A000032 -id:A000032 - cp's OEIS frontend

A064723 (L(p)-1)/p where L() are the Lucas numbers (A000032) and p runs through the primes.

1, 1, 2, 4, 18, 40, 210, 492, 2786, 39650, 97108, 1459960, 9030450, 22542396, 141358274, 2249412290, 36259245522, 91815545800, 1500020153484, 9702063416738, 24704432285040, 409634464205812, 2672366681180466, 44720842390302450, 1927655270098608960

Offset: 0

Shane Findley, Oct 13 2001

nonn

			a(0) = (Lucas(2) - 1)/2 = (3 - 1)/2 = 1; a(3) = (Lucas(7) - 1)/7 = (29 - 1)/7 = 4.

Harry J. Smith, Table of n, a(n) for n=0..100
Larry Ericksen, Primality Testing and Prime Constellations, Šiauliai Mathematical Seminar, Vol. 3 (11), 2008. Mentions this sequence.
S. Litsyn and V. Shevelev, Irrational Factors Satisfying the Little Fermat Theorem, International Journal of Number Theory, vol.1, no.4 (2005), 499-512.
V. Shevelev, A property of n-bonacci constant, Seqfan (Mar 23 2014)

Cf. A000032, A006206.

[(Lucas(NthPrime(n))-1)/NthPrime(n): n in [1..40]]; // Vincenzo Librandi, Aug 22 2015

A064723 := proc(n)
    p := ithprime(1+n) ;
    (A000032(p)-1)/p ;
end proc: # R. J. Mathar, Jan 09 2017

Array[(LucasL@ Prime@ # - 1)/Prime@ # &, {23}] (* Michael De Vlieger, Aug 22 2015 *)

lucas(n) = if(n==0,2, if(n==1,1,fibonacci(n+1)+fibonacci(n-1)))
forprime(n=1,100,print1((lucas(n)-1)/n, ", "))

lucas(n)= { if(n==0, 2, if(n==1, 1, fibonacci(n + 1) + fibonacci(n - 1))) }
{ n=-1; forprime (p=2, prime(101), write("b064723.txt", n++, " ", (lucas(p) - 1)/p) ) } \\ Harry J. Smith, Sep 23 2009

a(n) = A006206(A000040(n+1)). - Creighton Dement, Nov 04 2005

a(n) = (round(phi^prime(n+1)) - 1)/prime(n+1), where phi is golden ratio (A001622). Indeed, L(p) = round(phi^p), and round(phi^p) == 1 (mod p) and, what is more, for p>=5, round(phi^p) == 1 (mod 2*p) (see Shevelev link). In particular, all terms >=2 are even. - Vladimir Shevelev, Mar 24 2014

More terms from James Sellers and Klaus Brockhaus, Oct 16 2001

A304523 Number of ordered ways to write n as the sum of a Lucas number (A000032) and a positive odd squarefree number.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 2, 3, 2, 2, 1, 3, 1, 4, 2, 3, 2, 4, 3, 3, 3, 4, 3, 4, 3, 3, 1, 2, 1, 4, 2, 4, 3, 4, 3, 4, 3, 3, 3, 5, 3, 5, 2, 5, 2, 4, 2, 5, 2, 5, 2, 4, 2, 5, 3, 2, 3, 6, 3, 5, 3, 6, 2, 5, 2, 5, 1, 6, 3, 5, 3, 5, 3, 3, 3, 5, 3, 4, 3, 6, 3, 4, 3, 5, 2, 5, 4, 5, 4, 6

Offset: 1

Zhi-Wei Sun, May 13 2018

nonn

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 11, 13, 27, 29, 67, 139, 193, 247, 851.
It has been verified that a(n) > 0 for all n = 2..5*10^9.
See also A304331, A304333 and A304522 for similar conjectures.

			a(3) = 1 since 3 = A000032(0) + 1 with 1 odd and squarefree.
a(27) = 1 since 27 = A000032(3) + 23 with 23 odd and squarefree.
a(29) = 1 since 29 = A000032(6) + 11 with 11 odd and squarefree.
a(67) = 1 since 67 = A000032(0) + 5*13 with 5*13 odd and squarefree.
a(247) = 1 since 247 = A000032(6) + 229 with 229 odd and squarefree.
a(851) = 1 since 851 = A000032(0) + 3*283 with 3*283 odd and squarefree.

