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A077028
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The rascal triangle, read by rows: T(n,k) (n >= 0, 0 <= k <= n) = k(n-k)+1.
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16
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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 1,5
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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 (amarnath_murthy(AT)yahoo.com), Jul 15 2005
Comments from Moshe Newman (mshnoiman(AT)hotmail.com), Apr 06 2008: (Start) 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)
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REFERENCES
| A. Anggoro, E. Liu and A. Tulloch, The Rascal Triangle, College Math. J., Vol. 41, No. 5, Nov. 2010, pp. 393-395.
L. McHugh, CMJ Article Shows Collaboration Is Not Limited by Geography ... or Age, MAA Focus (Magazine), Vol. 31, No. 1, 2011, p. 13.
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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 (pbarry(AT)wit.ie), 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]
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EXAMPLE
| Third diagonal (1,3,5,7,...) consists of the positive integers congruent to 1 mod 2.
Triangle begins:
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
...
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 ...
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PROG
| (PARI) {T(n, k) = if( k<0 | k>n, 0, k * (n - k) + 1)} /* Michael Somos Mar 20 2011 */
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CROSSREFS
| Cf. A077029, A003991.
The maximum value for each anti-diagonal is given by sequence A033638.
Equals A004247(n) + 1.
Sequence in context: A107430 A132892 A174448 * A114225 A193515 A072704
Adjacent sequences: A077025 A077026 A077027 * A077029 A077030 A077031
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KEYWORD
| nonn,tabl
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AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu), Oct 19 2002
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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.
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