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A130154 Triangle read by rows: T(n, k) = 1 + 2*(n-k)*(k-1) (1 <= k <= n).

%I A130154 #51 Jan 27 2025 02:28:31
%S A130154 1,1,1,1,3,1,1,5,5,1,1,7,9,7,1,1,9,13,13,9,1,1,11,17,19,17,11,1,1,13,
%T A130154 21,25,25,21,13,1,1,15,25,31,33,31,25,15,1,1,17,29,37,41,41,37,29,17,
%U A130154 1,1,19,33,43,49,51,49,43,33,19,1,1,21,37,49,57,61,61,57,49,37,21,1
%N A130154 Triangle read by rows: T(n, k) = 1 + 2*(n-k)*(k-1) (1 <= k <= n).
%C A130154 Column k, except for the initial k-1 0's, is an arithmetic progression with first term 1 and common difference 2(k-1). Row sums yield A116731. First column of the inverse matrix is A129779.
%C A130154 Studied by _Paul Curtz_ circa 1993.
%C A130154 From _Rogério Serôdio_, Dec 19 2017: (Start)
%C A130154 T(n, k) gives the number of distinct sums of 2(k-1) elements in {1,1,2,2,...,n-1,n-1}. For example, T(6, 2) = the number of distinct sums of 2 elements in {1,1,2,2,3,3,4,4,5,5}, and because each sum from the smallest 1 + 1 = 2 to the largest 5 + 5 = 10 appears, T(6, 2) = 10 - 1 = 9. [In general: 2*(Sum_{j=1..(k-1)} n-j) - (2*(Sum_{j=1..k-1} j) - 1) = 2*n*(k-1) - 4*(k-1)*k/2 + 1 = 2*(k-1)*(n-k) + 1 = T(n, k). - _Wolfdieter Lang_, Dec 20 2017]
%C A130154 T(n, k) is the number of lattice points with abscissa x = 2*(k-1) and even ordinate in the closed region bounded by the parabola y = x*(2*(n-1) - x) and the x axis. [That is, (1/2)*y(2*(k-1)) + 1 = T(n, k). - _Wolfdieter Lang_, Dec 20 2017]
%C A130154 Pascal's triangle (A007318, but with apex in the middle) is formed using the rule South = West + East; the rascal triangle A077028 uses the rule South = (West*East + 1)/North; the present triangle uses a similar rule: South = (West*East + 2)/North. See the formula section for this recurrence. (End)
%H A130154 G. C. Greubel, <a href="/A130154/b130154.txt">Rows n = 1..100 of triangle, flattened</a>
%F A130154 T(n, k) = 1 + 2*(n-k)*(k-1) (1 <= k <= n).
%F A130154 G.f.: G(t,z) = t*z*(3*t*z^2 - z - t*z + 1)/((1-t*z)*(1-z))^2.
%F A130154 Equals = 2 * A077028 - A000012 as infinite lower triangular matrices. - _Gary W. Adamson_, Oct 23 2007
%F A130154 T(n, 1) = 1 and  T(n, n) = 1 for n >= 1; T(n, k) = (T(n-1, k-1)*T(n-1, k) + 2)/T(n-2, k-1), for n > 2 and 1 < k < n. See a comment above. - _Rogério Serôdio_, Dec 19 2017
%F A130154 G.f. column k (with leading zeros): (x^k/(1-x)^2)*(1 + (2*k-3)*x), k >= 1. See the g.f. of the triangle G(t,z) above: (d/dt)^k G(t,x)/k!|_{t=0}. - _Wolfdieter Lang_, Dec 20 2017
%e A130154 The triangle T(n, k) starts:
%e A130154   n\k  1  2  3  4  5  6  7  8  9 10 ...
%e A130154   1:   1
%e A130154   2:   1  1
%e A130154   3:   1  3  1
%e A130154   4:   1  5  5  1
%e A130154   5:   1  7  9  7  1
%e A130154   6:   1  9 13 13  9  1
%e A130154   7:   1 11 17 19 17 11  1
%e A130154   8:   1 13 21 25 25 21 13  1
%e A130154   9:   1 15 25 31 33 31 25 15  1
%e A130154  10:   1 17 29 37 41 41 37 29 17  1
%e A130154  ... reformatted. - _Wolfdieter Lang_, Dec 19 2017
%p A130154 T:=proc(n,k) if k<=n then 2*(n-k)*(k-1)+1 else 0 fi end: for n from 1 to 14 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
%t A130154 Flatten[Table[1+2(n-k)(k-1),{n,0,20},{k,n}]] (* _Harvey P. Dale_, Jul 13 2013 *)
%o A130154 (PARI) T(n, k) = 1 + 2*(n-k)*(k-1) \\ _Iain Fox_, Dec 19 2017
%o A130154 (PARI) first(n) = my(res = vector(binomial(n+1,2)), i = 1); for(r=1, n, for(k=1, r, res[i] = 1 + 2*(r-k)*(k-1); i++)); res \\ _Iain Fox_, Dec 19 2017
%o A130154 (Magma) [1 + 2*(n-k)*(k-1): k in [1..n], n in [1..12]]; // _G. C. Greubel_, Nov 25 2019
%o A130154 (Sage) [[1 + 2*(n-k)*(k-1) for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, Nov 25 2019
%o A130154 (GAP) Flat(List([1..12], n-> List([1..n], k-> 1 + 2*(n-k)*(k-1) ))); # _G. C. Greubel_, Nov 25 2019
%Y A130154 Cf. A116731, A129779. A007318, A077028.
%Y A130154 Column sequences (no leading zeros): A000012, A016813, A016921, A017077, A017281, A017533, A131877, A158057, A161705, A215145.
%K A130154 nonn,tabl,easy
%O A130154 1,5
%A A130154 _Emeric Deutsch_, May 22 2007
%E A130154 Edited by _Wolfdieter Lang_, Dec 19 2017