This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A104722 #36 Jul 02 2025 16:02:02 %S A104722 1,2,3,4,7,10,19,28,56,84,174,264,561,858,1859,2860,6292,9724,21658, %T A104722 33592,75582,117572,266798,416024,950912,1485800,3417340,5348880, %U A104722 12369285,19389690,45052515,70715340,165002460,259289580,607283490 %N A104722 Self-convolution of repeated Catalan numbers. %C A104722 This is the same as A059348 after the first term. [Proof by _James Sellers_, seqfan 19 May 2008: The generating functions are the same, ignoring the constant terms which cause the difference between the two sequences. If the g.f. in the formula here is expanded, the constant term ignored, we obtain ( 1 + 2x - x^2 - 4x^3 - (x+1)^2*sqrt(1-4*x^2) )/(2*x^4) ]. %C A104722 From the Bernhart Reference in A059348 we see that A059348 originates from A000108 padded with zeros, 1 0 1 0 2 0 5 0 14 0 42 0 132 ... with g.f. C(x^2). Taking the sum of each pair of consecutive values we get the auxiliary sequence 1 1 1 2 2 5 5 14 14 42 42 132 132 ... with g.f. ((1+x)C(x^2) - 1)/x . Sum pairs of consecutive values once more to obtain 2 2 3 4 7 10 19 28 56 ... which is A059348. %C A104722 So this generating function is (1+x)*(((1+x)*C(x^2) - 1)/x -1) / x again ignoring the constant term. Straightforward algebraic manipulations show that this quantity equals (1+2x-...)/(2x^4) above, again ignoring the constant term. %H A104722 G. C. Greubel, <a href="/A104722/b104722.txt">Table of n, a(n) for n = 0..1000</a> %F A104722 G.f.: (1+x)^2*c(x^2)^2, c(x) the g.f. of the Catalan numbers A000108; %F A104722 Let b(n) = (binomial(n-1, (n-1)/2)/((n-1)/2+1))*(1-(-1)^n)/2 + (binomial(n, n/2)/(n/2+1))*(1+(-1)^n)/2, then a(n) = Sum_{k=0..n} b(k)*b(n-k). %F A104722 Conjecture: (n+4)*a(n) + (n+1)*a(n-1) - 4*(n+1)*a(n-2) + 4*(2-n)*a(n-3) = 0. - _R. J. Mathar_, Nov 09 2012 %F A104722 a(n) ~ 2^(n + 1/2) * (9 + (-1)^n) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Mar 10 2018 %t A104722 CoefficientList[Series[( (1 + x)*(1 - Sqrt[1 - 4*x^2])/(2*x^2))^2, {x, 0, 100}], x] (* _G. C. Greubel_, Jan 07 2017 *) %o A104722 (PARI) Vec( ((1 + x)*(1 - sqrt(1 - 4*x^2))/(2*x^2))^2 + O(x^20)) \\ _G. C. Greubel_, Jan 07 2017 %Y A104722 Cf. A000108, A059348, A104721. %K A104722 easy,nonn %O A104722 0,2 %A A104722 _Paul Barry_, Mar 20 2005