Zhi-Wei Sun, Table of n, a(n) for n = 1..100000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: D. Chudnovsky and G. Chudnovsky (eds.), Additive Number Theory, Springer, New York, 2010, pp. 341-353.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000032, A005117, A304034, A304081, A304331, A304333, A304522.

f[n_]:=f[n]=LucasL[n];
QQ[n_]:=QQ[n]=n>0&&Mod[n,2]==1&&SquareFreeQ[n];
tab={};Do[r=0;k=0;Label[bb];If[k>0&&f[k]>=n,Goto[aa]];If[QQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,90}];Print[tab]

A005247 a(n) = 3*a(n-2) - a(n-4), a(0)=2, a(1)=1, a(2)=3, a(3)=2. Alternates Lucas (A000032) and Fibonacci (A000045) sequences for even and odd n.

Original entry on oeis.org

2, 1, 3, 2, 7, 5, 18, 13, 47, 34, 123, 89, 322, 233, 843, 610, 2207, 1597, 5778, 4181, 15127, 10946, 39603, 28657, 103682, 75025, 271443, 196418, 710647, 514229, 1860498, 1346269, 4870847, 3524578, 12752043, 9227465, 33385282, 24157817

Offset: 0

N. J. A. Sloane

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
T. Crilly, Double sequences of positive integers, Math. Gaz., 69 (1985), 263-271.
R. K. Guy, Letter to N. J. A. Sloane, Feb 1986
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A000032, A000045, A005013, A005013, A005246, A005248.

a005247 n = a005247_list !! n
a005247_list = f a000032_list a000045_list where
   f (x::xs) (:y:ys) = x : y : f xs ys
-- Reinhard Zumkeller, Dec 27 2012

I:=[2,1,3,2]; [n le 4 select I[n] else 3*Self(n-2) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017

with(combinat): A005247 := n-> if n mod 2 = 1 then fibonacci(n) else fibonacci(n+1)+fibonacci(n-1); fi;
A005247:=-(z+1)*(3*z**2-z-1)/(z**2-z-1)/(z**2+z-1); # Simon Plouffe in his 1992 dissertation. Gives sequence with an additional leading 1.

CoefficientList[Series[(2 + x - 3x^2 - x^3)/(1 - 3x^2 + x^4), {x, 0, 40}], x]
LinearRecurrence[{0,3,0,-1},{2,1,3,2},50] (* Harvey P. Dale, Oct 10 2012 *)

a(n)=if(n%2,fibonacci(n),fibonacci(n+1)+fibonacci(n-1))

a(0)=2, a(1)=1, a(2)=3, a(n) = (1+a(n-1)a(n-2))/a(n-3), n >= 3. a(-n) = a(n).
G.f.: (2+x-3*x^2-x^3)/((1-x-x^2)*(1+x-x^2))
a(n) = F(n) if n odd, a(n) = L(n) if n even. a(n) = F(n+1)+(-1)^nF(n-1). - Mario Catalani (mario.catalani(AT)unito.it), Sep 20 2002
a(n) = ((5+sqrt(5))/10)*(((1+sqrt(5))/2)^n+((-1+sqrt(5))/2)^n)+((5-sqrt(5))/10)*(((1-sqrt(5))/2)^n+((-1-sqrt(5))/2)^n). With additional leading 1: a(n)=((sqrt(5))/5)*(((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n)+((5+3*sqrt(5))/10)*((-1+sqrt(5))/2)^n+((5-3*sqrt(5))/10)*((-1-sqrt(5))/2)^n. - Tim Monahan, Jul 25 2011
From Peter Bala, Jan 11 2013: (Start)
Let phi = 1/2*(sqrt(5) - 1). This sequence is the simple continued fraction expansion of the real number 1 + product {n >= 0} (1 + sqrt(5)*phi^(4*n+1))/(1 + sqrt(5)*phi^(4*n+3)) = 2.77616 23282 02325 23857 ... = 2 + 1/(1 + 1/(3 + 1/(2 + 1/(7 + ...)))). Cf. A005248.
Furthermore, for k = 0,1,2,... the simple continued fraction expansion of 1 + product {n >= 0} (1 + 1/5^k*sqrt(5)*phi^(4*n+1))/(1 + 1/5^k*sqrt(5)*phi^(4*n+3)) equals [2; 1*5^k, 3, 2*5^k, 7, 5*5^k, 18, 13*5^k, 47, ...]. (End)
a(n) = hypergeom([(1-n)/2, n mod 2 - n/2], [1 - n], -4) for n > 2. - Peter Luschny, Sep 03 2019
E.g.f.: 2*cosh(x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Mar 15 2022

