A029546 -id:A029546 - cp's OEIS frontend

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

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

Offset: 0

N. J. A. Sloane

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+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_{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
One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006
Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - 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 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 aMohamed 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
a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - 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
From Richard R. Forberg, Aug 30 2013: (Start)
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. (End)
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

			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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Index entries for sequences related to Chebyshev polynomials.
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Cf. A000217, A000290, A001108, A001542, A001653, A001850, A002315, A002965, A278310.
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).
Cf. A323182.

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

a001109 n = a001109_list !! n :: Integer
a001109_list = 0 : 1 : zipWith (-)
   (map (* 6) $ tail a001109_list) a001109_list
-- Reinhard Zumkeller, Dec 17 2011

[n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015

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

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 *)
Table[Fibonacci[2n, 2]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)

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

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

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

is(n)=ispolygonal(n^2,3) \\ Charles R Greathouse IV, Nov 03 2016

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

[chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019

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(A001110(n)).
a(n) = A001542(n)/2.
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) = 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
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 05 2002
a(n) = ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) / (4*sqrt(2)). - Gregory V. Richardson, Oct 13 2002. Corrected for offset 0, and rewritten. - Wolfdieter Lang, Feb 10 2015
a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
a(n) = floor((3+2*sqrt(2))^n/(4*sqrt(2))). - Lekraj Beedassy, Apr 23 2003
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(2*n) = A001653(n)^2 - A001653(n-1)^2. - Charlie Marion, Jul 16 2003
For n > 0, a(n) = Sum_{k = 0..n-1}((2*k+1)*A001652(n-1-k)) + A000217(n). - Charlie Marion, Jul 18 2003
a(2*n+1) = a(n+1)^2 - a(n)^2. - Charlie Marion, Jan 12 2004
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
From Paul Barry, Feb 06 2004: (Start)
a(n) = A000129(2*n)/2;
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
A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement, Sep 16 2004
a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004
a(n) = A001653(n+1) - A038723(n); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement, Nov 19 2004, modified by Davide Colazingari, Jun 24 2016
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) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003
a(n) = 5*(a(n-1) + a(n-2)) - a(n-3). - Mohamed Bouhamida, 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..n, a(2*n-k) - a(k) = 2*a(n-k)*A001541(n). Also, a(2*n+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*binomial(n+k,2*k+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..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+2*k) = A003499(k)*a(n+k); a(n) + a(n+2*k+1) = A001653(k+1)*A002315(n+k). - Charlie Marion, Nov 29 2012
From Peter Bala, Dec 23 2012: (Start)
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
a(n-k)*a(n+k) + a(k)^2 = a(n)^2, a(n+k) + a(n-k) = A003499(k)*a(n), for n >= k >= 0. - Alexander Samokrutov, Sep 30 2015
Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016
4*a(n)^2 - 1 = A278310(n) for n > 0. - Bruno Berselli, Nov 24 2016
From Klaus Purath, Jan 18 2020: (Start)
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) = A002965(2*n)*A002965(2*n+1). - Jon E. Schoenfield, Jan 08 2022
a(n) = A002965(4*n)/2. - Gerry Martens, Jul 14 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023

Additional comments from Wolfdieter Lang, Feb 10 2000

Duplication of a formula removed by Wolfdieter Lang, Feb 10 2015

A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

Original entry on oeis.org

0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550, 16863297918705209269200, 572855720093639278238256

Offset: 0

Don N. Page

Triangular numbers that are twice other triangular numbers. - Don N. Page
Triangular numbers that are also pronic numbers. These will be shown to have a Pythagorean connection in a paper in preparation. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Mar 09 2002
In other words, triangular numbers which are products of two consecutive numbers. E.g., a(2) = 210: 210 is a triangular number which is the product of two consecutive numbers: 14 * 15. - Shyam Sunder Gupta, Oct 26 2002
Coefficients of the series giving the best rational approximations to sqrt(8). The partial sums of the series 3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(8) = 2 sqrt(2), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2; 1, 4, 1], [2; 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1, 4, 1] and so forth. - Gene Ward Smith, Sep 30 2006
This sequence satisfy the same recurrence as A165518. - Ant King, Dec 13 2010
Intersection of A000217 and A002378.
This is the sequence of areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + 1,z(n)) with x(0) = 0, y(0) = 1, z(0) = 1, a(0) = 0 and x(1) = 3, y(1) = 4, z(1) = 5, a(1) = 6. - George F. Johnson, Aug 20 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..100
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Shyam Sunder Gupta Fascinating Triangular Numbers
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A123478, A123479, A123480, A123482, A075528, A082405 (first differences).
Cf. A000129, A001108, A002203, A009111, A245031.
Cf. A000217, A001109, A001652, A002315, A002378, A011900, A029546, A046090, A046729, A053141, A084159, A096979, A157259, A165518.

