A127750 Row sums of inverse of number triangle A(n,k) = 1/(2n+1) if k <= n <= 2k, 0 otherwise.
1, 3, 2, 5, 2, 4, 2, 7, 2, 4, 2, 6, 2, 4, 2, 9, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 2, 6, 2, 4, 2, 11, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 2, 6, 2, 4, 2, 10, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 2, 6, 2, 4, 2, 13, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 2, 6, 2, 4, 2, 10, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 2, 6, 2, 4, 2, 12, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4
Offset: 0
Keywords
Links
- Antti Karttunen, Table of n, a(n) for n = 0..2048
Programs
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Mathematica
A[n_, k_] := If[k <= n <= 2k, 1/(2n+1), 0]; Total /@ Inverse[Array[A, {128, 128}, {0, 0}]] (* Jean-François Alcover, Feb 10 2021 *) -
PARI
up_to = 128; A127750aux(n,k) = if(k<=n,if(n<=(2*k),1/(n+n+1),0),0); A127750list(up_to) = { my(m1=matrix(up_to,up_to,n,k,A127750aux(n-1,k-1)), m2 = matsolve(m1,matid(up_to)), v = vector(up_to)); for(n=1,up_to,v[n] = vecsum(m2[n,])); (v); }; v127750 = A127750list(1+up_to); A127750(n) = v127750[1+n]; \\ Antti Karttunen, Sep 29 2018
Extensions
More terms from Antti Karttunen, Sep 29 2018
Comments