cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A115636 A divide-and-conquer number triangle.

Original entry on oeis.org

1, 1, -1, 4, 0, 1, 4, 0, 1, -1, 4, -4, 0, 0, 1, 4, -4, 0, 0, 1, -1, 16, 0, 4, 0, 0, 0, 1, 16, 0, 4, 0, 0, 0, 1, -1, 16, 0, 4, -4, 0, 0, 0, 0, 1, 16, 0, 4, -4, 0, 0, 0, 0, 1, -1, 16, -16, 0, 0, 4, 0, 0, 0, 0, 0, 1, 16, -16, 0, 0, 4, 0, 0, 0, 0, 0, 1, -1, 16, -16, 0, 0, 4, -4, 0, 0, 0, 0, 0, 0, 1, 16, -16, 0, 0, 4, -4, 0, 0, 0, 0, 0, 0, 1, -1
Offset: 0

Views

Author

Paul Barry, Jan 27 2006

Keywords

Examples

			Triangle begins
   1;
   1,  -1;
   4,   0,  1;
   4,   0,  1, -1;
   4,  -4,  0,  0,  1;
   4,  -4,  0,  0,  1, -1;
  16,   0,  4,  0,  0,  0,  1;
  16,   0,  4,  0,  0,  0,  1, -1;
  16,   0,  4, -4,  0,  0,  0,  0,  1;
  16,   0,  4, -4,  0,  0,  0,  0,  1, -1;
  16, -16,  0,  0,  4,  0,  0,  0,  0,  0,  1;
  16, -16,  0,  0,  4,  0,  0,  0,  0,  0,  1, -1;
  16, -16,  0,  0,  4, -4,  0,  0,  0,  0,  0,  0,  1;
  16, -16,  0,  0,  4, -4,  0,  0,  0,  0,  0,  0,  1, -1;
  64,   0, 16,  0,  0,  0,  4,  0,  0,  0,  0,  0,  0,  0,  1;
  64,   0, 16,  0,  0,  0,  4,  0,  0,  0,  0,  0,  0,  0,  1, -1;
  64,   0, 16,  0,  0,  0,  4, -4,  0,  0,  0,  0,  0,  0,  0,  0,  1;
		

Crossrefs

Cf. A115633 (inverse), A115637 (row sums), A115639 (first column).

Programs

  • Mathematica
    A115633[n_, k_]:= If[k==n, (-1)^n, If[k==n-1, Mod[n,2], If[n==2*k+2, -4, 0]]];
    T[n_, k_]:= T[n, k]= (-1)^k*If[k==n, 1, -Sum[T[n, j]*A115633[j, k], {j,k+1,n}] ];
    Table[T[n, k], {n,0,18}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 24 2021 *)
  • Sage
    @CachedFunction
    def A115633(n, k):
        if (k==n): return (-1)^n
        elif (k==n-1): return n%2
        elif (n==2*k+2): return -4
        else: return 0
    def A115636(n,k):
        if (k==0): return 4^(floor(log(n+2, 2)) -1)
        elif (k==n): return (-1)^n
        elif (k==n-1): return (n%2)
        else: return (-1)^(k+1)*sum( A115636(n, j)*A115633(j, k) for j in (k+1..n) )
    flatten([[A115636(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Nov 24 2021

Formula

T(n, 0) = A115639(n).
Sum_{k=0..n} T(n, k) = A115637(n).
T(n, k) = (-1)^k*( 1 if k = n otherwise (-1)*Sum_{j=k+1..n} T(n, j)*A115633(j, k) ). - G. C. Greubel, Nov 24 2021

A115637 In the binary expansion of n+2, transform 0->1 and 1->0 then interpret as base 4.

Original entry on oeis.org

1, 0, 5, 4, 1, 0, 21, 20, 17, 16, 5, 4, 1, 0, 85, 84, 81, 80, 69, 68, 65, 64, 21, 20, 17, 16, 5, 4, 1, 0, 341, 340, 337, 336, 325, 324, 321, 320, 277, 276, 273, 272, 261, 260, 257, 256, 85, 84, 81, 80, 69, 68, 65, 64, 21, 20, 17, 16, 5, 4, 1, 0, 1365, 1364, 1361, 1360, 1349
Offset: 0

Views

Author

Paul Barry, Jan 27 2006

Keywords

Comments

Row sums of number triangle A115636. Partial sums of A115638.
Old name was "A divide and conquer sequence".

