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.
1, 4, 7, 13, 18, 27, 34, 46, 55, 70, 81, 99, 112, 133, 148, 172, 189, 216, 235, 265, 286, 319, 342, 378, 403, 442, 469, 511, 540, 585, 616, 664, 697, 748, 783, 837, 874, 931, 970, 1030, 1071, 1134, 1177, 1243, 1288, 1357, 1404, 1476, 1525, 1600, 1651, 1729
Offset: 1
Examples
The fourth row is 8,9,10 and 13,(8+9+10 +13)/4 = 10. Triangle begins: 1 2 4 5 6 7 8 9 10 13 14 15 16 17 18 19 20 21 22 23 27 28 29 30 31 32 33 34
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).
Crossrefs
Cf. A085787. - R. J. Mathar, Aug 15 2008
Programs
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Mathematica
LinearRecurrence[{1, 2, -2, -1, 1}, {1, 4, 7, 13, 18}, 100] (* Paolo Xausa, Feb 09 2024 *) -
PARI
Vec(x*(1+3*x+x^2)/((1-x)^3*(1+x)^2) + O(x^100)) \\ Colin Barker, Jan 26 2016
Formula
a(1)=1, a(2n) = a(2n-1)+3n, a(2n+1)=a(2n)+2n+1. - Franklin T. Adams-Watters, May 01 2006
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
From Colin Barker, Jan 26 2016: (Start)
a(n) = (10*n^2+2*(-1)^n*n+10*n+(-1)^n-1)/16.
a(n) = (5*n^2+6*n)/8 for n even.
a(n) = (5*n^2+4*n-1)/8 for n odd. (End)
Extensions
More terms from Franklin T. Adams-Watters, May 01 2006