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A330987 Alternatively add and half-multiply pairs of the nonnegative integers.

%I A330987 #30 Feb 28 2020 23:19:14
%S A330987 1,3,9,21,17,55,25,105,33,171,41,253,49,351,57,465,65,595,73,741,81,
%T A330987 903,89,1081,97,1275,105,1485,113,1711,121,1953,129,2211,137,2485,145,
%U A330987 2775,153,3081,161,3403,169,3741,177,4095,185,4465,193,4851,201,5253,209
%N A330987 Alternatively add and half-multiply pairs of the nonnegative integers.
%C A330987 In groups of two, add and half-multiply the integers: 0+1, (2*3)/2, 4+5, (6*7)/2, ....
%C A330987 From _Bernard Schott_, Jan 06 2020: (Start)
%C A330987 The bisection of this sequence gives:
%C A330987 For n odd = 2*k+1, k >= 0: a(2*k+1) = 8*k+1 = A017077(k),
%C A330987 For n even = 2*k, k >= 1: a(2*k) = T(4*k-2) = A000217(4*k-2) = (2*k-1)*(4*k-1) = A033567(k) where T(j) is the j-th triangular number. (End)
%H A330987 Colin Barker, <a href="/A330987/b330987.txt">Table of n, a(n) for n = 1..1000</a>
%H A330987 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-3,0,1).
%F A330987 From _Colin Barker_, Jan 05 2020: (Start)
%F A330987 G.f.: x*(1 + 3*x + 6*x^2 + 12*x^3 - 7*x^4 + x^5) / ((1 - x)^3*(1 + x)^3).
%F A330987 a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>6.
%F A330987 a(n) = -1 + 2*(-1)^n - (1/2)*(-1+7*(-1)^n)*n + (1+(-1)^n)*n^2.
%F A330987 (End)
%F A330987 E.g.f.: (1 + 4*x + 2*x^2)*cosh(x) - (3 + x)*sinh(x) - 1. - _Stefano Spezia_, Jan 05 2020 after _Colin Barker_
%t A330987 a[n_]:=If[OddQ[n],4n-3,(n-1)(2n-1)]; Array[a,53] (* _Stefano Spezia_, Jan 05 2020 *)
%o A330987 (PARI) Vec(x*(1 + 3*x + 6*x^2 + 12*x^3 - 7*x^4 + x^5) / ((1 - x)^3*(1 + x)^3) + O(x^50)) \\ _Colin Barker_, Jan 06 2020
%Y A330987 Cf. A330983.
%Y A330987 Interspersion of A017077 and A033567 (excluding first term). - _Michel Marcus_, Jan 06 2020
%K A330987 nonn,easy
%O A330987 1,2
%A A330987 _George E. Antoniou_, Jan 05 2020