Search: id:A077028
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%I A077028
%S A077028 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,
%T A077028 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,
%U A077028 26,25,22,17,10,1,1,11,19,25,29,31,31,29,25,19,11,1,1,12,21,28,33,36
%N A077028 Triangle with diagonal n congruent to 1 mod (n-1).
%C A077028 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.
%C A077028 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
%C A077028 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
%C A077028 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.)
%C A077028 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.
%C A077028 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)
%F A077028 t(i, j)=(i-j)(j-1)+1.
%F A077028 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
%e A077028 Third diagonal (1,3,5,7,...) consists of the positive integers congruent 
               to 1 mod 2.
%e A077028 Triangle begins:
%e A077028 1
%e A077028 1 1
%e A077028 1 2 1
%e A077028 1 3 3 1
%e A077028 1 4 5 4 1
%e A077028 1 5 7 7 5 1
%e A077028 1 6 9 10 9 6 1
%e A077028 ...
%e A077028 As a square array read by antidiagonals, the first rows are:
%e A077028 1 1 1. 1. 1. 1 ...
%e A077028 1 2 3. 4. 5. 6 ...
%e A077028 1 3 5. 7. 9 11 ...
%e A077028 1 4 7 10 13 16 ...
%e A077028 1 5 9 13 17 21 ...
%Y A077028 Cf. A077029, A003991.
%Y A077028 The maximum value for each anti-diagonal is sequence A033638.
%Y A077028 A004247(n) + 1.
%Y A077028 Sequence in context: A099573 A107430 A132892 this_sequence A114225 A072704 
               A038792
%Y A077028 Adjacent sequences: A077025 A077026 A077027 this_sequence A077029 A077030 
               A077031
%K A077028 nonn,tabl
%O A077028 1,5
%A A077028 Clark Kimberling (ck6(AT)evansville.edu), Oct 19 2002

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