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.

A117592 a(n) = a(3n) = a(3n+1) = a(3n+2)/2 with a(0)=1.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 2, 4, 1, 1, 2, 1, 1, 2, 2, 2, 4, 2, 2, 4, 2, 2, 4, 4, 4, 8, 1, 1, 2, 1, 1, 2, 2, 2, 4, 1, 1, 2, 1, 1, 2, 2, 2, 4, 2, 2, 4, 2, 2, 4, 4, 4, 8, 2, 2, 4, 2, 2, 4, 4, 4, 8, 2, 2, 4, 2, 2, 4, 4, 4
Offset: 0

Views

Author

Paul Barry, Apr 05 2006

Keywords

Comments

Row sums of number triangle A117944.
Product of the nonzero digits of (n written in base 3). - Ilya Gutkovskiy, Nov 15 2020
a(n) = 1, 2, 4, 8, 16, 32, 64 iff n is respectively in A005836, A023699, A023700, A023701, A023702, A023703, A023704. - Bernard Schott, Dec 04 2020

Links

Crossrefs

See A338882 for similar sequences.
Cf. A081603 (log_2), A117942 (signed), A117944.
Cf. A005836, A023699, A023700, A023701, A023702, A023703, A023704.

Programs

  • Mathematica
    Nest[ Join[#, #, 2#] &, {1}, 5] (* Robert G. Wilson v, Jul 27 2014 *)
  • PARI
    a(n) = 1 << hammingweight(digits(n,3)>>1); \\ Kevin Ryde, Nov 15 2020
  • Python
    from gmpy2 import digits
def A117592(n): return 1<Chai Wah Wu, Dec 05 2024

Formula

a(n) = a(3n)/a(0) = a(3n+1)/a(1) = a(3n+2)/a(2).
a(n) = abs(A117942(n)).
G.f. A(x) satisfies: A(x) = (1 + x + 2*x^2) * A(x^3). - Ilya Gutkovskiy, Nov 15 2020
a(n) = 2^A081603(n). - Kevin Ryde, Nov 15 2020

Extensions

a(0) = 1 added to the Name by Bernard Schott, Dec 04 2020

A117939 Triangle related to powers of 3 partitions of n.

Original entry on oeis.org

1, 2, 1, 1, -2, 1, 2, 0, 0, 1, 4, 2, 0, 2, 1, 2, -4, 2, 1, -2, 1, 1, 0, 0, -2, 0, 0, 1, 2, 1, 0, -4, -2, 0, 2, 1, 1, -2, 1, -2, 4, -2, 1, -2, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 0, 0, 0, 0, 0, 0, 0, 2, 1, 2, -4, 2, 0, 0, 0, 0, 0, 0, 1, -2, 1, 4, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 1, 8, 4, 0, 4, 2, 0, 0, 0, 0, 4, 2, 0, 2, 1
Offset: 0

Views

Author

Paul Barry, Apr 05 2006

Keywords

Examples

			Triangle begins
  1;
  2,  1;
  1, -2, 1;
  2,  0, 0,  1;
  4,  2, 0,  2,  1;
  2, -4, 2,  1, -2,  1;
  1,  0, 0, -2,  0,  0, 1;
  2,  1, 0, -4, -2,  0, 2,  1;
  1, -2, 1, -2,  4, -2, 1, -2, 1;

Links

Crossrefs

Cf. A059151, A117940, A117944, A117946.
Cf. A120854 (matrix log), A117941 (inverse), A117947 (matrix square-root).

