A092683 Triangle, read by rows, such that the convolution of each row with {1,1} produces a triangle which, when flattened, equals this flattened form of the original triangle.
1, 1, 1, 2, 1, 2, 3, 3, 2, 3, 6, 5, 5, 3, 6, 11, 10, 8, 9, 6, 11, 21, 18, 17, 15, 17, 11, 21, 39, 35, 32, 32, 28, 32, 21, 39, 74, 67, 64, 60, 60, 53, 60, 39, 74, 141, 131, 124, 120, 113, 113, 99, 113, 74, 141, 272, 255, 244, 233, 226, 212, 212, 187, 215, 141, 272, 527, 499
Offset: 0
Examples
Rows begin:
1;
1, 1;
2, 1, 2;
3, 3, 2, 3;
6, 5, 5, 3, 6;
11, 10, 8, 9, 6, 11;
21, 18, 17, 15, 17, 11, 21;
39, 35, 32, 32, 28, 32, 21, 39;
74, 67, 64, 60, 60, 53, 60, 39, 74;
141, 131, 124, 120, 113, 113, 99, 113, 74, 141;
272, 255, 244, 233, 226, 212, 212, 187, 215, 141, 272;
527, 499, 477, 459, 438, 424, 399, 402, 356, 413, 272, 527;
1026, 976, 936, 897, 862, 823, 801, 758, 769, 685, 799, 527, 1026; ...
The convolution of each row with {1,1} gives the triangle:
1, 1;
1, 2, 1;
2, 3, 3, 2;
3, 6, 5, 5, 3;
6, 11, 10, 8, 9, 6;
11, 21, 18, 17, 15, 17, 11;
21, 39, 35, 32, 32, 28, 32, 21;
39, 74, 67, 64, 60, 60, 53, 60, 39; ...
which, when flattened, equals the original triangle in flattened form.
Links
Programs
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PARI
T(n,k)=if(n<0 || k>n,0, if(n==0 && k==0,1, if(n==1 && k<=1,1, if(k==n,T(n,0), T(n-1,k)+T(n-1,k+1))))) for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print("")) -
PARI
/* Generate Triangle by G.F. where F=1+x: */ {T(n,k)=local(A,F=1+x,d=1,G=x,H=1+x,S=ceil(log(n+1)/log(d+1))); for(i=0,n,G=x*subst(F,x,G+x*O(x^n)));for(i=0,S,H=subst(H,x,x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H,x,x*y^d +x*O(x^n)))/(x*subst(F,x,y)-y); polcoeff(polcoeff(A,n,x),k,y)} for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Jul 17 2006
Formula
T(n, k) = T(n-1, k) + T(n-1, k+1) for 0<=k
G.f.: A(x,y) = ( x*H(x) - y*H(x*y) )/( x*(1+y) - y ), where H(x) satisfies: H(x) = H(x^2/(1-x))/(1-x) and H(x) is the g.f. of column 0 (A092684). - Paul D. Hanna, Jul 17 2006
A092684 First column and main diagonal of triangle A092683, in which the convolution of each row with {1,1} produces a triangle that, when flattened, equals the flattened form of A092683.
1, 1, 2, 3, 6, 11, 21, 39, 74, 141, 272, 527, 1026, 2002, 3914, 7659, 14996, 29369, 57531, 112727, 220963, 433342, 850386, 1670011, 3282259, 6456475, 12711413, 25047465, 49396116, 97490480, 192552549, 380565123, 752619506, 1489234257
Offset: 0
Keywords
Comments
The self-convolution forms A100938. - Paul D. Hanna, Nov 23 2004
The limit of the matrix power A011973^n, as n->inf, results in a single column vector equal to this sequence. - Paul D. Hanna, May 03 2006
Examples
a(8) = Sum_{k=0..[8/2]} C(n-k,k)*a(k)
= C(8,0)*a(0) +C(7,1)*a(1) +C(6,2)*a(2) +C(5,3)*a(3) +C(4,4)*a(4)
= 1*1 + 7*1 + 15*2 + 10*3 + 1*6 = 74.
Crossrefs
Programs
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PARI
{T(n,k)=if(n<0 || k>n,0, if(n==0 && k==0,1, if(n==1 && k<=1,1, if(k==n,T(n,0), T(n-1,k)+T(n-1,k+1)))))} a(n)=T(n,0) -
PARI
a(n)=if(n==0,1,sum(k=0,n\2,binomial(n-k,k)*a(k))) \\ Paul D. Hanna, May 03 2006
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PARI
{a(n)=local(A=1+x);for(i=0,n\2,A=subst(A,x,x^2/(1-x+x*O(x^n)))/(1-x));polcoeff(A,n)} \\ Paul D. Hanna, Jul 10 2006
Formula
Invariant under the transformation of Fibonacci triangle A011973(n,k)=C(n-k,k): a(n) = Sum_{k=0..[n/2]} C(n-k,k)*a(k). - Paul D. Hanna, May 03 2006
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k)*a(k). - Vladeta Jovovic, May 07 2006
G.f. satisfies: A(x) = A( x^2/(1-x) )/(1-x). - Paul D. Hanna, Jul 10 2006
Comments