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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%5]; say unless $e==0 || (lucas_sequence($ has authored 63 sequences. Here are the ten most recent ones:

A275789 Least k such that sigma(n) divides Fibonacci(k) (k > 0).

Original entry on oeis.org

1, 4, 6, 8, 12, 12, 6, 20, 7, 12, 12, 24, 24, 12, 12, 30, 12, 28, 30, 24, 24, 12, 12, 60, 30, 24, 30, 24, 60, 12, 24, 24, 12, 36, 12, 56, 18, 60, 24, 60, 24, 24, 30, 24, 84, 12, 12, 30, 36, 60, 12, 168, 36, 60, 12, 60, 60, 60, 60, 24, 30, 24, 42, 128, 24, 12, 18, 24, 24, 12, 12
Offset: 1

Author

Altug Alkan, Aug 09 2016

Keywords

Examples

			a(5) = 12 because Fibonacci(12) = 144 is divisible by sigma(5) = 6.
		

Crossrefs

Programs

  • Mathematica
    Table[k = 1; While[! Divisible[Fibonacci@k, DivisorSigma[1, n]], k++]; k, {n, 120}] (* Michael De Vlieger, Aug 11 2016 *)
  • PARI
    a(n)=my(k=1); while(fibonacci(k) % sigma(n), k++); k;
    
  • PARI
    a(n)=my(s=sigma(n),a=Mod(1,s),b=a,k=1); while(a, [a,b]=[b,a+b]; k++); k \\ Charles R Greathouse IV, Aug 12 2016
    
  • Perl
    use ntheory ":all"; sub a275789 { my($sigma,$k)=(divisor_sum(shift), 1); return 1 if $sigma==1; $k++ while (lucas_sequence($sigma,1,-1,$k))[0]; $k; } # Dana Jacobsen, Aug 15 2016

Formula

a(n) = A001177(A000203(n)). - Robert Israel, Aug 09 2016
log n << a(n) << n log log n. - Charles R Greathouse IV, Aug 12 2016

A273626 A fourth-order divisibility sequence: a(n) = (1/14)*(Pell(4*n) + Pell(2*n)).

Original entry on oeis.org

1, 30, 995, 33660, 1142629, 38810970, 1318402631, 44786716920, 1521429030985, 51683794848150, 1755727563817259, 59643053188493940, 2026108079758297261, 68828031652259981010, 2338126968060165944975, 79427488882178225107440, 2698196495024745460575889
Offset: 1

Author

Peter Bala, May 31 2016

Keywords

Comments

This is a divisibility sequence, that is, a(n) divides a(m) whenever n divides m. The sequence satisfies a linear recurrence of order 4.
A000129(n) = Pell(n) is the Lucas sequence U_n(2,-1). In general, if U(n) = U_n(P,Q) is the Lucas sequence with integer parameters P and Q then when Q = 1 or Q = -1 both U(4*n) + U(2*n) and U(4*n) - 2*U(2*n) are divisibility sequences of the fourth order. Cf. A127595, A215466 and A273627.

Programs

  • Magma
    I:=[1,30,995,33660]; [n le 4 select I[n] else 40*Self(n-1)-206*Self(n-2)+40*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 04 2016
  • Maple
    #A273626
    A000129 := proc (n) option remember;
    if n <= 1 then n else 2*A000129(n-1) + A000129(n-2) end if
    end proc:
    seq(1/14*(A000129(4*n) + A000129(2*n)), n = 1..20);
  • Mathematica
    LinearRecurrence[{40,-206,40,-1},{1,30,995,33660},100] (* G. C. Greubel, Jun 02 2016 *)

Formula

a(n) = sqrt(2)/56*( (sqrt(2) + 1)^(4*n) - (sqrt(2) - 1)^(4*n) + (sqrt(2) + 1)^(2*n) - (sqrt(2) - 1)^(2*n) ).
a(n) = (A082405(n) + A001109(n))/7 .
a(n) = 1/14*Pell(2*n)*A081555(n).
a(n) = -a(-n).
a(n) = 40*a(n-1) - 206*a(n-2) + 40*a(n-3) - a(n-4) for n>4.
O.g.f.: x*(x^2 - 10*x + 1)/((x^2 - 6*x + 1)*(x^2 - 34*x + 1)).

A260711 Numbers of the form x^2 - y^2 with x >= y; x and y are odd, x + y is a power of 2.

