A236109 - OEIS

%I #27 Mar 13 2015 23:51:54

%S 0,0,0,0,0,1,0,0,0,1,0,0,0,2,2,0,0,0,0,1,2,0,0,0,0,2,3,3,0,0,0,0,0,2,

%T 3,3,0,0,0,0,0,2,2,4,4,0,0,0,0,0,0,2,3,4,4,0,0,0,0,0,0,3,4,5,5,5,0,0,

%U 0,0,0,0,0,2,2,3,5,5,0,0,0,0,0,0,0,3

%N Triangle read by rows: another version of A048158, only here the representation of A004125 is symmetric, as in the representation of A024916 and A000203.

%C Row sums give A004125.

%C For more information see A236104, A237591, A237593, A237270.

%e Triangle begins:

%e 0;

%e 0, 0;

%e 0, 0, 1;

%e 0, 0, 0, 1;

%e 0, 0, 0, 2, 2;

%e 0, 0, 0, 0, 1, 2;

%e 0, 0, 0, 0, 2, 3, 3;

%e 0, 0, 0, 0, 0, 2, 3, 3;

%e 0, 0, 0, 0, 0, 2, 2, 4, 4;

%e 0, 0, 0, 0, 0, 0, 2, 3, 4, 4;

%e 0, 0, 0, 0, 0, 0, 3, 4, 5, 5, 5;

%e 0, 0, 0, 0, 0, 0, 0, 2, 2, 3, 5, 5;

%e ...

%e For the symmetric representation of A000203, A024916, A004125 in the fourth quadrant using a diagram which arises from the sequence A236104 see below:

%e --------------------------------------------------

%e n     A000203  A024916            Diagram

%e --------------------------------------------------

%e .                         _ _ _ _ _ _ _ _ _ _ _ _

%e 1        1        1      |_| | | | | | | | | | | |

%e 2        3        4      |_ _|_| | | | | | | | | |

%e 3        4        8      |_ _|  _|_| | | | | | | |

%e 4        7       15      |_ _ _|    _|_| | | | | |

%e 5        6       21      |_ _ _|  _|  _ _|_| | | |

%e 6       12       33      |_ _ _ _|  _| |  _ _|_| |

%e 7        8       41      |_ _ _ _| |_ _|_|    _ _|

%e 8       15       56      |_ _ _ _ _|  _|     |* *

%e 9       13       69      |_ _ _ _ _| |      _|* *

%e 10      18       87      |_ _ _ _ _ _|  _ _|* * *

%e 11      12       99      |_ _ _ _ _ _| |* * * * *

%e 12      28      127      |_ _ _ _ _ _ _|* * * * *

%e .

%e The 12th row is ........ 0,0,0,0,0,0,0,2,2,3,5,5

%e .

%e The total number of cells in the first n set of symmetric regions of the diagram equals A024916(n). It appears that the total number of cells in the n-th set of symmetric regions of the diagram equals sigma(n) = A000203(n). Example: for n = 12 the 12th row of triangle is 144, 25, 9, 1, hence the alternating sums is 144 - 25 + 9 - 1 = 127. On the other hand we have that A000290(12) - A004125(12) = 144 - 17 = A024916(12) = 127, equaling the total number of cells in the diagram after 12 stages. The number of cells in the 12th set of symmetric regions of the diagram is sigma(12) = A000203(12) = 28. Note that in this case there is only one region. The number of "*"'s is A004125(12) = 17.

%Y Cf. A000203, A004125, A024916, A048158, A196020, A235799, A236104, A236630, A236631, A237591, A237593, A237270.

%K nonn,tabl

%O 1,14

%A _Omar E. Pol_, Jan 26 2014