A266195 -id:A266195 - cp's OEIS frontend

A266191 Sequence beginning with a(1)=1, a(2)=2, a(3)=3 and then after each new term a(n) is selected as the least unused number for which a(n)a(n-1)a(n-2) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

1, 2, 3, 6, 4, 7, 12, 8, 11, 15, 16, 22, 24, 5, 39, 14, 10, 59, 28, 20, 67, 31, 9, 63, 19, 18, 27, 38, 17, 54, 47, 34, 44, 51, 33, 88, 30, 32, 43, 48, 21, 86, 23, 42, 35, 46, 84, 70, 91, 168, 71, 55, 36, 75, 99, 40, 135, 110, 41, 60, 111, 80, 120, 62, 160, 134, 56, 141, 76, 112, 64, 78, 119, 113, 103, 183, 37, 167, 366, 73

Offset: 1

Antti Karttunen, Dec 23 2015

It is conjectured that this sequence is not only injective, but also surjective on N, i.e., that it is a true permutation of natural numbers.

A266351 Start with a(1) = 1, then always choose for a(n) the least unused number such that A057889(a(n)a(n-1)) = A057889(a(n)) A057889(a(n-1)), where A057889 is a bijective base-2 reverse.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 11, 17, 13, 32, 15, 24, 18, 20, 19, 33, 21, 36, 28, 27, 56, 34, 22, 64, 23, 65, 25, 40, 35, 72, 42, 48, 30, 51, 60, 66, 26, 68, 37, 96, 31, 99, 62, 128, 29, 129, 38, 80, 49, 73, 70, 130, 39, 256, 41, 131, 74, 136, 44, 132, 46, 257, 43, 258, 45, 260, 47, 512, 50, 133, 76, 160, 67, 84, 97, 137, 112, 54

Offset: 1

Antti Karttunen, Dec 28 2015

Equally: always choose for a(n) the least unused number such that a(n)*a(n-1) = A057889(A057889(a(n)) * A057889(a(n-1))).
Note that the adjacent terms of permutation A266195 satisfy the same condition, except that permutation is not the lexicographically earliest sequence of this kind (because it has a more restrictive condition). See A266194.
This is a bijection for the same reason that A266195 is. Any high enough 2^k will always save the permutation of being stuck, and will also immediately pick up as its succeeding pair the least term unused so far.

Antti Karttunen, Table of n, a(n) for n = 1..1450
Index entries for sequences that are permutations of the natural numbers

Inverse: A266352.
Cf. A057889, A266194.
Cf. A266195, A265405, A266405 (similar sequences).

A266117 Lexicographically first injection of positive integers beginning with a(1) such that a(n)*a(n+1) is a term of A265349, i.e., has no multiple occurrences of any nonzero digit when viewed in factorial base (A007623).

Original entry on oeis.org

1, 2, 3, 4, 5, 10, 11, 6, 7, 12, 8, 9, 24, 13, 18, 19, 26, 14, 17, 22, 21, 16, 15, 32, 30, 20, 23, 29, 42, 28, 25, 48, 34, 53, 41, 54, 27, 40, 33, 36, 37, 61, 65, 44, 49, 72, 39, 38, 51, 52, 55, 59, 47, 46, 58, 50, 60, 62, 31, 66, 56, 57, 64, 45, 80, 63, 74, 69, 68, 35, 79, 100, 73, 43, 67, 70, 71, 78, 84, 87, 92, 90, 76, 96, 75

Offset: 1

Antti Karttunen, Dec 22 2015

After a(1) = 1, always choose for a(n+1) the least unused k such that in factorial base representation (A007623), the product a(n)*a(n+1) will not show any of the nonzero digits present twice (or more times), regardless of the positions of the digits.

			For n = 6, we start searching from the least not yet used number in range a(1) .. a(5) [which is 6, because all the previous terms are fixed] for the first number whose product with a(5) = 5 results a number in A265349.
Multiplying 5 (in factorial base "21") with 6 (in factorial base "100") results 30, which in factorial base is "1100", containing digit "1" twice, thus 6 is disqualified.
Similarly, products 5*7, 5*8 and 5*9 result 35 = "1121", 40 = "1220" and 45 = "1311", where in all cases one of the nonzero digits occur more than once, so 7, 8 and 9 are also all disqualified.
But 5*10 = 50, which has a factorial base representation ("2010") that matches the criterion, thus a(6) = 10.

