|
| |
|
|
A077028
|
|
The rascal triangle, read by rows: T(n,k) (n >= 0, 0 <= k <= n) = k(n-k) + 1.
|
|
41
|
|
|
|
1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 7, 5, 1, 1, 6, 9, 10, 9, 6, 1, 1, 7, 11, 13, 13, 11, 7, 1, 1, 8, 13, 16, 17, 16, 13, 8, 1, 1, 9, 15, 19, 21, 21, 19, 15, 9, 1, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10, 1, 1, 11, 19, 25, 29, 31, 31, 29, 25, 19, 11, 1, 1, 12, 21, 28, 33, 36, 37, 36, 33, 28, 21, 12, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,5
|
|
|
COMMENTS
|
Pascal's triangle is formed using the rule South = West + East, whereas the rascal triangle uses the rule South = (West*East+1)/North. [Anggoro et al.]
The n-th diagonal is congruent to 1 mod n-1.
Row sums are the cake numbers, A000125. Alternating sum of row n is 0 if n even and (3-n)/2 if n odd. Rows are symmetric, beginning and ending with 1. The number of occurrences of k in this triangle is the number of divisors of k-1, given by A000005.
The triangle can be generated by numbers of the form k*(n-k) + 1 for k = 0 to n. Conjecture: except for n = 0,1 and 6 every row contains a prime. - Amarnath Murthy, Jul 15 2005
Above conjecture needs more exceptions, rows 30 and 54 do not contain primes. - Alois P. Heinz, Aug 31 2017
Consider the semigroup of words in x,y,q subject to the relationships: yx = xyq, qx = xq, qy = yq.
Now take words of length n in x and y, with exactly k y's. If there had been no relationships, the number of different words of this type would be n choose k, sequence A007318. Thanks to the relationships, the number of words of this type is the k-th entry in the n-th row of this sequence (read as a triangle, with the first row indexed by zero and likewise the first entry in each row.)
For example: with three letters and one y, we have three possibilities: xxy, xyx = xxyq, yxx = xxyqq. No two of them are equal, so this entry is still 3, as in Pascal's triangle.
With four letters, two y's, we have the first reduction: xyyx = yxxy = xxyyqq and this is the only reduction for 4 letters. So the middle entry of the fourth row is 5 instead of 6, as in the Pascal triangle. (End)
Main diagonals of this triangle sum to polygonal numbers. See A057145. - Raphie Frank, Oct 30 2012
T(n,k) gives the number of distinct sums of k elements in {1,2,...,n}, e.g., T(5,4) = the number of distinct sums of 4 elements in {1,2,3,4,5}, which is (5+4+3+2) - (4+3+2+1) + 1 = 5. - Derek Orr, Nov 26 2014
Conjecture: excluding the starting and ending 1's in each row, those that contain only prime numbers are n = 2, 3, 5, 7, 13, and 17. Tested up to row 10^9. - Rogério Serôdio, Sep 20 2017
The rascal triangle also uses the rule South = (West+East+1)-North. [Abstracts of AMS, Winter 2019, p. 526, 1145-VS-280, refers to Julian Fleron] - Michael Somos, Jan 12 2019
As a square array read by antidiagonals, selecting terms that give a remainder of 1 when divided by a prime gives evenly sized squares. For example, when each term is divided by 2, showing the remainder looks like:
1 1 1 1 1
1 0 1 0 1
1 1 1 1 1
1 0 1 0 1
T(n,k) is the number of binary words of length n which contain exactly k 1s and have at most 1 ascent. T(n,k) is also the number of ascent sequences avoiding 001 and 210 with length n+1 and exactly k ascents. - Amelia Gibbs, May 21 2024
|
|
|
LINKS
|
A. Anggoro, E. Liu and A. Tulloch, The Rascal Triangle, College Math. J., Vol. 41, No. 5, Nov. 2010, pp. 393-395.
|
|
|
FORMULA
|
As a square array read by antidiagonals, a(n, k) = 1 + n*k. a(n, k) = a(n-1, k) + k. Row n has g.f. (1+(n-1)x)/(1-x)^2, n >= 0. - Paul Barry, Feb 22 2003
Still thinking of square arrays. Let f:N->Z and g:N->Z be given and I an integer, then define a(n, k) = I + f(n)*g(k). Then a(n, k)*a(n-1, k-1) = a(n-1, k)*a(n, k-1) + I*(f(n) - f(n-1))*(g(k) - g(k-1)) for suitable n and k. S= (E*W + 1)/N. arises with I = 1, and f = g = id. - Terry Lindgren, Apr 10 2011
Using the above: Having just read J. Fleron's nice article in Discovering the Art of Mathematics on the rascal triangle, it is neat to note and straightforward to show that when I = 1, a(n, k) + a(n-1, k-1) = a(n-1, k) + a(n, k-1) + (f(n) - f(n-1))*(g(k) - g(k-1)), so with I = 1, and f = g = id, we have S+N = E+W + 1, as his students discovered. - Terry Lindgren, Nov 28 2016
O.g.f. (1 - x*(1 + t) + 2*t*x^2)/((1 - x)^2*(1 - t*x)^2) = 1 + (1 + t)*x + (1 + 2*t + t^2)*x^2 + .... Cf. A105851. - Peter Bala, Jul 26 2015
T(n, k) = 0 if n < k, T(n, 0) = 1, T(n,n) = 1, for n >= 0, and T(n, k) = (T(n-1, k-1)*T(n-1, k) + 1)/(T(n-2, k-1)) for 0 < k < n. See the first comment referring to the triangle with its apex in the middle. - Wolfdieter Lang, Dec 19 2017
|
|
|
EXAMPLE
|
Third diagonal (1,3,5,7,...) consists of the positive integers congruent to 1 mod 2.
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 1 1
2: 1 2 1
3: 1 3 3 1
4: 1 4 5 4 1
5: 1 5 7 7 5 1
6: 1 6 9 10 9 6 1
7: 1 7 11 13 13 11 7 1
8: 1 8 13 16 17 16 13 8 1
9: 1 9 15 19 21 21 19 15 9 1
10: 1 10 17 22 25 26 25 22 17 10 1
As a square array read by antidiagonals, the first rows are:
1, 1, 1, 1, 1, 1, ...
1, 2, 3, 4, 5, 6, ...
1, 3, 5, 7, 9, 11, ...
1, 4, 7, 10, 13, 16, ...
1, 5, 9, 13, 17, 21, ...
|
|
|
MAPLE
|
if n <0 or k<0 or k > n then
0;
else
k*(n-k)+1 ;
end if;
|
|
|
MATHEMATICA
|
t[n_, k_] := k (n - k) + 1; t[0, 0] = 1; Table[ t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Robert G. Wilson v, Jul 06 2012 *)
|
|
|
PROG
|
(PARI) {T(n, k) = if( k<0 || k>n, 0, k * (n - k) + 1)}; /* Michael Somos, Mar 20 2011 */
|
|
|
CROSSREFS
|
The maximum value for each antidiagonal is given by sequence A033638.
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
Better definition based on Murthy's comment of Jul 15 2005 and the Anggoro et al. paper. - N. J. A. Sloane, Mar 05 2011
|
|
|
STATUS
|
approved
|
| |
|
|
