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%I #156 Mar 21 2024 17:45:22
%S 1,4,21,120,697,4060,23661,137904,803761,4684660,27304197,159140520,
%T 927538921,5406093004,31509019101,183648021600,1070379110497,
%U 6238626641380,36361380737781,211929657785304,1235216565974041,7199369738058940,41961001862379597,244566641436218640
%N Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.
%C Solution to a*(a-1) = 2b*(b-1) in natural numbers: a = a(n), b = b(n) = A011900(n).
%C n such that n^2 = (1/2)*(n+floor(sqrt(2)*n*floor(sqrt(2)*n))). - _Benoit Cloitre_, Apr 15 2003
%C Place a(n) balls in an urn, of which b(n) = A011900(n) are red; draw 2 balls without replacement; 2*Probability(2 red balls) = Probability(2 balls); this is equivalent to the Pell equation A(n)^2-2*B(n)^2 = -1 with a(n) = (A(n)+1)/2; b(n) = (B(n)+1)/2; and the fundamental solution (7;5) and the solution (3;2) for the unit form. - _Paul Weisenhorn_, Aug 03 2010
%C Find base x in which repdigit yy has a square that is repdigit zzzz, corresponding to Diophantine equation zzzz_x = (yy_x)^2; then, solution z = a(n) with x = A002315(n) and y = A001653(n+1) for n >= 1 (see Maurice Protat reference). - _Bernard Schott_, Dec 21 2022
%D A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
%D Maurice Protat, Des Olympiades à l'Agrégation, De zzzz_x = (yy_x)^2 à Pell-Fermat, Problème 23, pp. 52-54, Ellipses, Paris, 1997.
%H Reinhard Zumkeller, <a href="/A046090/b046090.txt">Table of n, a(n) for n = 0..1000</a>
%H T. W. Forget and T. A. Larkin, <a href="http://www.fq.math.ca/Scanned/6-3/forget.pdf">Pythagorean triads of the form X, X+1, Z described by recurrence sequences</a>, Fib. Quart., 6 (No. 3, 1968), 94-104.
%H L. J. Gerstein, <a href="http://www.jstor.org/stable/30044157">Pythagorean triples and inner products</a>, Math. Mag., 78 (2005), 205-213.
%H H. J. Hindin, <a href="/A006062/a006062.pdf">Stars, hexes, triangular numbers and Pythagorean triples</a>, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%H S. Northshield, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Northshield/north4.html">An Analogue of Stern's Sequence for Z[sqrt(2)]</a>, Journal of Integer Sequences, 18 (2015), #15.11.6.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PythagoreanTriple.html">Pythagorean Triple</a>
%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F a(n) = (-1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n). [corrected by _Michel Marcus_, Dec 23 2022]
%F a(n) = 6*a(n-1) - a(n-2) - 2, n >= 2, a(0) = 1, a(1) = 4.
%F a(n) = (A(n+1) - 3*A(n) + 2)/4 with A(n) = A001653(n).
%F A001652(n) = -a(-1-n).
%F From _Barry E. Williams_, May 03 2000: (Start)
%F G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)).
%F a(n) = partial sums of A001541(n). (End)
%F From _Charlie Marion_, Jul 01 2003: (Start)
%F A001652(n)*A001652(n+1) + a(n)*a(n+1) = A001542(n+1)^2 = A084703(n+1).
%F Let a(n) = A001652(n), b(n) = this sequence and c(n) = A001653(n). Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. (End)
%F a(n) = 1/2 + ((1-2^(1/2))/4)*(3 - 2^(3/2))^n + ((1+2^(1/2))/4)*(3 + 2^(3/2))^n. - _Antonio Alberto Olivares_, Oct 13 2003
%F 2*a(n) = 2*A084159(n) + 1 + (-1)^(n+1) = 2*A046729(n) + 1 - (-1)^(n+1). - _Lekraj Beedassy_, Jul 16 2004
%F a(n) = A001109(n+1) - A053141(n). - _Manuel Valdivia_, Apr 03 2010
%F From _Paul Weisenhorn_, Aug 03 2010: (Start)
%F a(n+1) = round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^n)/2);
%F b(n+1) = round((2+(10+7*sqrt(2))*(3+2*sqrt(2))^n)/4) = A011900(n+1).
%F (End)
%F a(n)*(a(n)-1)/2 = b(n)*b(n+1) and 2*a(n) - 1 = b(n) + b(n+1), where b(n) = A001109. - _Kenneth J Ramsey_, Apr 24 2011
%F T(a(n)) = A011900(n)^2 + A001109(n), where T(n) is the n-th triangular number. See also A001653. - _Charlie Marion_, Apr 25 2011
%F a(0)=1, a(1)=4, a(2)=21, a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - _Harvey P. Dale_, Apr 13 2012
%F Limit_{n->oo} a(n+1)/a(n) = 3 + 2*sqrt(2) = A156035. - _Ilya Gutkovskiy_, Jul 10 2016
%F a(n) = A001652(n)+1. - _Dimitri Papadopoulos_, Jul 06 2017
%F a(n) = (A002315(n) + 1)/2. - _Bernard Schott_, Dec 21 2022
%F E.g.f.: (exp(x) + exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2. - _Stefano Spezia_, Mar 16 2024
%e For n=4: a(4)=697; b(4)=493; 2*binomial(493,2)=485112=binomial(697,2). - _Paul Weisenhorn_, Aug 03 2010
%p Digits:=100: seq(round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^(n-1))/2)/2, n=0..20); # _Paul Weisenhorn_, Aug 03 2010
%t Join[{1},#+1&/@With[{c=3+2Sqrt[2]},NestList[Floor[c #]+3&,3,20]]] (* _Harvey P. Dale_, Aug 19 2011 *)
%t LinearRecurrence[{7,-7,1},{1,4,21},25] (* _Harvey P. Dale_, Apr 13 2012 *)
%t a[n_] := (2-ChebyshevT[n, 3]+ChebyshevT[n+1, 3])/4; Array[a, 21, 0] (* _Jean-François Alcover_, Jul 10 2016, adapted from PARI *)
%o (PARI) a(n)=(2-subst(poltchebi(abs(n))-poltchebi(abs(n+1)),x,3))/4
%o (PARI) x='x+O('x^30); Vec((1-3*x)/((1-6*x+x^2)*(1-x))) \\ _G. C. Greubel_, Jul 15 2018
%o (Haskell)
%o a046090 n = a046090_list !! n
%o a046090_list = 1 : 4 : map (subtract 2)
%o (zipWith (-) (map (* 6) (tail a046090_list)) a046090_list)
%o -- _Reinhard Zumkeller_, Jan 10 2012
%o (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-3*x)/((1-6*x+x^2)*(1-x)))); // _G. C. Greubel_, Jul 15 2018
%Y Other 2 sides are A001652 and A001653.
%Y Cf. A001009, A001541, A001542, A011900, A046729, A053141, A084159, A084703, A156035.
%Y See comments in A301383.
%Y Cf. A001653, A002315.
%K nonn,easy,nice
%O 0,2
%A _Eric W. Weisstein_
%E Additional comments from _Wolfdieter Lang_
%E Comment moved to A001653 by _Claude Morin_, Sep 22 2023
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