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 41-47 of 47 results.

A114882 Transposition sequence of A114881.

Original entry on oeis.org

1, 3, 2, 5, 4, 7, 6, 8, 10, 9, 12, 11, 16, 24, 18, 13, 22, 15, 28, 48, 30, 17, 36, 14, 40, 120
Offset: 1

Views

Author

Clark Kimberling, Jan 03 2006

Keywords

Comments

A self-inverse permutation of the natural numbers.

Examples

			Start with the northwest corner of A114881:
1 3 5 7 9
2 8 14 20 26
4 24 34 54 64
6 48 76 90 118
a(1)=1 because 1=T(1,1) and T(1,1)=1.
a(2)=3 because 2=T(2,1) and T(1,2)=3.
a(3)=2 because 3=T(1,2) and T(2,1)=2.
a(20)=48 because 20=T(2,4) and T(4,2)=48.
		

Crossrefs

Formula

(See A114538 for definition of transposition sequence.)

A249743 Main diagonal of square arrays A114881 and A249741.

Original entry on oeis.org

1, 8, 34, 90, 208, 376, 628, 816, 1218, 1768, 2200, 2922, 3648, 4342, 5028, 5988, 7728, 8478, 10116, 11572, 12628, 14298, 16018, 17710, 21630, 23128, 24616, 26856, 28666, 30622, 35686, 38382, 42606, 44062, 50212, 52698, 56362, 60798, 63960, 68680, 73210, 76200, 82702, 85498, 90028, 92136, 101068, 109492, 114180, 119308, 126052, 133122
Offset: 1

Views

Author

Antti Karttunen, Nov 15 2014

Keywords

Comments

One less than the main diagonal of square arrays A083140 and A083221 formed from the sieve of Eratosthenes.

Crossrefs

One less than A083141.

Programs

Formula

a(1) = 1, a(n) = (A000040(n) * A000040(2*(n-1))) - 1. [Where A000040(n) gives the n-th prime, p_n].
a(n) = A083140(n,n) - 1 = A083221(n,n) - 1.
a(n) = A083141(n+1)-1. [With the current starting offset 2 of A083141].

A251758 Let n>=2 be a positive integer with divisors 1 = d_1 < d_2 < ... < d_k = n, and s = d_1*d_2 + d_2*d_3 + ... + d_(k-1)*d_k. The sequence lists the values a(n) = floor(n^2/s).

Original entry on oeis.org

2, 3, 1, 5, 1, 7, 1, 2, 1, 11, 1, 13, 1, 2, 1, 17, 1, 19, 1, 2, 1, 23, 1, 4, 1, 2, 1, 29, 1, 31, 1, 2, 1, 4, 1, 37, 1, 2, 1, 41, 1, 43, 1, 2, 1, 47, 1, 6, 1, 2, 1, 53, 1, 4, 1, 2, 1, 59, 1, 61, 1, 2, 1, 4, 1, 67, 1, 2, 1, 71, 1, 73, 1, 2, 1, 6, 1, 79, 1, 2, 1
Offset: 2

