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A065113 Sum of the squares of the n-th and the (n+1)st triangular numbers (A000217) is a perfect square. 4
6, 40, 238, 1392, 8118, 47320, 275806, 1607520, 9369318, 54608392, 318281038, 1855077840, 10812186006, 63018038200, 367296043198, 2140758220992, 12477253282758, 72722761475560, 423859315570606, 2470433131948080, 14398739476117878, 83922003724759192 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

The sequence of square roots of the sum of the squares of the n-th and the (n+1)st triangular numbers is A046176.

a(n)^2 + (a(n)+2)^2 = A075870(n+1)^2 = A165518(n+1). - Joerg Arndt, Feb 15 2012

LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Index entries for two-way infinite sequences

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

FORMULA

2 * A001652 = -1 + A002315 = a(n+1) - A003499(n+1).

From Michael Somos, Apr 07 2003: (Start)

G.f.: 2*x*(3-x)/((1-6*x+x^2)*(1-x)).

a(n) = 6*a(n-1) - a(n-2) + 4.

a(-1-n) = -a(n) - 2. (End)

a(1)=6, a(2)=40, a(3)=238, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3). - Harvey P. Dale, Dec 27 2011

a(n) = (-2-(3-2*sqrt(2))^n*(-1+sqrt(2))+(1+sqrt(2))*(3+2*sqrt(2))^n)/2. - Colin Barker, Mar 05 2016

EXAMPLE

T6 = 21 and T7 = 28, 21^2 + 28^2 = 441 + 784 = 1225 = 35^2.

MATHEMATICA

CoefficientList[ Series[2*(x - 3)/(-1 + 7x - 7x^2 + x^3), {x, 0, 24} ], x]

LinearRecurrence[{7, -7, 1}, {6, 40, 238}, 41] (* Harvey P. Dale, Dec 27 2011 *)

PROG

(PARI) a(n)=-1+subst(poltchebi(abs(n+1))-poltchebi(abs(n)), x, 3)/2

(PARI) Vec(2*x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^40)) \\ Colin Barker, Mar 05 2016

CROSSREFS

a(n) = 2 A001652(n) = A002315(n)-1. A003499(n)=a(n)-a(n-1). Cf. A065651.

Sequence in context: A229580 A254945 A026077 * A052518 A135032 A122074

Adjacent sequences:  A065110 A065111 A065112 * A065114 A065115 A065116

KEYWORD

nonn,easy

AUTHOR

Robert G. Wilson v, Nov 12 2001

STATUS

approved

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Last modified June 26 04:11 EDT 2019. Contains 324369 sequences. (Running on oeis4.)