Antti Karttunen, Table of n, a(n) for n = 1..14641
Eric Angelini, a(n)*a(n+1) shows at least twice the same digit, Posting on SeqFan-list Dec 21 2015. [Source of inspiration for this sequence.]
Index entries for sequences related to factorial base representation
Index entries for sequences that are permutations of the natural numbers

Left inverse: A266118 (also the right inverse if this sequence is a permutation of the positive integers).
Cf. A264990, A265349.
Cf. also A266121, A266191 and A266195 for similar permutations.

A266405 Start with a(1) = 1, then always choose for a(n) the least unused number such that A002487(a(n)a(n-1)) = A002487(a(n)) A002487(a(n-1)), where A002487 is Stern-Brocot sequence.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 6, 9, 10, 14, 16, 11, 17, 13, 12, 18, 15, 26, 24, 23, 19, 32, 20, 28, 34, 22, 27, 33, 29, 31, 25, 45, 49, 36, 30, 52, 48, 35, 64, 21, 69, 42, 128, 37, 256, 38, 46, 66, 54, 41, 83, 82, 108, 44, 39, 88, 68, 56, 40, 55, 65, 47, 130, 59, 96, 51, 192, 70, 72, 60, 104, 71, 80, 57, 63, 61, 126, 98, 90, 50, 62, 58, 124, 100, 121, 127

Offset: 1

Antti Karttunen, Dec 29 2015

nonn

This is a permutation of natural numbers for the same reason that A266195 and A266351 are. If nothing else works for the value of next a(n), then at least the next unused power of 2 will save the sequence from dying, and will also immediately pick up as its succeeding pair the least term not used so far. This follows because A002487(2^m) = 1 and A002487(2^m * n) = A002487(n) for all n and m.

Still, it would be nice to know when 149 will appear in the sequence.

Antti Karttunen, Table of n, a(n) for n = 1..2016
Index entries for sequences that are permutations of the natural numbers

Inverse: A266406.
Cf. A002487.
Cf. A266195, A266351, A265405 (for sequences with similar definitions).

A265405 Start with a(1) = 1, then always choose for a(n) the least unused number such that A193231(a(n)a(n-1)) = A193231(a(n)) A193231(a(n-1)), where A193231 is an involution of natural numbers called Blue code.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 16, 7, 17, 8, 15, 32, 12, 10, 18, 20, 19, 256, 9, 14, 34, 48, 40, 50, 33, 60, 257, 11, 97, 258, 13, 101, 209, 65536, 21, 259, 64, 30, 65, 51, 80, 24, 84, 36, 85, 66, 260, 22, 4352, 26, 4368, 28, 4369, 37, 768, 41, 770, 42, 771, 68, 90, 272, 45, 273, 56, 1200, 952, 4096, 23, 4097, 27, 4098, 86, 512, 54

Offset: 1

Antti Karttunen, Dec 29 2015

Does this sequence die after a(144) = 46 ?
No, a(145) = 16777216, but whether the sequence is finished remains open. - Rémy Sigrist, Feb 15 2019
The next unused number of the form 2^2^k is always a valid choice, so this sequence is infinite. - Charlie Neder, Apr 14 2019

Rémy Sigrist, Table of n, a(n) for n = 1..183 (first 144 terms from Antti Karttunen) [More terms needed for b-file.]
Rémy Sigrist, PARI program for A265405

Inverse: A265406.
Cf. A193231.
Cf. A266195, A266351, A266405 (sequences with similar definitions, of which at least the first two are known to be infinite and also bijective).

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A306465 Lexicographically earliest sequence of distinct positive terms such that the product of any two consecutive terms can be computed without carry by long multiplication in base 10.

