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 31-40 of 76 results. Next

A250245 Permutation of natural numbers: a(1) = 1, a(n) = A083221(A055396(n),a(A246277(n))).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 27, 22, 23, 24, 25, 26, 21, 28, 29, 30, 31, 32, 39, 34, 35, 36, 37, 38, 63, 40, 41, 54, 43, 44, 33, 46, 47, 48, 49, 50, 75, 52, 53, 42, 65, 56, 99, 58, 59, 60, 61, 62, 57, 64, 95, 78, 67, 68, 111, 70, 71, 72, 73, 74, 51, 76, 77, 126, 79, 80, 45, 82
Offset: 1

Views

Author

Antti Karttunen, Nov 17 2014

Keywords

Comments

The first 7-cycle occurs at: (33 39 63 57 99 81 45) which is mirrored by the cycle (66 78 126 114 198 162 90) with double-size terms.
The cycle which contains 55 as its smallest term, goes as: 55, 65, 95, 185, 425, 325, 205, 455, 395, 1055, 2945, 6035, 30845, ...
while to the other direction (A250246) it goes as: 55, 125, 245, 115, 625, 8575, 40375, ...
The cycle which contains 69 as its smallest term, goes as: 69, 111, 183, 351, 261, 273, 387, 489, 939, 1863, 909, 1161, 981, 1281, 4167, ...
while to the other direction (A250246) it goes as: 69, 135, 87, 105, 225, 207, 231, 195, 525, 1053, 3159, 24909, ...

Crossrefs

Inverse: A250246.
Other similar permutations: A250244, A250247, A250249, A243071, A252755.
Differs from the "vanilla version" A249817 for the first time at n=42, where a(42) = 54, while A249817(42) = 42.
Differs from A250246 for the first time at n = 33, where a(33) = 39, while A250246(33) = 45.
Differs from A250249 for the first time at n=73, where a(73) = 73, while A250249(73) = 103.

Formula

a(1) = 1, a(n) = A083221(A055396(n), a(A246277(n))).
a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A250469(a(A064989(2n+1))). - Antti Karttunen, Jan 18 2015
As a composition of related permutations:
a(n) = A252755(A243071(n)).
Other identities. For all n >= 1:
a(n) = a(2n)/2. [The even bisection halved gives the sequence back.]
A020639(a(n)) = A020639(n) and A055396(a(n)) = A055396(n). [Preserves the smallest prime factor of n].

A091202 Factorization-preserving isomorphism from nonnegative integers to binary codes for polynomials over GF(2).

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 6, 11, 8, 5, 14, 13, 12, 19, 22, 9, 16, 25, 10, 31, 28, 29, 26, 37, 24, 21, 38, 15, 44, 41, 18, 47, 32, 23, 50, 49, 20, 55, 62, 53, 56, 59, 58, 61, 52, 27, 74, 67, 48, 69, 42, 43, 76, 73, 30, 35, 88, 33, 82, 87, 36, 91, 94, 39, 64, 121, 46, 97, 100, 111, 98
Offset: 0

Views

Author

Antti Karttunen, Jan 03 2004

Keywords

Comments

E.g. we have the following identities: A000005(n) = A091220(a(n)), A001221(n) = A091221(a(n)), A001222(n) = A091222(a(n)), A008683(n) = A091219(a(n)), A014580(n) = a(A000040(n)), A049084(n) = A091227(a(n)).

Crossrefs

Inverse: A091203.
Several variants exist: A235041, A091204, A106442, A106444, A106446.
Cf. also A302023, A302025, A305417, A305427 for other similar permutations.

Programs

  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A091225(n) = polisirreducible(Pol(binary(n))*Mod(1, 2));
    A305420(n) = { my(k=1+n); while(!A091225(k),k++); (k); };
    A305421(n) = { my(f = subst(lift(factor(Pol(binary(n))*Mod(1, 2))),x,2)); for(i=1,#f~,f[i,1] = Pol(binary(A305420(f[i,1])))); fromdigits(Vec(factorback(f))%2,2); };
    A091202(n) = if(n<=1,n,if(!(n%2),2*A091202(n/2),A305421(A091202(A064989(n))))); \\ Antti Karttunen, Jun 10 2018