Original entry on oeis.org

0, 8, 16, 32, 48, 64, 96, 128, 160, 192, 224, 256, 320, 384, 448, 512, 576, 640, 704, 768, 832, 896, 960, 1024, 1152, 1280, 1408, 1536, 1664, 1792, 1920, 2048, 2176, 2304, 2432, 2560, 2688, 2816, 2944, 3072, 3200, 3328, 3456, 3584, 3712, 3840, 3968, 4096, 4352, 4608, 4864
Offset: 1

Author

Arkadiusz Wesolowski, Jul 30 2015

Keywords

Comments

These binomials do not have primitive factors.

Crossrefs

Subsequence of A008590.

Programs

  • Magma
    lst:=[0]; r:=4864; t:=1; d:=func; while d(t) gt t do s:=d(t); if s mod 2 eq 1 and t mod 2 eq 1 then s-:=1; end if; if s mod 2 eq 0 and t mod 2 eq 0 then s-:=1; end if; repeat; m:=(s+t)^2-(s-t)^2; if PrimeDivisors(s) eq [2] then Append(~lst, m); end if; s-:=2; until s-t lt 1; t+:=1; end while; Sort(lst); // Arkadiusz Wesolowski, Dec 19 2020

A238601 A sixth-order linear divisibility sequence related to the Fibonacci numbers: a(n) := (1/10)*Fibonacci(3*n)*Fibonacci(5*n)/Fibonacci(n).

Original entry on oeis.org

1, 44, 1037, 32472, 915305, 26874892, 776952553, 22595381424, 655633561309, 19040507781020, 552780012054689, 16050219184005336, 466002944275859873, 13530204273746536948, 392841165312292809085, 11405932444267712654688, 331164788382150547106857, 9615185834308570310716196
Offset: 1

Author

Peter Bala, Mar 06 2014

Keywords

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.
It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.
Here we take p = 3 and q = 5 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, and normalize the sequence {U(3*n)*U(5*n)/U(n)}n>=1 to have the initial term 1.
For other sequences of this type see A238600, A238602 and A238603. See also A238536.
Since Fibonacci(n) can be defined for all n, so can this sequence. - N. J. A. Sloane, May 07 2017

Examples

			G.f. = x + 44*x^2 + 1037*x^3 + 32472*x^4 + 915305*x^5 + 26874892*x^6 + ... - _Michael Somos_, May 07 2017
		

Programs

  • Magma
    [(Fibonacci(3*n) + (-1)^n*Fibonacci(5*n) + Fibonacci(7*n))/10: n in [1..30]]; // G. C. Greubel, Aug 07 2018
  • Maple
    with(combinat):
    seq(1/10*fibonacci(3*n)*fibonacci(5*n)/fibonacci(n), n = 1..20);
  • Mathematica
    Table[(1/10)*(Fibonacci[3*n] + (-1)^n*Fibonacci[5*n] + Fibonacci[7*n]), {n, 0, 50}] (* G. C. Greubel, Aug 07 2018 *)
  • PARI
    {a(n) = if(n, fibonacci(3*n) * fibonacci(5*n) / (10 * fibonacci(n)), 0)} /* Michael Somos, May 07 2017 */
    

Formula

a(n) = (1/10)*(Fibonacci(3*n) + (-1)^n*Fibonacci(5*n) + Fibonacci(7*n)).
The sequence can be extended to negative indices using a(-n) = (-1)^(n+1)*a(n).
O.g.f. x*(1 + 22*x - 181*x^2 - 22*x^3 + x^4)/( (1 - 4*x - x^2)*(1 + 11*x - x^2)*(1 - 29*x - x^2) ).
Recurrence equation: a(n) = 22*a(n-1) + 250*a(n-2) - 1320*a(n-3) - 250*a(n-4) + 22*a(n-5) + a(n-6).

A238600 A sixth-order linear divisibility sequence related to the Fibonacci numbers: a(n) := (1/6)*Fibonacci(3*n)*Fibonacci(4*n)/Fibonacci(n).

Original entry on oeis.org

1, 28, 408, 7896, 137555, 2496144, 44599477, 801617712, 14375440584, 258018516140, 4629531440711, 83076469908768, 1490726895438793, 26750144944686436, 480010941060482040, 8613453244178393184, 154562103244937408987, 2773504708179098411952
Offset: 1

Author

Peter Bala, Mar 01 2014

Keywords

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.
It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.
Here we take p = 3 and q = 4 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, and normalize the sequence to have the initial term 1.
For other sequences of this type see A238601, A238602 and A238603. See also A238536.

