cp's OEIS Frontend

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

Conjecture: A175181(n)/A214027(n) = a(n). This says that the zeros appear somewhat uniformly in a period. The second zero in a period is exactly where n divides the first Lucas number, so this relationship is not really surprising.

From Jianing Song, Aug 29 2018: (Start)

The comment above is correct, since n divides A000129(k*a(n)) for all integers k and clearly a(n) divides A175181(n), so the zeros appear uniformly.

a(n) <= 4*n/3 for all n, where the equality holds iff n is a power of 3.

(End)

The condition for an odd prime p to be a member of this sequence is that p^2 divides A001353(p - (3/p)).

Neither this quotient, nor the Lucas sequence U(4, 1) on which it is based, has a common name; but its fundamental discriminant of 3 places it between the quotient based on the Pell sequence U(2, -1) with discriminant 2 (A000129), and that based on the Fibonacci sequence U(1, -1) with discriminant 5 (A000045). Values of p dividing the Pell quotient will be found under A238736, while for the Fibonacci quotient it is known that there is no such p < 9.7*10^14.

The interest in this family of number-theoretic quotients derives from H. C. Williams, "Some formulas concerning the fundamental unit of a real quadratic field," p. 440, which proves a formula connecting the present quotient with the Fermat quotient base 2 (A007663), the Fermat quotient base 3 (A146211), and the harmonic number H(floor(p/12)) (see the Formula section below). As is well known, the vanishing of each of these Fermat quotients is a necessary condition for the failure of the first case of Fermat's Last Theorem (see discussions under A001220 and A014127); and a corresponding result concerning this type of harmonic number was proved by Dilcher and Skula. Thus, the vanishing mod p of the quotient based on U(4, 1) is also a necessary condition for the failure of the first case of Fermat's Last Theorem.

The pioneering computation for this quotient appears to be that of Elsenhans and Jahnel, "The Fibonacci sequence modulo p^2," p. 5, who report 103 as the only value of a(n) < 10^9. Extending the search to p < 2.5*10^10 has found only one further solution, 2297860813.

Let LucasQuotient(p) = A001353(p - (3/p))/p, q_2 = (2^(p-1) - 1)/p = A007663(p) be the corresponding Fermat quotient of base 2, q_3 = (3^(p-1) - 1)/p = A146211(p) be the corresponding Fermat quotient of base 3, H(floor(p/12)) be a harmonic number. Then Williams (1991) shows that 6*(3/p)*LucasQuotient(p) == -6*q_2 - 3*q_3 - 2*H(floor(p/12)) (mod p).

Also with an initial 2, primes p such that p^2 divides A001353(p - Kronecker(12,p)) (note that 12 is the discriminant of the characteristic polynomial of A001353, x^2 - 4x + 1). - Jianing Song, Jul 28 2018

a(n) is squarefree unless n is of the form A214028(A238736(k)) = {7, 30, 1546462, ...}. The terms in A238736 are called 2-Wall-Sun-Sun primes.

p always divides A316269(k, p-Kronecker(k^2-4, p)), so it's interesting to see when p^2 also divides A316269(k, p-Kronecker(k^2-4, p)).

In the following comments, let p = prime(n). Note that A316269(0, m) and A316269(1, m) is not defined, so here k must be understood as a remainder modulo p^2. because A316269(k+s*p^2, m) == A316269(k, m) (mod p^2).

Let p = prime(n). Every row contains 0, 1 and p^2 - 1. For n >= 3, the n-th row contains p - 2 numbers, whose remainders modulo p form a permutation of {0, 1, 3, 4, ..., p - 3, p - 1}.

Every row is antisymmetric, that is, k is a member iff p^2 - k is, k > 0. As a result, the sum of the n-th row is prime(n)^2*(prime(n) - 3)/2.

Equivalently, for n >= 2, row n lists 0 <= k < p^2 such that p^2 divides A316269(k, (p-Kronecker(k^2-4, p))/2), p = prime(n).

p always divides A172236(k, p-Kronecker(k^2+4, p)), so it's interesting to see when p^2 also divides A172236(k, p-Kronecker(k^2+4, p)). If this holds, then p is called a k-Wall-Sun-Sun prime (and thus being a (k + s*p^2)-Wall-Sun-Sun prime for all integer s). Specially, there is no Wall-Sun-Sun prime below 10^14 implies that there is no 1 in the first pi(10^14) rows.

Note that A172236(0, m) is not defined, so here k must be understood as a remainder modulo p^2. because A172236(k+s*p^2, m) == A172236(k, m) (mod p^2).

Let p = prime(n). Every row contains 0. For n >= 2, if p == 3 (mod 4), then the n-th row contains p numbers, whose remainders modulo p form a permutation of {0, 1, 2, 3, ..., p - 2, p - 1}. If p == 1 (mod 4), then the n-th row contains p - 2 numbers, whose remainders modulo p form a permutation of {0, 1, 2, 3, ..., p - 2, p - 1} \ {+-2*((p - 1)/2)! mod p}.

Every row is antisymmetric, that is, k is a member iff p^2 - k is, k > 0. As a result, the sum of the n-th row is prime(n)^2*(prime(n) - 1)/2 if prime(n) == 3 (mod 4) and prime(n)^2*(prime(n) - 3)/2 if prime(n) == 1 (mod 4).

Equivalently, if p = prime(n) == 1 (mod 4), then row n lists 0 <= k < p^2 such that p^2 divides A172236(k, (p-Kronecker(k^2+4, p))/2). - Jianing Song, Jul 06 2019