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A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.
(Formerly M4217 N1760)
127
0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

8*a(n)^2 + 1 is a perfect square. - Gregory V. Richardson, Oct 05 2002

For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if sum_{1<=j<k} a_j = sum_{k<j<=m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel (auela(AT)math.upenn.edu), Jan 12 2006

(a(n),b(n)) where b(n)=A082291(n) are the integer solutions of the equation 2*binomial(b,a)=binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n)=A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s=a(n+1)={A001652(n)+A046090(n)+A001653(n)}/2 and area rs=A029549(n)=6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]

n such that 8*n^2=floor(sqrt(8)*n*ceil(sqrt(8)*n)). - Benoit Cloitre, May 10 2003

For n>0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003

a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series) - Creighton Dement, Dec 16 2004

Kekule numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005

Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005

Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n>0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005

Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance t(20)=2t(14)=210, so 6 is in the sequence. - Floor van Lamoen (fvlamoen(AT)hotmail.com), Oct 13 2005

One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006

Pell trapezoids (cf. A084158); for n>0, a(n)=(A000129(n-1)+A000129(n+1))*A000129(n)/2; e.g. 204=(5+29)*12/2 - Charlie Marion, Apr 01 2006

Tested for 2<p<27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006

If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007

If 8n+5 and 8n+7 are twin primes then their product divides a(4n+3). - Kenneth J Ramsey, Jun 08 2006

If p is an odd prime, then if p == 1 or 7 mod 8, then a((p-1)/2) == 0 mod p and a((p+1)/2) == 1 mod p; if p == 3 or 5 mod 8, then a((p-1)/2) == 1 mod p and a((p+1)/2) == 0 mod p. Kenneth Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007

a(n)*[a(n+b) - a(b-2)] = [a(n+1)+1]*[a(n+b-1) - a(b-1)] This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007

For n<0, let a(n)=-a(-n). Then [a(n+j)+a(k+j)] * [a(n+b+k+j)-a(b-j-2)] = [a(n+j+1)+a(k+j+1)] * [a(n+b+k+j-1)-a(b-j-1)]. - Charlie Marion, Mar 04 2011

The remainder of the division of a(n) by 5 is: 0, 1 or 4. The remainder of the division of a(n) by 7 is: 0, 1 or 6. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 26 2009

Number of units of a(n) belongs to a periodic sequence: 0, 1, 6, 5, 4, 9. The remainder of the division of a(n) by 5 belongs to a periodic sequence: 0, 1, 1, 0, 4, 4. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 01 2009

Sequence gives y values of the Diophantine equation: 0+1+2+...+x=y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x=y^2 with a<c then a+b=c-d and ((d+b)^2,d^2-b^2) is a solution too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation: 0+1+2+...+x=y^2 with a<c<e then (8*d^2,d*(f-b)) is a solution too. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 29 2009

If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x=y^2 with p<r then r=3p+4q+1 and s=2p+3q+1. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 02 2009

a(n)/A002315(n) converges to cos^2(Pi/8) = 1/2 + 2^(1/2)/4. - Gary Detlefs, Nov 25 2009

Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010

If x=a(n), y=A055997(n+1) and z=x^2+y, then x^4+y^3=z^2. - Bruno Berselli, Aug 24 2010

In general, if b(0)=1, b(1)=k and for n>1, b(n)=6*b(n-1)-b(n-2), then

  for n>0, b(n)=a(n)*k-a(n-1); e.g.,

  for k=2, when b(n)=A038725(n), 2=1*2-0, 11=6*2-1, 64=35*2-6, 373=204*2-35;

  for k=3, when b(n)=A001541(n), 3=1*3-0, 17=6*3-1; 99=35*3-6; 577=204*3-35;

  for k=4, when b(n)=A038723(n), 4=1*4-0, 23=6*4-1; 134=35*4-6; 781=204*4-35;

  for k=5, when b(n)=A001653(n), 5=1*5-0, 29=6*5-1; 169=35*5-6; 985=204*5-35.

  See also A002315, A054488, A038761, A054489, A054490.

  - Charlie Marion, Dec 08 2010

See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012

a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012

a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012

The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:

a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;

a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2;. -  Richard R. Forberg, Aug 30 2013

REFERENCES

I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.

A. Auel, MSRI Emissary, Fall 2005, Jan 12 (2006), p. 1.

Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.

Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, Fall 2005. Problem 1.

D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.

S. J. Cyvin and I. Gutman, Kekule structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 301, 302, P_{13}).

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.

Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607.

H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.

P. Franklin, E. F. Beckenbach, H. S. M Coxeter, N. H. McCoy, K. Menger and J. L. Synge, The Carus Mathematical Monographs, The Mathematical Association of America, (1967), pp. 144-146 [Title of book?]

H. Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).

B. Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.

Michael A. Jones, Proof Without Words: The Square of a Balancing Number Is a Triangular Number, The College Mathematics Journal, Vol. 43, No. 3 (May 2012), p. 212.

Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.4

P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.

Poo-Sung Park, Ramanujan's Continued Fraction for a Puzzle, College Mathematics Journal, 2005, 363-365.

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

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

R. A. Sulanke, Bijective recurrences concerning Schroeder paths, Electron. J. Combin. 5 (1998), Research Paper 47, 11 pp.

A. Tekcan, M. Tayat, M. E. Ozbek, The diophantine equation 8x^2-y^2+8x(1+t)+(2t+1)^2=0 and t-balancing numbers, ISRN Combinatorics, Volume 2014, Article ID 897834, 5 pages, http://dx.doi.org/10.1155/2014/897834.

LINKS

T. D. Noe, Table of n, a(n) for n=0..200

Cf. Dario Alpern for Diophantine equation a^4+b^3=c^2. [From Bruno Berselli, Aug 24 2010]

A. Bogomolny, There exist triangular numbers that are also squares

John C. Butcher, On Ramanujan, continued Fractions and an interesting number

L. Euler, De solutione problematum diophanteorum per numeros integros, Par. 19

Tanya Khovanova, Recursive Sequences

Kalman Liptai, Fibionacci Balancing Numbers, Fib. Quart. 42 (4) (2004) 330-340.

Madras College, St Andrews, Square Triangular Numbers

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

Rajesh Ram, Triangle Numbers that are Perfect Squares

K. J. Ramsey, Relation of Mersenne Primes To Square Triangular Numbers [edited by K. J. Ramsey, May 14 2011]

Kenneth Ramsey, Generalized Proof re Square Triangular Numbers

A. Sandhya, Puzzle 4: A problem Srinivasa Ramanujan, the famous 20th century Indian Mathematician Solved

Sci.math Newsgroup, Square numbers which are triangular

R. A. Sulanke, Moments, Narayana numbers and the cut and paste for lattice paths

Eric Weisstein's World of Mathematics, Binomial coefficient, MathWorld

Eric Weisstein's World of Mathematics, Square Triangular Number, MathWorld.

Eric Weisstein's World of Mathematics, Triangular Number, MathWorld.

Wikipedia, Triangular square number

Rick Young, Relevant quotation from biography of Ramanujan

Index entries for sequences related to Chebyshev polynomials.

Index entries for two-way infinite sequences

Index entries for sequences related to linear recurrences with constant coefficients, signature (6,-1)

FORMULA

a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x).

a(n) = 3*a(n-1)+sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000

a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)). - Henry Bottomley, Apr 19 2000

a(n) ~ 1/8*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002

Lim n -> inf. a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 05 2002

a(n) = [(3 + sqrt(8))^(n-1) - [(3 - sqrt(8))^(n-1)] / (2*sqrt(8)). - Gregory V. Richardson, Oct 13 2002

a(n)=((3+2sqrt(2))^n-(3-2sqrt(2))^n)/(4sqrt(2)). a(2n)=a(n)*A003499(n). 4a(n)=A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003

a(n) = floor((3+2sqrt(2))^n/(4sqrt(2))). - Lekraj Beedassy, Apr 23 2003

G.f.: x / (1 - 6*x + x^2). a(-n) = -a(n). - Michael Somos, Apr 07 2003

For n>=1, a(n) = Sum_{k=0...n-1} A001653(k). - Charlie Marion, Jul 01 2003

For n > 0, 4*a(2n) = A001653(n)^2 - A001653(n-1)^2. - Charlie Marion, Jul 16 2003

For n>0, a(n)=sum_{k = 0...n-1}((2k+1)*A001652(n-1-k))+A000217(n). - Charlie Marion, Jul 18 2003

a(2n+1)=a(n+1)^2-a(n)^2. - Charlie Marion, Jan 12 2004

a(k)*a(2n+k)=a(n+k)^2-a(n)^2; e.g., 204*7997214=40391^2-35^2. - Charlie Marion, Jan 15 2004

For j<n+1, a(k+j)*a(2n+k-j)-sum_{i = 0...j-1}a(2n-(2i+1)) = a(n+k)^2-a(n)^2. - Charlie Marion, Jan 18 2004

a(n)=A000129(2n)/2; a(n) := ((1+sqrt(2))^(2n)-(1-sqrt(2))^(2n))sqrt(2)/8; a(n) := sum{i=0..n, sum{j=0..n, A000129(i+j)*n!/(i!j!(n-i-j)!)/2}}. - Paul Barry, Feb 06 2004

E.g.f. : exp(3x)sinh(2sqrt(2)x)/(2sqrt(2)). - Paul Barry, Apr 21 2004

A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement, Sep 16 2004

a(n)=sum{k=0..n, binomial(2n, 2k+1)2^(k-1)}. - Paul Barry, Oct 01 2004

a(n+1) = A001653(n+1) - A038723(n+1) (conjecture); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement, Nov 19 2004

a(n+1)=sum_{k=0...n} A001850(k)*A001850(n-k), self convolution of central Delannoy numbers. - Benoit Cloitre, Sep 28 2005

a_n = 7(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = [ (1 + sqrt(2) )^2n - (1 - sqrt(2) )^2n ] / [4*sqrt(2)]. - Antonio Alberto Olivares, Oct 23 2003

a(n) = 5*(a(n-1)+a(n-2))-a(n-3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 20 2006

