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%I #73 Feb 27 2024 12:56:50
%S 0,1,1,9,8,25,9,49,32,81,25,121,72,169,49,225,128,289,81,361,200,441,
%T 121,529,288,625,169,729,392,841,225,961,512,1089,289,1225,648,1369,
%U 361,1521,800,1681,441,1849,968,2025,529,2209,1152,2401,625,2601,1352
%N a(n) = n^2 / gcd(n+2, 4).
%C A061038(n), which appears in 4*a(n) formula, is a permutation of n^2.
%C Origin. In December 2010, I wrote in my 192-page Exercise Book no. 5, page 41, the array (difference table of the first row):
%C 1 0, 1/3, 1, 9/5, 8/3, 25/7, 9/2, 49/9, ...
%C -1, 1/3, 2/3, 4/5, 13/15, 19/21, 13/14, 17/18, 43/45, ...
%C Numerators are listed in A176126, denominators are in A064038, and denominator - numerator = 2, 2, 1, 1,... (A014695).
%C 4/3, 1/3, 2/15, 1/15, 4/105, 1/42, 1/63, 1/90, 4/495, ...
%C -1, -1/5, -1/15, -1/35, -1/70, -1/126, -1/210, -1/330, -1/495, ...
%C where the denominators of the second row are listed in A000332.
%C Also for those of the inverse binomial transform
%C 1, -1, 4/3, -1, 4/5, -2/3, 4/7, -1/2, 4/9, -2/5, 4/11, -1/3, ... ?
%C a(n) is the (n+1)-th term of the numerators of the first row.
%H Colin Barker, <a href="/A251091/b251091.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,3,0,0,0,-3,0,0,0,1).
%F a(n) = n^2/(period 4: repeat 2, 1, 4, 1).
%F a(4n) = 8*n^2, a(2n+1) = a(4n+2) = (2*n+1)^2.
%F a(n+4) = a(n) + 8*A060819(n).
%F a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11.
%F 4*a(n) = (period 4: repeat 2, 1, 4, 1) * A061038(n).
%F G.f.: -x*(x^10+x^9+9*x^8+8*x^7+22*x^6+6*x^5+22*x^4+8*x^3+9*x^2+x+1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - _Colin Barker_, May 14 2015
%F a(2n) = A181900(n), a(2n+1) = A016754(n). [_Bruno Berselli_, May 14 2015]
%F a(n) = ( 1 - (1/16)*(1+(-1)^n)*(5-(-1)^(n/2)) )*n^2. - _Bruno Berselli_, May 14 2015
%F Sum_{n>=1} 1/a(n) = 13*Pi^2/48. - _Amiram Eldar_, Aug 12 2022
%e a(0) = 0/2, a(1) = 1/1, a(2) = 4/4, a(3) = 9/1.
%p seq(seq((4*i+j-1)^2/[2,1,4,1][j],j=1..4),i=0..30); # _Robert Israel_, May 14 2015
%t f[n_] := Switch[ Mod[n, 4], 0, n^2/2, 1, n^2, 2, n^2/4, 3, n^2]; Array[f, 50, 0] (* or *) Table[(4 i + j - 1)^2/{2, 1, 4, 1}[[j]], {i, 0, 12}, {j, 4}] // Flatten (* after _Robert Israel_ *) (* or *) LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121}, 53] (* or *) CoefficientList[ Series[-((x (1 + x (1 + x (9 + x (8 + x (22 + x (6 + x (22 + x (8 + x (9 + x + x^2))))))))))/(-1 + x^4)^3), {x, 0, 52}], x] (* _Robert G. Wilson v_, May 19 2015 *)
%o (PARI) concat(0, Vec(-x*(x^10 + x^9 + 9*x^8 + 8*x^7 + 22*x^6 + 6*x^5 + 22*x^4 + 8*x^3 + 9*x^2 + x + 1) / ((x-1)^3*(x+1)^3*(x^2+1)^3) + O(x^100))) \\ _Colin Barker_, May 14 2015
%o (Magma) [(1-(1/16)*(1+(-1)^n)*(5-(-1)^(n div 2)) )*n^2: n in [0..60]]; // _Vincenzo Librandi_, Jun 12 2015
%Y Cf. A000290, A000332, A016754, A026741, A060819, A061038, A109008, A139098, A168077, A176895, A181900.
%K nonn,easy,mult
%O 0,4
%A _Paul Curtz_, May 08 2015
%E Missing term (1521) inserted in the sequence by _Colin Barker_, May 14 2015
%E Definition uses a formula by _Jean-François Alcover_, Jul 01 2015
%E Keyword:mult added by _Andrew Howroyd_, Aug 06 2018
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