Formula

a(0)=0, a(1)=1, a(p_i) = A014580(i) for primes p_i with index i and for composites a(p_i * p_j * ...) = a(p_i) X a(p_j) X ..., where X stands for carryless multiplication of GF(2)[X] polynomials (A048720).
Other identities. For all n >= 1, the following holds:
A091225(a(n)) = A010051(n). [Maps primes to binary representations of irreducible GF(2) polynomials, A014580, and nonprimes to union of {1} and the binary representations of corresponding reducible polynomials, A091242. The permutations A091204, A106442, A106444, A106446, A235041 and A245703 have the same property.]
From Antti Karttunen, Jun 10 2018: (Start)
For n <= 1, a(n) = n, for n > 1, a(n) = 2*a(n/2) if n is even, and if n is odd, then a(n) = A305421(a(A064989(n))).
a(n) = A305417(A156552(n)) = A305427(A243071(n)).
(End)

A246377 Permutation of natural numbers: a(1) = 1, a(p_n) = 2*a(n)+1, a(c_n) = 2*a(n), where p_n = n-th prime = A000040(n), c_n = n-th composite number = A002808(n).

Original entry on oeis.org

1, 3, 7, 2, 15, 6, 5, 14, 4, 30, 31, 12, 13, 10, 28, 8, 11, 60, 29, 62, 24, 26, 9, 20, 56, 16, 22, 120, 61, 58, 63, 124, 48, 52, 18, 40, 25, 112, 32, 44, 27, 240, 21, 122, 116, 126, 57, 248, 96, 104, 36, 80, 17, 50, 224, 64, 88, 54, 23, 480, 121, 42, 244, 232, 252, 114, 59, 496, 192, 208, 125, 72, 49, 160, 34, 100
Offset: 1

Views

Author

Antti Karttunen, Aug 27 2014

Keywords

Comments

This permutation is otherwise like Katarzyna Matylla's A135141, except that the role of even and odd numbers (or alternatively: primes and composites) has been swapped.
Because 2 is the only even prime, it implies that, apart from a(2)=3, odd numbers occur in odd positions only (along with many even numbers that also occur in odd positions).
This also implies that for each odd composite (A071904) there exists a separate infinite cycle in this permutation, apart from 9 and 15 which are in the same infinite cycle: (..., 23, 9, 4, 2, 3, 7, 5, 15, 28, 120, 82, 46, ...).

Crossrefs

Inverse: A246378.
Other related or similar permutations: A135141, A054429, A246201, A245703, A246376, A246379, A243071, A246681, A236854.
Differs from A237427 for the first time at n=19, where a(19) = 29, while A237427(19) = 62.

Formula

a(1) = 1, and for n > 1, if A010051(n) = 1 [i.e. when n is a prime], a(n) = 1+(2*a(A000720(n))), otherwise a(n) = 2*a(A065855(n)).
As a composition of related permutations:
a(n) = A054429(A135141(n)).
a(n) = A135141(A236854(n)).
a(n) = A246376(A246379(n)).
a(n) = A246201(A245703(n)).
a(n) = A243071(A246681(n)). [For n >= 1].
Other identities.
For all n > 1 the following holds:
A000035(a(n)) = A010051(n). [Maps primes to odd numbers > 1, and composites to even numbers, in some order. Permutations A246379 & A246681 have the same property].

A252756 Permutation of nonnegative integers: a(1) = 0, a(2) = 1, a(2n) = 2*a(n), a(2n+1) = 1 + 2*a(A250470(2n+1)).

Original entry on oeis.org

0, 1, 3, 2, 7, 6, 15, 4, 5, 14, 31, 12, 63, 30, 13, 8, 127, 10, 255, 28, 9, 62, 511, 24, 11, 126, 29, 60, 1023, 26, 2047, 16, 25, 254, 27, 20, 4095, 510, 61, 56, 8191, 18, 16383, 124, 17, 1022, 32767, 48, 23, 22, 21, 252, 65535, 58, 19, 120, 57, 2046, 131071, 52, 262143, 4094, 125, 32, 59, 50, 524287, 508, 49, 54, 1048575, 40
Offset: 1

Views

Author

Antti Karttunen, Jan 02 2015

Keywords

Crossrefs

Inverse: A252755.
Similar permutations: A243071, A252754, A054429, A250246.
Cf. also A250470, A253556 - A253559.
Differs from A243071 for the first time at n=21, where a(21) = 9, while A243071(21) = 29.

