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A111710 Consider the triangle shown below in which the n-th row contains the n smallest numbers greater than those in the previous row such that the arithmetic mean is an integer. Sequence contains the leading diagonal.

%I A111710 #27 Feb 09 2024 08:40:57
%S A111710 1,4,7,13,18,27,34,46,55,70,81,99,112,133,148,172,189,216,235,265,286,
%T A111710 319,342,378,403,442,469,511,540,585,616,664,697,748,783,837,874,931,
%U A111710 970,1030,1071,1134,1177,1243,1288,1357,1404,1476,1525,1600,1651,1729
%N A111710 Consider the triangle shown below in which the n-th row contains the n smallest numbers greater than those in the previous row such that the arithmetic mean is an integer. Sequence contains the leading diagonal.
%H A111710 Colin Barker, <a href="/A111710/b111710.txt">Table of n, a(n) for n = 1..1000</a>
%H A111710 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F A111710 a(1)=1, a(2n) = a(2n-1)+3n, a(2n+1)=a(2n)+2n+1. - _Franklin T. Adams-Watters_, May 01 2006
%F A111710 G.f.: -x*(1+3*x+x^2) / ( (1+x)^2*(x-1)^3 ). a(n+1)-a(n) = A080512(n+1). - _R. J. Mathar_, May 02 2013
%F A111710 From _Colin Barker_, Jan 26 2016: (Start)
%F A111710 a(n) = (10*n^2+2*(-1)^n*n+10*n+(-1)^n-1)/16.
%F A111710 a(n) = (5*n^2+6*n)/8 for n even.
%F A111710 a(n) = (5*n^2+4*n-1)/8 for n odd. (End)
%e A111710 The fourth row is 8,9,10 and 13,(8+9+10 +13)/4 = 10.
%e A111710 Triangle begins:
%e A111710 1
%e A111710 2 4
%e A111710 5 6 7
%e A111710 8 9 10 13
%e A111710 14 15 16 17 18
%e A111710 19 20 21 22 23 27
%e A111710 28 29 30 31 32 33 34
%t A111710 LinearRecurrence[{1, 2, -2, -1, 1}, {1, 4, 7, 13, 18}, 100] (* _Paolo Xausa_, Feb 09 2024 *)
%o A111710 (PARI) Vec(x*(1+3*x+x^2)/((1-x)^3*(1+x)^2) + O(x^100)) \\ _Colin Barker_, Jan 26 2016
%Y A111710 Cf. A111711, A111712.
%Y A111710 Cf. A085787. - _R. J. Mathar_, Aug 15 2008
%K A111710 easy,nonn
%O A111710 1,2
%A A111710 _Amarnath Murthy_, Aug 24 2005
%E A111710 More terms from _Franklin T. Adams-Watters_, May 01 2006