Views

Author

Michel Lagneau, Dec 08 2014

Keywords

Comments

s is always less than n^2 and if n is a prime number then s divides n^2.
For n >= 2, the sequence has the following properties:
a(n) = n if n is prime.
a(n) = 1 if n is in A005843 and > 2;
a(n) <= 2 if n is in A016945 and > 3;
a(n) <= 4 if n is in A084967 and > 5;
a(n) <= 6 if n is in A084968 and > 7;
a(n) = 8: <= 35336848261, ...;
a(n) <= 10 if n is in A084969 and > 11;
a(n) <= 12 if n is in A084970 and > 13;
a(n) = 14: 6678671, ...;
This is different from A250480 (a(n) = n for all prime n, and a(n) = A020639(n) - 1 for all composite n), which thus satisfies the above conditions exactly, while with this sequence A020639(n)-1 gives only the guaranteed upper limit for a(n) at composite n. Note that the first different term does not occur until at n = 2431 = 11*13*17, for which a(n) = 9. (See the example below.)
Conjecture: Terms x, where a(x)=n, x=p#k/p#j, p#i is the i-th primorial, k>j is suitable large k and j is the number of primes less than n. As an example, n=9, x = p#7/p#4 = 2431. For n=10, x = p#6/p#4 = 143 although 121 = 11^2 is the least x where a(x)=10 (see formula section). For n=8, x = p#12/p#4, p#13/p#4, p#14/p#4, p#15/p#4, p#16/p#4, etc. But is p#12/p#4 the least such x? - Robert G. Wilson v, Dec 18 2014
n^2/s is only an integer iff n is prime. - Robert G. Wilson v, Dec 18 2014
First occurrence of n >= 1: 4, 2, 3, 25, 5, 49, 7, ??? <= 35336848261, 2431, 121, 11, 169, 13, 6678671, 7429, 289, 17, 361, 19, 31367009, 20677, 529, 23, ..., . - Robert G. Wilson v, Dec 18 2014

Examples

			For n = 2431 = 11*13*17, we have (as the eight divisors of 2431 are [1, 11, 13, 17, 143, 187, 221, 2431]) a(n) = floor((2431*2431) / ((1*11)+(11*13)+(13*17)+(17*143)+(143*187)+(187*221)+(221*2431))) = floor(5909761/608125) = floor(9.718) = 9.
		

Crossrefs

Cf. A000040 (prime numbers), A005843 (even numbers), A016945 (6n+3), A084967 (GCD( 5k, 6) =1), A084968 (GCD( 7k, 30) =1), A084969 (GCD( 11k, 30) =1), A084970 (Numbers whose smallest prime factor is 13).
Cf. also A020639 (the smallest prime divisor), A055396 (its index) and arrays A083140 and A083221 (Sieve of Eratosthenes).
Differs from A250480 for the first time at n = 2431, where a(2431) = 9, while A250480(2431) = 10.
Cf. A078730 (sum of products of two successive divisors of n).

Programs

  • Maple
    with(numtheory):nn:=100:
    for n from 2 to nn do:
       x:=divisors(n):n0:=nops(x):s:=sum('x[i]*x[i+1]','i'=1..n0-1):
       z:=floor(n^2/s):printf(`%d, `,z):
    od:
  • Mathematica
    f[n_] := Floor[ n^2/Plus @@ Times @@@ Partition[ Divisors@ n, 2, 1]]; Array[f, 81, 2] (* Robert G. Wilson v, Dec 18 2014 *)

Formula

a(n) <= A250480(n), and especially, for all composite n, a(n) < A020639(n). [Cf. the Comments section above.] - Antti Karttunen, Dec 09 2014
From Robert G. Wilson v, Dec 18 2014: (Start)
a(n) = floor(n^2/A078730(n));
a(n) = n iff n is prime. (End)

Extensions

Comments section edited by Antti Karttunen, Dec 09 2014
Instances of n for which a(n) = 8 and 14 found by Robert G. Wilson v, Dec 18 2014

A298268 a(1) = 1, and for any n > 1, if n is the k-th number with greatest prime factor p, then a(n) is the k-th number with least prime factor p.

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 7, 6, 15, 25, 11, 21, 13, 49, 35, 8, 17, 27, 19, 55, 77, 121, 23, 33, 65, 169, 39, 91, 29, 85, 31, 10, 143, 289, 119, 45, 37, 361, 221, 95, 41, 133, 43, 187, 115, 529, 47, 51, 161, 125, 323, 247, 53, 57, 209, 203, 437, 841, 59, 145, 61, 961
Offset: 1

Views

Author

Rémy Sigrist, Jan 27 2018

Keywords

Comments

This sequence is a permutation of the natural numbers, with inverse A298882.
For any prime p and k > 0:
- if s_p(k) is the k-th p-smooth number and r_p(k) is the k-th p-rough number,
- then a(p * s_p(k)) = p * r_p(k),
- for example: a(11 * A051038(k)) = 11 * A008364(k).