Additional comments from Michael Somos, May 01 2000

A058036 Smallest primitive prime factor of the n-th Lucas number (A000032); i.e., L(n), L(0) = 2, L(1) = 1 and L(n) = L(n-1) + L(n-2).

Original entry on oeis.org

2, 1, 3, 1, 7, 11, 1, 29, 47, 19, 41, 199, 23, 521, 281, 31, 2207, 3571, 107, 9349, 2161, 211, 43, 139, 1103, 101, 90481, 5779, 14503, 59, 2521, 3010349, 1087, 9901, 67, 71, 103681, 54018521, 29134601, 79, 1601, 370248451, 83, 6709, 263, 181, 4969

Offset: 0

Robert G. Wilson v, Nov 16 2000

nonn

A Lucas number can have more than one primitive factor; the primitive factors of L(22) are 43 and 307.

T. D. Noe, Table of n, a(n) for n = 0..1000 (using Blair Kelly's data).
Mansur S. Boase, A Result About the Primes Dividing Fibonacci Numbers, The Fibonacci Quarterly, 39.5 (2001) 386.
J. Brillhart, P. L. Montgomery and R. D. Silverman, Tables of Fibonacci and Lucas factorizations, Math. Comp. 50 (1988), 251-260, S1-S15. Math. Rev. 89h:11002.
Blair Kelly, Fibonacci and Lucas Factorizations

Cf. A000032, A086600 (number of primitive prime factors in L(n)).

Cf. A001578 (analog for Fibonacci).

a=3; b=-1; prms={}; Table[c=a+b; a=b; b=c; f=First/@FactorInteger[c]; p=Complement[f, prms]; prms=Join[prms, p]; If[p=={}, 1, First[p]], {47}]

lucas(n) = fibonacci(n+1)+fibonacci(n-1); \\ A000032
a(n) = {n++; my(v = vector(n, k, k--; lucas(k))); my(vf = vector(n, k, factor(v[k])[,1]~)); for (k=1, n-1, vf[n] = setminus(vf[n], vf[k]);); if (#vf[n], vecmin(vf[n]), 1);} \\ Michel Marcus, May 11 2021

A079451 Largest prime dividing the n-th Lucas number (A000032); 1 when no such prime exists.

Original entry on oeis.org

2, 1, 3, 2, 7, 11, 3, 29, 47, 19, 41, 199, 23, 521, 281, 31, 2207, 3571, 107, 9349, 2161, 211, 307, 461, 1103, 151, 90481, 5779, 14503, 19489, 2521, 3010349, 4481, 9901, 63443, 911, 103681, 54018521, 29134601, 859, 3041, 370248451, 1427, 144481, 967, 541, 275449

Offset: 0

Lekraj Beedassy, Jan 13 2003

nonn

Amiram Eldar, Table of n, a(n) for n = 0..1000 (using Blair Kelly's data)
J. Brillhart, P. L. Montgomery and R. D. Silverman, Tables of Fibonacci and Lucas factorizations, Math. Comp. 50 (1988), 251-260, S1-S15. Math. Rev. 89h:11002.
Blair Kelly, Factorizations of first 1000 Lucas numbers.

