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

%I A121722 #12 Sep 08 2022 08:45:27
%S A121722 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,
%T A121722 106,127,1,29,57,85,113,141,169,197,1,37,73,109,145,181,217,253,289,1,
%U A121722 46,91,136,181,226,271,316,361,406,1,56,111,166,221,276,331,386,441,496,551
%N A121722 Triangle T(n,k) = 1 + k*n*(n+1)/2, read by rows.
%C A121722 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.
%C A121722 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}.
%H A121722 G. C. Greubel, <a href="/A121722/b121722.txt">Rows n = 0..100 of triangle, flattened</a>
%F A121722 T(n, k) = 1 + k*binomial(n+1,2).
%e A121722 Triangle begins as:
%e A121722   1;
%e A121722   1,  2;
%e A121722   1,  4,  7;
%e A121722   1,  7, 13, 19;
%e A121722   1, 11, 21, 31, 41;
%e A121722   1, 16, 31, 46, 61, 76;
%p A121722 seq(seq( 1 + k*binomial(n+1,2), k=0..n), n=0..10); # _G. C. Greubel_, Nov 21 2019
%t A121722 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 *)
%t A121722 Table[Table[1+k*n*(1+n)/2, {k,0,n}], {n,0,10}]//Flatten
%o A121722 (PARI) T(n, k) = 1 + k*binomial(n+1,2); \\ _G. C. Greubel_, Nov 21 2019
%o A121722 (Magma) [1+k*Binomial(n+1,2): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Nov 21 2019
%o A121722 (Sage) [[1+k*binomial(n+1,2) for k in (0..n)] for n in (0..10)] # _G. C. Greubel_, Nov 21 2019
%o A121722 (GAP) Flat(List([0..10], n-> List([0..n], k-> 1 + k*Binomial(n+1,2) ))); # _G. C. Greubel_, Nov 21 2019
%Y A121722 Cf. A001844, A002061, A003215, A005448, A005891.
%K A121722 nonn,tabl
%O A121722 0,3
%A A121722 _Roger L. Bagula_, Sep 08 2006
%E A121722 Edited by _G. C. Greubel_, Nov 21 2019