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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A203861 G.f.: Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^3 where Lucas(n) = A000204(n).

Original entry on oeis.org

1, -3, -9, 20, 45, 0, -151, -231, 0, 140, 1107, 2052, 49, -1305, 0, -15004, -28260, 0, 17710, 0, 81, 324040, 589953, 0, -375570, -1089, 0, -124124, -10659705, -19764180, -121, 12605358, 0, 0, 4158315, 0, 567552368, 1052295189, -780030, -669901660, 0, 0, -221399431, -85965, 0
Offset: 0

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Author

Paul D. Hanna, Jan 07 2012

Keywords

Comments

a(A020757(n)) = 0 where A020757 lists numbers that are not the sum of two triangular numbers.

Examples

			G.f.: A(x) = 1 - 3*x - 9*x^2 + 20*x^3 + 45*x^4 - 151*x^6 - 231*x^7 +...
-log(A(x))/3 = x + 3*3*x^2/2 + 4*4*x^3/3 + 7*7*x^4/4 + 6*11*x^5/5 + 12*18*x^6/6 +...+ sigma(n)*A000204(n)*x^n/n +...
The g.f. equals the product:
A(x) = (1-x-x^2)^3 * (1-3*x^2+x^4)^3 * (1-4*x^3-x^6)^3 * (1-7*x^4+x^8)^3 * (1-11*x^5-x^10)^3 * (1-18*x^6+x^12)^3 *...* (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^3 *...
Positions of zeros form A020757:
[5,8,14,17,19,23,26,32,33,35,40,41,44,47,50,52,53,54,59,62,63,...]
which are numbers that are not the sum of two triangular numbers.
		

Crossrefs

Programs

  • PARI
    /* Subroutine used in PARI programs below: */
    {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
    
  • PARI
    {a(n)=polcoeff(exp(sum(k=1, n, -3*sigma(k)*Lucas(k)*x^k/k)+x*O(x^n)), n)}
    
  • PARI
    {a(n)=polcoeff(prod(m=1, n, 1 - Lucas(m)*x^m + (-1)^m*x^(2*m) +x*O(x^n))^3, n)}

Formula

G.f.: exp( Sum_{n>=1} -3 * sigma(n) * A000204(n) * x^n/n ).

A022801 n-th Lucas number (A000204(n)) + n-th non-Lucas number (A090946(n+1)).

Original entry on oeis.org

3, 8, 10, 15, 20, 28, 41, 60, 90, 138, 215, 339, 540, 863, 1385, 2229, 3594, 5802, 9374, 15153, 24503, 39631, 64109, 103713, 167793, 271476, 439238, 710682, 1149887, 1860535, 3010387, 4870886, 7881236, 12752084, 20633281, 33385325, 54018565
Offset: 1

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Author

Keywords

Programs

  • Maple
    with(combinat): lucas:= n->fibonacci(n+1)+fibonacci(n-1): n:=1: for k from 1 to 7 do for nonlucas from lucas(k)+1 to lucas(k+1)-1 do printf("%d, ",nonlucas+lucas(n)) :n:=n+1 od od: # C. Ronaldo (aga_new_ac(AT)hotmail.com)
  • Mathematica
    Module[{nn=40,luc,nluc},luc=LucasL[Range[nn]];nluc=Complement[Range[ Last[ luc]],luc];Total/@Thread[{luc,Take[nluc,Length[luc]]}]] (* Harvey P. Dale, May 02 2019 *)
  • Python
    from sympy import lucas
    def A022801(n):
        def f(x):
            if x<=2: return n+1
            a, b, c = 1, 3, 0
            while b<=x:
                a, b = b, a+b
                c += 1
            return n+1+c
        m, k = n+1, f(n+1)
        while m != k: m, k = k, f(k)
        return m+lucas(n) # Chai Wah Wu, Sep 10 2024

Formula

a(n) = A000204(n) + A090946(n + 1). - Sean A. Irvine, May 21 2019

Extensions

Thanks to Karima MOUSSAOUI (bouyao(AT)wanadoo.fr), who noticed that there were two versions of this sequence, differing at about the 22nd term, Feb 28 2004
More terms from Emeric Deutsch, Jan 14 2005

A065156 Numbers n such that some Lucas number (A000204) is divisible by n.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 11, 14, 18, 19, 22, 23, 27, 29, 31, 38, 41, 43, 44, 46, 47, 49, 54, 58, 59, 62, 67, 71, 76, 79, 81, 82, 83, 86, 94, 98, 101, 103, 107, 116, 118, 121, 123, 124, 127, 129, 131, 134, 139, 142, 151, 158, 161, 162, 163, 166, 167, 179, 181, 191, 199
Offset: 1

Views

Author

Dean Hickerson, Oct 18 2001

Keywords

Comments

From A.H.M. Smeets, Sep 20 2020 (Start)
For the Fibonacci numbers, each natural number divides some Fibonacci number (see A001177).
If, for some number m, m divides some Lucas number L_i (=A000204(i)), then, the smallest i satisfies i <= m. (End)

Crossrefs

Complement of A064362. Cf. A000204.

