login
Triangle {H(n,k)} similar to Pascal's with sides of 1's, but interior entries are obtained by the rule: H(n,k) = |H(n-1,k)+(-1)^m(n,k)*H(n-1,k-1)|, where m(n,k) = H(n-1,k) + H(n-1,k-1).
5

%I #28 Nov 14 2014 13:02:56

%S 1,1,1,1,2,1,1,1,1,1,1,2,2,2,1,1,1,4,4,1,1,1,2,3,8,3,2,1,1,1,1,5,5,1,

%T 1,1,1,2,2,6,10,6,2,2,1,1,1,4,8,16,16,8,4,1,1,1,2,3,12,24,32,24,12,3,

%U 2,1,1,1,1,9,36,56,56,36,9,1,1,1,1,2,2,10

%N Triangle {H(n,k)} similar to Pascal's with sides of 1's, but interior entries are obtained by the rule: H(n,k) = |H(n-1,k)+(-1)^m(n,k)*H(n-1,k-1)|, where m(n,k) = H(n-1,k) + H(n-1,k-1).

%C Let us consider the operation <+> over integers such that k<+>m = |k+(-1)^(k+m)*m|. Then H(n,k) = H(n-1,k)<+>H(n-1,k-1).

%C This is an analog of the formula binomial(n,k) = binomial(n-1,k) + binomial(n-1,k-1).

%H Peter J. C. Moses, <a href="/A245618/b245618.txt">First 50 rows.</a>

%e Triangle begins

%e 1

%e 1 1

%e 1 2 1

%e 1 1 1 1

%e 1 2 2 2 1

%e 1 1 4 4 1 1

%e 1 2 3 8 3 2 1

%e ....................

%t parityAdd[a_, b_] := Abs[a + b (-1)^(a + b)];

%t triangleHP[n_, 0] := 1;

%t triangleHP[n_, n_] := 1;

%t triangleHP[n_, k_] := triangleHP[n, k] = parityAdd[triangleHP[n - 1, k - 1], triangleHP[n - 1, k]];

%t Flatten[Table[triangleHP[n, k], {n, 0, 15}, {k, 0, n}]] (* _Peter J. C. Moses_, Nov 05 2014 *)

%Y Cf. A007318, row sums in A245619, row "sums", using <+>, in A249388.

%K nonn,tabl

%O 0,5

%A _Vladimir Shevelev_, Nov 05 2014

%E More terms from _Peter J. C. Moses_, Nov 05 2014