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A117178 Riordan array (c(x^2)/sqrt(1-4*x^2), x*c(x^2)), c(x) the g.f. of A000108. 3
1, 0, 1, 3, 0, 1, 0, 4, 0, 1, 10, 0, 5, 0, 1, 0, 15, 0, 6, 0, 1, 35, 0, 21, 0, 7, 0, 1, 0, 56, 0, 28, 0, 8, 0, 1, 126, 0, 84, 0, 36, 0, 9, 0, 1, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1, 462, 0, 330, 0, 165, 0, 55, 0, 11, 0, 1, 0, 792, 0, 495, 0, 220, 0, 66, 0, 12, 0, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,4
COMMENTS
Row sums are A058622(n+1). Diagonal sums are A001791(n+1), with interpolated zeros. Inverse is A117179.
De-aerated and rows reversed, this matrix apparently becomes A014462. The nonzero antidiagonals are embedded in several entries and apparently contain partial sums of previous nonzero antidiagonals. - Tom Copeland, May 30 2017
LINKS
FORMULA
T(n,k) = C(n+1, (n-k)/2) * (1 + (-1)^(n-k))/2.
Column k has e.g.f. Bessel_I(k,2x) + Bessel_I(k+2, 2x).
From G. C. Greubel, Aug 08 2022: (Start)
Sum_{k=0..n} T(n, k) = A058622(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = ((1+(-1)^n)/2) * A001791((n+2)/2).
T(2*n, n) = ((1+(-1)^n)/2) * A052203(n/2).
T(2*n+1, n) = ((1-(-1)^n)/2) * A224274((n+1)/2).
T(2*n-1, n-1) = ((1+(-1)^n)/2) * A224274(n/2). (End)
EXAMPLE
Triangle begins
1;
0, 1;
3, 0, 1;
0, 4, 0, 1;
10, 0, 5, 0, 1;
0, 15, 0, 6, 0, 1;
35, 0, 21, 0, 7, 0, 1;
0, 56, 0, 28, 0, 8, 0, 1;
126, 0, 84, 0, 36, 0, 9, 0, 1;
MATHEMATICA
T[n_, k_]:= Binomial[n+1, (n-k)/2]*(1+(-1)^(n-k))/2;
Table[T[n, k], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Aug 08 2022 *)
PROG
(Magma) [(1+(-1)^(n-k))*Binomial(n+1, Floor((n-k)/2))/2: k in [0..n], n in [0..15]]; // G. C. Greubel, Aug 08 2022
(SageMath)
def A117178(n, k): return (1 + (-1)^(n-k))*binomial(n+1, (n-k)//2)/2
flatten([[A117178(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Aug 08 2022
CROSSREFS
Sequence in context: A049769 A117179 A111526 * A111527 A035695 A100257
KEYWORD
easy,nonn,tabl
AUTHOR
Paul Barry, Mar 01 2006
STATUS
approved

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Last modified March 19 03:33 EDT 2024. Contains 370952 sequences. (Running on oeis4.)