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A130753
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A folded-back triangular sequence based on symmetry of CnH2*n+2 straight chain alkanes and the number of hydrogen atoms of a given symmetry type: Besides methane at 4 there are only three symmetry types: two CH3's->6: a single CH2->2, two CH2's->4.
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0
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4, 6, 6, 2, 6, 4, 6, 4, 2, 6, 4, 4, 6, 4, 4, 2, 6, 4, 4, 4, 6, 4, 4, 4, 2, 6, 4, 4, 4, 4, 6, 4, 4, 4, 4, 2, 6, 4, 4, 4, 4, 4, 6, 4, 4, 4, 4, 4, 2, 6, 4, 4, 4, 4, 4, 4, 6, 4, 4, 4, 4, 4, 4, 2, 6, 4, 4, 4, 4, 4, 4, 4, 6, 4, 4, 4, 4, 4, 4, 4, 2, 6, 4, 4, 4, 4, 4, 4, 4, 4, 6, 4, 4, 4, 4, 4, 4, 4, 4, 2, 6, 4, 4, 4, 4
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OFFSET
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1,1
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LINKS
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FORMULA
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a(n,m) = If[n == m == 1, 4, If[n == 1 && m >1, 6, If[Mod[m, 2] == 1 && n == Floor[m/2] + 1, 2, 4]]]
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EXAMPLE
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{4},->Ch4
{6}, ->CH3CH3
{6, 2}, ->CH3CH2CH3
{6, 4}, ->CH3CH2CH2CH3
{6, 4, 2}, ->CH3CH2CH2CH2CH3
{6, 4, 4}, ->CH3CH2CH2CH2CH2CH3
{6, 4, 4, 2}, ->CH3CH2CH2CH2CH2CH2CH3
{6, 4, 4, 4}, ->CH3CH2CH2CH2CH2CH2CH2CH3
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MATHEMATICA
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f[n_, m_] = If[n == m ==1, 4, If[n == 1 && m > 1, 6, If[Mod[m, 2] == 1 && n == Floor[m/2] + 1, 2, 4]]]; Table[Table[f[n, m], {n, 1, If[m == 1, 1, If[Mod[m, 2] == 0, Floor[m/2], Floor[m/2] + 1]]}], {m, 1, 20}]; Flatten[%]
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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