A233775 Number of vertices in the n-th row of the Sierpinski gasket (cf. A047999).
1, 2, 3, 4, 5, 4, 6, 8, 9, 4, 6, 8, 10, 8, 12, 16, 17, 4, 6, 8, 10, 8, 12, 16, 18, 8, 12, 16, 20, 16, 24, 32, 33, 4, 6, 8, 10, 8, 12, 16, 18, 8, 12, 16, 20, 16, 24, 32, 34, 8, 12, 16, 20, 16, 24, 32, 36, 16, 24, 32, 40, 32, 48, 64, 65, 4, 6, 8, 10, 8, 12
Offset: 0
Examples
Illustration of initial terms:
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Diagram n a(n) A233774(n)
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* 0 1 1
/T\
*---* 1 2 3
/T\ /T\
*---*---* 2 3 6
/T\ /T\
*---* *---* 3 4 10
/T\ /T\ /T\ /T\
*---*---*---*---* 4 5 15
/T\ /T\
*---* *---* 5 4 19
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After five stages the number of "black" triangles in the structure is A006046(5) = 11 and the number of "black" triangles in row 5 is A001316(5-1) = 2. The number of vertices in row 5 is equal to 4, so a(5) = 4.
Written as an irregular triangle the sequence begins:
1;
2;
3;
4,5;
4,6,8,9;
4,6,8,10,8,12,16,17;
4,6,8,10,8,12,16,18,8,12,16,20,16,24,32,33;
...
Links
- N. J. A. Sloane, Table of n, a(n) for n = 0..10000
- N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
- Eric Weisstein's World of Mathematics, Sierpiński Gasket Graph.
- Wikipedia, Sierpinski triangle.
- Index entries for coordination sequences
Programs
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Maple
A000120 := n -> add(i, i=convert(n, base, 2)): A007814 := n -> padic[ordp](n, 2): A233775 := n->(2^A007814(n)+1)*(2^(A000120(n)-1); # N. J. A. Sloane, Sep 19 2020
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Mathematica
A233775[n_] := If[n == 0, 1, (2^IntegerExponent[n, 2]+1)*2^(DigitSum[n, 2]-1)]; Array[A233775, 100, 0] (* Paolo Xausa, Aug 05 2024 *)
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PARI
print1("1, "); for(k=1,70, print1((2^valuation(k,2)+1) *2^(hammingweight(k)-1),", ")) \\ Hugo Pfoertner, Jul 27 2020
Formula
a(0)=1, a(n) = (2^t(n) + 1) * 2^(c(n) - 1) where t(n) = A007814(n) is the number of trailing zeros in the binary representation of n and c(n) = A000120(n) is the total number of ones in the binary representation of n. - Johan Falk, Jun 24 2020
Comments