Cf. A000032 (Lucas numbers), A006530 (greatest prime factor).
Cf. A001606 (indices of prime Lucas numbers <=> where a(n) = A000032(n)), subsequence of A076697 (indices of record values in this sequence).
Cf. A280104 (same for smallest prime factor).

[2,1] cat [Maximum(PrimeDivisors(Lucas(n))): n in [2..60]]; // Vincenzo Librandi, Dec 26 2016

A079451 := proc(n)
        A006530(A000032(n)) ;
end proc:
seq(A079451(n),n=0..30) ; # R. J. Mathar, Oct 26 2013
# second Maple program:
a:= n-> max(ifactors((<<1|1>, <1|0>>^n. <<2, -1>>)[1, 1])[2][..., 1][], 1):
seq(a(n), n=0..46);  # Alois P. Heinz, Aug 04 2025

Join[{2,1},f[n_]:=(FactorInteger@LucasL@n)[[-1,1]];Array[f,60,2]] (* Vincenzo Librandi, Dec 26 2016 *)

a(n) = my(f = factor(fibonacci(n+1)+fibonacci(n-1))); if (om = #f~, f[om, 1], 1); \\ Michel Marcus, Oct 26 2013

A079451(n) = A006530(A000032(n)) \\ M. F. Hasler, Apr 10 2025

def A079451(n): return A006530(A000032(n)) # M. F. Hasler, Apr 10 2025

a(n) = A006530(A000032(n)). - Felix Fröhlich, Dec 26 2016

More terms from Michel Marcus, Oct 26 2013

Modified b-file and scripts so that a(1)=1. - David Radcliffe, Aug 03 2025

A082762 Trinomial transform of Lucas numbers (A000032).

Original entry on oeis.org

1, 8, 44, 232, 1216, 6368, 33344, 174592, 914176, 4786688, 25063424, 131233792, 687149056, 3597959168, 18839158784, 98643116032, 516502061056, 2704439902208, 14160631169024, 74146027405312, 388233639755776, 2032817728913408, 10643971814457344

Offset: 0

Emanuele Munarini, May 21 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
Yuhan Jiang, The doubly asymmetric simple exclusion process, the colored Boolean process, and the restricted random growth model, arXiv:2312.09427 [math.CO], 2023.
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A000032, A027907, A091042, A292277.

I:=[1, 8]; [n le 2 select I[n] else 6*Self(n-1)-4*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017

a[n_]:=(MatrixPower[{{2,2},{2,4}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
f[n_] := Block[{s = Sqrt@ 5}, Simplify[((1 + s)(3 + s)^n + (1 - s)(3 - s)^n)/2]]; Array[f, 21, 0] (* Robert G. Wilson v, Mar 07 2011 *)
LinearRecurrence[{6,-4}, {1, 8}, 30] (* G. C. Greubel, Dec 21 2017 *)

x='x+O('x^30); Vec((1 + 2*x)/(1 - 6*x + 4*x^2)) \\ G. C. Greubel, Dec 21 2017

a(n) = Sum_{k=0..2*n} Trinomial(n,k)*Lucas(k+1), where Trinomial(n,k) = trinomial coefficients (A027907).
a(n) = 2^n*Lucas(2*n+1), where Lucas = A000032.
From Philippe Deléham, Mar 01 2004: (Start)
a(n) = 2^n*A002878(n) = 2^(-n)*Sum_{k>=0} C(2*n+1,2*k)*5^k; see A091042.
a(0) = 1, a(1) = 8, a(n+1) = 6*a(n) - 4*a(n-1). (End)
From Al Hakanson (hawkuu(AT)gmail.com), Jul 13 2009: (Start)
a(n) = ((1+sqrt(5))*(3+sqrt(5))^n + (1-sqrt(5))*(3-sqrt(5))^n)/2.
Third binomial transform of 1, 5, 5, 25, 25, 125. (End)
G.f.: (1 + 2*x)/(1 - 6*x + 4*x^2). - Colin Barker, Mar 23 2012

A099923 Fourth powers of Lucas numbers A000032.

