A127750 -id:A127750 - cp's OEIS frontend

A127752 Row sums of inverse of number triangle A(n,k) = 1/(3n+1) if k <= n <= 2k, 0 otherwise.

1, 4, 3, 7, 3, 6, 3, 10, 3, 6, 3, 9, 3, 6, 3, 13, 3, 6, 3, 9, 3, 6, 3, 12, 3, 6, 3, 9, 3, 6, 3, 16, 3, 6, 3, 9, 3, 6, 3, 12, 3, 6, 3, 9, 3, 6, 3, 15, 3, 6, 3, 9, 3, 6, 3, 12, 3, 6, 3, 9, 3, 6, 3, 19, 3, 6, 3, 9, 3, 6, 3, 12, 3, 6, 3, 9, 3, 6, 3, 15, 3, 6, 3, 9, 3, 6, 3, 12, 3, 6, 3, 9, 3, 6, 3, 18, 3, 6, 3, 9, 3, 6, 3, 12, 3, 6

Offset: 0

Paul Barry, Jan 28 2007

nonn

Row sums of number triangle A127751.
a(n) mod 2 is first Feigenbaum symbolic sequence A035263 (conjecture).
The conjecture is true at least up to 2048 first terms. (But please note the different indexing, here 0-based.) - Antti Karttunen, Sep 29 2018

Antti Karttunen, Table of n, a(n) for n = 0..2048

Cf. A127750, A127751.

A[n_, k_] := If[k <= n <= 2k, 1/(3n+1), 0];
Total /@ Inverse[Array[A, {128, 128}, {0, 0}]] (* Jean-François Alcover, Feb 11 2021 *)

up_to = 128;
A127752aux(n,k) = if(k<=n,if(n<=(2*k),1/((3*n)+1),0),0);
A127752list(up_to) = { my(m1=matrix(up_to,up_to,n,k,A127752aux(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); };
v127752 = A127752list(1+up_to);
A127752(n) = v127752[1+n]; \\ Antti Karttunen, Sep 29 2018

More terms from Antti Karttunen, Sep 29 2018

A127749 Inverse of number triangle A(n,k) = 1/(2n+1) if k <= n <= 2k, 0 otherwise.

Original entry on oeis.org

1, 0, 3, 0, -3, 5, 0, 3, -5, 7, 0, 0, 0, -7, 9, 0, -3, 5, 0, -9, 11, 0, 0, 0, 0, 0, -11, 13, 0, 3, -5, 7, 0, 0, -13, 15, 0, 0, 0, 0, 0, 0, 0, -15, 17, 0, 0, 0, -7, 9, 0, 0, 0, -17, 19, 0, 0, 0, 0, 0, 0, 0, 0, 0, -19, 21, 0, -3, 5

Offset: 0

Paul Barry, Jan 28 2007

Conjectures: row sums modulo 2 are the Fredholm-Rueppel sequence A036987; row sums of triangle modulo 2 are A111982. Row sums are A127750.

The first conjecture is equivalent to the row sums conjecture in A111967. - R. J. Mathar, Apr 21 2021

			Triangle begins
  1;
  0,  3;
  0, -3,  5;
  0,  3, -5,  7;
  0,  0,  0, -7,  9;
  0, -3,  5,  0, -9,  11;
  0,  0,  0,  0,  0, -11,  13;
  0,  3, -5,  7,  0,   0, -13,  15;
  0,  0,  0,  0,  0,   0,   0, -15,  17;
  0,  0,  0, -7,  9,   0,   0,   0, -17,  19;
  0,  0,  0,  0,  0,   0,   0,   0,   0, -19,  21;
  0, -3,  5,  0, -9,  11,   0,   0,   0,   0, -21,  23;
  0,  0,  0,  0,  0,   0,   0,   0,   0,   0,   0, -23, 25;
Inverse of triangle
  1;
  0, 1/3;
  0, 1/5, 1/5;
  0,  0,  1/7, 1/7;
  0,  0,  1/9, 1/9,  1/9;
  0,  0,   0,  1/11, 1/11, 1/11;
  0,  0,   0,  1/13, 1/13, 1/13, 1/13;
  0,  0,   0,   0,   1/15, 1/15, 1/15, 1/15;
  0,  0,   0,   0,   1/17, 1/17, 1/17, 1/17, 1/17;
  0,  0,   0,   0,    0,   1/19, 1/19, 1/19, 1/19, 1/19;
  0,  0,   0,   0,    0,   1/21, 1/21, 1/21, 1/21, 1/21, 1/21;

Cf. A111967.

A127749 := proc(n,k)
    option remember ;
    if k > n then
        0 ;
    elif k = n then
        2*n+1 ;
    else
        -(2*k+1)*add( procname(n,i)/(2*i+1),i=k+1..min(n,2*k)) ;
    end if;
end proc:
seq(seq( A127749(n,k),k=0..n),n=0..20) ; # R. J. Mathar, Feb 09 2021

nmax = 10;
A[n_, k_] := If[k <= n <= 2k, 1/(2n+1), 0];
invA = Inverse[Table[A[n, k], {n, 0, nmax}, {k, 0, nmax}]];
T[n_, k_] := invA[[n+1, k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 05 2020 *)

T(n,k) = (2*k+1)*A111967(n,k). - R. J. Mathar, Apr 21 2021

cp's OEIS Frontend

A127752 Row sums of inverse of number triangle A(n,k) = 1/(3n+1) if k <= n <= 2k, 0 otherwise.

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