List([0..20], n-> (Lucas(2,-1, 4*n+2)[2] -6)/32 ); # G. C. Greubel, Jan 13 2020

a029549 n = a029549_list !! n
a029549_list = [0,6,210] ++
   zipWith (+) a029549_list
               (map (* 35) $ tail delta)
   where delta = zipWith (-) (tail a029549_list) a029549_list
-- Reinhard Zumkeller, Sep 19 2011

(makelist(binom(n,2),n,1,999999),intersection(%%,2*%%)) /* Bill Gosper, Feb 07 2010 */

R:=PowerSeriesRing(Integers(), 25); [0] cat Coefficients(R!(6/(1-35*x+35*x^2-x^3))); // G. C. Greubel, Jul 15 2018

A029549 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,6]) ;
    else
        34*procname(n-1)-procname(n-2)+6 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

Table[Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Original program from author, corrected by Ray Chandler, Jul 09 2015 *)
CoefficientList[Series[6/(1 - 35x + 35x^2 - x^3), {x, 0, 14}], x]
Intersection[#, 2#] &@ Table[Binomial[n, 2], {n, 999999}] (* Bill Gosper, Feb 07 2010 *)
LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* Harvey P. Dale, Jun 06 2011 *)
(LucasL[4Range[20] - 2, 2] -6)/32 (* G. C. Greubel, Jan 13 2020 *)

concat(0,Vec(6/(1-35*x+35*x^2-x^3)+O(x^25))) \\ Charles R Greathouse IV, Jun 13 2013

[(lucas_number2(4*n+2, 2, -1) -6)/32 for n in (0..20)] # G. C. Greubel, Jan 13 2020

val triNums = (0 to 39999).map(n => (n * n + n)/2)
triNums.filter( % 2 == 0).filter(n => (triNums.contains(n/2))) // _Alonso del Arte, Jan 12 2020

G.f.: 6*x/(1 - 35*x + 35*x^2 - x^3) = 6*x /( (1-x)*(1 - 34*x + x^2) ).
a(n) = 6*A029546(n-1) = 2*A075528(n).
a(n) = -3/16 + ((3+2*sqrt(2))/32) *(17 + 12*sqrt(2))^n + ((3-2*sqrt(2))/32) *(17 - 12*sqrt(2))^n. - Gene Ward Smith, Sep 30 2006
From Bill Gosper, Feb 07 2010: (Start)
a(n) = (cosh((4*n + 2)*log(1 + sqrt(2))) - 3)/16.
a(n) = binomial(A001652(n) + 1, 2) = 2*binomial(A053141(n) + 1, 2). (End)
a(n) = binomial(A046090(n), 2) = A000217(A001652(n)). - Mitch Harris, Apr 19 2007, R. J. Mathar, Jun 26 2009
a(n) = ceiling((3 + 2*sqrt(2))^(2n + 1) - 6)/32 = floor((1/32) (1+sqrt(2))^(4n+2)). - Ant King, Dec 13 2010
Sum_{n >= 1} 1/a(n) = 3 - 2*sqrt(2) = A157259 - 4. - Ant King, Dec 13 2010
a(n) = a(n - 1) + A001109(2n). - Charlie Marion, Feb 10 2011
a(n+2) = 34*a(n + 1) - a(n) + 6. - Charlie Marion, Feb 11 2011
From George F. Johnson, Aug 20 2012: (Start)
a(n) = ((3 + 2*sqrt(2))^(2*n + 1) + (3 - 2*sqrt(2))^(2*n + 1) - 6)/32.
8*a(n) + 1 = (A002315(n))^2, 4*a(n) + 1 = (A000129(2*n + 1))^2, 32*a(n)^2 + 12*a(n) + 1 are perfect squares.
a(n + 1) = 17*a(n) + 3 + 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1) = 17*a(n) + 3 - 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1)*a(n + 1) = a(n)*(a(n) - 6), a(n) = A096979(2*n).
a(n) = (1/2)*A084159(n)*A046729(n) = (1/2)*A001652(n)*A046090(n).
Limit_{n->infinity} a(n)/a(n - 1) = 17 + 12*sqrt(2).
Limit_{n->infinity} a(n)/a(n - 2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Limit_{n->infinity} a(n)/a(n - r) = (17 + 12*sqrt(2))^r.
Limit_{n->infinity} a(n - r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r. (End)
a(n) = 3 * T( b(n) ) + (2*b(n) + 1)*sqrt( T( b(n) ) ) where b(n) = A001108(n) (indices of the square triangular numbers), T(n) = A000217(n) (the n-th triangular number). - Dimitri Papadopoulos, Jul 07 2017
a(n) = (Pell(2*n + 1)^2 - 1)/4 = (Q(4*n + 2) - 6)/32, where Q(n) are the Pell-Lucas numbers (A002203). - G. C. Greubel, Jan 13 2020
a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - Pontus von Brömssen, Sep 11 2024

