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.

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A300834 a(n) = Product_{d|n, dA019565(A003714(d)), where A003714(n) is the n-th Fibbinary number.

Original entry on oeis.org

1, 2, 2, 6, 2, 30, 2, 60, 10, 42, 2, 4200, 2, 126, 70, 660, 2, 9240, 2, 13860, 210, 330, 2, 5082000, 14, 78, 220, 32760, 2, 3783780, 2, 42900, 550, 780, 294, 924924000, 2, 1092, 130, 41621580, 2, 3898440, 2, 112200, 60060, 306, 2, 28078050000, 42, 235620, 1300, 92820, 2, 200119920, 770, 128648520, 1820, 1122, 2, 424964656116000, 2, 3366
Offset: 1

Views

Author

Antti Karttunen, Mar 18 2018

Keywords

Crossrefs

Cf. A003714, A019565, A300835 (rgs-transform of this sequence), A300836.

Programs

  • PARI
    A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
    A003714(n) = { my(s=0,w); while(n>2, w = A072649(n); s += 2^(w-1); n -= fibonacci(w+1)); (s+n); }
    A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
    A300834(n) = { my(m=1); fordiv(n,d,if(d < n,m *= A019565(A003714(d)))); m; };

Formula

a(n) = Product_{d|n, dA019565(A003714(d)).
For n >= 1, A001222(a(n)) = A300836(n).

A345201 Bit-reverse the odd part of the Zeckendorf representation of n: a(n) = A022290(A057889(A003714(n))).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 19, 18, 17, 20, 21, 22, 23, 24, 30, 26, 27, 31, 29, 25, 28, 32, 33, 34, 35, 36, 37, 48, 39, 43, 49, 42, 40, 44, 50, 53, 47, 38, 41, 45, 51, 52, 46, 54, 55, 56, 57, 58, 77, 60, 69, 78, 63, 64, 70, 79, 85
Offset: 0

Views

Author

Rémy Sigrist, Jun 10 2021

Keywords

Comments

This sequence is a self-inverse permutation of the nonnegative integers.
This sequence is similar to A343150 and to A344682.

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(n) < A000045(k) for any n < A000045(k).

A129761 First differences of Fibbinary numbers (A003714).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 11, 1, 1, 2, 1, 3, 1, 1, 22, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 43, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 11, 1, 1, 2, 1, 3, 1, 1, 86, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 11, 1, 1, 2, 1, 3, 1, 1, 22, 1, 1, 2, 1, 3, 1, 1, 6, 1, 1, 2, 1, 171, 1, 1, 2, 1, 3, 1, 1, 6
Offset: 0

Views

Author

Ralf Stephan, May 14 2007

Keywords

Comments

Theorem: If the Zeckendorf representation of M ends with exactly k >= 0 zeros, ...10^k, then a(n) = ceiling(2^k/3). Also, if the Zeckendorf representation of n (A014417(n)) is even then a(n) is given by A319952, otherwise a(n) = 1. - Jeffrey Shallit and N. J. A. Sloane, Oct 03 2018

Crossrefs

Programs

  • Maple
    with(combinat): F:=fibonacci:
    A072649:= proc(n) local j; global F; for j from ilog[(1+sqrt(5))/2](n)
           while F(j+1)<=n do od; (j-1); end:
    A003714 := proc(n) global F; option remember; if(n < 3) then RETURN(n); else RETURN((2^(A072649(n)-1))+A003714(n-F(1+A072649(n)))); fi; end:
    A129761 := n -> A003714(n+1)-A003714(n):
    [seq(A129761(n),n=0..120)]; # N. J. A. Sloane, Oct 03 2018, borrowing programs from other sequences
  • Mathematica
    Differences[Select[Range[600], !MemberQ[Partition[IntegerDigits[#, 2], 2, 1], {1, 1}] &]] (* Harvey P. Dale, Jul 17 2011 *)