Programs

  • Mathematica
    test[ n_ ] := For[ a=1; b=3, True, t=b; b=Mod[ a+b, n ]; a=t, If[ b==0, Return[ True ] ]; If[ a==2&&b==1, Return[ False ] ] ]; Select[ Range[ 200 ], test ]
    Take[Flatten[Divisors/@LucasL[Range[200]]]//Union,70] (* Harvey P. Dale, Jun 07 2020 *)
  • Python
    a, n = 0, 0
    while n < 1000:
        a, f0, f1, i = a+1, 1, 2, 1
        if f1%a == 0:
            n = n+1
            print(n,a)
        else:
            while f0%a != 0 and i <= a:
                f0, f1, i = f0+f1, f0, i+1
            if i <= a:
                n = n+1
                print(n,a) # A.H.M. Smeets, Sep 20 2020

Formula

Equals {1,2,4} union {p^e | p in A140409 and e > 0} union {2*p^e | p in A140409 and e > 0} union {4*p | p in A053032} union {4*p*q | p, q in A053032}. - A.H.M. Smeets, Sep 20 2020

A130774 Cumulative concatenation of A000204 Lucas numbers (beginning at 1).

Original entry on oeis.org

1, 13, 134, 1347, 134711, 13471118, 1347111829, 134711182947, 13471118294776, 13471118294776123, 13471118294776123199, 13471118294776123199322, 13471118294776123199322521, 13471118294776123199322521843
Offset: 1

Views

Author

Jonathan Vos Post, Aug 19 2007

Keywords

Comments

a(2) = 13 and a(10) = 13471118294776123 are prime. What is the next prime?
The next prime is a(31), a number of 120 digits. - Giovanni Resta, Jun 18 2016

Crossrefs

Cf. A000204.

Programs

  • Mathematica
    Module[{nn=20,ll},ll=LucasL[Range[nn]];Table[FromDigits[Flatten[IntegerDigits/@Take[ll,n]]],{n,nn}]] (* Harvey P. Dale, May 07 2023 *)

Formula

a(1) = 1; a(n+1) = Concatenate(a(n),A000204(n+1)).

A233360 Primes of the form L(k) + q(m) with k > 0 and m > 0, where L(k) is the k-th Lucas number (A000204), and q(.) is the strict partition function (A000009).

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 101, 103, 107, 127, 131, 149, 151, 193, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 307, 337, 347, 349, 379, 397, 401, 419, 421, 449, 463, 487, 523, 541, 571, 643, 647, 661
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 08 2013

Keywords

Comments

Conjecture: The sequence has infinitely many terms.
This follows from the conjecture in A233359.

Examples

			a(1) = 2 since L(1) + q(1) = 1 + 1 = 2.
a(2) = 3 since L(1) + q(3) = 1 + 2 = 3.
a(3) = 5 since L(2) + q(3) = 3 + 2 = 5.
		

Crossrefs

Programs

  • Mathematica
    n=0
    Do[Do[If[LucasL[j]>=Prime[m],Goto[aa],
    Do[If[PartitionsQ[k]==Prime[m]-LucasL[j],
    n=n+1;Print[n," ",Prime[m]];Goto[aa]];If[PartitionsQ[k]>Prime[m]-LucasL[j],Goto[bb]];Continue,{k,1,2*(Prime[m]-LucasL[j])}]];
    Label[bb];Continue,{j,1,2*Log[2,Prime[m]]}];
    Label[aa];Continue,{m,1,125}]

A288039 Number of compositions (ordered partitions) of n into Lucas numbers (beginning with 1) (A000204).

Original entry on oeis.org

1, 1, 1, 2, 4, 6, 9, 16, 27, 43, 70, 118, 195, 318, 524, 868, 1430, 2351, 3878, 6399, 10542, 17367, 28634, 47206, 77793, 128212, 211346, 348360, 574153, 946342, 1559849, 2571016, 4237616, 6984659, 11512526, 18975464, 31276187, 51550993, 84968944, 140049801, 230836734, 380476447, 627119783, 1033648857
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 04 2017

Keywords

Examples

			a(4) = 4 because we have [4], [3, 1], [1, 3] and [1, 1, 1, 1].
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/(1 - Sum[x^LucasL[k], {k, 1, 15}]), {x, 0, 43}], x]

Formula

G.f.: 1/(1 - Sum_{k>=1} x^A000204(k)).