Additional comments from Christian G. Bower, Sep 19 2002; T. D. Noe, Nov 07 2006; and others

Edited by N. J. A. Sloane, Apr 18 2007, following suggestions from Andrew S. Plewe and Tanya Khovanova

A075528 Triangular numbers that are half other triangular numbers.

Original entry on oeis.org

0, 3, 105, 3570, 121278, 4119885, 139954815, 4754343828, 161507735340, 5486508657735, 186379786627653, 6331426236682470, 215082112260576330, 7306460390622912753, 248204571168918457275, 8431648959352604634600, 286427860046819639119128

Offset: 0

Christian G. Bower, Sep 19 2002

This is the sequence of 1/2 the areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n)=x(n)+1, z(n)) with x(0)=0, y(0)=1, z(0)=1, a(0)=0 and x(1)=3, y(1)=4, z(1)=5, a(1)=3. - George F. Johnson, Aug 24 2012

Colin Barker, Table of n, a(n) for n = 0..653
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A000217, A001652, A002315, A029546, A029549, A046090, A046729, A053141, A084159, A096979.

CoefficientList[ Series[ 3x/(1 - 35 x + 35 x^2 - x^3), {x, 0, 15}], x] (* Robert G. Wilson v, Jun 24 2011 *)

concat(0, Vec(3*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 18 2015

a(n) = 3*A029546(n-1) = A029549(n)/2.
G.f.: 3*x/((1-x)*(1-34*x+x^2)).
From George F. Johnson, Aug 24 2012: (Start)
a(n) = ((3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1) - 6)/64.
8*a(n)+1 = A000129(2*n+1)^2.
16*a(n)+1 = A002315(n)^2.
128*a(n)^2 + 24*a(n) + 1 is a perfect square.
a(n+1) = 17*a(n) + 3/2 + 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1) = 17*a(n) + 3/2 - 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1)*a(n+1) = a(n)*(a(n)-3); a(n+1) = 34*a(n) - a(n-1) + 3.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2); a(n) = A096979(2*n)/2.
a(n) = A084159(n)*A046729(n)/4 = A001652(n)*A046090(n)/4.
Lim_{n->infinity} a(n)/a(n-1) =  17 + 12*sqrt(2).
Lim_{n->infinity} a(n)/a(n-2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Lim_{n->infinity} a(n)/a(n-r) = (17 + 12*sqrt(2))^r.
Lim_{n->infinity} a(n-r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r.
(End)
a(n) = 34*a(n-1) - a(n-2) + 3, n >= 2. - R. J. Mathar, Nov 07 2015
a(n) = A000217(A053141(n)). - R. J. Mathar, Aug 16 2019
a(n) = (a(n-1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
Sum_{n>=1} 1/a(n) = 2*(3 - 2*sqrt(2)). - Amiram Eldar, Dec 04 2024

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

Original entry on oeis.org

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024

Offset: 1

Joseph L. Pe, Jan 07 2003

Equivalently, both k+1 and 2*k+1 are perfect squares.
The square roots of (k+1)*(2*k+1) are in A046176.
Also numbers k such that 3*A000217(k) + A000217(k+1) is a perfect square. - Bruno Berselli, Nov 17 2016
From Sergey Pavlov, Mar 14 2017: (Start)
The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).
Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section).
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.

Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020

I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019

seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020

RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)

concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016

def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

From Benoit Cloitre, Jan 19 2003: (Start)
a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.
a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).
a(n) = A008844(n) - 1. (End)
From R. J. Mathar, Sep 21 2011: (Start)
G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).
a(n) = 24*A029546(n-2). (End)
a(n) = (A001653(n)^2 - 1)/2 = A002315(n-1)^2 - 1. - Tomohiro Yamada, Sep 15 2019
a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020
From Amiram Eldar, Dec 02 2024: (Start)
a(n) = A001542(n-1)*A001542(n).
Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)

Edited by Bruno Berselli, Nov 17 2016

A245031 Numbers m such that 3m+1 and 8m+1 are both squares.

Original entry on oeis.org

0, 1, 21, 120, 2080, 11781, 203841, 1154440, 19974360, 113123361, 1957283461, 11084934960, 191793804840, 1086210502741, 18793835590881, 106437544333680, 1841604094101520, 10429793134197921, 180458407386358101, 1022013289607062600

Offset: 1

Bruno Berselli, Jul 15 2014

Naturally, all terms are triangular numbers.
Numbers m such that k*m+1 and 8*m+1 are both squares:
k=1: A006454;
k=3: this sequence;
k=4: A029549;
k=5: 0, 3, 231, 4560, 333336, 6575751, ...
k=6: A200999;
k=7: A157879.
Numbers m such that 3*m+1 and k*m+1 are both squares:
k=1: A045899;
k=2: A045502;
k=4: A059989;
k=5: A159683;
k=6: 8*A029546;
k=7: A160695;
k=8: this sequence.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000217.

Cf. A006454, A029546, A029549, A045502, A045899, A059989, A157879, A159683, A160695, A200999.

I:=[0,1,21,120,2080]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..20]];

LinearRecurrence[{1, 98, -98, -1, 1}, {0, 1, 21, 120, 2080}, 20] (* or *) CoefficientList[Series[x (1 + 20 x + x^2)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 20}], x]

a[1]:0$ a[2]:1$ a[3]:21$ a[4]:120$ a[5]:2080$ a[n]:=a[n-1]+98*a[n-2]-98*a[n-3]-a[n-4]+a[n-5]$ makelist(a[n], n, 1, 20);

a=vector(20); a[1]=0; a[2]=1; a[3]=21; a[4]=120; a[5]=2080; for(i=6, #a, a[i]=a[i-1]+98*a[i-2]-98*a[i-3]-a[i-4]+a[i-5]); a

G.f.: x^2*(1 + 20*x + x^2)/((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)).
a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
G.f. of the quadrisections:
a(4k+1):   40*x*(52 + 3*x)/((1 - x)*(1 - 9602*x + x^2));
a(4k+2): (1 + 2178*x + 21*x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+3): (21 + 2178*x + x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+4): 40*(3 + 52*x)/((1 - x)*(1 - 9602*x + x^2)).

Changed offset from 0 to 1 and adapted formulas by Bruno Berselli, Mar 03 2016

A278310 Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.

Original entry on oeis.org

3, 143, 4899, 166463, 5654883, 192099599, 6525731523, 221682772223, 7530688524099, 255821727047183, 8690408031080163, 295218051329678399, 10028723337177985443, 340681375412721826703, 11573138040695364122499, 393146012008229658338303, 13355391270239113019379843

Offset: 1

Bruno Berselli, Nov 17 2016

Equivalently, both m+1 and 2*m+3 are squares for nonnegative m.
Corresponding triangular numbers T(m): 6, 10296, 12002550, 13855048416, 15988853699286, 18451128064030200, 21292585958400815526, ...
Square roots of T(m) + 3*T(m+1) are listed by A082405 (after 0).
Negative values of m for which T(m) + 3*T(m+1) is a square: -1, -2, -26, -842, -28562, -970226, -32959082, ...

			3 is in the sequence because T(3) + 3*T(4) = 6 + 3*10 = 6^2.
For n=5 is a(5) = 5654883, therefore floor(sqrt(5654883)) = 2377 = A182189(5) - 2 = 2379 - 2.

