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 1-2 of 2 results.

A379176 Inverse permutation to A379175.

Original entry on oeis.org

0, 1, 2, 4, 3, 5, 9, 7, 6, 8, 10, 18, 16, 19, 14, 12, 11, 13, 15, 20, 17, 21, 39, 35, 33, 36, 40, 29, 27, 30, 25, 23, 22, 24, 26, 31, 28, 32, 41, 37, 34, 38, 42, 78, 74, 79, 70, 66, 64, 67, 71, 80, 75, 81, 60, 56, 54, 57, 61, 50, 48, 51, 46, 44, 43, 45, 47, 52
Offset: 0

Views

Author

Rémy Sigrist, Dec 17 2024

Keywords

Examples

			A379175(67) = 49, so a(49) = 67.
		

Crossrefs

Cf. A379175.

Programs

  • PARI
    \\ See Links section.

A379147 Irregular triangle T(n, k), n >= 0, k = 1..2^A007895(n), read by rows; the n-th row lists the integers m such that A184617(abs(m)) = A003714(n).

Original entry on oeis.org

0, -1, 1, -2, 2, -4, 4, -5, -3, 3, 5, -8, 8, -9, -7, 7, 9, -10, -6, 6, 10, -16, 16, -17, -15, 15, 17, -18, -14, 14, 18, -20, -12, 12, 20, -21, -19, -13, -11, 11, 13, 19, 21, -32, 32, -33, -31, 31, 33, -34, -30, 30, 34, -36, -28, 28, 36
Offset: 0

Views

Author

Rémy Sigrist, Dec 16 2024

Keywords

Comments

A permutation of the integers (Z).
For any n >= 0:
- in the Zeckendorf expansion of n,
- replace each Fibonacci number, say A000045(2+i) with i >= 0, by 2^i or -2^i,
- the various values obtained make up the n-th row.

Examples

			Triangle T(n, k) begins:
  n   n-th row
  --  ----------------------------------
   0  0
   1  -1, 1
   2  -2, 2
   3  -4, 4
   4  -5, -3, 3, 5
   5  -8, 8
   6  -9, -7, 7, 9
   7  -10, -6, 6, 10
   8  -16, 16
   9  -17, -15, 15, 17
  10  -18, -14, 14, 18
  11  -20, -12, 12, 20
  12  -21, -19, -13, -11, 11, 13, 19, 21
  13  -32, 32
  14  -33, -31, 31, 33
  15  -34, -30, 30, 34
		

Crossrefs

Programs

  • PARI
    tozeck(n) = { for (i=0, oo, if (n<=fibonacci(2+i), my (v=0, f); forstep (j=i, 0, -1, if (n>=f=fibonacci(2+j), n-=f; v+=2^j;); if (n==0, return (v););););); }
    row(n) = { my (z = tozeck(n), r = [0], b); while (z, z -= b = 2^valuation(z, 2); r = concat([v - b | v <- r], [v + b | v <- r]);); return (r); }

Formula

T(n, 1) = -A003714(n).
T(n, 2^A007895(n)) = A003714(n).
T(n, k) = -T(n, 2^A007895(n)+1-k) for k = 1..2^A007895(n).
Showing 1-2 of 2 results.