Formula

a(n) = A005578(A035614(n)). - Alan Michael Gómez Calderón, Nov 01 2023

Extensions

a(0)=1 added by N. J. A. Sloane, Oct 02 2018

A300835 Restricted growth sequence transform of A300834, product_{d|n, dA019565(A003714(d)); Filter sequence related to Zeckendorf-representations of proper divisors of n.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 6, 7, 2, 8, 2, 9, 10, 11, 2, 12, 2, 13, 14, 15, 2, 16, 17, 18, 19, 20, 2, 21, 2, 22, 23, 24, 25, 26, 2, 27, 28, 29, 2, 30, 2, 31, 32, 33, 2, 34, 7, 35, 36, 37, 2, 38, 39, 40, 41, 42, 2, 43, 2, 44, 45, 46, 47, 48, 2, 49, 50, 51, 2, 52, 2, 53, 54, 55, 56, 57, 2, 58, 59, 60, 2, 61, 41, 62, 63, 64, 2, 65, 66, 67, 68, 69
Offset: 1

Views

Author

Antti Karttunen, Mar 18 2018

Keywords

Comments

For all i, j: a(i) = a(j) => A001065(i) = A001065(j).
For all i, j: a(i) = a(j) => A300836(i) = A300836(j).

Examples

			For cases n=10 and 49, we see that 10 has proper divisors 1, 2 and 5 and these have Zeckendorf-representations (A014417) 1, 10 and 1000, while 49 has proper divisors 1 and 7 and these have Zeckendorf-representations 1 and 1010. When these Zeckendorf-representations are summed (columnwise without carries), result in both cases is 1011, thus a(10) = a(49).
		

Crossrefs

Cf. also A293215, A293217, A293223, A293224, A293232, A300833 for similar filtering sequences.

Programs

  • PARI
    up_to = 65537;
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
    A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
    A003714(n) = { my(s=0,w); while(n>2, w = A072649(n); s += 2^(w-1); n -= fibonacci(w+1)); (s+n); }
    A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
    A300834(n) = { my(m=1); fordiv(n,d,if(d < n,m *= A019565(A003714(d)))); m; };
    write_to_bfile(1,rgs_transform(vector(up_to,n,A300834(n))),"b300835.txt");

A318469 Multiplicative with a(p^e) = A019565(A003714(e)).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 10, 2, 6, 2, 6, 4, 4, 2, 10, 3, 4, 5, 6, 2, 8, 2, 7, 4, 4, 4, 9, 2, 4, 4, 10, 2, 8, 2, 6, 6, 4, 2, 20, 3, 6, 4, 6, 2, 10, 4, 10, 4, 4, 2, 12, 2, 4, 6, 14, 4, 8, 2, 6, 4, 8, 2, 15, 2, 4, 6, 6, 4, 8, 2, 20, 10, 4, 2, 12, 4, 4, 4, 10, 2, 12, 4, 6, 4, 4, 4, 14, 2, 6, 6, 9, 2, 8, 2, 10, 8
Offset: 1

Views

Author

Antti Karttunen, Aug 30 2018

Keywords

Crossrefs

Programs

  • PARI
    A003714(n) = { my(s=0,w); while(n>2, w = A072649(n); s += 2^(w-1); n -= fibonacci(w+1)); (s+n); }
    A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
    A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
    A318469(n) = factorback(apply(e -> A019565(A003714(e)),factor(n)[,2]));

Formula

For all n >= 1, A001222(a(n)) = A318464(n).

A374354 Irregular table T(n, k), n >= 0, 0 <= k < A277561(n), read by rows; the n-th row lists the fibbinary numbers f <= n such that n - f is also a fibbinary number whose binary expansion has no common 1's with that of f (where fibbinary numbers correspond to A003714).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 2, 0, 4, 0, 1, 4, 5, 2, 4, 2, 5, 0, 8, 0, 1, 8, 9, 0, 2, 8, 10, 1, 2, 9, 10, 4, 8, 4, 5, 8, 9, 4, 10, 5, 10, 0, 16, 0, 1, 16, 17, 0, 2, 16, 18, 1, 2, 17, 18, 0, 4, 16, 20, 0, 1, 4, 5, 16, 17, 20, 21, 2, 4, 18, 20, 2, 5, 18, 21, 8, 16, 8, 9, 16, 17
Offset: 0

Views

Author

Rémy Sigrist, Jul 06 2024

Keywords

Comments

In other words, we partition n into pairs of fibbinary numbers whose binary expansions have no common 1's and list the corresponding fibbinary numbers to get the n-th row.

