A121722 Triangle T(n,k) = 1 + k*n*(n+1)/2, read by rows.

1, 1, 2, 1, 4, 7, 1, 7, 13, 19, 1, 11, 21, 31, 41, 1, 16, 31, 46, 61, 76, 1, 22, 43, 64, 85, 106, 127, 1, 29, 57, 85, 113, 141, 169, 197, 1, 37, 73, 109, 145, 181, 217, 253, 289, 1, 46, 91, 136, 181, 226, 271, 316, 361, 406, 1, 56, 111, 166, 221, 276, 331, 386, 441, 496, 551

Offset: 0

Author

Roger L. Bagula, Sep 08 2006

Keywords

• nonn
• tabl

Comments

A triangular form based on the Hex number recursion: a(n) = 2*a(n-1) - a(n-1) + 6: A003215 form as generalized to Integer m.

A solution for the general type for m held constant: a(n) = 2*a(n-1) - a(n-2) + m, with first two values as {1, 1+m}.

Examples

			Triangle begins as:
  1;
  1,  2;
  1,  4,  7;
  1,  7, 13, 19;
  1, 11, 21, 31, 41;
  1, 16, 31, 46, 61, 76;
		

Links

• G. C. Greubel, Rows n = 0..100 of triangle, flattened

Crossrefs

Cf. A001844, A002061, A003215, A005448, A005891.

Programs

GAP

Flat(List([0..10], n-> List([0..n], k-> 1 + k*Binomial(n+1,2) ))); # G. C. Greubel, Nov 21 2019

Magma

[1+k*Binomial(n+1,2): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 21 2019

Maple

seq(seq( 1 + k*binomial(n+1,2), k=0..n), n=0..10); # G. C. Greubel, Nov 21 2019

Mathematica

f[n_Integer] = Module[{a}, a[n]/.RSolve[{a[n]==2*a[n-1]-a[n-2]+m, a[0] ==1, a[1]==1+m}, a[n], n][[1]]//FullSimplify] (* formula of triangle *)
Table[Table[1+k*n*(1+n)/2, {k,0,n}], {n,0,10}]//Flatten

PARI

T(n, k) = 1 + k*binomial(n+1,2); \\ G. C. Greubel, Nov 21 2019

Sage

[[1+k*binomial(n+1,2) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Nov 21 2019

Formula

T(n, k) = 1 + k*binomial(n+1,2).

Extensions

Edited by G. C. Greubel, Nov 21 2019