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Number of partitions of n into 3 parts which form the sides of a nondegenerate isosceles triangle.
23

%I #103 Dec 30 2023 23:52:03

%S 0,0,1,0,1,1,2,1,2,2,3,2,3,3,4,3,4,4,5,4,5,5,6,5,6,6,7,6,7,7,8,7,8,8,

%T 9,8,9,9,10,9,10,10,11,10,11,11,12,11,12,12,13,12,13,13,14,13,14,14,

%U 15,14,15,15,16,15,16,16,17,16,17,17,18,17,18,18,19,18,19,19,20,19,20,20

%N Number of partitions of n into 3 parts which form the sides of a nondegenerate isosceles triangle.

%C Also number of 0's in n-th row of triangle in A071026. - _Hans Havermann_, May 26 2002

%C Exponent of 2 in factorization of A030436(n-1) and A026655(n-1). First differences of A001971. - _Ralf Stephan_, Mar 21 2004

%C Conjecture: this is 0 followed by A026922. - _R. J. Mathar_, Oct 05 2008 [See the g.f. given there by Michael Somos and the one given below for the proof. - _Wolfdieter Lang_, May 10 2017]

%C a(n+1) is for n >= 0 the number of integers k in the left-sided open interval ((n+1)/4, floor(n/2)]. This is needed for the number of zeros of Chebyshev S polynomials in the open interval (-sqrt(2), sqrt(2)) given in A285869. - _Wolfdieter Lang_, May 10 2017

%H Vincenzo Librandi, <a href="/A059169/b059169.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).

%F a(2*n + 2) = a(2*n - 1) = A004526(n).

%F a(n) = A005044(n) - A005044(n-6).

%F From _Vladeta Jovovic_, Dec 29 2001: (Start)

%F a(n) = a(n-1) + a(n-4) - a(n-5).

%F G.f.: x^3*(1 - x + x^2)/(1 - x - x^4 + x^5). (End)

%F The g.f. can also be written as x^3 * (1 + x^3) / ((1 - x^2) * (1 - x^4)). - _Michael Somos_, May 05 2015

%F Euler transform of length 6 sequence [0, 1, 1, 1, 0, -1]. - _Michael Somos_, Oct 14 2008

%F a(n) = -a(3 - n) for all n in Z. - _Michael Somos_. Oct 14 2008

%F a(n) = abs(floor((n-1)*(-1)^n/4)). - _Wesley Ivan Hurt_, Oct 22 2013

%F a(n) = abs(A178804(n+1) - A178804(n)). - _Reinhard Zumkeller_, Nov 15 2014

%F a(n) = floor(n/2) - floor(n/4) - (1 if n even). - _David Pasino_, Jun 17 2016

%F E.g.f.: (4 - sin(x) - cos(x) + x*sinh(x) + (x - 3)*cosh(x))/4. - _Ilya Gutkovskiy_, Jun 21 2016

%F a(n) = floor((n-1)/2) - floor(n/4), n >= 0 (from the preceding a(n) formula). - _Wolfdieter Lang_, May 08 2017

%F a(n) = (2*n - 3 - 2*cos(n*Pi/2) - 3*cos(n*Pi) - 2*sin(n*Pi/2))/8. - _Wesley Ivan Hurt_, Oct 01 2017

%F a(n) = Sum_{i=1..floor((n-1)/2)} (n-i-1) mod 2. - _Wesley Ivan Hurt_, Nov 17 2017

%e Consider the number 13. The following partitions give a nondegenerate triangle: 4 4 5; 3 5 5; 1 6 6; 2 5 6; 3 4 6. Since the first three partitions represent isosceles triangles, we have A059169(13) = 3.

%e G.f. = x^3 + x^5 + x^6 + 2*x^7 + x^8 + 2*x^9 + 2*x^10 + 3*x^11 + 2*x^12 + ...

%p a[1] := 0: a[2] := 0: a[3] := 1: a[4] := 0: a[5] := 1: for n from 6 to 300 do a[n] := a[n-1] + a[n-4] - a[n-5]: end do: seq(a[n], n=1..82);

%p a := n -> A005044(n) - A005044(n-6): A005044 := n-> floor((1/48)*(n^2 + 3*n + 21 + (-1)^(n-1)*3*n)): seq(a(n), n = 1..82); # _Johannes W. Meijer_, Oct 10 2013

%t CoefficientList[Series[x^2 (1 - x + x^2)/(1 - x - x^4 + x^5), {x, 0, 100}], x] (* _Vincenzo Librandi_, Aug 15 2013 *)

%t LinearRecurrence[{1,0,0,1,-1},{0,0,1,0,1},100] (* _Harvey P. Dale_, Feb 09 2015 *)

%t a[ n_] := Quotient[ n - 1, 2] - Quotient[ n, 4]; (* _Michael Somos_, May 05 2015 *)

%o (PARI) {a(n) = (n - 1) \ 2 - (n \ 4)}; /* _Michael Somos_, Oct 14 2008 */

%o (PARI) {a(n) = if( n<1, -a(3 - n), polcoeff( x^3 * (1 - x + x^2) / (1 - x - x^4 + x^5) + x * O(x^n), n))}; /* _Michael Somos_, Oct 14 2008 */

%o (Haskell)

%o a059169 n = a059169_list !! (n-1)

%o a059169_list = map abs $ zipWith (-) (tail a178804_list) a178804_list

%o -- _Reinhard Zumkeller_, Nov 15 2014

%o (Magma) [Floor((n-1)/2) - Floor(n/4): n in [1..80]]; // _G. C. Greubel_, Mar 08 2018

%Y Essentially the same as A008624.

%Y Cf. A178804.

%K nonn,easy

%O 1,7

%A _Floor van Lamoen_, Jan 13 2001

%E More terms from _Sascha Kurz_, Mar 25 2002