Examples

			Triangle T(n, k) begins:
  n   n-th row
  --  -----------
   0  0
   1  0, 1
   2  0, 2
   3  1, 2
   4  0, 4
   5  0, 1, 4, 5
   6  2, 4
   7  2, 5
   8  0, 8
   9  0, 1, 8, 9
  10  0, 2, 8, 10
  11  1, 2, 9, 10
  12  4, 8
  13  4, 5, 8, 9
  14  4, 10
  15  5, 10
  16  0, 16
		

Crossrefs

See A295989 and A374361 for similar sequences.

Programs

  • PARI
    row(n) = { my (r = [0], e, x, y, b); while (n, x = y = 0; e = valuation(n, 2); for (k = 0, oo, if (bittest(n, e+k), n -= b = 2^(e+k); [x, y] = [y + b, x], r = concat([v + y | v <- r], [v + x | v <- r]); break;););); return (r); }

Formula

T(n, 0) = 0 iff n is a fibbinary number.
T(n, k) + T(n, A277561(n)-1-k) = n.
T(n, 0) = A374355(n).
T(n, A277561(n)-1) = A374356(n).
Sum_{k = 0..A277561(n)-1} T(n, k) = n * 2^A037800(n).

A374355 a(n) is the least fibbinary number f <= n such that n - f is also a fibbinary number whose binary expansion has no common 1's with that of f (where fibbinary numbers correspond to A003714).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 0, 1, 4, 4, 4, 5, 0, 0, 0, 1, 0, 0, 2, 2, 8, 8, 8, 9, 8, 8, 10, 10, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 0, 1, 4, 4, 4, 5, 16, 16, 16, 17, 16, 16, 18, 18, 16, 16, 16, 17, 20, 20, 20, 21, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 0, 1, 4, 4, 4, 5, 0
Offset: 0

Views

Author

Rémy Sigrist, Jul 06 2024

Keywords

Comments

To compute a(n): replace every other bit with zero (starting with the first bit) in each run of consecutive 1's in the binary expansion of n.

Examples

			The first terms, in binary and in decimal, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     0       1          0
   2     0      10          0
   3     1      11          1
   4     0     100          0
   5     0     101          0
   6     2     110         10
   7     2     111         10
   8     0    1000          0
   9     0    1001          0
  10     0    1010          0
  11     1    1011          1
  12     4    1100        100
  13     4    1101        100
  14     4    1110        100
  15     5    1111        101
  16     0   10000          0
		

Crossrefs

Programs

  • PARI
    a(n) = { my (v = 0, e, x, y, b); while (n, x = y = 0; e = valuation(n, 2); for (k = 0, oo, if (bittest(n, e+k), n -= b = 2^(e+k); [x, y] = [y + b, x], v += y; break;););); return (v); }

Formula

a(n) = A374354(n, 0).
a(n) = n - A374356(n).
a(n) >= 0 with equality iff n is a fibbinary number.

A277332 a(n) = A253565(A003714(n)).

Original entry on oeis.org

1, 2, 3, 5, 9, 7, 25, 15, 11, 49, 35, 21, 75, 13, 121, 77, 55, 245, 33, 147, 105, 17, 169, 143, 91, 847, 65, 605, 385, 39, 363, 231, 165, 735, 19, 289, 221, 187, 1859, 119, 1183, 1001, 85, 845, 715, 455, 4235, 51, 507, 429, 273, 2541, 195, 1815, 1155, 23, 361, 323, 247, 3757, 209, 3179, 2431, 133, 2023, 1547, 1309, 13013, 95
Offset: 0

Views

Author

Antti Karttunen, Oct 12 2016

Keywords

Comments

After the initial terms 1, 2 and 3, all other terms can be inductively generated by applying any finite composition-combination of A253560 and A253550 to 3, but with such a restriction that A253560 may not be applied twice in succession.
A permutation of A277334.
Note how A253565(A022340(n)) = A253565(2*A003714(n)) yields a permutation of A056911, odd squarefree numbers.