Original entry on oeis.org

1, 2, 3, 10, 4, 11, 5, 100, 6, 101, 7, 110, 8, 111, 9, 1000, 12, 13, 20, 14, 21, 22, 30, 23, 102, 24, 112, 31, 32, 103, 33, 120, 40, 121, 41, 200, 34, 201, 42, 202, 43, 1001, 15, 1010, 16, 1011, 17, 1100, 18, 1101, 25, 1110, 26, 1111, 27, 10000, 19, 10001, 28

Offset: 1

Rémy Sigrist, Feb 17 2019

This sequence is the variant of A266195 in base 10.
This sequence is a permutation of the natural numbers, with inverse A306466. Proof:
- we can always extend the sequence with a power of ten not yet in the sequence, hence the sequence is well defined and infinite,
- for any k > 0, 10^(k-1) is the first k-digit number appearing in the sequence,
- all powers of ten appear in the sequence, in increasing order,
- a power of ten is always followed by the least number unused so far,
hence every number eventually appears. QED

			The first terms, alongside their digital sum and the digital sum of the product with the next term, are:
  n   a(n)  ds(a(n))  ds(a(n)*a(n+1))
  --  ----  --------  ---------------
   1     1         1                2
   2     2         2                6
   3     3         3                3
   4    10         1                4
   5     4         4                8
   6    11         2               10
   7     5         5                5
   8   100         1                6
   9     6         6               12
  10   101         2               14
  11     7         7               14
  12   110         2               16
  13     8         8               24
  14   111         3               27
  15     9         9                9
  16  1000         1                3
  17    12         3               12

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program for A306465
Rémy Sigrist, Colored logarithmic scatterplot of the sequence for n = 1..200000 (where the color is function of A054055(a(n)))
Index entries for sequences that are permutations of the natural numbers

Cf. A007953, A054055, A266195, A306466 (inverse).

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A007953(a(n) * a(n+1)) = A007953(a(n)) * A007953(a(n+1)).

A054055(a(n)) * A054055(a(n+1)) <= 9.

A379126 a(1) = 1; for n > 1, a(n) is the least number k such that A325567(k) = n, or 0 if no such number exists.

Original entry on oeis.org

1, 4, 9, 8, 35, 18, 49, 16, 135, 70, 33, 36, 65, 98, 225, 32, 527, 270, 133, 140, 651, 66, 161, 72, 775, 130, 837, 196, 899, 450, 961, 64, 2079, 1054, 525, 540, 259, 266, 273, 280, 2583, 1302, 129, 132, 2835, 322, 705, 144, 3087, 1550, 3213, 260, 3339, 1674, 385, 392, 1539, 1798, 3717, 900, 3843, 1922, 3969, 128

Offset: 1

Antti Karttunen, Dec 21 2024

nonn

By definition, sequence is injective (apart from possible 0's) and each a(n) is a multiple of n.

Cf. A048720, A065621, A277320, A325567, A379128 (odd bisection), A379228 [= a(n)/n].

Cf. also A115872, A266195, A266351.

A048720(b, c) = fromdigits(Vec(Pol(binary(b))*Pol(binary(c)))%2, 2);
A065621(n) = bitxor(n-1, n+n-1);
memoA325567 = Map();
A325567(n) = if(1==n,1,my(v); if(mapisdefined(memoA325567,n,&v), v, fordiv(n, d, if((d>1)&&A048720(A065621(n/d), d)==n, v = (n/d); break)); mapput(memoA325567,n,v); (v)));
A379126(n) = for(k=1,oo,if(A325567(k)==n, return(k)));

a(n) = n * A379228(n).

A289726 Lexicographically earliest sequence of distinct positive terms such that, for any n > 0, multiplying n by a(n) does not produce any carries when performed in base 2.

Original entry on oeis.org

1, 2, 4, 3, 6, 5, 8, 7, 10, 9, 16, 17, 32, 18, 33, 11, 12, 14, 20, 19, 24, 34, 64, 21, 40, 65, 66, 36, 128, 68, 129, 13, 15, 22, 72, 28, 48, 80, 130, 25, 96, 67, 256, 132, 136, 257, 258, 37, 73, 133, 160, 260, 512, 264, 513, 137, 514, 516, 520, 272, 1024, 528

Offset: 1

Rémy Sigrist, Jul 10 2017

This sequence is a self-inverse permutation of the natural numbers.
A priori, there are only two fixed points: 1 and 2 (note that a fixed point must be a power of 2).
This sequence is related to A266195: here n * a(n), there a(n) * a(n+1), does not produce any carries when performed in base 2.

