A269448 The first of 26 consecutive positive integers the sum of the squares of which is a square.

25, 301, 454, 3850, 31966, 47569, 393925, 3261481, 4852834, 40177750, 332640346, 494942749, 4097737825, 33926055061, 50479308814, 417929081650, 3460124977126, 5148394557529, 42624668591725, 352898821613041, 525085765560394, 4347298267275550

Offset

1,1

Comments

Positive integers y in the solutions to 2*x^2-52*y^2-1300*y-11050 = 0.

All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...25 for this sequence, k=0...22 for A269447, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,0,102,-102,0,-1,1).

Formula

G.f.: x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6) / ((1-x)*(1-102*x^3+x^6)).

a(1)=25, a(2)=301, a(3)=454, a(4)=3850, a(5)=31966, a(6)=47569, a(n)=102*a(n-3) - a(n-6) + 1250. - Daniel Mondot, Aug 05 2016

Example

25 is in the sequence because sum(k=25, 50, k^2) = 38025 = 195^2.

Mathematica

Rest@ CoefficientList[Series[x (25 + 276 x + 153 x^2 + 846 x^3 - 36 x^4 - 3 x^5 - 11 x^6)/((1 - x) (1 - 102 x^3 + x^6)), {x, 0, 22}], x] (* Michael De Vlieger, Aug 07 2016 *)

Prog

(PARI) Vec(x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6)/((1-x)*(1-102*x^3+x^6)) + O(x^30))

Crossrefs

Cf. A001032, A001652, A094196, A106521, A257765, A269447, A269449, A269451.

Sequence in context: A022620 A367457 A042206 * A128377 A171445 A359097

Adjacent sequences: A269445 A269446 A269447 * A269449 A269450 A269451

Keyword

nonn,easy

Author

Colin Barker, Feb 27 2016

Status

approved