OFFSET
0,1
COMMENTS
Cf. A000204 for Lucas numbers beginning with 1.
Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014
Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017
This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003
For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.
a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).
Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007
From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)
The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.
The first six multiplication formulas are:
L(2n) = L(n)^2 - 2*(-1)^n;
L(3n) = L(n)^3 - 3*(-1)^n*L(n);
L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;
L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);
L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.
Generally, L(n) | L(mn) if and only if m is odd.
In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:
For a >= b and odd b, L(a + b) + L(a - b) = 5*F(a)*F(b).
For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).
For a >= b and odd b, L(a + b) - L(a - b) = L(a)*L(b).
For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).
A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:
L(n) - L(n - 3) = 2*L(n - 2);
L(n) - L(n - 4) = 5*F(n - 2);
L(n) - L(n - 6) = 4*L(n - 3);
L(n) - L(n - 12) = 40*F(n - 6);
L(n) - L(n - 60) = 4160200*F(n - 30).
These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).
The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)
Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009
The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014
If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015
A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015
A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017
Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017
For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Klaus Purath, Apr 19 2019: (Start)
While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:
First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.
Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.
Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)
L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019
If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019
From Kai Wang, Dec 18 2019: (Start)
L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).
Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251.
(End)
From Peter Bala, Dec 23 2021: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)
For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024
REFERENCES
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Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
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S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 83-84.
LINKS
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Taras Goy and Mark Shattuck, Fibonacci and Lucas identities from Toeplitz-Hessenberg matrices, Applications and Applied Mathematics: An International Journal, 14:2 (2019), 699-715.
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Index entries for linear recurrences with constant coefficients, signature (1,1).
FORMULA
G.f.: (2 - x)/(1 - x - x^2).
L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n = phi^n + (1-phi)^n.
L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000
a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]
a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003
L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004
a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).
L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).
L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).
L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)
Inverse: floor(log_phi(a(n)) + 1/2) = n, for n>1. Also for n >= 0, floor((1/2)*log_phi(a(n)*a(n+1))) = n. Extension valid for all integers n: floor((1/2)*sign(a(n)*a(n+1))*log_phi|a(n)*a(n+1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007
Let f(n) = phi^n + phi^(-n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k>=0} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide Colazingari, Jul 01 2016
Starting (1, 3, 4, 7, 11, ...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007
a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).
a(n) = round(1/min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
E.g.f.: exp(phi*x) + exp(-x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010
L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010
a(n) = 2*a(n-2) + a(n-3), n>2. - Gary Detlefs, Sep 09 2010
L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010
L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010
L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs, Nov 29 2010
L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)), n != 0. - Gary Detlefs, Dec 13 2010
L(n) = sqrt(A001254(n)) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n+1)). - Gary Detlefs, Dec 26 2010
L(n) = floor(phi^n) + ((-1)^n + 1)/2 = A014217(n) +((-1)^n+1)/2, where phi = A001622. - Gary Detlefs, Jan 20 2011
L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs, May 19 2011
For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012
a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) == - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012
From Gary Detlefs, Dec 21 2012: (Start)
L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.
L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.
where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
(End)
L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H. W. Gould]. - Gary Detlefs, Jan 20 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013
L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014
L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014
L(i+j+1) = L(i)*F(j) + L(i+1)*F(j+1) with F(n)=A000045(n). - J. M. Bergot, Feb 12 2016
a(n) = (L(n+1)^2 + 5*(-1)^n)/L(n+2). - J. M. Bergot, Apr 06 2016
Dirichlet g.f.: PolyLog(s,-1/phi) + PolyLog(s,phi), where phi is the golden ratio. - Ilya Gutkovskiy, Jul 01 2016
L(n) = F(n+2) - F(n-2). - Yuchun Ji, Feb 14 2016
L(n+1) = A087131(n+1)/2^(n+1) = 2^(-n)*Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2). - Tony Foster III, Oct 14 2017
L(2*n) = (F(k+2*n) + F(k-2*n))/F(k); n >= 1, k >= 2*n. - David James Sycamore, May 04 2018
From Greg Dresden and Shaoxiong Yuan, Jul 16 2019: (Start)
L(3n + 4)/L(3n + 1) has continued fraction: n 4's followed by a single 7.
