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A299146 Modified Pascal's triangle read by rows: T(n,k) = C(n+1,k) - n, 1 <= k <= n.

%I A299146 #63 Sep 08 2022 08:46:20
%S A299146 1,1,1,1,3,1,1,6,6,1,1,10,15,10,1,1,15,29,29,15,1,1,21,49,63,49,21,1,
%T A299146 1,28,76,118,118,76,28,1,1,36,111,201,243,201,111,36,1,1,45,155,320,
%U A299146 452,452,320,155,45,1,1,55,209,484,781,913,781,484,209,55,1
%N A299146 Modified Pascal's triangle read by rows: T(n,k) = C(n+1,k) - n, 1 <= k <= n.
%C A299146 If we define T_m(n, k) = binomial(n+m,k) - m*n where m <= k <= n, then T_0 is Pascal's triangle A007318 and T_1 is the current triangle sequence.
%C A299146 This modified Pascal's triangle is symmetric: C(n+m, k) - m*n = C(n+m, n-k+1) - m*n for any nonnegative integer m.
%F A299146 T(n, k) = T_1(n, k) = binomial(n+1, k) - n, for 1 <= k <= n.
%e A299146 The triangle T(n, k) begins:
%e A299146 n\k  1    2    3    4    5    6    7    8    9   10
%e A299146 1    1;
%e A299146 2    1,   1;
%e A299146 3    1,   3,   1;
%e A299146 4    1,   6,   6,   1;
%e A299146 5    1,  10,  15,  10,   1;
%e A299146 6    1,  15,  29,  29,  15,   1;
%e A299146 7    1,  21,  49,  63,  49,  21,   1;
%e A299146 8    1,  28,  76, 118, 118,  76,  28,   1;
%e A299146 9    1,  36, 111, 201, 243, 201, 111,  36,   1;
%e A299146 10   1,  45, 155, 320, 452, 452, 320, 155,  45,   1; etc.
%p A299146 seq(seq(binomial(n+1,k)-n, k=1..n), n=1..10); # _Muniru A Asiru_, Feb 05 2018
%t A299146 Table[Binomial[n + 1, k] - n, {n, 11}, {k, n}] // Flatten (* _Michael De Vlieger_, Feb 05 2018 *)
%o A299146 (Magma) [[Binomial(n+1, k)- 1*n: k in [1..n]]: n in [1..10]];
%o A299146 (GAP) Flat(List([1..100],n->List([1..n],k->Binomial(n+1,k)-n))); # _Muniru A Asiru_, Feb 05 2018
%o A299146 (PARI) T(n, k) = binomial(n+1,k) - n;
%o A299146 tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Mar 01 2018
%Y A299146 Cf. A000012, A000217, A005286, A083858, A007318, A276472.
%K A299146 nonn,tabl,easy
%O A299146 1,5
%A A299146 _Juri-Stepan Gerasimov_, Feb 03 2018