%I #53 Sep 18 2023 14:05:07
%S 0,0,2,1,6,2,12,3,20,4,30,5,42,6,56,7,72,8,90,9,110,10,132,11,156,12,
%T 182,13,210,14,240,15,272,16,306,17,342,18,380,19,420,20,462,21,506,
%U 22,552,23,600,24,650,25,702,26,756,27,812,28,870,29,930,30,992,31,1056,32,1122,33,1190
%N a(n) is the numerator of Cesàro means sequence c(n) of A237420 when the denominator is A141310(n).
%C So, we get c(n) = a(n) / A141310(n) for n >= 0 (see Formula and Example section).
%C Cesàro mean theorem: when the series u(n) has a limit (finite or infinite) in the usual sense, then c(n) = (u(0)+...+u(n))/(n+1) has the same Cesàro limit, but the converse is false.
%C A237420 is such a counterexample in the case of an infinite limit.
%C Proof: A237420 is not convergent in the usual sense because a(2n+1) = 0, while a(2n) -> oo when n -> oo. Now, the successive arithmetic means c(n) of the first n terms of the sequence are 0/1, 0/2, 2/3, 2/4, 6/5, 6/6, 12/7, 12/8, 20/9, 20/10, ... so c(2n)= (n*(n+1))/(2*n+1) ~ n/2 and c(2n+1) = n/2, hence the Cesàro limit is infinity because c(n) -> oo as n -> oo (Arnaudiès et al.), QED.
%C The first few irreducible fractions c(n) are in the last row of the Example section. The differences between row 4 and last row exist only when n = 4*k+1, k>0, then respectively c(n) = 2k/2 = k/1.
%C This sequence consists of the oblong numbers (A002378) interlaced with the natural numbers (A001477)
%C Note that A033999 is a counterexample in the case of a finite Cesàro limit.
%C Also, the converse of the Cesàro mean theorem is true iff u(n) is monotonic.
%D J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.
%H ProofWiki, <a href="https://proofwiki.org/wiki/Cesàro_Mean">Cesàro mean</a>.
%H The MacTutor History of Mathematics archive, <a href="https://mathshistory.st-andrews.ac.uk/Biographies/Cesaro/">Ernesto Cesàro</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ernesto_Cesàro">Ernesto Cesàro</a>.
%H Wikipédia, <a href="https://fr.wikipedia.org/wiki/Lemme_de_Cesàro">Lemme de Cesàro </a> (in French).
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-3,0,1).
%F a(n) = (A141310(n)/(n+1)) * Sum_{k=0..n} A237420(k).
%F For n >= 0, a(2n) = n*(n+1) = A002378(n), a(2n+1) = n = A001477(n).
%F G.f.: x^2*(2 + x - x^3)/(1 - x^2)^3. - _Stefano Spezia_, May 23 2022
%e Table with the first few terms:
%e Indices n : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
%e A237420(n) : 0, 0, 2, 0, 4, 0, 6, 0, 8, 0, ...
%e Partial sums : 0, 0, 2, 2, 6, 6, 12, 12, 20, 20, ...
%e Cesàro means c(n) : 0/1, 0/2, 2/3, 1/2, 6/5, 2/2, 12/7, 3/2, 20/9, 4/2, ...
%e Numerator a(n) : 0, 0, 2, 1, 6, 2, 12, 3, 20, 4, ...
%e Denominator A141310(n) : 1, 2, 3, 2, 5, 2, 7, 2, 9, 2, ...
%e Irreducible Cesàro mean : 0/1, 0/2, 2/3, 1/2, 6/5, 1/1, 12/7, 3/2, 20/9, 2/1, ...
%t m = 50; Accumulate[Table[If[OddQ[n], 0, n], {n, 0, 2*m - 1}]] * Flatten[Table[{2*n - 1, 2}, {n, 1, m}]] / Range[2*m] (* _Amiram Eldar_, Jun 05 2022 *)
%o (PARI) c(n) = sum(k=0, n, if (k%2, 0, k))/(n+1);
%o f(n) = if(n%2, 2, 1+n); \\ A141310
%o a(n) = c(n)*f(n); \\ _Michel Marcus_, Jun 06 2022
%Y Cf. A001477, A002378, A033999, A141310 (denominators), A237420.
%K nonn,easy
%O 0,3
%A _Bernard Schott_, May 22 2022