			1 * 1 can be computed without carry in base 2, hence a(1) = 1.
2 * 2 can be computed without carry in base 2, hence a(2) = 2.
3 * 3 cannot be computed without carry in base 2.
3 * 4 can be computed without carry in base 2, hence a(3) = 4.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program for A289726
Index entries for sequences that are permutations of the natural numbers

Cf. A266195.

PARI
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See Links section.
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A343650 a(n) is the number of divisors d of n such that the product d * (n/d) can be computed without carries in binary.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 2, 4, 2, 6, 2, 4, 4, 5, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 4, 6, 2, 8, 2, 6, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 2, 6, 4, 4, 2, 10, 2, 4, 4, 6, 2, 8, 2, 8, 2, 4, 2, 12, 2, 4, 6, 7, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 2, 6, 2, 4, 2, 10, 2, 4, 2, 6, 4, 4

Offset: 1

Rémy Sigrist, Apr 24 2021

See A343651 for the corresponding divisors.

			For n = 18:
- we have the following divisors:
     d   18/d  bin(d)  bin(18/d)  Requires carries?
     --  ----  ------  ---------  -----------------
      1    18       1      10010  No
      2     9      10       1001  No
      3     6      11        110  Yes
      6     3     110         11  Yes
      9     2    1001         10  No
     18     1   10010          1  No
- so a(18) = #{1, 2, 9, 18} = 4.

Cf. A000005, A001511, A067824, A266195, A343651.

a(n, h=hammingweight) = my (hn=h(n)); sumdiv(n, d, hn==h(d)*h(n/d))

a(n) <= A000005(n).
a(2^n) = n + 1 for any n >= 0.
a(2^n - 1) = A067824(n) for any n > 0.
A001511(n) divides a(n).

cp's OEIS Frontend

A266191 Sequence beginning with a(1)=1, a(2)=2, a(3)=3 and then after each new term a(n) is selected as the least unused number for which a(n)*a(n-1)*a(n-2) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

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A266351 Start with a(1) = 1, then always choose for a(n) the least unused number such that A057889(a(n)*a(n-1)) = A057889(a(n)) * A057889(a(n-1)), where A057889 is a bijective base-2 reverse.

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A266117 Lexicographically first injection of positive integers beginning with a(1) such that a(n)*a(n+1) is a term of A265349, i.e., has no multiple occurrences of any nonzero digit when viewed in factorial base (A007623).

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A266405 Start with a(1) = 1, then always choose for a(n) the least unused number such that A002487(a(n)*a(n-1)) = A002487(a(n)) * A002487(a(n-1)), where A002487 is Stern-Brocot sequence.

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A265405 Start with a(1) = 1, then always choose for a(n) the least unused number such that A193231(a(n)*a(n-1)) = A193231(a(n)) * A193231(a(n-1)), where A193231 is an involution of natural numbers called Blue code.

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A306465 Lexicographically earliest sequence of distinct positive terms such that the product of any two consecutive terms can be computed without carry by long multiplication in base 10.

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A379126 a(1) = 1; for n > 1, a(n) is the least number k such that A325567(k) = n, or 0 if no such number exists.

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A289726 Lexicographically earliest sequence of distinct positive terms such that, for any n > 0, multiplying n by a(n) does not produce any carries when performed in base 2.

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A343650 a(n) is the number of divisors d of n such that the product d * (n/d) can be computed without carries in binary.

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A266191 Sequence beginning with a(1)=1, a(2)=2, a(3)=3 and then after each new term a(n) is selected as the least unused number for which a(n)a(n-1)a(n-2) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

A266351 Start with a(1) = 1, then always choose for a(n) the least unused number such that A057889(a(n)a(n-1)) = A057889(a(n)) A057889(a(n-1)), where A057889 is a bijective base-2 reverse.

A266405 Start with a(1) = 1, then always choose for a(n) the least unused number such that A002487(a(n)a(n-1)) = A002487(a(n)) A002487(a(n-1)), where A002487 is Stern-Brocot sequence.

A265405 Start with a(1) = 1, then always choose for a(n) the least unused number such that A193231(a(n)a(n-1)) = A193231(a(n)) A193231(a(n-1)), where A193231 is an involution of natural numbers called Blue code.