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Revision History for A153281

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Showing entries 1-10 | older changes
Triangle read by rows, A130321 * A127647. Also, number of subsets of [n+2] with consecutive integers that start at k.
(history; published version)
#17 by Harvey P. Dale at Sun Apr 26 11:39:55 EDT 2020
STATUS

editing

approved

#16 by Harvey P. Dale at Sun Apr 26 11:39:52 EDT 2020
MATHEMATICA

Table[2^(n+1-k) Fibonacci[k], {n, 0, 10}, {k, n+1}]//Flatten (* Harvey P. Dale, Apr 26 2020 *)

STATUS

approved

editing

#15 by Joerg Arndt at Sat Aug 12 10:39:04 EDT 2017
STATUS

editing

approved

#14 by Joerg Arndt at Sat Aug 12 10:38:23 EDT 2017
MAPLE

>with(combinat, fibonacci):

>seq(seq(2^(n+1-k)*fibonacci(k), k=1..(n+1)), n=0..10);

STATUS

proposed

editing

Discussion
Sat Aug 12
10:39
Joerg Arndt: Remove prompts.
#13 by Jon E. Schoenfield at Sat Aug 12 10:34:58 EDT 2017
STATUS

editing

proposed

Discussion
Sat Aug 12
10:35
Michel Marcus: keep the > in maple ??
10:37
Jon E. Schoenfield: I don't know -- are those prompts?
#12 by Jon E. Schoenfield at Sat Aug 12 10:33:35 EDT 2017
COMMENTS

Row sums = A008466(k-2): (1, 3, 8, 19, 43, 94, ...).

T(n,k) is the number of subsets of {1,...,n+2} that contain consecutive integers and that have k as the first integer in the first consecutive string. (See the example below.) Hence rows sums of T(n,k) give the number of subsets of {1,...,n+2} that contain consecutive integers. Also, T(n,k) = F(k)*2^(n+1-k), where F(k) is the k-th Fibonacci number, since there are F(k) subsets of {1,...,k-2} that contain no consecutive integers and there are 2^(n+1-k) subsets of {k+2,...,n+2}. [From _Dennis P. Walsh, _, Dec 21 2011]

FORMULA

Triangle read by rows, A130321 * A127647. A130321 = an infinite lower triangular matrix with powers of 2: (A000079) in every column: (1, 2, 4, 8, ...).

matrix with powers of 2: (A000079) in every column: (1, 2, 4, 8,...).

A127647 = an infinite lower triangular matrix with the Fibonacci numbers, A000045 as the main diagonal and the rest zeros.

A000045 as the main diagonal and T(n,k)=2^(n+1-k)*F(k) where F(k) is the rest zerosk-th Fibonacci number. [_Dennis Walsh_, Dec 21 2011]

T(n,k)=2^(n+1-k)*F(k) where F(k) is the k-th Fibonacci number. [From Dennis Walsh, Dec 21 2011]

EXAMPLE

First few rows of the triangle =:

1;

2, 1;

4, 2, 2;

8, 4, 4, 3;

16, 8, 8, 6, 5;

32, 16, 16, 12, 10, 8;

64, 32, 32, 24, 20, 16, 13;

128, 64, 64, 48, 40, 32, 26, 21;

256, 128, 128, 96, 80, 64, 52, 42, 34;

512, 256, 256, 192, 160, 128, 104, 84, 68, 55;

...

...

{1,3,4}, {1,3,4,5}, {1,3,4,6}, {1,3,4,7}, {1,3,4,5,6}, {1,3,4,5,7}, {1,3,4,6,7}, {1,3,4,5,6,7}, {3,4}, {3,4,5}, {3,4,6}, {3,4,7}, {3,4,5,6}, {3,4,5,7}, {3,4,6,7}, and {3,4,5,6,7}. [_Dennis P. Walsh_, Dec 21 2011]

[From Dennis Walsh, Dec 21 2011]

CROSSREFS
STATUS

approved

editing

#11 by Joerg Arndt at Mon Dec 24 02:07:17 EST 2012
STATUS

proposed

approved

#10 by Michel Marcus at Sun Dec 23 09:47:20 EST 2012
STATUS

editing

proposed

#9 by Michel Marcus at Sun Dec 23 09:44:35 EST 2012
COMMENTS

T(n,k) is the number of subsets of {1,...,n+2} that contain consecutive integers and that have k as the first integer in the first consecutive string. (See the example below.) Hence rows sums of T(n,k)give the number of subsets of {1,...,n+2} that contain consecutive integers. Also, T(n,k)=F(k)*2^(n+1-k), where F(k) is the k-th Fibonacci number, since there are F(k) subsets of {1,...,k-2} that contain no consective consecutive integers and there are 2^(n+1-k) subsets of {k+2,...,n+2}. [From Dennis P. Walsh, Dec 21 2011]

STATUS

approved

editing

#8 by Russ Cox at Fri Mar 30 17:25:33 EDT 2012
AUTHOR

_Gary W. Adamson (qntmpkt(AT)yahoo.com), _, Dec 23 2008

Discussion
Fri Mar 30
17:25
OEIS Server: https://oeis.org/edit/global/135