul 08 2022

A275695 a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1; for n>6, a(n) = ( a(n-1)+a(n-3)+a(n-5) )*( a(n-2)+a(n-4)+a(n-6) ) / a(n-7).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 9, 33, 385, 13825, 5474305, 75873853441, 415386585427968001, 3501887406773528570406162401, 44079910680970588907541344275243042224979209, 400942556117903539711475671972145122347091674105174721165559627509313

Offset: 0

Bruno Langlois, Aug 21 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..21

Cf. A006723, A276130.

I:=[1,1,1,1,1,1,1]; [n le 7 select I[n] else (Self(n-1) + Self(n-3) + Self(n-5))*(Self(n-2) + Self(n-4) + Self(n-6))/Self(n-7): n in [1..17]]; // G. C. Greubel, Feb 21 2018

RecurrenceTable[{a[n] == (a[n - 1] + a[n - 3] + a[n - 5]) (a[n - 2] + a[n - 4] + a[n - 6])/a[n - 7], a[0] == a[1] == a[2] == a[3] == a[4] == a[5] == a[6] == 1}, a, {n, 0, 16}] (* Michael De Vlieger, Aug 25 2016 *)
nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,((g+e+c)(f+d+b))/a}; NestList[ nxt,{1,1,1,1,1,1,1},20][[All,1]] (* Harvey P. Dale, May 04 2019 *)

a(n) = if (n <=6, 1, (a(n-1)+a(n-3)+a(n-5))*(a(n-2)+a(n-4)+a(n-6))/a(n-7)); \\ Michel Marcus, Aug 25 2016

def A(m, n)
  a = Array.new(2 * m + 1, 1)
  ary = [1]
  while ary.size < n + 1
    i = (1..m).inject(0){|s, i| s + a[2 * i - 1]} * (1..m).inject(0){|s, i| s + a[2 * i]}
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A275695(n)
  A(3, n)
end # Seiichi Manyama, Aug 27 2016

a(n) = (8-4*(-1)^n)*a(n-1)*a(n-3)*a(n-5) - a(n-2) - a(n-4) - a(n-6).

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

A276130 a(0) = a(1) = a(2) = a(3) = a(4) = 1; for n>4, a(n) = ( a(n-1) +a(n-3) )*( a(n-2)+a(n-4) ) / a(n-5).

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 10, 55, 649, 38881, 6414706, 24978826228, 2913605221297249, 112139525368095766797655, 8403341152380185679389503620974065, 146904111947501701959735285821948223340424963459227

Offset: 0

Bruno Langlois, Aug 21 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..22

Cf. A006721.

a(n) = if (n<=4, 1, (9-3*(-1)^n)/2*a(n-1)*a(n-3)-a(n-2)-a(n-4)); \\ Michel Marcus, Aug 27 2016

a(n) = (9-3*(-1)^n)/2*a(n-1)*a(n-3)-a(n-2)-a(n-4).

A276124 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)a(n-2)a(n-3))/a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 4, 22, 589, 399253, 41144206447, 77387327118194895379, 10169897514576967837097322386922878932, 259050897146323086186965020577200627526185475088368701480903471601830

Offset: 0

Bruno Langlois, Aug 21 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..15

Cf. A165903, A072878.

RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 3]^2 + a[n - 1] a[n - 2] a[n - 3])/a[n - 4], a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 11}] (* Michael De Vlieger, Aug 21 2016 *)

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(0){|s, i| s + i * i} + a[1..-1].inject(:*)
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276124(n)
  A(4, n)
end # Seiichi Manyama, Aug 21 2016

a(n) = 8*a(n-1)*a(n-2)*a(n-3)-a(n-1)*a(n-2)-a(n-1)*a(n-3)-a(n-2)*a(n-3)-a(n-4).

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

A276122 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1)^2+a(n-2)^2+a(n-1)+a(n-2))/a(n-3).

Original entry on oeis.org

1, 1, 1, 4, 22, 526, 69427, 219111589, 91273561736491, 119994570874632853695766, 65713991236617279734602790963627271046, 47311933073383646516067037755547920981262829886906923065810924

Offset: 0

Bruno Langlois, Aug 21 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..16

Cf. A064098, A072881, A101879, A165903.

RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 1] + a[n - 2])/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 11}] (* Michael De Vlieger, Aug 21 2016 *)

a(n) = 6*a(n-1)*a(n-2)-a(n-3)-1.

a(n) ~ 1/6 * c^(phi^n), where c = 2.059783590102273... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 20 2017

a(10) corrected by Seiichi Manyama, Aug 21 2016

cp's OEIS Frontend

User: Bruno Langlois

A354597 a(n) is the smallest number k>0 such that -n is not a quadratic residue modulo k.

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

A275695 a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1; for n>6, a(n) = ( a(n-1)+a(n-3)+a(n-5) )*( a(n-2)+a(n-4)+a(n-6) ) / a(n-7).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Ruby

Formula

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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Ruby

Formula

A276130 a(0) = a(1) = a(2) = a(3) = a(4) = 1; for n>4, a(n) = ( a(n-1) +a(n-3) )*( a(n-2)+a(n-4) ) / a(n-5).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

A276124 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)*a(n-2)*a(n-3))/a(n-4).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Ruby

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A276122 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1)^2+a(n-2)^2+a(n-1)+a(n-2))/a(n-3).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

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.

A276124 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)a(n-2)a(n-3))/a(n-4).