A378578 G.f. A(x) equals the series obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3.
1, 1, 1, 2, 13, 19, 52, 412, 73, 1405, 11735, 20000, 7300, 388606, 664316, 2325118, 20832709, 11815463, 95438089, 861817318, 1495813613, 5231996647, 47291366710, 3025568936, 199838851432, 1828302724054, 3320026962314, 439614522008, 73390614310810, 131344935434920, 55179693272894, 3321671735661494
Offset: 0
Keywords
Examples
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 + ... The expansion of A(x)^3 begins 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 + ... where g.f. A(x) is obtained by removing all factors of 3 from the coefficients in 1 + x*A(x)^3. SPECIFIC VALUES. A(t) = 8/5 at t = 0.257842038645833456558... A(t) = 3/2 at t = 0.24869526467110689667648213094860932113462559982219... A(t) = 4/3 at t = 0.21321674572378383093755902318049913517774115880785... A(t) = 5/4 at t = 0.18151790234803008203317827057063199289923020437324... A(t) = 6/5 at t = 0.15638236848650043639095127985605468995430265567872... A(1/4) = 1.5104498750225954401497052152244291483533940069402... A(1/5) = 1.2948177731384287040434619555644894329636242990640... A(1/6) = 1.2192544950152148905144159115908573870687883121699... A(1/8) = 1.1487089332444818621810139499703458742589412625833... A(1/9) = 1.1286969902151862545955480786537992451685724113531... PARITY OF TERMS. The run lengths of the odd terms, which starts [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, ...], appears to equal A282162 (offset 1), the first differences of the upper Wythoff sequence (A001950). The run lengths of the even terms, which starts [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, ...], 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.
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..800
Programs
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PARI
N = 30; A=vector(N+1); A[1]=1; \\ N = number of terms {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) )} for(n=0, N, print1(a(n), ", "))
Comments