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Subsequence of A000466.
Cf. A000217, A001109, A029546, A046176, A082405, A156164, A182189.
Cf. A278438: numbers m such that T(m) + 2*T(m+1) is a square.
Cf. A078522: numbers m such that 3*T(m) + T(m+1) is a square.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: A084703 (k=-1), A076218 (k=3), this sequence (k=-5).

Iv:=[3,143]; [n le 2 select Iv[n] else 34*Self(n-1)-Self(n-2)+40: n in [1..20]];

P:=proc(q) local n; for n from 3 to q do if type(sqrt(2*n^2+5*n+3),integer) then print(n); fi; od; end: P(10^9); # Paolo P. Lava, Nov 18 2016

Table[((1 + Sqrt[2])^(4 n) + (1 - Sqrt[2])^(4 n))/8 - 5/4, {n, 1, 20}]
RecurrenceTable[{a[1] == 3, a[2] == 143, a[n] == 34 a[n - 1] - a[n - 2] + 40}, a, {n, 1, 20}]
LinearRecurrence[{35, -35, 1}, {3, 143, 4899}, 50] (* G. C. Greubel, Nov 20 2016 *)

Vec(x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)) + O(x^50)) \\ G. C. Greubel, Nov 20 2016

def A278310():
    a, b = 3, 143
    yield a
    while True:
        yield b
        a, b = b, 34*b - a + 40
a = A278310(); print([next(a) for  in range(18)]) # _Peter Luschny, Nov 18 2016

O.g.f.: x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)).
E.g.f.: (exp((1-sqrt(2))^4*x) + exp((1+sqrt(2))^4*x) - 10*exp(x))/8 + 1.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3.
a(n) = 34*a(n-1) - a(n-2) + 40 for n>2.
a(n) = a(-n) = ((1 + sqrt(2))^(4*n) + (1 - sqrt(2))^(4*n))/8 - 5/4.
a(n) = 4*A001109(n)^2 - 1.
a(n) = -A029546(n) + 38*A029546(n-1) + 3*A029546(n-2) for n>1.
Lim_{n -> infinity} a(n)/a(n-1) = A156164.
Floor(sqrt(a(n))) = A182189(n) - 2.
a(n) - a(n-1) = 4*A046176(n) for n>1.

A220185 Numbers n such that n^2 + n(n+1) is an oblong number (A002378).

Original entry on oeis.org

0, 10, 348, 11830, 401880, 13652098, 463769460, 15754509550, 535189555248, 18180690368890, 617608282987020, 20980500931189798, 712719423377466120, 24211479893902658290, 822477596969312915748, 27940026817062736477150, 949138434183163727307360

Offset: 1

Alex Ratushnyak, Apr 12 2013

Also numbers n such that the sum of the hexagonal numbers H(n) and H(n+1) is equal to m^2 + (m+1)^2 for some m. - Colin Barker, Dec 10 2014

Also nonnegative integers x in the solutions to 4*x^2-2*y^2+2*x-2*y = 0, the corresponding values of y being A251867. - Colin Barker, Dec 10 2014

			a(3) = A089928(7) = 348.

Colin Barker, Table of n, a(n) for n = 1..654
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A002378, A014105 (n^2 + n(n+1)), A029546, A084703 (numbers n such that n^2 + n(n+1) is a square).

Cf. A000290, A000384, A251867.

#include 
#include 
typedef unsigned long long U64;
U64 rootPronic(U64 a) {
    U64 sr = 1L<<31, s, b;
    if (a < sr*(sr+1)) {
          sr>>=1;
          while (a < sr*(sr+1))  sr>>=1;
    }
    for (b = sr>>1; b; b>>=1) {
            s = sr+b;
            if (a >= s*(s+1))  sr = s;
    }
    return sr;
}
int main() {
  U64 a, n, r, t;
  for (n=0; n < (1L<<31); n++) {
    a = (n*(n+1)) + n*n;
    t = rootPronic(a);
    if (a == t*(t+1))  printf("%llu\n", n);
  }
}

[Floor(((1+Sqrt(2))^(4*n-3)+(1-Sqrt(2))^(4*n-3)-2)/8): n in [1..20]]; // Vincenzo Librandi, Sep 08 2015

f:= gfun:-rectoproc({a(n)=35*(a(n-1)-a(n-2))+a(n-3),a(1)=0,a(2)=10,a(3)=348},a(n),remember):
map(f, [$1..50]); # Robert Israel, Sep 06 2015

