

A124801


Triangle, row sums = Fibonacci numbers in two ways.


1



1, 1, 0, 1, 0, 1, 1, 0, 3, 1, 1, 0, 6, 4, 1, 1, 0, 10, 10, 10, 3, 1, 0, 15, 20, 30, 18, 5, 1, 0, 21, 35, 70, 63, 35, 8, 1, 0, 28, 56, 140, 168, 140, 64, 13
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,9


COMMENTS

nth row sum of signed terms = Fn; nth row sum of unsigned terms = F(2n3).


LINKS

Table of n, a(n) for n=1..45.


FORMULA

Diagonalize the inverse binomial transform of the Fibonacci sequence as an infinite matrix, M; and P = Pascal's triangle as an infinite lower triangular matrix. The triangle A124802 = P*M, with the zeros deleted.


EXAMPLE

A039834, (1, 0, 1, 1, 2, 3, 5, 8, 13...) = the diagonal of M, then the first few rows of P*M =
1;
1, 0;
1, 0, 1;
1, 0, 3, 1;
1, 0, 6, 4, 2;
1, 0, 10, 10, 10, 3;
1, 0, 15, 20, 30, 18, 5;
1, 0, 21, 35, 70, 63, 35, 8;
...
Row 7 terms (signed) = F7 = 13 = (1 + 15 20 + 30  18 + 5).
Row 7 terms (unsigned) = F11 = 28 = (1 + 15 + 20 + 30 + 18 + 5).


CROSSREFS

Cf. A039834, A124802.
Sequence in context: A287822 A162169 A216954 * A124926 A175946 A115378
Adjacent sequences: A124798 A124799 A124800 * A124802 A124803 A124804


KEYWORD

tabl,sign


AUTHOR

Gary W. Adamson, Nov 08 2006


STATUS

approved