L(3n + 3)/L(3n) has continued fraction: n 4's followed by a single 2.
L(3n + 2)/L(3n - 1) has continued fraction: n 4's followed by a single -3. (End)
From Klaus Purath, Sep 15 2019: (Start)
All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
L(n) = (2*L(n+2) - L(n-3))/5.
L(n) = (2*L(n-2) + L(n+3))/5.
L(n) = F(n-3) + 2*F(n).
L(n) = 2*F(n+2) - 3*F(n).
L(n) = (3*F(n-1) + F(n+2))/2.
L(n) = 3*F(n-3) + 4*F(n-2).
L(n) = 4*F(n+1) - F(n+3).
L(n) = (F(n-k) + F(n+k))/F(k) with odd k>0.
L(n) = (F(n+k) - F(n-k))/F(k) with even k>0.
L(n) = A001060(n-1) - F(n+1).
L(n) = (A022121(n-1) - F(n+1))/2.
L(n) = (A022131(n-1) - F(n+1))/3.
L(n) = (A022139(n-1) - F(n+1))/4.
L(n) = (A166025(n-1) - F(n+1))/5.
The following two formulas apply for all sequences of the Fibonacci type.
(a(n-2*k) + a(n+2*k))/a(n) = L(2*k).
(a(n+2*k+1) - a(n-2*k-1))/a(n) = L(2*k+1). (End)
L(n) = F(n-k)*L(k+1) + F(n-k-1)*L(k), for all k >= 0, where F(n) = A000045(n). - Michael Tulskikh, Dec 06 2019
F(n+2*m) = L(m)*F(n+m) + (-1)^(m-1)*F(n) for all n >= 0 and m >= 0. - Alexander Burstein, Mar 31 2022
a(n) = i^(n-1)*cos(n*c)/cos(c) = i^(n-1)*cos(c*n)*sec(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
From Yike Li and Greg Dresden, Aug 25 20022: (Start)
L(2*n) = 5*binomial(2*n-1,n) - 2^(2*n-1) + 5*Sum_{j=1..n/5} binomial(2*n,n+5*j) for n>0.
L(2*n+1) = 2^(2n) - 5*Sum_{j=0..n/5} binomial(2*n+1,n+5*j+3). (End)
From Andrea Pinos, Jul 04 2023: (Start)
L(n) ~ Gamma(1/phi^n) + gamma.
L(n) = Re(phi^n + e^(i*Pi*n)/phi^n). (End)
EXAMPLE
G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + ...
MAPLE
with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)
MATHEMATICA
a[0] := 2; a[n] := Nest[{Last[#], First[#] + Last[#]} &, {2, 1}, n] // Last
Array[2 Fibonacci[# + 1] - Fibonacci[#] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
Table[LucasL[n], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)
LucasL[Range[0, 20]] (* Eric W. Weisstein, Aug 07 2017 *)
CoefficientList[Series[(-2 + x)/(-1 + x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
PROG
(Magma) [Lucas(n): n in [0..120]];
(PARI) {a(n) = if(n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};
(PARI) {a(n) = if(n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};
(PARI) {a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};
(PARI) a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011
(PARI) polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011
(Sage) [lucas_number2(n, 1, -1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008
(Haskell)
a000032 n = a000032_list !! n
a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
-- Reinhard Zumkeller, Aug 20 2011
(Python)
def A000032_gen(): # generator of terms
a, b = 2, 1
while True:
yield a
a, b = b, a+b
it = A000032_gen()
A000032_list = [next(it) for _ in range(50)] # Cole Dykstra, Aug 02 2022
(Python)
from sympy import lucas
def A000032(n): return lucas(n) # Chai Wah Wu, Sep 23 2023
CROSSREFS
First row of array A103324.
a(n) = A101220(2, 0, n), for n > 0.
Cf. A131774, A001622, A002878 (L(2n+1)), A005248 (L(2n)), A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods), A057854 (complement), A354265 (generalized Lucas numbers).
Cf. sequences with formula Fibonacci(n+k)+Fibonacci(n-k) listed in A280154.
Subsequence of A047201.
KEYWORD
nonn,nice,easy,core
AUTHOR
N. J. A. Sloane, May 24 1994
STATUS
approved