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A378577 G.f. A(x) equals the series obtained by removing all factors of 2 from the coefficients in 1 + x*A(x)^2.

%I A378577 #36 Jan 03 2025 13:46:33
%S A378577 1,1,1,3,1,9,13,55,5,201,309,1467,541,7009,10905,53103,2493,232713,
%T A378577 369973,1895875,711901,10057761,15917809,80998215,7682147,389278901,
%U A378577 625035439,3285433779,1245382467,18187624005,29024837139,150736553103,7191395811,735853765941,1190542570455,6387028801323,2429801651419
%N A378577 G.f. A(x) equals the series obtained by removing all factors of 2 from the coefficients in 1 + x*A(x)^2.
%H A378577 Paul D. Hanna, <a href="/A378577/b378577.txt">Table of n, a(n) for n = 0..1030</a>
%F A378577 a(n) = A000265( Sum_{k=0..n-1} a(k)*a(n-k-1) ) for n > 0 with a(0) = 1, where A000265(m) = odd part of m.
%e A378577 G.f.: A(x) = 1 + x + x^2 + 3*x^3 + x^4 + 9*x^5 + 13*x^6 + 55*x^7 + 5*x^8 + 201*x^9 + 309*x^10 + 1467*x^11 + 541*x^12 + 7009*x^13 + 10905*x^14 + 53103*x^15 + 2493*x^16 + ...
%e A378577 The expansion of A(x)^2 begins
%e A378577 A(x)^2 = 1 + 2*x + 3*x^2 + 8*x^3 + 9*x^4 + 26*x^5 + 55*x^6 + 160*x^7 + 201*x^8 + 618*x^9 + 1467*x^10 + 4328*x^11 + 7009*x^12 + 21810*x^13 + 53103*x^14 + 159552*x^15 + ...
%e A378577 where g.f. A(x) is obtained by removing all factors of 2 from the coefficients in 1 + x*A(x)^2.
%e A378577 Notice that the powers of 2 that divide the coefficients in 1 + x*A(x)^2 to give the respective coefficients in A(x) form the period 16 sequence [1, 2, 1, 8, 1, 2, 1, 32, 1, 2, 1, 8, 1, 2, 1, 64], ignoring the initial constant terms.
%e A378577 Given A000265(n) = odd part of n, we may also use the following formula
%e A378577   a(n) = A000265( Sum_{k=0..n-1} a(k)*a(n-k-1) ) for n > 0
%e A378577 as illustrated by
%e A378577   a(1) = 1;
%e A378577   a(2) = A000265(1*1 + 1*1) = 1;
%e A378577   a(3) = A000265(1*1 + 1*1 + 1*1) = 3;
%e A378577   a(4) = A000265(1*3 + 1*1 + 1*1 + 3*1) = 1;
%e A378577   a(5) = A000265(1*1 + 1*3 + 1*1 + 3*1 + 1*1) = 9;
%e A378577   a(6) = A000265(1*9 + 1*1 + 1*3 + 3*1 + 1*1 + 9*1) = 13;
%e A378577   a(7) = A000265(1*13 + 1*9 + 1*1 + 3*3 + 1*1 + 9*1 + 13*1) = 55;
%e A378577   ...
%e A378577 RATIOS OF TERMS.
%e A378577 The ratios a(n+1)/a(n) seem to tend to a period 16 sequence of reals as a limit.
%e A378577 For example, for n = 9968..9983 the values are approximately
%e A378577 [1.648278, 5.66786, 0.386617, 16.11420, 1.628799, 5.52564, 0.096966, 57.3175, 1.645484, 5.64340, 0.3869028, 16.0584, 1.629124, 5.52096, 0.0484979, 112.4854].
%e A378577 And for n = 16370..16385 the 16 values are approximately
%e A378577 [1.648318, 5.66829, 0.386625, 16.11592, 1.628837, 5.52605, 0.096968, 57.3244, 1.645524, 5.64383, 0.3869115, 16.0601, 1.629162, 5.52138, 0.0484989, 112.4990].
%e A378577 What is the limit of each of these 16 values?
%e A378577 SPECIFIC VALUES.
%e A378577 A(t) = 2 at t = 0.36423557025609119319840508971928979325085050158297...
%e A378577 A(t) = 7/4 at t = 0.34185808023756575598995230721363143324449415040840...
%e A378577 A(t) = 5/3 at t = 0.32872963418965851479105080085792251790114332592550...
%e A378577 A(t) = 3/2 at t = 0.29042932434410374379568213857247308991788862102621...
%e A378577 A(t) = 4/3 at t = 0.23035252030969595871280245310990008598072787136152...
%e A378577 A(t) = 5/4 at t = 0.18922043486502922451746312802717681803115006233322...
%e A378577 A(1/3) = 1.6934776520525683546148601762295937105641618330534...
%e A378577 A(1/4) = 1.3803475910136768240697468770816757080230324208973...
%e A378577 A(1/5) = 1.2702054444980255497491030452212140699102670480027...
%e A378577 A(1/6) = 1.2107705896242485243544918479672582628536284699829...
%e A378577 A(1/8) = 1.1470812698447982797390188599870804830531241984640...
%o A378577 (PARI) N = 40; A=vector(N+1);  \\ N = number of terms
%o A378577 {a(n) = A[n+1] = if(n==0,1, sum(k=0,n-1, A[k+1]*A[n-k]) ); A[n+1] = A[n+1]/2^valuation(A[n+1],2)}
%o A378577 for(n=0,N, print1(a(n),", "))
%Y A378577 Cf. A378578, A000108, A000265.
%K A378577 nonn
%O A378577 0,4
%A A378577 _Paul D. Hanna_, Dec 31 2024