login
A046177
Squares (A000290) which are also hexagonal numbers (A000384).
8
1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481
OFFSET
1,2
COMMENTS
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
REFERENCES
M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.
LINKS
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Eric Weisstein's World of Mathematics, Square Triangular Number.
FORMULA
a(n) = A001110(2n-1). - Alexander Adamchuk, Nov 06 2007
a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009
a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - Richard Choulet, May 01 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
(End)
a(n) = A096979(4*n - 3). - Ivan N. Ianakiev, Sep 05 2016
a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - Ivan N. Ianakiev, Sep 05 2016
MATHEMATICA
LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)
PROG
(PARI) Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015
CROSSREFS
Cf. A001110 (Numbers that are both triangular and square).
Sequence in context: A267297 A151657 A218273 * A031748 A031658 A031533
KEYWORD
nonn,easy
STATUS
approved