LinearRecurrence[{35, - 35, 1}, {0, 10, 348}, 20] (* Vincenzo Librandi, Sep 06 2015 *)

concat(0, Vec(2*x^2*(5-x)/((1-x)*(1-34*x+x^2))+O(x^100))) \\ Colin Barker, Dec 10 2014

For n>1, a(n) = A089928(n*4-5).
From Bruno Berselli, Apr 12 2013: (Start)
G.f.: 2*x^2*(5-x)/((1-x)*(1-34*x+x^2)).
a(n) = ((1+sqrt(2))^(4n-3)+(1-sqrt(2))^(4n-3)-2)/8.
a(n+2) = 10*A029546(n)-2*A029546(n-1). (End)
a(n) = 35*a(n-1)-35*a(n-2)+a(n-3). - Colin Barker, Dec 10 2014
a(n) = A251867(n) - A001542(n-1)^2. - Alexander Samokrutov, Sep 05 2015

More terms from Bruno Berselli, Apr 12 2013

A305539 a(n) is a generalized pentagonal number such that 2*a(n) is also a generalized pentagonal number.

Original entry on oeis.org

0, 1, 35, 1190, 40426, 1373295, 46651605, 1584781276, 53835911780, 1828836219245, 62126595542551, 2110475412227490, 71694037420192110, 2435486796874304251, 82734857056306152425, 2810549653117534878200, 95475953348939879706376, 3243371864210838375138585

Offset: 1

Peter Luschny, Jun 04 2018

Enge, Hart and Johansson prove that every generalized pentagonal number c >= 5 is the sum of a smaller one and twice a smaller one, that is, there are generalized pentagonal numbers a, b < c such that c = 2a + b (see link, theorem 5). We look here at those c >= 0 which have b = 0. A305538 lists the smallest b > 0 for a given c.

			For n = 56 and k = -80 there is k*(3*k + 2) - 6*n^2 - 4*n = 0, hence A001318(56) = 1190 is in this sequence. And indeed also 2380 is a generalized pentagonal number, A001318(79).

G. C. Greubel, Table of n, a(n) for n = 1..650
Andreas Enge, William Hart and Fredrik Johansson, Short addition sequences for theta functions, Journal of Integer Sequences, Vol. 21 (2018), Article 18.2.4. Also available as arXiv:1608.06810 [math.NT], 2016-2018.
Simon Plouffe Conjectures of the OEIS, as of June 20, 2018.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Essentially A029546.

Cf. A001318, A305538.

I:=[0,1,35]; [n le 3 select I[n] else 35*Self(n-1) -35*Self(n-2) +Self(n-3): n in [1..41]]; // G. C. Greubel, Jun 05 2023

a := proc(searchlimit) local L, g, n, s; L := NULL;
g := n -> ((6*n^2+6*n+1)-(2*n+1)*(-1)^n)/16;
for n from 0 to searchlimit do
    s := isolve(k*(3*k+2)-6*n^2-4*n = irem(n,2)*(4*n+2));
    if s <> NULL then L:=L,g(n); fi
od: L end:
a(12000);

LinearRecurrence[{35,-35,1}, {0,1,35}, 18] (* Jean-François Alcover, Jul 14 2019, after A029546 *)
(Fibonacci[2*Range[40]-1,2]^2 -1)/24 (* G. C. Greubel, Jun 05 2023 *)

[(lucas_number1(2*n-1,2,-1)^2 -1)/24 for n in range(1,41)] # G. C. Greubel, Jun 05 2023

If for given n there is an integer k such that k*(3*k + 2) - 6*n^2 - 4*n = (n mod 2)*(4*n + 2) then A001318(n) is in this sequence.
G.f.: x^2/(1 - 35*x + 35*x^2 -x^3). - Simon Plouffe, Jun 20 2018
a(n) = (Pell(2*n-1)^2 - 1)/24, n > 0. - G. C. Greubel, Jun 05 2023

A008845 Numbers k such that k+1 and k/2+1 are squares.

