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 1-6 of 6 results.

A266343 Positions of records in A266342.

Original entry on oeis.org

1, 120, 1980, 7920, 30240, 109200, 120120, 480480, 491400, 1853280, 1884960, 1965600, 2106720, 7207200, 8168160, 8482320, 8648640, 31600800, 121080960, 122522400, 129729600, 136936800, 465585120, 490089600, 497296800, 514594080, 537213600, 551350800
Offset: 1

Views

Author

Antti Karttunen, Dec 28 2015

Keywords

Comments

The corresponding record values A266342(a(n)) are 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 21, 25, 26, ...

Crossrefs

Cf. A266342.

Programs

  • PARI
    A000523(n) = if(n<1,0,#binary(n) - 1);
    A266342(n) = sumdiv(n, d, ((d <= (n/d)) && (A000523(d)==A000523(n/d))));
    m=0; i=0; for(n=1, 2^32, k = A266342(n); if(k > m, m = k; i++; write("b266343.txt", i, " ", n)));

Extensions

a(20)-a(28) from Hiroaki Yamanouchi, Jan 02 2016

A266344 a(n) = number of ways n can be divided into two factors that have the same number of digits in factorial base representation (the two different orders for unequal factors are counted only once).

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 2, 0, 0, 0, 1, 0, 2, 1, 0, 0, 0, 0, 2, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 2, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 1, 3
Offset: 1

Views

Author

Antti Karttunen, Dec 28 2015

Keywords

Examples

			1 can be factored just one way, as 1*1, and thus a(1) = 1.
4 can be factored as 2*2, and thus also a(4) = 1, and generally for all perfect squares k, a(k) >= 1.
14 can be factored as 2*7, but as A007623(2) = 2 and A007623(7) = 101, with different number of digits in factorial base (and 1*14 fares even less well), a(14) = 0.
72 can be factored to two divisors so that the factorial base representations are of equal length as 6*12 or 8*9 (where the corresponding factorial base representations are "100" * "200" and "110" * "111"), thus a(72) = 2.
120 can be similarly factored as 6*20 ("100" * "310"), 8*15 ("110" * "211") and 10*12 ("120" * "200"), thus a(120) = 3.
		

Crossrefs

Cf. A084558.
Cf. A266345 (positions of records).
Cf. also A078781, A266342.

Programs

  • Mathematica
    r = Most@ Reverse@ Range@ 10; Map[Length, Table[Flatten@ Map[Differences@ IntegerLength[#, MixedRadix@ r] &, Transpose@ {#, n/#}] &@ TakeWhile[Divisors@ n, # <= Sqrt@ n &], {n, 120}] /. k_ /; k > 0 -> Nothing] (* Michael De Vlieger, Dec 30 2015, Version 10.2 *)
  • PARI
    A084558(n) = { my(m=1); if(0==n,n,while(m!<=n,m++);return(m-1)); }
    A266344(n) = sumdiv(n, d, ((d <= (n/d)) && (A084558(d)==A084558(n/d))));
    for(n=1, 14161, write("b266344.txt", n, " ", A266344(n)));

Formula

a(n) = Sum_{d|n} [(d <= (n/d)) and (A084558(d) = A084558(n/d))].
(In the above formula [ ] stands for Iverson bracket, giving as its result 1 only if d is less than or equal to n/d and in factorial base representation d and n/d require equal number of digits, and 0 otherwise.)

A266346 Numbers that can be represented as a product of two numbers with an equal number of significant digits (bits) in binary system.

Original entry on oeis.org

0, 1, 4, 6, 9, 16, 20, 24, 25, 28, 30, 35, 36, 42, 49, 64, 72, 80, 81, 88, 90, 96, 99, 100, 104, 108, 110, 112, 117, 120, 121, 126, 130, 132, 135, 140, 143, 144, 150, 154, 156, 165, 168, 169, 180, 182, 195, 196, 210, 225, 256, 272, 288, 289, 304, 306, 320, 323, 324, 336, 340, 342, 352, 357, 360, 361, 368, 374, 378, 380, 384, 391
Offset: 0

Views

Author

Antti Karttunen, Dec 28 2015

Keywords

Comments

Indexing starts from zero as a(0) = 0 is a special case in this sequence.