Programs

  • Mathematica
    T[n_, k_]:= Sum[JacobiSymbol[Binomial[n, j], 3]*JacobiSymbol[Binomial[n-j, k], 3], {j, 0, n}]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 29 2021 *)
  • PARI
    T(n,k)=(matrix(n+1,n+1,r,c,(binomial(r-1,c-1)+1)%3-1)^2)[n+1,k+1] \\ Paul D. Hanna, Jul 08 2006
  • Sage
    def A117939(n, k): return sum(jacobi_symbol(binomial(n, j), 3)*jacobi_symbol(binomial(n-j, k), 3) for j in (0..n))
flatten([[A117939(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Oct 29 2021

Formula

Triangle T(n,k) = Sum_{j=0..n} L(C(n,j)/3)*L(C(n-j,k)/3) where L(j/p) is the Legendre symbol of j and p.
T(n, k) mod 2 = A117944(n,k).
T(n, 0) = A059151(n).
T(n, 1) = A117946(n).
Sum_{k=0..n} T(n, k) = A117940(n).
Matrix square of triangle A117947. Matrix log is the integer triangle A120854. - Paul D. Hanna, Jul 08 2006

A117941 Inverse of number triangle A117939.

Original entry on oeis.org

1, -2, 1, -5, 2, 1, -2, 0, 0, 1, 4, -2, 0, -2, 1, 10, -4, -2, -5, 2, 1, -5, 0, 0, 2, 0, 0, 1, 10, -5, 0, -4, 2, 0, -2, 1, 25, -10, -5, -10, 4, 2, -5, 2, 1, -2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, -2, 0, 0, 0, 0, 0, 0, 0, -2, 1, 10, -4, -2, 0, 0, 0, 0, 0, 0, -5, 2, 1, 4, 0, 0, -2, 0, 0, 0, 0, 0, -2, 0, 0, 1, -8, 4, 0, 4, -2, 0, 0, 0, 0, 4, -2, 0, -2, 1
Offset: 0

Views

Author

Paul Barry, Apr 05 2006

Keywords

Comments

Row sums are A117942.
T(n, k) mod 2 = A117944(n,k).

Examples

			Triangle begins
   1;
  -2,   1;
  -5,   2,  1;
  -2,   0,  0,   1;
   4,  -2,  0,  -2, 1;
  10,  -4, -2,  -5, 2, 1;
  -5,   0,  0,   2, 0, 0,  1;
  10,  -5,  0,  -4, 2, 0, -2, 1;
  25, -10, -5, -10, 4, 2, -5, 2, 1;

Links

Crossrefs

Cf. A117939, A117942, A117944.

Programs

  • Mathematica
    M[n_, k_]:= M[n, k]= If[k>n, 0, Sum[JacobiSymbol[Binomial[n, j], 3]*JacobiSymbol[Binomial[n-j, k], 3], {j,0,n}], 0];
m:= m= With[{q = 60}, Table[M[n, k], {n,0,q}, {k,0,q}]];
T[n_, k_]:= Inverse[m][[n+1, k+1]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 29 2021 *)

A117945 Triangle related to powers of 3 partitions of n.

Original entry on oeis.org

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

Views

Author

Paul Barry, Apr 05 2006

Keywords

Comments

Row sums are A039966.
Inverse of A117944.

Examples

			Triangle begins
   1;
   0,  1;
  -1,  0,  1;
   0,  0,  0,  1;
   0,  0,  0,  0, 1;
   0,  0,  0, -1, 0, 1;
  -1,  0,  0,  0, 0, 0,  1;
   0, -1,  0,  0, 0, 0,  0, 1;
   1,  0, -1,  0, 0, 0, -1, 0, 1;
   0,  0,  0,  0, 0, 0,  0, 0, 0,  1;
   0,  0,  0,  0, 0, 0,  0, 0, 0,  0, 1;
   0,  0,  0,  0, 0, 0,  0, 0, 0, -1, 0, 1;

Links

Crossrefs

Cf. A039966, A117944.

Programs

  • Mathematica
    M[n_, k_]:= M[n, k] = If[k>n, 0, Mod[Sum[JacobiSymbol[Binomial[n, j], 3]*JacobiSymbol[Binomial[n-j, k], 3], {j,0,n}], 2], 0];
m:= m= With[{q=60}, Table[M[n, k], {n,0,q}, {k,0,q}]];
T[n_, k_]:= Inverse[m][[n+1, k+1]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 29 2021 *)
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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