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A floretion-generated sequence relating NSW and Pell numbers.

Elements of odd index in the sequence gives A002315. a(n+2) - a(n) = A002203(n+2).

Floretion Algebra Multiplication Program, FAMP Code: 2tesseq[B*C} with B = - .25'i + .25'j + .5'k - .25i' + .25j' + .5k' - .5'kk' - .25'ik' - .25'jk' - .25'ki' - .25'kj' - .5e and C = + .5'i - .25'j + .25'k + .5i' - .25j' + .25k' - .5'ii' - .25'ij' - .25'ik' - .25'ji' - .25'ki' - .5e

This is the row reversed version of A180870. See also A157751 and A228565.

The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.

This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.

The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)

The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by

[1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)

The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.

The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.

The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

Bisection (even part) of Chebyshev sequence with Diophantine property.

b(n)^2 - 8*a(n)^2 = 17, with the companion sequence b(n)= A077240(n).

The odd part is A077413(n) with Diophantine companion A077239(n).

a(p)=0 if p is an odd prime. If n is an odd composite number, then a(n) is a square; see A001110 for numbers that are both triangular and square. - Victoria A Sapko (vsapko(AT)frc.mass.edu), Sep 28 2007

From Jon E. Schoenfield, May 25 2014: (Start)

If n is an odd semiprime, then a triangular number t having exactly n divisors must be of the form t = p^(2r) * q^(2s) = (p^r * q^s)^2, where p and q are distinct primes (p < q) and r and s are positive integers such that (2r+1)*(2s+1) = n.

If t is the k-th triangular number k(k+1)/2, it can be factored as t = u * v where

u = k and v = (k+1)/2 if k is odd, or

u = k/2 and v = k+1 if k is even.

Since neither p nor q (each of which is greater than 1) can divide both k and (k+1)/2, or both k/2 and k+1, only four cases need to be considered:

Case 1: k is even, q^(2s) = k/2, p^(2r) = k+1

Case 2: k is even, p^(2r) = k/2, q^(2s) = k+1

Case 3: k is odd, q^(2s) = k, p^(2r) = (k+1)/2

Case 4: k is odd, p^(2r) = k, q^(2s) = (k+1)/2

These yield the following equations:

Case 1: 2 * q^(2s) + 1 = p^(2r)

Case 2: 2 * p^(2r) + 1 = q^(2s)

Case 3: 2 * p^(2r) - 1 = q^(2s)

Case 4: 2 * q^(2s) - 1 = p^(2r)

Case 1 can be ruled out: since q > p, q is odd, so 2 * q^(2s) + 1 == 3 (mod 16), but p^(2r) cannot be 3 (mod 16).

For a Case 2 solution, since q is odd, 2 * p^(2r) + 1 = q^(2s) == 1 (mod 8), so p^(2r) == 0 (mod 4), so p must be even. Therefore, p = 2, and we must satisfy the equation 2 * 2^(2r) + 1 = q^(2s), whose left-hand side, which is divisible by 3 for every nonnegative integer r, is thus the square of a prime iff it is 3^2. So r=1, q=3, and s=1, yielding t = 2^2 * 3^2 = 36, which is the smallest triangular number with exactly 9 divisors, so a(9)=36.

In Case 3, both p and q must be odd, and p^r must be a number w having the property that 2*w^2 - 1 is square (i.e., (q^s)^2); every such number w is in A001653 (1, 5, 29, 169, 985, ...), and the corresponding value of q^s = sqrt(2*w^2 - 1) is in A002315 (1, 7, 41, 239, 1393, ...). (Note that A001653 and A002315 are defined with offsets of 1 and 0, respectively, so A001653(j) corresponds to A002315(j-1).) However, for odd semiprime n > 9, we need r > 1 and/or s > 1. The only nontrivial power (i.e., number of the form x^m, where both x and m are integers greater than 2) in A001653 is A001653(4) = 169 = 13^2 [Pethö], so the only Case 3 solution with r > 1 is 2 * 13^4 - 1 = 239^2, which yields the 15-divisor triangular number 13^4 * 239^2 = 1631432881 = a(15). A Case 3 solution with r = 1 and s > 1 would require 2 * p^2 - 1 = q^(2s), which is impossible since p < q.