Formula

a(1) = 0, a(2) = 1, a(2n) = 2*a(n), a(2n+1) = 1 + 2*a(A250470(2n+1)).
As a composition of related permutations:
a(n) = A054429(A252754(n)).
a(n) = A243071(A250246(n)).

A292385 a(1) = 0, a(2) = 1, and for n > 2, a(n) = 2*a(A252463(n)) + [n == 1 (mod 4)].

Original entry on oeis.org

0, 1, 2, 2, 5, 4, 10, 4, 5, 10, 20, 8, 41, 20, 8, 8, 83, 10, 166, 20, 21, 40, 332, 16, 11, 82, 8, 40, 665, 16, 1330, 16, 41, 166, 16, 20, 2661, 332, 80, 40, 5323, 42, 10646, 80, 17, 664, 21292, 32, 23, 22, 164, 164, 42585, 16, 42, 80, 333, 1330, 85170, 32, 170341, 2660, 40, 32, 83, 82, 340682, 332, 665, 32, 681364, 40, 1362729, 5322, 20, 664, 33, 160
Offset: 1

Views

Author

Antti Karttunen, Sep 16 2017

Keywords

Comments

Variant of A292381. Here the most significant 1-bit is at the one step smaller position.

Crossrefs

Formula

a(1) = 0, a(2) = 1, and for n > 2, a(n) = 2*a(A252463(n)) + [n == 1 (mod 4)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 4k+1, and 0 otherwise.
For n >= 1, a(n) + A292383(n) = A243071(n); a(A163511(n)) = A292271(n).
For n >= 2, A004754(a(n)) = A292381(n).

A245606 Permutation of natural numbers: a(1) = 1, a(2n) = 1 + A003961(a(n)), a(2n+1) = A003961(1+a(n)). [Where A003961(n) shifts the prime factorization of n one step left].

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 10, 7, 8, 15, 16, 11, 26, 21, 22, 13, 12, 27, 28, 25, 36, 81, 82, 19, 14, 45, 52, 125, 56, 39, 40, 29, 18, 33, 46, 17, 126, 99, 100, 31, 50, 51, 226, 41, 626, 129, 130, 89, 24, 63, 34, 35, 176, 87, 154, 59, 344, 825, 298, 115, 86, 189, 190, 43, 32, 105, 76, 23, 66, 57, 88, 53, 20
Offset: 1

Views

Author

Antti Karttunen, Jul 29 2014

Keywords

Comments

The even bisection halved gives A245608. The odd bisection incremented by one and halved gives A245708.

Crossrefs

Programs

Formula

a(1) = 1, a(2n) = A243501(a(n)), a(2n+1) = A003961(1+a(n)).
As a composition of related permutations:
a(n) = A064216(A245608(n)).

A249824 Permutation of natural numbers: a(n) = A078898(A003961(A003961(2*n))).

Original entry on oeis.org

1, 2, 3, 9, 4, 12, 5, 42, 17, 19, 6, 59, 7, 22, 26, 209, 8, 82, 10, 92, 31, 29, 11, 292, 41, 32, 115, 109, 13, 129, 14, 1042, 40, 39, 48, 409, 15, 49, 45, 459, 16, 152, 18, 142, 180, 52, 20, 1459, 57, 202, 54, 159, 21, 572, 63, 542, 68, 62, 23, 642, 24, 69, 213
Offset: 1

Views

Author

Antti Karttunen, Nov 06 2014

Keywords

Examples

			a(4) = 9 because of the following. 2n = 2*4 = 8 = 2^3. We replace the prime factor 2 of 8 with the next prime 3 to get 3^3, then replace 3 with 5 to get 5^3 = 125. The smallest prime factor of 125 is 5. 125 is the 9th term of A084967: 5, 25, 35, 55, 65, 85, 95, 115, 125, ..., thus a(4) = 9.
		

