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.

Previous Showing 61-70 of 70 results.

A266643 Permutation of nonnegative integers: a(n) = A264965(3*n) / 3.

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 19, 12, 11, 22, 21, 16, 25, 18, 13, 20, 15, 14, 27, 24, 73, 58, 23, 28, 39, 30, 37, 64, 57, 46, 17, 36, 67, 26, 55, 40, 31, 66, 49, 76, 51, 54, 65, 48, 41, 50, 75, 44, 163, 34, 29, 56, 77, 78, 61, 60, 53, 74, 45, 32, 59, 42, 43, 68, 81, 70, 33, 72, 35, 82, 63, 52, 71, 38, 47, 80, 69
Offset: 0

Views

Author

Antti Karttunen, Jan 04 2016

Keywords

Crossrefs

Formula

a(n) = A264965(3*n) / 3.
As a composition of related permutations:
a(n) = A263273(A246200(n)).

A266644 Permutation of nonnegative integers: a(n) = A264966(3*n) / 3.

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 13, 12, 19, 22, 21, 16, 35, 18, 11, 20, 15, 14, 27, 24, 17, 38, 23, 28, 55, 30, 41, 64, 71, 54, 73, 36, 31, 78, 29, 40, 49, 66, 67, 52, 63, 34, 79, 48, 43, 50, 45, 76, 61, 46, 39, 56, 33, 26, 65, 60, 59, 82, 75, 32, 47, 42, 37, 68, 81, 70, 77, 72, 25, 62, 51, 44, 57, 58, 83, 80, 69
Offset: 0

Views

Author

Antti Karttunen, Jan 04 2016

Keywords

Crossrefs

Inverse: A266643.
Differs from A264965 for the first time at n=17, where a(17) = 35, while A264965(17) = 25.

Programs

Formula

a(n) = A264966(3*n) / 3.
As a composition of related permutations:
a(n) = A246200(A263273(n)).

A264980 Base-3 reversal of 2^n: a(n) = A030102(A000079(n)).

Original entry on oeis.org

1, 2, 4, 8, 16, 64, 32, 184, 352, 704, 1408, 1880, 2824, 14032, 10328, 56128, 100576, 145784, 189472, 370304, 731752, 4388248, 2924096, 11175712, 15965704, 31930448, 63861880, 383165344, 255439712, 1021772344, 510875648, 2550188248, 5619691648, 9689861048, 17830350904, 79068724264, 34109913224, 192259976368, 133338241880
Offset: 0

Views

Author

Antti Karttunen, Dec 05 2015

Keywords

Examples

			2^5 = 32 in base 3 = "1012" (= A007089(32)) as 1*27 + 1*3 + 2*1 = 32. 2^6 = 64 in base 3 = "2101" as 2*27 + 1*9 + 1*1 = 64. "1012" reversed is "2101" and vice versa, thus a(5) = 64 and a(6) = 32.
		

Crossrefs

Leftmost column of A265345.
Cf. also A036215.

Programs

  • PARI
    base(n) = {my(a=[n%3]); while(0Altug Alkan, Dec 29 2015

Formula

a(n) = A030102(A000079(n)) = A263273(A000079(n)).
a(0) = 1, for n >= 1, a(n) = A265342(a(n-1)).

A265340 Number of iterations of A265339 needed to reach zero; a(0) = 0; for n >= 1, a(n) = 1 + a(A265339(n)).

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 4, 4, 5, 5, 5, 5, 5, 4, 5, 4, 4, 5, 5, 5, 5, 5, 5, 6, 5, 6, 7, 6, 6, 7, 6, 5, 6, 6, 6, 7, 7, 6, 7, 6, 6, 6, 6, 6, 7, 7, 6, 7, 6, 5, 6, 6, 6, 7, 6, 6, 7, 6, 6, 6, 6, 6, 7, 7, 6, 7, 7, 6, 6, 6, 6, 7, 7, 6, 7, 7
Offset: 0

Views

Author

Antti Karttunen, Dec 18 2015

Keywords

Comments

Also the number of significant digits in the binary representation of A263273(n).

Crossrefs

Formula

a(0) = 0; for n >= 1, a(n) = 1 + a(A265339(n)).
a(0) = 0; for n >= 1, a(n) = A070939(A263273(n)).
a(n) = A265336(n) + A265337(n).

A265342 Permutation of even numbers: a(n) = 2 * A265351(n).

Original entry on oeis.org

0, 2, 4, 6, 8, 22, 12, 10, 16, 18, 20, 58, 24, 26, 76, 66, 64, 70, 36, 14, 40, 30, 28, 34, 48, 46, 52, 54, 56, 166, 60, 62, 184, 174, 172, 178, 72, 74, 220, 78, 80, 238, 228, 226, 232, 198, 68, 202, 192, 190, 196, 210, 208, 214, 108, 38, 112, 42, 44, 130, 120, 118, 124, 90, 32, 94, 84, 82, 88, 102, 100, 106, 144
Offset: 0

Views

Author

Antti Karttunen, Dec 07 2015

Keywords

Comments

Iterating this sequence as 1, a(1), a(a(1)), a(a(a(1))), ... yields A264980.

Crossrefs

Cf. A265351.
Cf. also A265341, A263273, A264980.

Programs

  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a263272(n): return a263273(2*n)/2
    def a(n): return 2*a263272(a263273(n)) # Indranil Ghosh, May 25 2017
  • Scheme
    (define (A265342 n) (* 2 (A265351 n)))
    

Formula

a(n) = 2 * A265351(n).

A264986 Even bisection of A263272; terms of A264974 doubled.

