[math-fun] Square hex numbers?
Just happened to notice that in the centered hexagonal numbers H_n = 1 + 6 T_n = 3n^2 + 3n + 1 we have H_7 = 13^2. Also H_0 = 1^2. So I tried to find all such cases of H_n = K^2, but am not sure of the best way to proceed. Suggestions? —Dan
Maybe work out the first few terms and look it up somewhere?!! The OEIS has hundreds of sequences mentioning those words - for example, %N A001922 Numbers n such that 3*n^2-3*n+1 is both a square (A000290) and a centered hexagonal number (A003215). and %N A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215). Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Tue, Apr 23, 2019 at 11:10 PM Dan Asimov <dasimov@earthlink.net> wrote:
Just happened to notice that in the centered hexagonal numbers
H_n = 1 + 6 T_n = 3n^2 + 3n + 1
we have H_7 = 13^2. Also H_0 = 1^2.
So I tried to find all such cases of
H_n = K^2,
but am not sure of the best way to proceed. Suggestions?
—Dan
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See: http://oeis.org/A006051 On Tue, Apr 23, 2019 at 11:10 PM Dan Asimov <dasimov@earthlink.net> wrote:
Just happened to notice that in the centered hexagonal numbers
H_n = 1 + 6 T_n = 3n^2 + 3n + 1
we have H_7 = 13^2. Also H_0 = 1^2.
So I tried to find all such cases of
H_n = K^2,
but am not sure of the best way to proceed. Suggestions?
—Dan
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Your equation is of the form: (quadratic in n) = (quadratic in K) so you can complete the square on both sides and multiply both sides by a constant factor to obtain the Pell-like equation: x^2 - d y^2 = c where x is a linear function of n, and y is a linear function of K (both with integer coefficients). Each pair (x,y) in Z^2 can be identified with the algebraic integer: x + y sqrt(d) and it's a solution if and only if the (number-theoretic) norm x^2 - d y^2 is equal to c. But norms are multiplicative, so if you find a single solution (which you have done) together with a primitive unit (e.g. the smallest norm-1 element with x,y > 0) then you can generate all of the solutions to the Pell-like equation. Then the solutions to your original problem will be the subset of solutions which satisfy some congruence condition. Best wishes, Adam P. Goucher
Sent: Wednesday, April 24, 2019 at 4:10 AM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Square hex numbers?
Just happened to notice that in the centered hexagonal numbers
H_n = 1 + 6 T_n = 3n^2 + 3n + 1
we have H_7 = 13^2. Also H_0 = 1^2.
So I tried to find all such cases of
H_n = K^2,
but am not sure of the best way to proceed. Suggestions?
—Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hi Dan There is a general way to get all solutions to your problem. You are looking for 3n^2 + 3n + 1 = K^2. First transform this to 3(2n + 1)^2 + 1 = 4K^2, which further simplifies to x^2 - 3y^2 = 1, where x = 2K and y = 2n + 1. This is a form of Pell's equation in x and y with solutions x = [(2 + sqrt(3))^m + (2 - sqrt(3))^m]/2 and y = [(2 + sqrt(3))^m - (2 - sqrt(3))^m]/sqrt(12) for m = 0, 1, 2,.... This leads to K = [(2 + sqrt(3))^m + (2 - sqrt(3))^m]/4 and n = [(2 + sqrt(3))^m - (2 - sqrt(3))^m]/sqrt(12), where m = 1, 3, 5,... to produce integer solutions. The first few solutions are (n, K) = (0, 1), (7, 13), (104, 181), (1455, 2521), (20272, 35113), (282359, 489061), (3932760, 6811741). Note that K_(i+1) = 14K_(i) + K_(i-1) and n_(i+1) = 14n_(i) + n_(i-1) + 6, so the solution string can be easily extended. This method also works for the more general Fermat equation, x^2 - Cy^2 = +H or -H. Multiple families of solutions are possible depending on C and H. -Dick Hess On 4/23/2019 8:10 PM, Dan Asimov wrote:
Just happened to notice that in the centered hexagonal numbers
H_n = 1 + 6 T_n = 3n^2 + 3n + 1
we have H_7 = 13^2. Also H_0 = 1^2.
So I tried to find all such cases of
H_n = K^2,
but am not sure of the best way to proceed. Suggestions?
—Dan
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participants (5)
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Adam P. Goucher -
Dan Asimov -
Neil Sloane -
Richard -
W. Edwin Clark