Examples

			1 can be represented as 1*1 (1 being "1" also in base-2 system), thus it is included.
4 can be represented as 2*2, and like any square, is included.
6 can be represented as 2*3, and both "10" and "11" require two bits in binary system, thus 6 is included.
		

Crossrefs

Positions of nonzeros in A266342.
Cf. A266347 (complement).
Cf. A000290, A085721, A261073, A261074, A261075 (subsequences).
Cf. also A266342.

Programs

  • Mathematica
    {0}~Join~Flatten[Position[#, k_ /; k > 0] &@ Table[Length@ DeleteCases[Flatten@ Map[Differences@ IntegerLength[#, 2] &, Transpose@ {#, n/#}] &@ TakeWhile[Divisors@ n, # <= Sqrt@ n &], k_ /; k > 0], {n, 400}]] (* Michael De Vlieger, Dec 30 2015 *)

A266347 Numbers that cannot be represented as the product of two numbers with an equal number of significant digits (bits) in their binary representations.

Original entry on oeis.org

2, 3, 5, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 21, 22, 23, 26, 27, 29, 31, 32, 33, 34, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 82, 83, 84, 85, 86, 87, 89, 91, 92, 93, 94, 95, 97, 98, 101, 102, 103, 105
Offset: 1

Views

Author

Antti Karttunen, Dec 28 2015

Keywords

Comments

All primes p are in the sequence since the only pair of divisors of p is {1, p} and since the smallest p = 2 has more bits than 1; all larger primes written in binary will require at least 2 bits to represent p. Thus A000040 is a subsequence of this sequence. - Michael De Vlieger, Dec 30 2015

Examples

			From _Michael De Vlieger_, Dec 30 2015: (Start)
Consider pairs of divisors {d, d'} of n, both integers such that d * d' = n:
2 is a term, since the only pair of divisors of 2 written in binary are {1, 10}, with unequal numbers of bits.
3 is a term, since the only pair of divisors of 3 written in binary are {1, 11}, with unequal numbers of bits.
8 is a term, since the pair of divisors of 8 written in binary are {1, 100} and {10, 100}, both with unequal numbers of bits.
12 is a term, since the elements of {1, 1100}, {10, 110}, and {11, 100} are both unequal in length in all cases.
...
(End)
		

Crossrefs

Positions of zeros in A266342.
Cf. A266346 (complement).
Cf. A000040 (a subsequence).

Programs

  • Mathematica
    Position[#, k_ /; k == 0] &@ Map[Length, Table[Flatten@ Map[Differences@ IntegerLength[#, 2] &, Transpose@ {#, n/#}] &@ TakeWhile[Divisors@ n, # <= Sqrt@ n &], {n, 100}] /. k_ /; k > 0 -> Nothing] // Flatten (* Michael De Vlieger, Dec 30 2015 *)

A324392 a(n) is the number of divisors d of n such that A000120(d) divides n, where A000120(d) gives the binary weight of d.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Mar 05 2019

Keywords

Comments

Number of such positive integers k that both k and A000120(k) [the Hamming weight of k] divide n.

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[n, 1 &, Divisible[n, DigitCount[#, 2, 1]] &]; Array[a, 100] (* Amiram Eldar, Dec 04 2020 *)
  • PARI
    A324392(n) = sumdiv(n, d, !(n%hammingweight(d)));

Formula

a(n) = Sum_{d|n} [A000120(d) does divide n], where [ ] is the Iverson bracket.
a(n) = A000005(n) - A324393(n).
a(p) = 1 for all odd primes p.

A324393 a(n) is the number of such divisors d of n that A000120(d) does not divide n, where A000120(d) gives the binary weight of d.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Mar 05 2019

Keywords

Comments

Number of such positive integers k that divide n but A000120(k) [the Hamming weight of k] does not divide n.

Crossrefs

Cf. A000005, A000120, A324392, A306263 (positions of zeros).

Programs

  • Mathematica
    a[n_] := DivisorSum[n, 1 &, !Divisible[n, DigitCount[#, 2, 1]] &]; Array[a, 100] (* Amiram Eldar, Dec 04 2020 *)
  • PARI
    A324393(n) = sumdiv(n, d, !!(n%hammingweight(d)));

Formula

a(n) = Sum_{d|n} [A000120(d) does not divide n], where [ ] is the Iverson bracket.
a(n) = A000005(n) - A324392(n).
a(p) = 1 for all odd primes p.
Showing 1-6 of 6 results.