Examples

			55 = A253550(A253550(A253560(A253550(3)))), 55 is in this sequence.
		

Crossrefs

Cf. A277334 (same sequence sorted into ascending order).
Cf. also A056911, A277006, A277331.

Programs

Formula

a(n) = A253565(A003714(n)).

A300867 a(n) is the least positive k such that k * n is a Fibbinary number (A003714).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 3, 3, 5, 3, 11, 1, 1, 1, 7, 1, 1, 3, 3, 3, 13, 5, 3, 3, 5, 11, 11, 1, 1, 1, 39, 1, 1, 7, 7, 1, 1, 1, 3, 3, 13, 3, 7, 3, 21, 13, 23, 5, 5, 3, 3, 3, 9, 5, 11, 11, 9, 11, 43, 1, 1, 1, 35, 1, 1, 39, 15, 1, 1, 1, 31, 7, 57, 7, 7, 1
Offset: 0

Views

Author

Rémy Sigrist, Mar 14 2018

Keywords

Comments

This sequence is well defined: for any positive n, according to the pigeonhole principle, A195156(i) mod n = A195156(j) mod n for some distinct i and j, hence n divides f = abs(A195156(i) - A195156(j)), and as f is a Fibbinary number, a(n) <= f/n.
All terms are odd.

Examples

			The first terms, alongside the binary representation of n * a(n), are:
  n  a(n)   bin(n * a(n))
  -- ----   -------------
   0    1           0
   1    1           1
   2    1          10
   3    3        1001
   4    1         100
   5    1         101
   6    3       10010
   7    3       10101
   8    1        1000
   9    1        1001
  10    1        1010
  11    3      100001
  12    3      100100
  13    5     1000001
  14    3      101010
  15   11    10100101
  16    1       10000
  17    1       10001
  18    1       10010
  19    7    10000101
  20    1       10100
		

Crossrefs

Programs

  • PARI
    a(n) = forstep (k=1, oo, 2, if (bitand(k*n, 2*k*n)==0, return (k)))

Formula

a(n) = A300889(n) / n for any n > 0.
a(2*n) = a(n).
a(n) = 1 iff n belongs to A003714.

A300889 a(n) is the least positive multiple of n which is a Fibbinary number (A003714).

Original entry on oeis.org

1, 2, 9, 4, 5, 18, 21, 8, 9, 10, 33, 36, 65, 42, 165, 16, 17, 18, 133, 20, 21, 66, 69, 72, 325, 130, 81, 84, 145, 330, 341, 32, 33, 34, 1365, 36, 37, 266, 273, 40, 41, 42, 129, 132, 585, 138, 329, 144, 1029, 650, 1173, 260, 265, 162, 165, 168, 513, 290, 649
Offset: 1

Views

Author

Rémy Sigrist, Mar 14 2018

Keywords

Examples

			The first terms, alongside their binary representation, are:
  n  a(n)    binary(a(n))
  -- ----    ------------
   1    1           1
   2    2          10
   3    9        1001
   4    4         100
   5    5         101
   6   18       10010
   7   21       10101
   8    8        1000
   9    9        1001
  10   10        1010
  11   33      100001
  12   36      100100
  13   65     1000001
  14   42      101010
  15  165    10100101
  16   16       10000
  17   17       10001
  18   18       10010
  19  133    10000101
  20   20       10100
		

Crossrefs

Programs

  • PARI
    a(n) = forstep (k=1, oo, 2, if (bitand(k*n, 2*k*n)==0, return (k*n)))

Formula

a(n) = n * A300867(n).
a(2*n) = 2*a(n).
a(n) = n iff n belongs to A003714.
Previous Showing 11-20 of 236 results. Next