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A001109
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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)
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0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720
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OFFSET
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0,3
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COMMENTS
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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_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, 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*ceiling(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
Kekulé 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)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005
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 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+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 J 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
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, 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 = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009
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 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:
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015
Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015
The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and 2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016
a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016
Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017
In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021
Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021
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REFERENCES
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Julio R. Bastida, 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, pp. 193, 197.
D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.
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.
P. Franklin, E. F. Beckenbach, H. S. M Coxeter, N. H. McCoy, K. Menger, and J. L. Synge, Rings And Ideals, No 8, The Carus Mathematical Monographs, The Mathematical Association of America, (1967), pp. 144-146.
A. Patra, G. K. Panda, and T. Khemaratchatakumthorn. "Exact divisibility by powers of the balancing and Lucas-balancing numbers." Fibonacci Quart., 59:1 (2021), 57-64; see B(n).
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).
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022
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LINKS
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Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E. Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See v_n in Lemma 7.9 p. 21.
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Jeremiah Bartz, Bruce Dearden, Joel Iiams, and Julia Peterson, Powers of Two as Sums of Two Balancing Numbers, Combinatorics, Graph Theory and Computing (SEICCGTC 2021) Springer Proc. Math. Stat., Vol 448, pp. 383-392. See p. 384.
Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, Fall 2005. Problem 1, (6 MB).
Brian Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.
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FORMULA
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G.f.: x / (1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation.
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) = sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).
a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000
a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
a(k)*a(2*n+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(2*n+k-j) - Sum_{i = 0..j-1} a(2*n-(2*i+1)) = a(n+k)^2 - a(n)^2. - Charlie Marion, Jan 18 2004
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End)
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Apr 21 2004
a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004
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) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003
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
a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - Paul Barry, Apr 20 2009
Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
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
a(n) = (a(n-1)^2 - a(n-3)^2)/a(n-2) + a(n-4) for n > 3. - Patrick J. McNab, Jul 24 2015
Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016
a(n) = (a(n-3) + a(n+3))/198.
a(n) = Sum_{i=1..n} A001653(i), n>=1.
a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - Federico Provvedi, Feb 01 2021
(End)
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023
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EXAMPLE
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G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
6 is in the sequence since 6^2 = 36 is a triangular number: 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. - Michael B. Porter, Jul 02 2016
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MAPLE
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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
with (combinat):seq(fibonacci(2*n, 2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008
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MATHEMATICA
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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 *)
TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)
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PROG
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(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
(Sage) [chebyshev_U(n-1, 3) for n in (0..20)] # G. C. Greubel, Dec 23 2019
(Haskell)
a001109 n = a001109_list !! n :: Integer
a001109_list = 0 : 1 : zipWith (-)
(map (* 6) $ tail a001109_list) a001109_list
(Magma) [n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015
(GAP) a:=[0, 1];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Dec 18 2018
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CROSSREFS
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Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), this sequence (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
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KEYWORD
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nonn,easy,nice,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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