A223544 T(n, k) = n*k - 1.

0, 1, 3, 2, 5, 8, 3, 7, 11, 15, 4, 9, 14, 19, 24, 5, 11, 17, 23, 29, 35, 6, 13, 20, 27, 34, 41, 48, 7, 15, 23, 31, 39, 47, 55, 63, 8, 17, 26, 35, 44, 53, 62, 71, 80, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99, 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, 11, 23, 35, 47, 59, 71, 83, 95, 107, 119, 131, 143

Offset: 1

Author

Richard R. Forberg, Jul 19 2013

Keywords

• nonn
• tabl

Comments

Previous name was: Triangle T(n,k), 0 < k <= n, T(n,1) = n - 1, T(n,k) = T(n,k-1) + n; read by rows.

This simple triangle arose analyzing f(x) = x/(n + e^(c/x)), for n <> 0. f(x) converges towards a rational number for large values of x, if x is rational. T(n+1,k)/(n+1)^2 equals the fractional portion of f(x) if x is large and restricted to the positive integers, c = 1 and n>=1, whereby the value of the fractional portion changes on a cycle with period n+1 (as k goes from 1 to n+1) for each n in the denominator of f(x). Other, somewhat similar triangles (or repeating fractional patterns) arise with other rational values of n or c, or other rational increments of x (even if a large irrational initial value of x is used).

Let S(n) = row sums = Sum(k>=1, T(n,k)), then:

S(n) = A077414(n); S(n)/(n+2) = A000217(n); S(n)/n = A000096(n);

Let Sq(n) = sum of squares of row elements = Sum(k>=1, T(n,k)^2), then:

Sq(n)/n^2 - 1/n = A058373(n)

Let D(n) = diagonal sums = Sum(k>=1, T(n-k+1, k)) then:

D(2n) = A131423(n); D(2n-1) = 2/3*n^3 + 1/2*n^2 - 7/6*n;

D(2n) - D(2n-1) = A000217(n); D(2n+1) - D(2n) = A115067(n);

D(2n+2) - D(2n)= A056220(n+1); D(2n+1) - D(2n -1) = A014106(n).

Equals A144204 with the first column of negative ones removed. - Georg Fischer, Jul 26 2023

Examples

			Triangle begins as:
0;
1,  3;
2,  5,  8;
3,  7, 11, 15;
4,  9  14, 19, 24;
5, 11, 17, 23, 29, 35;
6, 13, 20, 27, 34, 41, 48;
7, 15, 23, 31, 39, 47, 55, 63;
8, 17, 26, 35, 44, 53, 62, 71, 80;
		

Links

• Boris Putievskiy, Rows n = 1..140 of triangle, flattened

Crossrefs

Cf. A075362, A144204.

Formula

Also note: T(n+1,k) = T(n,k)+ k, and T(n,n) = n^2 - 1.

a(n) = A075362(n)-1; a(n)=i(t+1)-1, where i=n-t*(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Jul 24 2013

T(n, k) = n*k - 1. - Georg Fischer, Jul 26 2023

Extensions

Simpler name from Georg Fischer, Jul 26 2023