Programs

  • Magma
    [(Fibonacci(2*n) + (-1)^n*Fibonacci(4*n) + Fibonacci(6*n))/6: n in [1..30]]; // G. C. Greubel, Aug 07 2018
  • Maple
    with(combinat):
    seq(1/6*fibonacci(3*n)*fibonacci(4*n)/fibonacci(n), n = 1..20);
  • Mathematica
    Table[(1/6)*(Fibonacci[2*n] + (-1)^n*Fibonacci[4*n] + Fibonacci[6*n]), {n, 1, 500}] (* G. C. Greubel, Aug 07 2018 *)
    LinearRecurrence[{14,90,-350,90,14,-1},{1,28,408,7896,137555,2496144},20] (* Harvey P. Dale, Aug 26 2020 *)
  • PARI
    vector(30, n, (fibonacci(2*n) + (-1)^n*fibonacci(4*n) + fibonacci(6*n))/6) \\ G. C. Greubel, Aug 07 2018
    

Formula

a(n) = (1/6)*Fibonacci(3*n)*Fibonacci(4*n)/Fibonacci(n).
a(n) = (1/6)*( Fibonacci(2*n) + (-1)^n*Fibonacci(4*n) + Fibonacci(6*n) ).
The sequence can be extended to negative indices when a(-n) = -a(n).
O.g.f. x*(1 + 14*x - 74*x^2 + 14*x^3 + x^4)/( (1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) ).
Recurrence equation: a(n) = 14*a(n-1) + 90*a(n-2) - 350*a(n-3) + 90*a(n-4) + 14*a(n-5) - a(n-6).

A238603 A sixth-order linear divisibility sequence related to A000225: a(n) := (1/105)*(2^(3*n) - 1)*(2^(4*n) - 1)/(2^n - 1).

Original entry on oeis.org

1, 51, 2847, 170391, 10555655, 664857063, 42215949223, 2691226507047, 171901443816999, 10990938133564455, 703076406514657319, 44985901769992495143, 2878746218051469266983, 184228512166784552153127, 11790264946382521291370535, 754565442462197107544125479
Offset: 1

Author

Peter Bala, Mar 06 2014

Keywords

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.
It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.
Here we take p = 3 and q = 4 with P = 3 and Q = 2, for which U(n) is the sequence A000225 (sometimes called the Mersenne numbers), and normalize the sequence {U(3*n)*U(4*n)/U(n)}n>=1 to have the initial term 1.
For other sequences of this type see A238600, A238601 and A238602. See also A238536.

Examples

			G.f. = x + 51*x^2 + 2847*x^3 + 170391*x^4 + 10555655*x^5 + 664857063*x^6 + ... - _Michael Somos_, May 07 2017
		

Crossrefs

Programs

  • Magma
    [(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1): n in [1..50]]; // G. C. Greubel, Aug 07 2018
  • Maple
    seq(1/105*(2^(3*n)-1)*(2^(4*n)-1)/(2^n-1), n = 1..20);
  • Mathematica
    Table[(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1), {n, 1, 50}] (* G. C. Greubel, Aug 07 2018 *)
  • PARI
    {a(n) = if( n, (8^n - 1) * (16^n - 1) / (105 * (2^n - 1)), 0)}; /* Michael Somos, May 07 2017 */
    

Formula

a(n) = (1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1).
O.g.f.: x*(4096*x^4 - 4352*x^3 + 1160*x^2 - 68*x + 1 )/( (1-x)*(1-2*x)(1-4*x)*(1-16*x)*(1-32*x)*(1-64*x) ).
The formula for a(n) may be used to define it for all n in Z, and then we have a(n) = -(64)^n * a(-n). - Michael Somos, May 07 2017

A238602 A sixth-order linear divisibility sequence related to the Pell numbers: a(n) := (1/60)*Pell(3*n)*Pell(4*n)/Pell(n).

Original entry on oeis.org

1, 238, 45507, 9063516, 1792708805, 355009117386, 70287911575687, 13916722851826872, 2755438412296182921, 545562971271797876390, 108018710075587599558731, 21387159127038457710621972, 4234549485214861760195346253, 838419411023095574089504928386
Offset: 1

Author

Peter Bala, Mar 06 2014

Keywords

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.
It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.
Here we take p = 3 and q = 4 with P = 2 and Q = -1, for which U(n) is the sequence of Pell numbers, A000129, and normalize the sequence {U(3*n)*U(4*n)/U(n)}n>=1 to have the initial term 1.
For other sequences of this type see A238600, A238601 and A238603. See also A238536.