Define f[x,s] = s x + Sqrt[(s^2-1)x^2+1]; f[0,s]=0. a(n) = f[a(n-1),3], see second formula. - Marcos Carreira, Dec 27 2006

The perfect median m(n) can be expressed in terms of the Pell numbers P() = A000129() by m(n) = P( n + 2) * ( P ( n + 2) + (P (n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007

For k = 0,1,...,n, a(2n-k)-a(k)=2*a(n-k)*A001541(n). Also, a(2n+1-k)-a(k)=A002315(n-k)*A001653(n). - Charlie Marion, Jul 18 2007

[A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008

a(n)=sum{k=0..n-1, 4^k*C(n+k,2k+1)}. - Paul Barry, Apr 20 2009

a(n+1)^2-6*a(n+1)*a(n)+a(n)^2 = 1. - Charlie Marion, Dec 14 2010

a(n) = A002315(m)*A011900(n-m-1) + A001653(m)*A001652(n-m-1)-a(m) = A002315(m)*A053141(n-m-1) + A001653(m)*A046090(n-m-1) + a(m) with m<n; otherwise a(n) = A002315(m)*A053141(m-n) - A001653(m)*A011900(m-n) + a(m) = A002315(m)*A053141(m-n) - A001653(m)*A046090(m-n) - a(m) =(A002315(n)-A001653(n))/2. - Kenneth J Ramsey, Oct 12 2011

16*a(n)^2 + 1 = A056771(n). - James R. Buddenhagen, Dec 09 2011

A010054(A000290(a(n))) = 1. - Reinhard Zumkeller, Dec 17 2011

In general, a(n+k)^2 - A003499(k)*a(n+k)*a(n) + a(n)^2 = a(k)^2. - Charlie Marion, Jan 11 2012

a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*5^k. - Philippe Deléham, Feb 10 2012

PSUM transform of a(n+1) is A053142. PSUMSIGN transform of a(n+1) is A084158. BINOMIAL transform of a(n+1) is A164591. BINOMIAL transform of A086347 is a(n+1). BINOMIAL transform of A057087(n-1). - Michael Somos, May 11 2012

a(n+k) = A001541(k)*a(n) + sqrt(A132592(k)*a(n)^2 + a(k)^2). Generalizes formula dated Oct 09 2000. - Charlie Marion, Nov 27 2012

a(n) + a(n+2k) = A003499(k)*a(n+k); a(n) + a(n+2k+1) = A001653(k+1)*A002315(n+k). - Charlie Marion, Nov 29 2012

Product {n >= 1} (1 + 1/a(n)) = 1 + sqrt(2). - Peter Bala, Dec 23 2012

Product {n >= 2} (1 - 1/a(n)) = 1/3*(1 + sqrt(2)). - Peter Bala, Dec 23 2012

G.f.: G(0)*x/(2-6*x), where G(k)= 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013

G.f.: H(0)*x/2, where H(k) = 1 + 1/( 1 - x*(6-x)/(x*(6-x) + 1/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 18 2014

EXAMPLE

G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...

MAPLE

a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..26); # Emeric Deutsch

A001109:=1/(z**2-6*z+1); # Simon Plouffe in his 1992 dissertation

with (combinat):seq(fibonacci(2*n, 2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008

MATHEMATICA

Transpose[NestList[Flatten[{Rest[#], ListCorrelate[{-1, 6}, #]}]&, {0, 1}, 30]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)

CoefficientList[Series[x/(1-6x+x^2), {x, 0, 30}], x]  (* Harvey P. Dale, Mar 23 2011 *)

LinearRecurrence[{6, -1}, {0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)

a[ n_] := ChebyshevU[ n - 1, 3]; (* Michael Somos, Sep 02 2012 *)

PROG

(PARI) {a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */

(PARI) {a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */

(PARI) {a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */

(Sage) [lucas_number1(n, 6, 1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

(Haskell)

a001109 n = a001109_list !! n :: Integer

a001109_list = 0 : 1 : zipWith (-)

   (map (* 6) $ tail a001109_list) a001109_list

-- Reinhard Zumkeller, Dec 17 2011

CROSSREFS

sqrt(A001110). Cf. A001108, A002315. a(n)=sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).

2*a(n) = A001542.

Cf. A001653, A001850.

Sequence in context: A161727 A121838 A242629 * A180033 A144638 A117671

Adjacent sequences:  A001106 A001107 A001108 * A001110 A001111 A001112

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

EXTENSIONS

Additional comments from Wolfdieter Lang, Feb 10 2000

More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000

STATUS

approved

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Last modified October 21 11:38 EDT 2014. Contains 248377 sequences.