Finally, in Case 4, both p and q must be odd, q^s must be in A001653, and p^r must be the corresponding term in A002315. However, using the only nontrivial power in A001653 (i.e., 169 = 13^2) as q^s would not yield a valid solution here because it would mean p = 239 and q = 13 (contradicting p < q). Thus, if a Case 4 solution exists for odd semiprime n > 9, we must have s = 1 and r > 1, so n = (2r+1)*(2s+1) = (2r+1)*3, where 2r+1 is prime. Such a solution requires an index j satisfying two conditions: (1) A001653(j) = q^1 = q is prime, and (2) the corresponding term A002315(j-1) = p^r is a nontrivial prime power. There are no nontrivial powers (whether of primes or composites) among the terms in A002315 below 10^5000. Moreover, the terms in A001653 are the odd-indexed terms from A000129 (Pell numbers), so condition (1) requires that j satisfy A000129(2j-1) = q. A096650 lists the indices of all prime or probable prime Pell numbers up to 100000. A check of the value A002315(j-1) corresponding to each prime or probable prime among the odd-indexed Pell numbers A000129(2j-1) up to j=50000 determined that none were nontrivial powers, so if any Case 4 solution with n > 9 exists, it will yield a triangular number t = p^(2r) * q^2 = (2 * q^2 - 1) * q^2, where q >= A000129(100001) = 3.16...*10^38277, so t > 10^153110. Since there are no nontrivial powers at all in A002315 below 10^5000, and since prime Pell numbers above A000129(50000) seem so scarce, it seems extremely unlikely that any such solution exists.

Thus, a(n) = 0 for every odd semiprime n that is not divisible by 3, and assuming that no Case 4 solution for odd semiprime n > 9 exists, the only nonzero a(n) where n is an odd semiprime greater than 9 is a(15) = 13^4 * 239^2 = 1631432881.

If j is prime and n=2j, then a(n) (if nonzero) must be of the form p^r * q, where p and q are distinct primes, r = j-1, and q is one of 3 functions of p^r:

q = f1(p^r) = 2*p^r - 1

q = f2(p^r) = 2*p^r + 1

q = f3(p^r) = (p^r - 1)/2

Of these, q = f1(p^r) for all but two cases among n < 1000:

at n=362, q = f2(p^r), with p=3;

at n=514, q = f3(p^r), with p=331.

Conjecture: a(2j) > 0 for all j > 1. (This conjecture holds at least through n = 2j = 1000. The largest a(n) for even n <= 1000 is a(898) = 20599^448 * (2 * 20599^448 - 1) = 3.21...*10^3865.) (End)

For more known terms, and information about unknown terms, see Links. - Jon E. Schoenfield, May 26 2014

If d(k*(k+1)/2) = 21, note that 2*q^2 = p^6 + 1 = (p^2 + 1)*(p^4 - p^2 + 1) has no prime solutions, so then k = p^2 and k+1 = 2*q^6, where p and q are distinct primes. We can prove 2*x^3 - y^2 = 1 has only one positive solution (1, 1) which shows that p^2 + 1 = 2*q^6 has no prime solutions. In the ring of Gaussian integers, x^3 = (1+y*i)*(1-y*i)/((1+i)*(1-i)) and (1+y*i)/(1+i) is coprime to (1-y*i)/(1-i), thus (1+y*i)/(1+i) = (u+v*i)^3 and (1-y*i)/(1-i) = (u-v*i)^3 for some integers u and v. Note that 1+y*i = (1+i)*(u+v*i)^3 = (u+v)*(u^2+v^2-4*u*v) + (u-v)*(u^2+v^2+4*u*v)*i, we have (u+v)*(u^2+v^2-4*u*v) = 1. Therefore, u = 1 and v = 0 if u > v, which means y = (u-v)*(u^2+v^2+4*u*v) = 1. This implies that a(21) = 0. - Jinyuan Wang, Aug 22 2020

a(n) is a perfect number for all n such that n/2 is in A000043. - J. Lowell, Mar 16 2024

