A092686 Triangle, read by rows, such that the convolution of each row with {1,2} produces a triangle which, when flattened, equals this flattened form of the original triangle.
1, 2, 2, 6, 4, 6, 16, 14, 12, 16, 46, 40, 40, 32, 46, 132, 120, 112, 110, 92, 132, 384, 352, 334, 312, 316, 264, 384, 1120, 1038, 980, 940, 896, 912, 768, 1120, 3278, 3056, 2900, 2776, 2704, 2592, 2656, 2240, 3278, 9612, 9012, 8576, 8256, 8000, 7840, 7552, 7758
Offset: 0
Examples
Rows begin:
1;
2, 2;
6, 4, 6;
16, 14, 12, 16;
46, 40, 40, 32, 46;
132, 120, 112, 110, 92, 132;
384, 352, 334, 312, 316, 264, 384;
1120, 1038, 980, 940, 896, 912, 768, 1120;
3278, 3056, 2900, 2776, 2704, 2592, 2656, 2240, 3278;
9612, 9012, 8576, 8256, 8000, 7840, 7552, 7758, 6556, 9612;
28236, 26600, 25408, 24512, 23840, 23232, 22862, 22072, 22724, 19224, 28236; ...
Convolution of each row with {1,2} results in the triangle:
1, 2;
2, 6, 4;
6, 16, 14, 12;
16, 46, 40, 40, 32;
46, 132, 120, 112, 110, 92;
132, 384, 352, 334, 312, 316, 264;
384, 1120, 1038, 980, 940, 896, 912, 768; ...
which, when flattened, equals the original triangle in flattened form.
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..495
Programs
-
PARI
T(n,k)=if(n<0 || k>n,0, if(n==0 && k==0,1, if(n==1 && k<=1,2, if(k==n,T(n,0), 2*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 the G.F.: */ {T(n,k)=local(A,F=1+2*x,d=1,G=x,H=1+2*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) = 2*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+2y) - y ), where H(x) satisfies: H(x) = H(x^2/(1-2x))/(1-2x) and H(x) is the g.f. of column 0 (A092687). - Paul D. Hanna, Jul 17 2006
A092687 First column and main diagonal of triangle A092686, in which the convolution of each row with {1,2} produces a triangle that, when flattened, equals the flattened form of A092686.
1, 2, 6, 16, 46, 132, 384, 1120, 3278, 9612, 28236, 83072, 244752, 722048, 2132704, 6306304, 18666190, 55300732, 163968612, 486528288, 1444571068, 4291629384, 12756459936, 37934818112, 112855778768, 335867740704, 999895548736
Offset: 0
Keywords
Comments
Conjecture: Limit n->infinity a(n)^(1/n) = 3. - Vaclav Kotesovec, Jun 29 2015
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..850
Programs
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Mathematica
m = 27; A[] = 1; Do[A[x] = A[x^2/(1-2x)]/(1-2x) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Nov 03 2019 *)
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PARI
T(n,k)=if(n<0||k>n,0, if(n==0&k==0,1, if(n==1&k<=1,2, if(k==n,T(n,0), 2*T(n-1,k)+T(n-1,k+1))))) a(n)=T(n,0) for(n=0,30,print1(a(n),", "))
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PARI
a(n)=local(A=1+x);for(i=0,n\2,A=subst(A,x,x^2/(1-2*x+x*O(x^n)))/(1-2*x));polcoeff(A,n) \\ Paul D. Hanna, Jul 10 2006
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PARI
/* Using Recurrence: */ a(n)=if(n==0, 1, sum(k=0, n\2, binomial(n-k, k)*2^(n-2*k)*a(k))) for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Jul 10 2006
Formula
G.f. satisfies: A(x) = A( x^2/(1-2x) )/(1-2x). Recurrence: a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*2^(n-2k)*a(k). - Paul D. Hanna, Jul 10 2006
Comments