Original entry on oeis.org

16, 1, 81, 256, 2401, 14641, 104976, 707281, 4879681, 33362176, 228886641, 1568239201, 10750371856, 73680216481, 505022001201, 3461445366016, 23725169980801, 162614549665681, 1114577187760656, 7639424429247601

Offset: 0

Ralf Stephan, Nov 01 2004

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 56.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Pridon Davlianidze, Problem B-1270, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; Four Telescopic Infinite Products, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
Toufik Mansour, A formula for the generating functions of powers of Horadam's sequence, Australas. J. Combin. 30 (2004) 207-212.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000032, A000045, A001254, A001622, A075515.

Fourth row of array A103324.

[ Lucas(n)^4 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011

LucasL[Range[0,20]]^4 (* or *) LinearRecurrence[{5,15,-15,-5,1},{16,1,81,256,2401},21] (* Harvey P. Dale, Jul 04 2014 *)
CoefficientList[Series[(16 - 79 x - 164 x^2 + 76 x^3 + x^4)/((1 - x) (1 + 3*x+x^2)*(1-7*x+x^2)), {x,0,50}], x] (* G. C. Greubel, Dec 21 2017 *)

for(n=0, 30, print1( (fibonacci(n+1) + fibonacci(n-1))^4, ", ")) \\ G. C. Greubel, Dec 21 2017

x='x+O('x^30); Vec((16-79*x-164*x^2+76*x^3+x^4)/((1-x)*(1+3*x+x^2)*(1-7*x+x^2))) \\ G. C. Greubel, Dec 21 2017

a(n) = A000032(n)^4 = A001254(n)^2.
a(n) = L(4*n) + 4*(-1)^n*L(2*n) + 6.
a(n) = L(n-2)*L(n-1)*L(n+1)*L(n+2) + 25, for n >=1.
G.f.: (16-79*x-164*x^2+76*x^3+x^4)/((1-x)*(1+3*x+x^2)*(1-7*x+x^2)). [See Mansour p. 207] - R. J. Mathar, Oct 26 2008
a(0)=16, a(1)=1, a(2)=81, a(3)=256, a(4)=2401, a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Jul 04 2014
Sum_{i=0..n} a(i) = 11 + 6*n + 4*(-1)^n*F(2*n+1) + F(4*n+2), for F = A000045. - Adam Mohamed and Greg Dresden, Jul 02 2024
Product_{n>=2} (1 - 25/a(n)) = phi^5/18, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 04 2024

A100107 Inverse Moebius transform of Lucas numbers (A000032) 1,3,4,7,11,..

Original entry on oeis.org

1, 4, 5, 11, 12, 26, 30, 58, 81, 138, 200, 355, 522, 876, 1380, 2265, 3572, 5880, 9350, 15272, 24510, 39806, 64080, 104084, 167773, 271968, 439285, 711530, 1149852, 1862022, 3010350, 4873112, 7881400, 12755618, 20633280, 33391491, 54018522, 87413156

Offset: 1

Jonathan Vos Post, Dec 26 2004

nonn

			a(2) = 4 because the prime 2 is divisible only by 1 and 2, so L(1) + L(2) = 1 + 3 = 4.
a(3) = 5 because the prime 3 is divisible only by 1 and 3, so L(1) + L(3) = 1 + 4 = 5.
a(4) = 11 because the semiprime 4 is divisible only by 1, 2, 4, so L(1) + L(2) + L(4) = 1 + 3 + 7 = 11.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A000032, A007435, A100279.

with(numtheory): with(combinat): a:=proc(n) local div: div:=divisors(n): sum(2*fibonacci(div[j]+1)-fibonacci(div[j]),j=1..tau(n)) end: seq(a(n),n=1..42); # Emeric Deutsch, Jul 31 2005

Table[Plus @@ Map[Function[d, LucasL[d]], Divisors[n]], {n, 100}] (* T. D. Noe, Aug 14 2012 *)

a(n) = Sum_{d|n} Lucas(d) = Sum_{d|n} A000032(d).
G.f.: Sum_{k>=1} Lucas(k) * x^k/(1 - x^k) = Sum_{k>=1} x^k * (1 + 2*x^k)/(1 - x^k - x^(2*k)). - Ilya Gutkovskiy, Aug 14 2019
a(n) ~ phi^n, where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 10 2021

More terms from Emeric Deutsch, Jul 31 2005

A111958 Lucas numbers (A000032) mod 8.