Crossrefs

Programs

  • Mathematica
    t = PositionIndex[FactorInteger[#][[1, 1]] & /@ Range[10^4]]; f[n_] := Times @@ Power[If[# == 1, 1, NextPrime@ #] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n; Flatten@ Table[Position[Lookup[t, FactorInteger[#][[1, 1]] ], #] &[f@ f[2 n]], {n, 120}] (* Michael De Vlieger, Jul 25 2016, Version 10 *)
  • Scheme
    (define (A249824 n) (A078898 (A003961 (A003961 (* 2 n)))))

Formula

a(n) = A078898(A246278(3,n)).
As a composition of other permutations:
a(n) = A249746(A048673(n)).
a(n) = A250475(A249826(n)).
a(n) = A275716(A243071(n)).
Other identities. For all n >= 1:
a(2n) = A273669(a(n)) and a(A003961(n)) = A273664(a(n)). -- Antti Karttunen, Aug 07 2016

A292274 a(n) = A292383(A163511(n)).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 2, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 8, 8, 10, 11, 8, 8, 10, 11, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 16, 17, 16, 17, 20, 20, 22, 22, 16, 17, 16, 17, 20, 20, 22, 22, 0, 0, 2, 2, 0, 1, 0, 0, 8, 8, 10, 11, 8, 8, 10, 11, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 32, 32, 34, 35, 32, 32, 34, 35, 40, 41, 40, 40, 44
Offset: 0

Views

Author

Antti Karttunen, Sep 16 2017

Keywords

Comments

Because A292383(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate which numbers are of the form 4k+3 in binary tree A005940 on that trajectory which leads from the root of the tree to the node containing A163511(n). This works because A243071(n) = A054429(A156552(n)), a bit-flipped variant of Leonid Broukhis's unary-binary encoded compressed factorization of natural numbers, A156552(n) being an inverse of Doudna map f(n) = A005940(1+n).

Examples

			A163511(18) = 54, that is, at "node address" 18 in binary tree A163511 (which is the mirror image of A005940) sits number 54. 18 in binary is "10010", which when read from left to right (after the most significant bit which is always 1) gives the directions to follow in either tree when starting from the root, so that we land in number 54. (E.g. in A005940-tree, turn right from 2, turn right from 4, turn left from 8 and then turn right from 27 and one lands in 54, this corresponds with the four lowermost bits of the code, "0010". In A163511 the sense of direction is just reversed). When one selects the numbers of the form 4k+3 from this path 1 -> 2 -> 4 -> 8 -> 27 -> 54, one sees that only one is 27, which corresponds with the second rightmost bit (which also is the only 1-bit) in the code, which can be masked with 2 (binary "10"), thus a(18) = 2.
A163511(15) = 7, that is, at "node address" 15 in binary tree A163511 sits number 7. 15 in binary is "1111", which tells that 7 can be located in mirror-image tree A005940 by going (after the initial root 1 and 2) three steps towards left from 2: 1 -> 2 -> 3 -> 5 -> 7. Of these numbers, only 3 and 7 are of the form 4k+3, thus the mask with which to obtain the corresponding bits from "1111" is "00101" (5 in binary), thus a(15) = 5.
A163511(31) = 11, that is, at "node address" 31 in binary tree A163511 sits number 11. 31 in binary is "11111", which tells that 11 can be located in mirror-image tree A005940 by going (after the initial root 1 and 2) four steps towards left from 2: 1 -> 2 -> 3 -> 5 -> 7 -> 11. Of these numbers, only 3, 7 and 11 are of the form 4k+3, thus the mask with which to obtain the corresponding bits from "11111" is "001011" (11 in binary), thus a(31) = 11.
		

Crossrefs

Differs from related A292592 for the first time at n=31, where a(31) = 11, while A292592(31) = 10. Compare also the scatter plots.

Programs

  • Mathematica
    f[n_] := Reverse@ Map[Ceiling[(Length@ # - 1)/2] &, DeleteCases[Split@ Join[Riffle[IntegerDigits[n, 2], 0], {0}], {k__} /; k == 1]]; Map[FromDigits[Reverse@ NestWhileList[Function[k, Which[k == 1, 1, EvenQ@ k, k/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ k]], #, # > 1 &] /. k_ /; IntegerQ@ k :> If[Mod[k, 4] == 3, 1, 0], 2] &, {1}~Join~Table[Function[t, Prime[t] Product[Prime[m]^(f[n][[m]]), {m, t}]][DigitCount[n, 2, 1]], {n, 120}]] (* Michael De Vlieger, Sep 22 2017 *)

Formula

a(n) = A292383(A163511(n)).
a(n) + A292271(n) = n, a(n) AND A292271(n) = 0.
a(n) AND n = a(n), where AND is bitwise-AND (A004198).