Examples

			The first terms, alongside A006530(n), are:
  n     a(n)   gpf(n)
  --    ----   ------
   1      1      1
   2      2      2
   3      3      3
   4      4      2
   5      5      5
   6      9      3
   7      7      7
   8      6      2
   9     15      3
  10     25      5
  11     11     11
  12     21      3
  13     13     13
  14     49      7
  15     35      5
  16      8      2
  17     17     17
  18     27      3
  19     19     19
  20     55      5
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(1) = 1.
a(A125624(n, k)) = A083140(n, k) for any n > 0 and k > 0.
a(n) = A083140(A061395(n), A078899(n)) for any n > 1.
Empirically:
- a(n) = n iff n belongs to A046022,
- a(2^k) = 2 * k for any k > 0,
- a(2 * p) = p^2 for any prime p,
- a(3 * p) = p * A151800(p) for any odd prime p.

A216240 Composite numbers arising in Eratosthenes sieve with removing the multiples of every other remaining numbers after 2 (see comment).

Original entry on oeis.org

9, 21, 33, 49, 51, 77, 87, 119, 121, 123, 141, 177, 187, 201, 203, 219, 237, 287, 289, 291, 309, 319, 327, 329, 357, 393, 413, 417, 447, 451, 469, 471, 493, 501, 511, 517, 543, 553, 573, 591, 633, 649, 669, 679, 687, 697, 721, 723, 737, 763, 771, 799, 803, 807
Offset: 1

Views

Author

Vladimir Shevelev, Mar 14 2013

Keywords

Comments

We remove even numbers except for 2. The first two remaining numbers are 3,5. Further we remove all remaining numbers multiple of 5,except for 5. The following two remaining numbers are 7,9. Now we remove all remaining numbers multiple of 9, except for 9, etc. The sequence lists the remaining composite numbers.
Conjecture. There exists x_0 such that for every x>=x_0, the number of a(n)<=x is more than pi(x).

Crossrefs

Programs

  • Mathematica
    Module[{a=Insert[Range[1,1000,2], 2, 2], k=4}, While[Length[a] >= 2k, a = Flatten[{Take[a,k], Select[Take[a,-Length[a]+k], Mod[#,a[[k]]] != 0 &]}]; k+=2]; Rest[Select[a,!PrimeQ[#]&]]] (* Peter J. C. Moses, Mar 27 2013 *)

A298882 a(1) = 1, and for any n > 1, if n is the k-th number with least prime factor p, then a(n) is the k-th number with greatest prime factor p.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 7, 16, 6, 32, 11, 64, 13, 128, 9, 256, 17, 512, 19, 1024, 12, 2048, 23, 4096, 10, 8192, 18, 16384, 29, 32768, 31, 65536, 24, 131072, 15, 262144, 37, 524288, 27, 1048576, 41, 2097152, 43, 4194304, 36, 8388608, 47, 16777216, 14, 33554432, 48
Offset: 1

Views

Author

Rémy Sigrist, Jan 28 2018

Keywords

Comments

This sequence is a permutation of the natural numbers, with inverse A298268.
For any prime p and k > 0:
- if s_p(k) is the k-th p-smooth number and r_p(k) is the k-th p-rough number,
- then a(p * r_p(k)) = p * s_p(k),
- for example: a(11 * A008364(k)) = 11 * A051038(k).

Examples

			The first terms, alongside A020639(n), are:
  n     a(n)    lpf(n)
  --    ----    ------
   1       1      1
   2       2      2
   3       3      3
   4       4      2
   5       5      5
   6       8      2
   7       7      7
   8      16      2
   9       6      3
  10      32      2
  11      11     11
  12      64      2
  13      13     13
  14     128      2
  15       9      3
  16     256      2
  17      17     17
  18     512      2
  19      19     19
  20    1024      2
		

Crossrefs

Formula

a(1) = 1.
a(A083140(n, k)) = A125624(n, k) for any n > 0 and k > 0.
a(n) = A125624(A055396(n), A078898(n)) for any n > 1.
Empirically:
- a(n) = n iff n belongs to A046022,
- a(2 * k) = 2^k for any k > 0,
- a(p^2) = 2 * p for any prime p,
- a(p * q) = 3 * p for any pair of consecutive odd primes (p, q).

