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%I #27 Aug 22 2022 14:29:24
%S 3,12,19,28,35,44,51,60,67,76,83,92,99,108,115,124,131,140,147,156,
%T 163,172,179,188,195,204,211,220,227,236,243,252,259,268,275,284,291,
%U 300,307,316,323,332,339,348,355,364,371,380,387,396,403,412,419,428
%N a(1)=3, a(n) = a(n-1) + 7 + 2*mod(n-1, 2) for n>=2.
%C Starting weights for pyramid game.
%C Numbers n such that the equation m(m + 1)/2 + 1 - n == 0 mod m has a solution.
%C Numbers congruent to {3, 12} mod 16. - _Philippe Deléham_, Nov 28 2016
%H Colin Barker, <a href="/A134965/b134965.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F From _R. J. Mathar_, Feb 05 2008: (Start)
%F G.f.: (3+9*x+4*x^2)/((1-x)^2*(x+1)).
%F a(n) - a(n-1) = A010729(n).
%F (End)
%F From _Colin Barker_, Nov 29 2016: (Start)
%F a(n) = 8*n - 4 for n even.
%F a(n) = 8*n - 5 for n odd.
%F a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.
%F (End)
%F E.g.f.: 4 + ((16*x - 9)*exp(x) + exp(-x))/2. - _David Lovler_, Aug 21 2022
%t Flatten[Table[If[ IntegerQ[2*Sqrt[ -7 + 8*n]] && Mod[n - 7, 8] == 0, f[n], {}], {n, 1, 10000}]]
%t LinearRecurrence[{1,1,-1},{3,12,19},60] (* _Harvey P. Dale_, Oct 05 2017 *)
%o (PARI) Vec(x*(3 + 9*x + 4*x^2) / ((1 - x)^2 * (1 + x)) + O(x^100)) \\ _Colin Barker_, Nov 29 2016
%o (PARI) a(n)=8*n - 4 - n%2 \\ _Charles R Greathouse IV_, Nov 29 2016
%K nonn,easy
%O 1,1
%A _Roger L. Bagula_, Jan 31 2008
%E Definition adapted to offset by _Georg Fischer_, Jun 19 2021
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