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 +68*x-698*x^2+68*x^3+x^4)/((1-6*x+x^2)*(1+34*x+x^2)*(1-198*x+x^2)))); // G. C. Greubel, Aug 07 2018
  • Mathematica
    Table[(1/60)*(Fibonacci[2*n, 2] + (-1)^n*Fibonacci[4*n, 2] + Fibonacci[6*n, 2]), {n, 1, 50}] (* G. C. Greubel, Aug 07 2018 *)
  • PARI
    x='x+O('x^30); Vec(x*(1+68*x-698*x^2+68*x^3+x^4)/((1-6*x+x^2)*(1 + 34*x+x^2)*(1-198*x+x^2))) \\ G. C. Greubel, Aug 07 2018
    

Formula

a(n) = (1/60)*( Pell(2*n) + (-1)^n*Pell(4*n) + Pell(6*n) ).
The sequence can be extended to negative indices using a(-n) = -a(n).
O.g.f. x*(1 + 68*x - 698*x^2 + 68*x^3 + x^4)/( (1 - 6*x + x^2)*(1 + 34*x + x^2)*(1 - 198*x + x^2) ).
Recurrence equation: a(n) = 170*a(n-1) + 5745*a(n-2) - 40052*a(n-3) + 5745*a(n-4) + 170*a(n-5) - a(n-6).

A238536 A fourth-order linear divisibility sequence related to the Fibonacci numbers: a(n) = (1/2)*Fibonacci(3*n)*Lucas(n).

Original entry on oeis.org

1, 12, 68, 504, 3355, 23256, 158717, 1089648, 7463884, 51170460, 350695511, 2403786672, 16475579353, 112925875764, 774003961940, 5305106018016, 36361727272627, 249227013404808, 1708227291909269, 11708364225400920, 80250321774226396, 550043889533755332, 3770056901455017263
Offset: 1

Author

Peter Bala, Feb 28 2014

Keywords

Comments

Let P and Q be integers. The Lucas sequences U(n) and V(n) (which depend on P and Q) are a pair of integer sequences that satisfy the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1 and V(0) = 2, V(1) = P, respectively. The sequence {U(n)} n >= 1 is a linear divisibility sequence of order 2, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0. In general, V(n) is not a divisibility sequence. However, it can be shown that if p >= 3 is an odd integer then the sequence {U(p*n)*V(n)} n >= 1 is a linear divisibility sequence of order 4. For a proof and a generalization of this result see the Bala link. Here we take p = 3 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, V(n) is the sequence of Lucas numbers, A000032, and normalize the sequence to have the initial term 1. For other sequences of this type see A238537 and A238538.

References

  • S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80.

Programs

  • Magma
    I:=[12, 68, 504, 3355]; [1] cat [n le 4 select I[n] else 4*Self(n-1) + 19*Self(n-2) + 4*Self(n-3) - Self(n-4): n in [1..30]]; // G. C. Greubel, Dec 25 2017
  • Maple
    with(combinat): lucas:= n->fibonacci(n+1)+ fibonacci(n-1):
    seq(1/2*lucas(n)*fibonacci(3*n), n = 1..24);
  • Mathematica
    Table[Fibonacci(3*n)*Lucas(n)/2, {n,1,30}] (* or *) Join[{1}, LinearRecurrence[{4,19,4,-1}, {12, 68, 504, 3355}, 30]] (* G. C. Greubel, Dec 25 2017 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,4,19,4]^(n-1)*[1;12;68;504])[1,1] \\ Charles R Greathouse IV, Oct 07 2016
    

Formula

a(n) = (1/2)*Fibonacci(3*n)*Lucas(n) = (1/2)*A000045(3*n)*A000032(n).
a(n) = (1/2)*Fibonacci(2*n)*Fibonacci(3*n)/Fibonacci(n).
a(n) = (1/(2*sqrt(5)))*( ((7 + 3*sqrt(5))/2)^n - ((7 - 3*sqrt(5))/2)^n + (-1)^n*((3 + sqrt(5))/2)^n - (-1)^n*((3 - sqrt(5))/2)^n ).
The sequence can be extended to negative indices by setting a(-n) = -a(n).
O.g.f. x*(1 + 8*x + x^2)/( (1 + 3*x + x^2)*(1 - 7*x + x^2) ).
Recurrence equation: a(n) = 4*a(n-1) + 19*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (1/2) * (Fibonacci(4*n) + (-1)^n*Fibonacci(2*n)). - Ralf Stephan, Mar 01 2014

A238538 A fourth-order linear divisibility sequence: a(n) = (2^n + 1)*(2^(3*n) - 1)/ ( (2 + 1)*(2^3 - 1) ).