A304093 a(n) is the number of the proper divisors of n that are Lucas numbers (A000204, with 2 excluded).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 4, 1, 1, 2, 2, 1, 3, 1, 3, 2, 1, 1, 3, 2, 1, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 1, 2, 2, 3, 2, 1, 2, 2, 1, 1, 4, 1, 1, 3, 3, 1, 3, 2, 2, 2, 2, 1, 3, 1, 2, 3, 2, 1, 2, 1, 2, 3
Offset: 1

Views

Author

Antti Karttunen, May 13 2018

Keywords

Crossrefs

Programs

  • PARI
    isA000204(n) = { my(u1=1,u2=3,old_u1); if(n<=2,(n%2),while(n>u2,old_u1=u1;u1=u2;u2=old_u1+u2);(u2==n)); };
    A304093(n) = sumdiv(n,d,(dA000204(d));

Formula

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A093540. - Amiram Eldar, Jul 05 2025

A304094 Number of Lucas numbers (A000204: 1, 3, 4, 7, 11, ... excluding 2) that divide n.

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 2, 2, 2, 1, 2, 3, 1, 2, 2, 2, 1, 3, 1, 2, 3, 2, 1, 3, 1, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 4, 1, 1, 2, 2, 1, 3, 1, 3, 2, 1, 2, 3, 2, 1, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 1, 2, 3, 3, 2, 1, 2, 2, 1, 1, 4, 1, 1, 3, 3, 1, 3, 2, 2, 2, 2, 1, 3, 1, 2, 3, 2, 1, 2, 1, 2, 3
Offset: 1

Views

Author

Antti Karttunen, May 13 2018

Keywords

Crossrefs

Programs

  • PARI
    isA000204(n) = { my(u1=1,u2=3,old_u1); if(n<=2,(n%2),while(n>u2,old_u1=u1;u1=u2;u2=old_u1+u2);(u2==n)); };
    A304094(n) = sumdiv(n,d,isA000204(d));

Formula

a(n) = A304092(n) - A059841(n).
a(n) = A304096(n) + A079978(n) + 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A093540 = 1.962858... . - Amiram Eldar, Dec 31 2023

A134072 Concatenation of A000204 Lucas numbers (beginning at 1) in reverse order.

Original entry on oeis.org

1, 31, 431, 7431, 117431, 18117431, 2918117431, 472918117431, 76472918117431, 12376472918117431, 19912376472918117431, 32219912376472918117431, 52132219912376472918117431, 84352132219912376472918117431
Offset: 1

Views

Author

Alexander Adamchuk, Oct 06 2007

Keywords

Comments

Indices of prime terms are {2, 3, 5, 11, ...}. Primes are listed in A134071 = {31, 431, 117431, 19912376472918117431, ...}.

Crossrefs

Cf. A000204 (Lucas numbers).
Cf. A130774 (concatenation of Lucas numbers).
Cf. A019523 (concatenation of Fibonacci numbers).
Cf. A038399 (concatenation of first n nonzero Fibonacci numbers in reverse order).
Cf. A134069 (primes in A038399).
Cf. A134070 (primes in A130774).
Cf. A134071 (primes in A134072).

Programs

  • Mathematica
    Module[{nn=20,lnos},lnos=LucasL[Range[nn]];Table[FromDigits[Flatten[ IntegerDigits/@ Reverse[Take[lnos,n]]]],{n,nn}]] (* Harvey P. Dale, Jul 27 2015 *)

Extensions

Edited by Charles R Greathouse IV, Apr 26 2010

A153180 a(n) = L(13n)/L(n) where L(n) = Lucas number A000204(n).

Original entry on oeis.org

521, 90481, 35355581, 10525900321, 3489827263001, 1111126318086721, 359316586176453881, 115509240442846111681, 37216910406644366498621, 11980863523543017476802001, 3858153294795970321295258921
Offset: 1

Views

Author

Artur Jasinski, Dec 20 2008

Keywords

Comments

All numbers in this sequence are:
congruent to 1 mod 10
congruent to 1 mod 100 (iff n is congruent to 0 mod 5),Q congruent to 21 mod 100 (iff n is congruent to 1 or 4 mod 5),
congruent to 81 mod 100 (iff n is congruent to 2 or 3 mod 5).Q

Crossrefs

Programs

  • Mathematica
    Table[LucasL[13 n]/LucasL[n], {n, 1, 150}]

Formula

a(n)= +233*a(n-1) +33552*a(n-2) -1493064*a(n-3) -27372840*a(n-4) +186135312*a(n-5) +488605194*a(n-6) -488605194*a(n-7) -186135312*a(n-8) +27372840*a(n-9) +1493064*a(n-10) -33552*a(n-11) -233*a(n-12) +a(n-13). G.f.: -1+ (-2-123*x)/(x^2+123*x+1) +(2-322*x)/(x^2-322*x+1) +(-2-3*x)/(x^2+3*x+1) +(2-7*x)/(x^2-7*x+1) +(2-47*x)/(x^2-47*x+1) -1/(x-1)+ (-2-18*x)/(x^2+18*x+1). [From R. J. Mathar, Oct 22 2010]
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