A139377 A Jacobsthal-Catalan triangle.
1, 1, 1, 3, 2, 1, 5, 6, 3, 1, 11, 15, 10, 4, 1, 21, 41, 30, 15, 5, 1, 43, 113, 92, 51, 21, 6, 1, 85, 327, 284, 171, 79, 28, 7, 1, 171, 982, 897, 570, 286, 115, 36, 8, 1, 341, 3066, 2895, 1913, 1016, 446, 160, 45, 9, 1
Offset: 0
Examples
Triangle begins
1;
1, 1;
3, 2, 1;
5, 6, 3, 1;
11, 15, 10, 4, 1;
21, 41, 30, 15, 5, 1;
43, 113, 92, 51, 21, 6, 1;
85, 327, 284, 171, 79, 28, 7, 1;
171, 982, 897, 570, 286, 115, 36, 8, 1;
The production matrix for this array is
1, 1,
2, 1, 1,
-2, 1, 1, 1,
0, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1
Links
Crossrefs
Programs
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Maple
#define auxiliary sequence with(combinat): b := proc (n) (2/3)*(1+I)^(n-1) + (2/3)*(1-I)^(n-1) - (4/3)*(1+I)^(n-2)-(4/3)*(1-I)^(n-2) + (1/3)*(-1)^n*fibonacci(n+1) + (2/3)*(-1)^n*fibonacci(n); end proc: A139377 := proc (n, k) add(b(j)*binomial(2*n-k-j, n), j = 0..n-k); end proc: #display sequence as a triangle for n from 0 to 10 do seq(A139377(n, k), k = 0..n); end do; # Peter Bala, Feb 20 2018
Formula
Riordan array (1/(1-x-2x^2), xc(x)) where c(x) is the g.f. of A000108
From Peter Bala, Feb 20 2018: (Start)
Define a(n) = floor(2^(n+2)/3) - floor(2^(n+1)/3) = A001045(n+1). Then T(n,0) = a(n) and T(n,k) = Sum_{j = 0..n-k} a(j)*k/(2*n-k-2*j)*binomial(2*n-k-2*j,n-k-j) for 1 <= k <= n.
Define b(n) = (2/3)*(1+i)^(n-1) + (2/3)*(1-i)^(n-1) - (4/3)*(1+i)^(n-2) - (4/3)*(1-i)^(n-2) + (1/3)*(-1)^n*Fibonacci(n+1) + (2/3)*(-1)^n*Fibonacci(n). Then T(n,k) = Sum_{j = 0..n-k} b(j)*binomial(2*n-k-j,n) for 0 <= k <= n.
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 - 2*x)/((1 - x)*(1 + x - x^2)*(1 - 2*x + 2*x^2)) * 1/(1 - x)^n about 0. For example, for n = 4, (1 - 2*x)/((1 - x)*(1 + x - x^2)*(1 - 2*x + 2*x^2)) * 1/(1 - x)^4 = (11*x^4 + 15*x^3 + 10*x^2 + 4*x + 1) + O(x^5). (End)
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