%I #23 Oct 06 2015 22:33:14
%S 1,1,2,3,5,9,10,20,22,40,51,67,114,126,203,230,354,468,571,885,908,
%T 1486,1674,2250,3045,3586,5418,5322,8186,9560,12234,16341,17976,27912,
%U 26970,38435,46383,57024,76794,80805,125205,116376,165914,201580,232352,322980,332388,508710,456154,645661,800495,886018,1241190,1226382,1916682,1693454,2290704,2935492
%N G.f. A(x) satisfies: a([n/r^2]) = [x^n] A(x)^2/x and a([n/r^3]) = [x^n] A(x)^3/x^2, for n>=1, where r^2 + r^3 = 1.
%C The integer floor values, [n/r^2] and [n/r^3] where r^2 + r^3 = 1, form Beatty sequences and thus together contain all the positive integers without repetition.
%C Here r = 6 / ( (108 + 12*sqrt(69))^(1/3) + (108 - 12*sqrt(69))^(1/3) ) = 0.75487766624669276.... satisfies r^2 + r^3 = 1.
%C Not equal to A090845.
%C What is the rate of growth of this sequence?
%H Paul D. Hanna, <a href="/A262990/b262990.txt">Table of n, a(n) for n = 1..1000</a>
%e G.f.: A(x) = x + x^2 + 2*x^3 + 3*x^4 + 5*x^5 + 9*x^6 + 10*x^7 + 20*x^8 + 22*x^9 + 40*x^10 + 51*x^11 + 67*x^12 + 114*x^13 + 126*x^14 + 203*x^15 +...
%e where the terms are formed from the union of coefficients in A(x)^2 and A(x)^3.
%e The coefficients of A(x)^2 begin:
%e A^2 = [1, 2, 5, 10, 20, 40, 67, 126, 203, 354, 571, 908, 1486, 2250, 3586, 5322, 8186, 12234, 17976, 26970, 38435, 57024, 80805, 116376, 165914, 232352,...]
%e and form the terms of this sequence at positions [n/r^2] for n>=1:
%e {[n/r^2]} = [1, 3, 5, 7, 8, 10, 12, 14, 15, 17, 19, 21, 22, 24, 26, 28, 29, 31, 33, 35, 36, 38, 40, 42, 43, 45, 47, 49, 50, 52, 54, 56, 57, 59, ...].
%e The coefficients of A(x)^3 begin:
%e A^3 = [1, 3, 9, 22, 51, 114, 230, 468, 885, 1674, 3045, 5418, 9560, 16341, 27912, 46383, 76794, 125205, 201580, 322980, 508710, 800495, 1241190, ...]
%e and form the terms of this sequence at positions [n/r^3] for n>=1:
%e {[n/r^3]} = [2, 4, 6, 9, 11, 13, 16, 18, 20, 23, 25, 27, 30, 32, 34, 37, 39, 41, 44, 46, 48, 51, 53, 55, 58, 60, 62, 65, 67, 69, 72, ...].
%o (PARI) {a(n) = local(A=vector(n+1),B=A,C=A,r=6/((108+12*sqrt(69))^(1/3)+(108-12*sqrt(69))^(1/3))); A[1]=1;A[2]=1;
%o for(i=1,ceil(log(#A)/log(1/r)),
%o B=vector(floor(#A/r^2));for(n=1,#A,m=floor(n/r^2);if(m<=#B,B[m]=Vec(Ser(A)^2)[n]));
%o C=vector(floor(#A/r^3));for(n=1,#A,m=floor(n/r^3);if(m<=#C,C[m]=Vec(Ser(A)^3)[n]));
%o A=vector(#A,n,if(C[n]==0,B[n],C[n]));); A[n]}
%o for(n=1,80,print1(a(n),", "))
%Y Cf. A075778.
%K nonn
%O 1,3
%A _Paul D. Hanna_, Oct 06 2015