A215674
a(1) = 1, a(n) = 2 if 1
Original entry on oeis.org
1, 2, 2, 4, 4, 3, 5, 5, 3, 7, 7, 5, 9, 9, 5, 8, 8, 4, 9, 9, 6, 11, 11, 6, 9, 9, 4, 11, 11, 8, 15, 15, 8, 13, 13, 6, 15, 15, 10, 19, 19, 10, 15, 15, 6, 14, 14, 9, 17, 17, 9, 13, 13, 5, 14, 14, 10, 19, 19, 10, 16, 16, 7, 18, 18, 12, 23, 23, 12, 18, 18, 7, 16
Offset: 1
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a:= proc(n) option remember; 1+ `if`(n=1, 0, `if`(n<=3, 1,
`if`(irem(n, 3, 'r')=0, a(r), a(r)+a(r+1))))
end:
seq (a(n), n=1..80); # Alois P. Heinz, Aug 23 2012
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a[n_] := a[n] = If[n < 3, n, {q, r} = QuotientRemainder[n, 3];
Switch[r, 0, a[q] + 1, 1|2, a[q] + a[q+1] + 1]];
Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Apr 24 2022 *)
A215675
a(1) = 1, a(n) = 2 if 1
Original entry on oeis.org
1, 2, 2, 4, 3, 5, 3, 7, 5, 8, 4, 9, 6, 9, 4, 11, 8, 13, 6, 14, 9, 13, 5, 14, 10, 16, 7, 16, 10, 14, 5, 16, 12, 20, 9, 22, 14, 20, 7, 21, 15, 24, 10, 23, 14, 19, 6, 20, 15, 25, 11, 27, 17, 24, 8, 24, 17, 27, 11, 25, 15, 20, 6, 22, 17, 29, 13, 33, 21, 30, 10
Offset: 1
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a:= proc(n) option remember; 1+ `if`(n=1, 0, `if`(n<=3, 1,
`if`(irem(n-1, 2, 'r')=0, a(r), a(r)+a(r+1))))
end:
seq (a(n), n=1..80); # Alois P. Heinz, Aug 23 2012
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a[n_] := a[n] = If[n < 3, n, {q, r} = QuotientRemainder[n, 2];
Switch[r, 1, a[q] + 1, 0, a[q-1] + a[q] + 1]];
Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Apr 24 2022 *)
A215676
a(1) = 1, a(n) = 2 if 1
Original entry on oeis.org
1, 2, 2, 2, 4, 4, 3, 5, 5, 3, 5, 5, 3, 7, 7, 5, 9, 9, 5, 8, 8, 4, 9, 9, 6, 11, 11, 6, 9, 9, 4, 9, 9, 6, 11, 11, 6, 9, 9, 4, 11, 11, 8, 15, 15, 8, 13, 13, 6, 15, 15, 10, 19, 19, 10, 15, 15, 6, 14, 14, 9, 17, 17, 9, 13, 13, 5, 14, 14, 10, 19, 19, 10, 16, 16, 7
Offset: 1
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a:= proc(n) option remember; 1+ `if`(n=1, 0, `if`(n<=4, 1,
`if`(irem(n-1, 3, 'r')=0, a(r), a(r)+a(r+1))))
end:
seq (a(n), n=1..80); # Alois P. Heinz, Aug 23 2012
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a[n_] := a[n] = Switch[n, 1, 0, 2|3|4, 1, _, {q, r} = QuotientRemainder[n-1, 3]; If[r == 0, a[q], a[q]+a[q+1]]]+1;
Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Sep 01 2022 *)
A294991
Let S be the sequence of integer sets defined by the following rules: S(0) = {0}, S(1) = {1} and for any k > 0, S(2*k) = {2*k} U S(k) and S(2*k+1) = {2*k+1} U S(k) U S(k+1) (where X U Y denotes the union of the sets X and Y); a(n) = the number of elements of S(n).
Original entry on oeis.org
1, 1, 2, 3, 3, 4, 4, 5, 4, 6, 5, 6, 5, 7, 6, 7, 5, 8, 7, 8, 6, 8, 7, 8, 6, 9, 8, 9, 7, 9, 8, 9, 6, 10, 9, 10, 8, 10, 9, 10, 7, 10, 9, 10, 8, 10, 9, 10, 7, 11, 10, 11, 9, 11, 10, 11, 8, 11, 10, 11, 9, 11, 10, 11, 7, 12, 11, 12, 10, 12, 11, 12, 9, 12, 11, 12, 10
Offset: 0
The first terms, alongside the corresponding set S(n), are:
n a(n) S(n)
-- ---- -----
0 1 { 0 }
1 1 { 1 }
2 2 { 1, 2 }
3 3 { 1, 2, 3 }
4 3 { 1, 2, 4 }
5 4 { 1, 2, 3, 5 }
6 4 { 1, 2, 3, 6 }
7 5 { 1, 2, 3, 4, 7 }
8 4 { 1, 2, 4, 8 }
9 6 { 1, 2, 3, 4, 5, 9 }
10 5 { 1, 2, 3, 5, 10 }
11 6 { 1, 2, 3, 5, 6, 11 }
12 5 { 1, 2, 3, 6, 12 }
13 7 { 1, 2, 3, 4, 6, 7, 13 }
14 6 { 1, 2, 3, 4, 7, 14 }
15 7 { 1, 2, 3, 4, 7, 8, 15 }
16 5 { 1, 2, 4, 8, 16 }
17 8 { 1, 2, 3, 4, 5, 8, 9, 17 }
18 7 { 1, 2, 3, 4, 5, 9, 18 }
19 8 { 1, 2, 3, 4, 5, 9, 10, 19 }
20 6 { 1, 2, 3, 5, 10, 20 }
See also illustration of the first terms in Links section.
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a(n) = my (S = Set(n), u = 1); while (u <= #S, my (v = S[#S-u+1]); if (v>1, if (v%2==0, S = setunion(S, Set(v/2)), S = setunion(S, Set([(v-1)/2, (v+1)/2])))); u++;); return (#S)
Showing 1-4 of 4 results.
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