Crossrefs

Cf. A000695, A035327, A115633, A115636, A115638 (first differences), A374625.

Programs

  • Maple
    b:= n-> 1-(n mod 2)+`if`(n<2, 0, b(iquo(n, 2))*4):
    a:= n-> b(n+2):
    seq(a(n), n=0..66);  # Alois P. Heinz, Jul 16 2024
  • Mathematica
    A115637[n_] := FromDigits[1 - IntegerDigits[n + 2, 2], 4];
    Array[A115637, 100, 0] (* Paolo Xausa, Jul 16 2024 *)
  • PARI
    up_to = 1024;
    A115633array(n, k) = (((-1)^n)*if(n==k,1, if((k+k+2)==n, -4, if((k+1)==n, -(1+(-1)^k)/2, 0))));
    A115637list(up_to) = { my(mA115633=matrix(up_to,up_to,n,k,A115633array(n-1,k-1)), mA115636 = matsolve(mA115633,matid(up_to)), v = vector(up_to)); for(n=1,up_to,v[n] = vecsum(mA115636[n,])); (v); };
    v115637 = A115637list(up_to+1);
    A115637(n) = v115637[1+n]; \\ Antti Karttunen, Nov 02 2018
    
  • PARI
    a(n) = fromdigits([!b |b<-binary(n+2)], 4); \\ Kevin Ryde, Jul 15 2024
    
  • Python
    def A115637(n): return int(bin((~(n+2))^(-1<<(n+2).bit_length()))[2:],4) # Chai Wah Wu, Jul 17 2024

Formula

G.f.: (1/(1-x))*Sum_{k>=0} 4^k*x^(2^(k+1)-2)/(1+x^(2^k)); the g.f. G(x) satisfies G(x) - 4(1+x)*x^2*G(x^2) = 1/(1-x^2).
a(n) = A000695(A035327(n+2)). - Kevin Ryde, Jul 15 2024

Extensions

New name from Kevin Ryde, Jul 15 2024

A115634 Expansion of (1-4*x^2)/(1-x^2).

Original entry on oeis.org

1, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0, -3, 0
Offset: 0

Views

Author

Paul Barry, Jan 27 2006

Keywords

Comments

Row sums of number triangle A115633.

Crossrefs

Programs

  • Magma
    [4*0^n -3*(1+(-1)^n)/2: n in [0..100]]; // G. C. Greubel, Nov 23 2021
    
  • Mathematica
    Join[{1}, -3*Mod[Range[100] -1, 2]] (* G. C. Greubel, Nov 23 2021 *)
    CoefficientList[Series[(1-4x^2)/(1-x^2),{x,0,100}],x] (* or *) LinearRecurrence[{0,1},{1,0,-3},100] (* or *) PadRight[{1},100,{-3,0}] (* Harvey P. Dale, Dec 06 2024 *)
  • Sage
    [1]+[-3*((n-1)%2) for n in (1..100)] # G. C. Greubel, Nov 23 2021

Formula

a(n) = 4*0^n - 3*(1 + (-1)^n)/2.
a(n) = Sum_{k=0..n} A115633(n, k).
From G. C. Greubel, Nov 23 2021: (Start)
a(n) = 1 if n = 0, otherwise a(n) = -A010674(n-1).
E.g.f.: 4 - 3*cosh(x). (End)

A115635 Periodic {1,1,-5,0,1,-3,-1,0,-3,1,-1,-4}.

Original entry on oeis.org

1, 1, -5, 0, 1, -3, -1, 0, -3, 1, -1, -4, 1, 1, -5, 0, 1, -3, -1, 0, -3, 1, -1, -4, 1, 1, -5, 0, 1, -3, -1, 0, -3, 1, -1, -4, 1, 1, -5, 0, 1, -3, -1, 0, -3, 1, -1, -4, 1, 1, -5, 0, 1, -3, -1, 0, -3, 1, -1, -4, 1, 1, -5, 0, 1, -3, -1, 0, -3, 1, -1, -4
Offset: 0

Views

Author

Paul Barry, Jan 27 2006

Keywords

Comments

Diagonal sums of number triangle A115633.

Formula

G.f.: (1+2x-3x^2-4x^3-5x^4-4x^5)/((1-x^3)*(1+x+x^2+x^3)); a(n)=sum{k=0..floor(n/2), A115633(n-k, k)}.
Showing 1-4 of 4 results.