Extensions

Comments and examples from Antti Karttunen, Sep 22 2017

A365808 Numbers k such that A163511(k) is a square.

Original entry on oeis.org

0, 2, 5, 8, 11, 17, 20, 23, 32, 35, 41, 44, 47, 65, 68, 71, 80, 83, 89, 92, 95, 128, 131, 137, 140, 143, 161, 164, 167, 176, 179, 185, 188, 191, 257, 260, 263, 272, 275, 281, 284, 287, 320, 323, 329, 332, 335, 353, 356, 359, 368, 371, 377, 380, 383, 512, 515, 521, 524, 527, 545, 548, 551, 560, 563, 569, 572, 575
Offset: 1

Views

Author

Antti Karttunen, Oct 01 2023

Keywords

Comments

The sequence is defined inductively as:
(a) it contains 0 and 2,
and
(b) for any nonzero term a(n), (2*a(n)) + 1 and 4*a(n) are also included as terms.
Because the inductive definition guarantees that all terms after 0 are of the form 3k+2 (A016789), and because for any n >= 0, n^2 == 0 or 1 (mod 3), (i.e., squares are in A032766), it follows that there are no squares in this sequence after the initial 0.

Crossrefs

Cf. A000290, A010052, A032766, A163511, A365807 (characteristic function).
Positions of even terms in A365805.
Sequence A243071(n^2), n >= 1, sorted into ascending order.
Subsequences: A004171, A055010, A365809 (odd terms).
Subsequence of A016789 (after the initial 0).

Programs

  • PARI
    A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
    isA365808v2(n) = issquare(A163511(n));
    
  • PARI
    isA365808(n) = if(n<=2, !(n%2), if(n%2, isA365808((n-1)/2), if(n%4, 0, isA365808(n/4))));
    
  • Python
    from itertools import count, islice
    def A365808_gen(): # generator of terms
        return map(lambda n:(3*(n+1)>>2)-1,filter(lambda n:n==1 or (n&3==3 and not '00' in bin(n)),count(1)))
    A365808_list = list(islice(A365808_gen(),20)) # Chai Wah Wu, Feb 12 2025

A279341 a(1) = 0, a(2) = 1, for n > 2, if A079559(n) = 0, a(n) = 2*a(A256992(n)), otherwise a(n) = 1 + 2*a(A256992(n)).

Original entry on oeis.org

0, 1, 3, 7, 2, 6, 15, 5, 14, 13, 31, 4, 12, 30, 11, 29, 10, 27, 63, 28, 26, 9, 25, 62, 61, 23, 8, 24, 60, 22, 59, 21, 58, 55, 127, 20, 54, 57, 53, 126, 19, 51, 56, 52, 18, 125, 123, 50, 47, 17, 124, 122, 49, 121, 46, 45, 119, 16, 48, 120, 44, 118, 43, 117, 42, 111, 255, 116, 110, 41, 109, 254, 115, 107, 40, 108, 114, 253, 39, 106, 103
Offset: 1

Views

Author

Antti Karttunen, Dec 10 2016

Keywords

Comments

Note the indexing: the domain starts from 1, while the range includes also zero.

Crossrefs

Inverse: A279342.
Related or similar permutations: A054429, A243071, A279338, A279343, A279347.

Programs

Formula

a(1) = 0, a(2) = 1, for n > 2, if A079559(n) = 0 [when n is a term of A055938], a(n) = 2*a(A256992(n)), otherwise a(n) = 1 + 2*a(A256992(n)).
As a composition of other permutations:
a(n) = A054429(A279343(n)).
a(n) = A279343(A279347(n)).
a(n) = A243071(A279338(n)).
Other identities. For all n >= 1:
A000120(a(n)) = A279345(n).
For all n >= 2, A070939(a(n)) = A256993(n).
Previous Showing 31-40 of 76 results. Next