|
|
A284168
|
|
Integers n such that sigma(binomial(n,k)) = sigma(binomial(n-1,k-1)) + sigma(binomial(n-1,k)) for some k.
|
|
1
|
|
|
3, 6, 15, 25, 27, 30, 35, 40, 48, 50, 54, 60, 63, 66, 78, 80, 100, 108, 112, 118, 120, 123, 124, 126, 140, 144, 158, 175, 192, 198, 200, 207, 216, 220, 224, 225, 232, 238, 243, 247, 304, 310, 316, 319, 341, 345, 348, 358, 364, 368, 375, 385, 391, 408, 416, 425, 432
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Consider the triangle formed by replacing each m in Pascal's triangle with sigma(m). Then this sequence consists of the row indices where there is a term that is equal to the sum of its NW and N neighbors as in a Pascal triangle.
|
|
LINKS
|
|
|
EXAMPLE
|
Here is the triangle also described in A074801.
1,
1, 1,
1, 3, 1,
1, 4, 4, 1,
1, 7, 12, 7, 1,
1, 6, 18, 18, 6, 1,
On row index 3, we have 4 which is the sum of 1 and 3 its NW and N neighbors.
So a(1)= 3, and its column index is 1 which will be corresponding value in A284169.
|
|
PROG
|
(PARI) T(n, k) = sigma(binomial(n, k));
isokT(n, k) = T(n-1, k-1) + T (n-1, k) == T(n, k);
isokn(n) = for (k=1, n-1, if (isokT(n, k), return(1)));
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|