From Herb Conn, Jan 28 2005: (Start)

"Letting x = 2 Cos 2A, we have the following trigonometric identities:

"Sin 3A = 3*Sin A - 4*Sin^3 A

"Sin 5A = 5*Sin A - 20*Sin^3 A + 16*Sin^5 A

"Sin 7A = 7*Sin A - 56*Sin^3 A + 112*Sin^5 A - 64*Sin^7 A

"Sin 9A = 9*Sin A - 120*Sin^3 A + 432*Sin^5 A - 576*Sin^7 A + 256*Sin^9 A, etc." (End)

Cayley (1876) states "Write sin u = x, then we have sin u = x, [...] sin 3u = 3x - 4x^3, [...] sin 5u = 5x - 20x^3 + 16 x^5, [...]". Since T_n(cos(u)) = cos(nu) for all integer n, sin(u) = cos(u - Pi/2), and sin(u + k*Pi) = (-1)^k sin(u) it follows that T_n(sin(u)) = (-1)^((n-1)/2) sin(nu) for all odd integer n. - Michael Somos, Jun 19 2012

From Wolfdieter Lang, Aug 05 2014: (Start)

The coefficient triangle t(n,k) for the row polynomials Todd(n, x) := T_{2*n+1}(sqrt(x))/sqrt(x) = sum(t(n,k)*x^k, k=0..n) is the Riordan triangle ((1-z)/(1+z)^2, 4*z/(1+z)^2) (rewrite the g.f. for the present triangle a(n,k) given in the formula section). The triangle entries t(n,k) = a(n,k), but the interpretation of the row polynomials is different for both cases.

From the relation Todd(n, x) = S(n, 2*(2*x-1)) - S(n-1, 2*(2*x-1)) with the Chebyshev S-polynomials (see A049310 and the formula section of A130777) follows the recurrence: Todd(n, x) = 2*(-1)^n*(1-x)*Todd(n-1, 1-x) + (2x-1)*Todd(n-1, x), n >= 1, Todd(0, x) = 1.

This corresponds to the triangle recurrence t(n,k) = (2*(k+1)*(-1)^(n-k) - 1)*t(n-1,k) + 2*(1 +(-1)^(n-k))*t(n-1,k-1) + 2*(-1)^(n-k)*sum(binomial(l+1,k)*t(n-1,l), l=k+1..n-1), n >= k >= 1, t(n,k) = 0 if n < k, t(n,0) = (-1)^n*(2*n+1). Compare this with the shorter recurrence involving the rational A-sequence for this Riordan triangle which has g.f. x^2/(2-x-2*sqrt(1-x)). t(n,k) = sum(A(j)*t(n-1,k-1+j), j=0..n-k), n >= k >= 1. The Z-sequence has g.f. -(1 + 2/sqrt(1-x)). For the A- and Z-sequence see a link under A006232. (End)

For each positive integer n, there exists a unique pair (j,k) of positive integers such that (j + k + 1)^2 - 4*k = 8*n^2; in fact, j(n) = A087056(n) and k(n) = A087059(n).

Suppose g >= 1 and let k = k(g). The numbers in row g of array D are among those n for which (j + k + 1)^2 - 4*k = 8*n^2 for some j; that is, k stays fixed and j and n vary - hence the name "fixed-k dispersion". (The fixed-j dispersion for Q = 8 is A120860.)

Every positive integer occurs exactly once in array D and every pair of rows are mutually interspersed. That is, beginning at the first term of any row having greater initial term than that of another row, all the following terms individually separate the individual terms of the other row.

Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007

Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009

As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011

Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - Colin Barker, Jan 16 2015

Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015

Row sums of triangle A135838. - Gary W. Adamson, Dec 01 2007

Row sums of triangle A152842. - Reinhard Zumkeller, May 01 2014

Positive integers n such that A000217(n) + A000217(n + 1) + A000217(n + 2) + A000217(n + 3) is a square (=A075870(n+1)^2).