Original entry on oeis.org

2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3, 7, 2, 1, 3, 4, 7, 3, 2, 5, 7

Offset: 0

N. J. A. Sloane, Nov 28 2005

nonn

This sequence has period-length 12.

G. C. Greubel, Table of n, a(n) for n = 0..9999
Paulo Ribenboim, FFF (Favorite Fibonacci Flowers), Fib. Quart. 43 (No. 1, 2005), 3-14.
Index entries for linear recurrences with constant coefficients, signature (1,-1,1,-1,1,-1,1,-1,1,-1,1).

Cf. A000032, A079344, A130893.

LinearRecurrence[{1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1}, {2, 1, 3, 4, 7, 3, 2, 5, 7, 4, 3}, 105] (* Ray Chandler, Aug 27 2015 *)
Mod[LucasL[Range[0, 99]], 8] (* Alonso del Arte, Dec 19 2015 *)

From G. C. Greubel, Feb 08 2016: (Start)
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6) + a(n-7) - a(n-8) + a(n-9) - a(n-10) + a(n-11).
a(n+12) = a(n). (End)

A130247 Inverse Lucas (A000032) numbers: index k of a Lucas number such that Lucas(k)=n; max(k|Lucas(k) < n), if there is no such index.

Original entry on oeis.org

1, 0, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9

Offset: 1

Hieronymus Fischer, May 19 2007, Jul 02 2007

nonn

Inverse of the Lucas sequence (A000032), since a(Lucas(n))=n for n >= 0 (see A130241 and A130242 for other versions). Same as A130241 except for n=1.

			a(2)=0, since Lucas(0)=2; a(10)=4, since Lucas(4) = 7 < 10 but Lucas(5) = 11 > 10.

G. C. Greubel, Table of n, a(n) for n = 1..10000

For partial sums see A130248. Other related sequences: A000032, A130241, A130242, A130245, A130249, A130255, A130259. Indicator sequence A102460. For Fibonacci inverse see A130233 - A130240, A104162.

Join[{1, 0}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 3, 50}]] (* G. C. Greubel, Dec 21 2017 *)

from itertools import islice, count
def A130247_gen(): # generator of terms
    yield from (1,0)
    a, b = 3, 4
    for i in count(2):
        yield from (i,)*(b-a)
        a, b = b, a+b
A130247_list = list(islice(A130247_gen(),40)) # Chai Wah Wu, Jun 08 2022

a(n)=c(n), if (n^2-4)/5 is a square number, a(n)=s(n), if (n^2+4)/5 is a square number and a(n)=floor(log_phi(n)) otherwise, where s(n)=floor(arcsinh(n/2)/log(phi)), c(n)=floor(arccosh(n/2)/log(phi)) and phi=(1+sqrt(5))/2.
a(n) = A130241(n) except for n=2.
G.f.: g(x) = (1/(1-x))*(Sum_{k>=1} x^Lucas(k)) - x^2.
a(n) = floor(log_phi(n+1/2)) for n >= 3, where phi is the golden ratio.

cp's OEIS Frontend

A064723 (L(p)-1)/p where L() are the Lucas numbers (A000032) and p runs through the primes.

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A304523 Number of ordered ways to write n as the sum of a Lucas number (A000032) and a positive odd squarefree number.

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A005247 a(n) = 3*a(n-2) - a(n-4), a(0)=2, a(1)=1, a(2)=3, a(3)=2. Alternates Lucas (A000032) and Fibonacci (A000045) sequences for even and odd n.

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A058036 Smallest primitive prime factor of the n-th Lucas number (A000032); i.e., L(n), L(0) = 2, L(1) = 1 and L(n) = L(n-1) + L(n-2).

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A079451 Largest prime dividing the n-th Lucas number (A000032); 1 when no such prime exists.

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A082762 Trinomial transform of Lucas numbers (A000032).

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A099923 Fourth powers of Lucas numbers A000032.

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