Original entry on oeis.org

0, 48, 1680, 57120, 1940448, 65918160, 2239277040, 76069501248, 2584123765440, 87784138523760, 2982076586042448, 101302819786919520, 3441313796169221280, 116903366249966604048, 3971273138702695316400, 134906383349641674153600, 4582845760749114225906048

Offset: 0

N. J. A. Sloane

			48+1 = 49 = 7^2 and 48/2+1 = 24+1 = 25 = 5^2.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Colin Barker, Table of n, a(n) for n = 0..650
Henry Ernest Dudeney, Amusements in Mathematics, 1917. See problem 114, "Curious numbers".
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

a:=[0,48,1680];; for n in [4..20] do a[n]:=35*a[n-1]-35*a[n-2] +a[n-3]; od; a; # G. C. Greubel, Sep 13 2019

I:=[0,48]; [n le 2 select I[n] else  34*Self(n-1) - Self(n-2)+48: n in [1..20]]; // Vincenzo Librandi, Mar 03 2016

seq(coeff(series(48*x/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 0..20); # G. C. Greubel, Sep 13 2019

LinearRecurrence[{35,-35,1},{0,48,1680},20] (* Harvey P. Dale, May 24 2014 *)

concat(0, Vec(48*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016

def A008845_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(48*x/((1-x)*(1-34*x+x^2))).list()
A008845_list(20) # G. C. Greubel, Sep 13 2019

a(n) = 2*(A008844(n)-1) = 16*A075528(n) = 48*A029546(n). - corrected by Sean A. Irvine, Apr 07 2018
a(0)=0, a(1)=48, a(2)=1680, a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). - Harvey P. Dale, May 24 2014
From Colin Barker, Mar 02 2016: (Start)
a(n) = (-6+(3-2*sqrt(2))*(17+12*sqrt(2))^(-n)+(3+2*sqrt(2))*(17+12*sqrt(2))^n)/4.
G.f.: 48*x / ((1-x)*(1-34*x+x^2)).
(End)
a(n) = 34*a(n-1) - a(n-2) + 48. - Vincenzo Librandi, Mar 03 2016

A247215 Integers k such that 3k+1 and 6k+1 are both squares.

Original entry on oeis.org

0, 8, 280, 9520, 323408, 10986360, 373212840, 12678250208, 430687294240, 14630689753960, 497012764340408, 16883803297819920, 573552299361536880, 19483894374994434008, 661878856450449219400, 22484397224940279025600, 763807626791519037651008

Offset: 1

Casey Leung, Nov 26 2014

			When n=1, a(1)=0, 3(0)+1=1, 6(0)+1=1.
When n=2, a(2)=8, 3(8)+1=25, 6(8)+1=49.
When n=3, a(3)=280, 3(280)+1=841=29^2, 6(280)+1=1681=41^2.
When n=4, a(4)=9520, 3(9520)+1=28560=169^2, 6(9520)+1=57121=239^2.

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

The common terms of A062717 and A001082.

LinearRecurrence[{35,-35,1},{0,8,280},20] (* Harvey P. Dale, Mar 25 2025 *)

concat(0, Vec(-8*x^2/((x-1)*(x^2-34*x+1)) + O(x^100))) \\ Colin Barker, Nov 26 2014

a(n) = (1/72)*(3*(3*(17-12*sqrt(2))^n+2*sqrt(2)*(17-12*sqrt(2))^n+3*(17+12*sqrt(2))^n-2*sqrt(2)*(17+12*sqrt(2))^n)-18).
From Colin Barker, Nov 26 2014: (Start)
a(n) = 8*A029546(n).
a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).
G.f.: -8*x^2 / ((x-1)*(x^2-34*x+1)).
(End)
Lim_{n -> infinity} a(n+1)/a(n) = 33.970562748... = (1+sqrt(2))^4 (the dominant root of x^2-34*x+1). - Joerg Arndt, Dec 01 2014

More terms from Colin Barker, Nov 26 2014

cp's OEIS Frontend

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

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A029549 a(n + 3) = 35*a(n + 2) - 35*a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

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A075528 Triangular numbers that are half other triangular numbers.

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A078522 Numbers k such that (k+1)*(2*k+1) is a perfect square.

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A245031 Numbers m such that 3*m+1 and 8*m+1 are both squares.

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A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

A245031 Numbers m such that 3m+1 and 8m+1 are both squares.