A306353 Number of composites among the first n composite numbers whose least prime factor p is that of the n-th composite number.

Original entry on oeis.org

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

Views

Author

Jamie Morken and Vincenzo Librandi, Feb 09 2019

Keywords

Comments

Composites with least prime factor p are on that row of A083140 which begins with p
Sequence with similar values: A122005.
Sequence written as a jagged array A with new row when a(n) > a(n+1):
1, 2, 3,
1, 4, 5, 6,
2, 7, 8, 9,
3, 10, 11,
1, 12,
4, 13, 14, 15,
5, 16,
2, 17, 18,
6, 19, 20, 21,
7, 22, 23,
1, 24,
8, 25, 26,
3, 27,
9, 28, 29, 30.
A153196 is the list B of the first values in successive rows with length 4.
B is given by the formula for A002808(x)=A256388(n+3), an(x)=A153196(n+2)
For example: A002808(26)=A256388(3+3), an(26)=A153196(3+2).
A243811 is the list of the second values in successive rows with length 4.
A047845 is the list of values in the second column and A104279 is the list of values in the third column of the jagged array starting on the second row.
Sequence written as an irregular triangle C with new row when a(n)=1:
1,2,3,
1,4,5,6,2,7,8,9,3,10,11,
1,12,4,13,14,15,5,16,2,17,18,6,19,20,21,7,22,23,
1,24,8,25,26,3,27,9,28,29,30,10,31,4,32,33,11,34,35,36,12,37,2,38,39,13,40,41,5,42,14,43,44,3,45,15,46,6,47,48,16,49,50,51,17,52,53,54,18,55,56,7,57,19,58,4,59.
A243887 is the last value in each row of C.
The second value D on the row n > 1 of the irregular triangle C is a(A053683(n)) or equivalently A084921(n). For example for row 3 of the irregular triangle:
D = a(A053683(3)) = a(16) = 12 or D = A084921(3) = 12. This is the number of composites < A066872(3) with the same least prime factor p as the A053683(3) = 16th composite, A066872(3) = 26.
The number of values in each row of the irregular triangle C begins: 3,11,18,57,39,98,61,141,265,104,351,268,...
The second row of the irregular triangle C is A117385(b) for 3 < b < 15.
The third row of the irregular triangle C has similar values as A117385 in different order.

Examples

			First composite 4, least prime factor is 2, first case for 2 so a(1)=1.
Next composite 6, least prime factor is 2, second case for 2 so a(2)=2.
Next composite 8, least prime factor is 2, third case for 2 so a(3)=3.
Next composite 9, least prime factor is 3, first case for 3 so a(4)=1.
Next composite 10, least prime factor is 2, fourth case for 2 so a(5)=4.
		

Crossrefs

Programs

  • Mathematica
    counts = {}
    values = {}
    For[i = 2, i < 130, i = i + 1,
    If[PrimeQ[i], ,
    x = PrimePi[FactorInteger[i][[1, 1]]];
      If[Length[counts] >= x,
       counts[[x]] = counts[[x]] + 1;
       AppendTo[values, counts[[x]]], AppendTo[counts, 1];
       AppendTo[values, 1]]]]
       (* Print[counts] *)
       Print[values]
  • PARI
    c(n) = for(k=0, primepi(n), isprime(n++)&&k--); n; \\ A002808
    a(n) = my(c=c(n), lpf = vecmin(factor(c)[,1]), nb=0); for(k=2, c, if (!isprime(k) && vecmin(factor(k)[,1])==lpf, nb++)); nb; \\ Michel Marcus, Feb 10 2019

Formula

a(n) is approximately equal to A002808(n)*(A038110(x)/A038111(x)), with A000040(x)=A020639(A002808(n)).
For example if n=325, a(325)~= A002808(325)*(A038110(2)/A038111(2)) with A000040(2)=A020639(A002808(325)).
This gives an estimate of 67.499... and the actual value of a(n)=67.
Previous Showing 41-47 of 47 results.