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A378578 G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.

%I A378578 #36 Jan 04 2025 07:38:42
%S A378578 1,1,1,2,13,19,52,412,73,1405,11735,20000,7300,388606,664316,2325118,
%T A378578 20832709,11815463,95438089,861817318,1495813613,5231996647,
%U A378578 47291366710,3025568936,199838851432,1828302724054,3320026962314,439614522008,73390614310810,131344935434920,55179693272894,3321671735661494
%N A378578 G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.
%C A378578 Conjecture: a(n) == binomial(3*n,n)/(2*n+1) (mod 2) for n >= 0.
%H A378578 Paul D. Hanna, <a href="/A378578/b378578.txt">Table of n, a(n) for n = 0..800</a>
%F A378578 a(n) = A038502( Sum_{k=0..n-1} a(k) * Sum_{j=0..n-1-k} a(j)*a(n-1-k-j) ) for n > 0 with a(0) = 1, where A038502(m) = m/3^A007949(m) and A007949(m) = 3-adic valuation of m.
%e A378578 G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 13*x^4 + 19*x^5 + 52*x^6 + 412*x^7 + 73*x^8 + 1405*x^9 + 11735*x^10 + 20000*x^11 + 7300*x^12 + ...
%e A378578 The expansion of A(x)^3 begins
%e A378578 A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 57*x^4 + 156*x^5 + 412*x^6 + 1971*x^7 + 4215*x^8 + 11735*x^9 + 60000*x^10 + 197100*x^11 + ...
%e A378578 where g.f. A(x) is obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.
%e A378578 SPECIFIC VALUES.
%e A378578 A(t) = 8/5 at t = 0.257842038645833456558...
%e A378578 A(t) = 3/2 at t = 0.24869526467110689667648213094860932113462559982219...
%e A378578 A(t) = 4/3 at t = 0.21321674572378383093755902318049913517774115880785...
%e A378578 A(t) = 5/4 at t = 0.18151790234803008203317827057063199289923020437324...
%e A378578 A(t) = 6/5 at t = 0.15638236848650043639095127985605468995430265567872...
%e A378578 A(1/4) = 1.5104498750225954401497052152244291483533940069402...
%e A378578 A(1/5) = 1.2948177731384287040434619555644894329636242990640...
%e A378578 A(1/6) = 1.2192544950152148905144159115908573870687883121699...
%e A378578 A(1/8) = 1.1487089332444818621810139499703458742589412625833...
%e A378578 A(1/9) = 1.1286969902151862545955480786537992451685724113531...
%e A378578 PARITY OF TERMS.
%e A378578 The run lengths of the odd terms, which starts
%e A378578 [3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, ...],
%e A378578 appears to equal A282162 (offset 1), the first differences of the upper Wythoff sequence (A001950).
%e A378578 The run lengths of the even terms, which starts
%e A378578 [1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 42, 1, 2, 5, 1, 10, 1, 2, 85, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 170, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 42, 1, 2, 5, 1, 10, 1, 2, 341, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 42, 1, 2, 5, 1, 10, 1, 2, 85, 1, 2, 5, 1, 10, 1, 2, 21, 1, 2, 5, 1, 682, 1, 2, 5, 1, 10, ...],
%e A378578 appears to equal A085358, the runs of zeros in binomial(3*n,n)/(2*n+1) (mod 2), the records of which are given by A000975 and occur at Fibonacci numbers.
%o A378578 (PARI) N = 30; A=vector(N+1); A[1]=1; \\ N = number of terms
%o A378578 {a(n) = if(n==0,1, A[n+1] = Vec(1 + x*Ser(A)^3)[n+1]; A[n+1] = A[n+1] / 3^valuation(A[n+1], 3) )}
%o A378578 for(n=0, N, print1(a(n), ", "))
%Y A378578 Cf. A378577, A085358, A001764, A282162.
%K A378578 nonn
%O A378578 0,4
%A A378578 _Paul D. Hanna_, Jan 03 2025