A090845
Let A denote the sequence; A is equal to the union of the self-convolutions A^2 and A^3, with terms in ascending order by size.
Original entry on oeis.org
1, 1, 2, 3, 5, 9, 10, 20, 22, 40, 51, 67, 114, 126, 203, 230, 354, 468, 571, 885, 908, 1486, 1674, 2250, 3045, 3586, 5322, 5418, 8186, 9560, 12234, 16341, 17976, 26970, 27912, 38435, 46383, 57024, 76794, 80805, 116376, 125205, 165914, 201580, 232352
Offset: 0
A={1,1,2,3,5,9,10,20,22,40,51,...} since A is the sorted union of:
A^2={1,2,5,10,20,40,67,126,203,354,571,908,1486,2250,3586,...} and
A^3={1,3,9,22,51,114,230,468,885,1674,3045,5418,9560,16341,...}.
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{a(n)=local(A=[1,1]);for(i=1,#binary(3*n+1),A=vecsort(concat(Vec(Ser(A)^2),Vec(Ser(A)^3))));A[n+1]}
for(n=0,60,print1(a(n),", "))
Original entry on oeis.org
1, 3, 9, 22, 51, 114, 230, 468, 885, 1674, 3045, 5418, 9560, 16341, 27912, 46383, 76794, 125205, 201580, 322980, 508710, 800495, 1241190, 1916682, 2935492, 4456617, 6747393, 10101532, 15105042, 22378362, 33035166, 48520809, 70776711, 103072393, 148899756
Offset: 0
G.f.: A(x) = 1 + 3*x + 9*x^2 + 22*x^3 + 51*x^4 + 114*x^5 + 230*x^6 +...
Let G(x) = A(x)^(1/3) denote the g.f. of A090845:
G(x) = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 9*x^5 + 10*x^6 + 20*x^7 + 22*x^8 + 40*x^9 + 51*x^10 + 67*x^11 + 114*x^12 + 126*x^13 + 203*x^14 +...
then the coefficients of G(x)^2 and G(x)^3 begin:
G(x)^2: [1, 2, 5, 10, 20, 40, 67, 126, 203, 354, 571, 908, 1486, ...];
G(x)^3: [1, 3, 9, 22, 51, 114, 230, 468, 885, 1674, 3045, 5418, ..];
where the sorted union of these coefficients yield sequence A090845.
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{a(n)=local(A=[1, 1]); for(i=1, #binary(3*n+1), A=vecsort(concat(Vec(Ser(A)^2), Vec(Ser(A)^3)))); Vec(Ser(A)^3)[n+1]}
for(n=0, 60, print1(a(n), ", "))
Showing 1-2 of 2 results.
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