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A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

%I A253475 #26 May 30 2025 10:10:30
%S A253475 1,6,55,540,5341,52866,523315,5180280,51279481,507614526,5024865775,
%T A253475 49741043220,492385566421,4874114620986,48248760643435,
%U A253475 477613491813360,4727886157490161,46801248083088246,463284594673392295,4586044698650834700,45397162391834954701
%N A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).
%C A253475 Also positive integers x in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0, the corresponding values of y being A054318.
%C A253475 Also indices of centered hexagonal numbers (A003215) which are also hexagonal numbers (A000384).
%C A253475 Also indices of terms in sequence A193218 which are the square root of a sum of 5th powers (A000539). - _Daniel Poveda Parrilla_, Jun 10 2017
%H A253475 Colin Barker, <a href="/A253475/b253475.txt">Table of n, a(n) for n = 1..1000</a>
%H A253475 Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
%H A253475 Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_1.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
%H A253475 Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_2.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
%H A253475 Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_3.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
%H A253475 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11,-11,1).
%F A253475 a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
%F A253475 G.f.: x*(5*x-1) / ((x-1)*(x^2-10*x+1)).
%F A253475 a(n) = sqrt((-2-(5-2*sqrt(6))^n-(5+2*sqrt(6))^n)*(2-(5-2*sqrt(6))^(1+n)-(5+2*sqrt(6))^(1+n)))/(4*sqrt(2)). - _Gerry Martens_, Jun 04 2015
%F A253475 2*a(n) = 1+A054320(n-1). - _R. J. Mathar_, Feb 07 2022
%e A253475 6 is in the sequence because the 6th centered square number is 61, which is also the 5th centered hexagonal number.
%t A253475 LinearRecurrence[{11, -11, 1}, {1, 6, 55}, 25] (* _Paolo Xausa_, May 30 2025 *)
%o A253475 (PARI) Vec(x*(5*x-1)/((x-1)*(x^2-10*x+1)) + O(x^100))
%Y A253475 Cf. A000384, A000539, A001844, A003215, A054318, A193218, A253175.
%K A253475 nonn,easy
%O A253475 1,2
%A A253475 _Colin Barker_, Jan 02 2015