Original entry on oeis.org

1, 15, 219, 3315, 51491, 811395, 12882499, 205321155, 3278747331, 52408827075, 838132189379, 13406842675395, 214483303960771, 3431523432591555, 54902699475185859, 878429788032676035, 14054769379960303811, 224875452250864496835, 3598000373385828511939
Offset: 1

Author

Peter Bala, Feb 28 2014

Keywords

Comments

This is a fourth-order linear divisibility sequence, that is, the sequence satisfies a linear recurrence of order 4 and if n | m then a(n) | a(m). This is a consequence of the following more general result: The polynomials P(n,x,y) := (x^n + y^n)*(x^(3*n) - y^(3*n)) form a fourth-order linear divisibility sequence in the polynomial ring Z[x,y]. See the Bala link.
Hence, for a fixed integers M and N, the normalized sequence (M^n + N^n)*(M^(3*n) - N^(3*n))/ ( (M + N)*(M^3 - N^3) ) for n = 1,2,3,... is a linear divisibility sequence of order 4. It has the rational o.g.f. x*(1 - 2*M*N*(M^2 - M*N + N^2)*x + (M*N)^4*x^2)/( (1 - M^4*x)*(1 - M^3*N*x)*(1 - M*N^3*x)*(1 - N^4*x) ). This is the case M = 2, N = 1. For other cases see A238539(M = 2, N = -1), A238540(M = 3, N = 1) and A238541(M = 3, N = 2). See also A238536, A238537 and A215466.
Note, these sequences do not belong to the family of linear divisibility sequences of the fourth order studied by Williams and Guy, which have o.g.f.s of the form x*(1 - q*x^2)/Q(x), Q(x) a quartic polynomial and q an integer parameter.

Programs

  • Maple
    seq(1/21*(2^n + 1)*(2^(3*n) - 1), n = 1..20);
  • Mathematica
    LinearRecurrence[{27,-202,432,-256},{1,15,219,3315},20] (* Harvey P. Dale, Jul 04 2019 *)

Formula

a(n) = (1/21)*(2^n + 1)*(2^(3*n) - 1) = A000051(n)*A024088(n)/21.
a(n) = (1/21)*(4^n - 1)*(8^n - 1)/(2^n - 1).
O.g.f.: x*(1 - 12*x + 16*x^2)/((1 - x)*(1 - 2*x)*(1 - 8*x)*(1 - 16*x)).
Recurrence equation: a(n) = 27*a(n-1) - 202*a(n-2) + 432*a(n-4) - 256*a(n-4).

A228440 Numbers n dividing u(n), where the Lucas sequence is defined u(i) = u(i-1) - 3*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 11, 121, 253, 1331, 2783, 5819, 11891, 14641, 29161, 30613, 64009, 130801, 133837, 161051, 273493, 320771, 336743, 558877, 640343, 670703, 704099, 895873, 1438811, 1472207, 1771561, 3008423, 3078251, 3528481, 3544453, 3704173, 6147647, 6290339, 7027801
Offset: 1

Author

Thomas M. Bridge, Nov 02 2013

Keywords

Comments

Since the absolute value of the discriminant of the characteristic polynomial is prime (=11), the sequence contains every nonnegative integer power of 11 (A001020 is subsequence). Other terms are formed on multiplication of 11^k by sporadic primes.

Examples

			u(1)=1 and u(11)=253. Clearly n divides u(n) for these terms.
		

Crossrefs

Cf. A214733 (Lucas sequence u(n) ignoring sign).
Cf. A001020 (powers of 11).

Programs

  • Mathematica
    nn = 10000; s = LinearRecurrence[{1, -3}, {1, 1}, nn]; t = {}; Do[
    If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 06 2013 *)

Extensions

a(27)-a(34) from Lars Blomberg, Feb 15 2016