Original entry on oeis.org

0, 2, 4, 6, 8, 10, 12, 14, 32, 18, 20, 38, 24, 26, 28, 30, 16, 34, 36, 22, 40, 42, 68, 86, 96, 50, 104, 54, 56, 110, 60, 74, 92, 114, 44, 98, 72, 62, 116, 78, 80, 82, 84, 46, 100, 90, 64, 118, 48, 70, 88, 102, 52, 106, 108, 58, 112, 66, 76, 94, 120, 122, 284, 126, 176, 338, 204, 230, 248, 258, 140, 302, 288
Offset: 0

Views

Author

Antti Karttunen, Dec 05 2015

Keywords

Crossrefs

Programs

  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a(n): return a263273(4*n)/2 # Indranil Ghosh, May 23 2017
  • Scheme
    (define (A264986 n) (A263272 (+ n n)))
    

Formula

a(n) = A263272(2*n).
a(n) = 2 * A264974(n).
a(n) = A263273(4*n)/2.

A264987 Odd bisection of A263272.

Original entry on oeis.org

1, 3, 5, 11, 9, 7, 13, 15, 23, 29, 33, 17, 35, 27, 19, 37, 21, 25, 31, 39, 41, 95, 45, 59, 113, 69, 77, 83, 87, 47, 101, 99, 65, 119, 51, 71, 89, 105, 53, 107, 81, 55, 109, 57, 73, 91, 111, 43, 97, 63, 61, 115, 75, 79, 85, 93, 49, 103, 117, 67, 121, 123, 203, 257, 285, 149, 311, 135, 167, 329, 177, 221, 275
Offset: 0

Views

Author

Antti Karttunen, Dec 05 2015

Keywords

Crossrefs

Programs

  • Python
    from sympy import factorint
    from sympy.ntheory.factor_ import digits
    from operator import mul
    def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
    def a038502(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
    def a038500(n): return n/a038502(n)
    def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
    def a(n): return a263273(2*(2*n + 1))/2 # Indranil Ghosh, May 23 2017
  • Scheme
    (define (A264987 n) (A263272 (+ 1 n n)))
    

Formula

a(n) = A263272((2*n)+1).

A264995 Bijective base-5 reverse: a(0) = 0; for n >= 1, a(n) = A030104(A132739(n)) * A060904(n).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 11, 16, 21, 10, 7, 12, 17, 22, 15, 8, 13, 18, 23, 20, 9, 14, 19, 24, 25, 26, 51, 76, 101, 30, 31, 56, 81, 106, 55, 36, 61, 86, 111, 80, 41, 66, 91, 116, 105, 46, 71, 96, 121, 50, 27, 52, 77, 102, 35, 32, 57, 82, 107, 60, 37, 62, 87, 112, 85, 42, 67, 92, 117, 110, 47, 72, 97, 122, 75, 28, 53, 78, 103, 40, 33
Offset: 0

Views

Author

Antti Karttunen, Dec 07 2015

Keywords

Comments

Self-inverse permutation of nonnegative integers.

Crossrefs

Cf. similar sequences A057889 (base-2), A263273 (base-3), A264994 (base-4), A264979 (base-9).

Programs

Formula

a(0) = 0; for n >= 1, a(n) = A030104(A132739(n)) * A060904(n).
Other identities. For all n >= 0:
a(5*n) = 5*a(n).
A010873(a(n)) = 0 if and only if A010873(n) = 0 and it also seems that A010873(a(n)) = A010873(n) for all n.

A265343 Permutation of nonnegative integers: a(n) = A264978(A263272(n)).

Original entry on oeis.org

0, 1, 2, 3, 8, 5, 6, 17, 4, 9, 10, 7, 24, 26, 14, 15, 44, 16, 18, 19, 20, 51, 53, 11, 12, 35, 13, 27, 28, 29, 30, 89, 23, 21, 62, 22, 72, 71, 25, 78, 80, 41, 42, 125, 43, 45, 46, 47, 132, 134, 50, 48, 143, 49, 54, 55, 56, 57, 170, 59, 60, 179, 58, 153, 152, 52, 159, 161, 32, 33, 98, 31, 36, 37, 34, 105, 107, 38, 39, 116, 40, 81
Offset: 0

Views

Author

Antti Karttunen, Dec 07 2015

Keywords

Comments

Composition of A263272 with the permutation obtained from its quadrisection. Equally: composition of permutations that are obtained from the bisection and octisection of A263273.

Crossrefs

Inverse: A265344.
Cf. also A265363.

Programs

Formula

a(n) = A264978(A263272(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).

A265344 Permutation of nonnegative integers: a(n) = A263272(A264978(n)).

Original entry on oeis.org

0, 1, 2, 3, 8, 5, 6, 11, 4, 9, 10, 23, 24, 26, 14, 15, 17, 7, 18, 19, 20, 33, 35, 32, 12, 38, 13, 27, 28, 29, 30, 71, 68, 69, 74, 25, 72, 73, 77, 78, 80, 41, 42, 44, 16, 45, 46, 47, 51, 53, 50, 21, 65, 22, 54, 55, 56, 57, 62, 59, 60, 101, 34, 99, 100, 104, 105, 107, 95, 96, 98, 37, 36, 109, 110, 114, 116, 113, 39, 119, 40, 81
Offset: 0

Views

Author

Antti Karttunen, Dec 07 2015

Keywords

Comments

Composition of A263272 with the permutation obtained from its quadrisection. Equally: composition of permutations that are obtained from the bisection and octisection of A263273.

Crossrefs

Inverse: A265343.
Cf. also A265364.

Programs

Formula

a(n) = A263272(A264978(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
Previous Showing 61-70 of 70 results.