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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A353954 a(0) = 1; a(n) = A019565(A109812(n)).

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 21, 11, 15, 14, 33, 35, 22, 105, 13, 30, 77, 26, 55, 42, 65, 66, 91, 110, 39, 70, 143, 210, 17, 165, 182, 51, 154, 195, 34, 231, 130, 119, 330, 221, 462, 85, 78, 385, 102, 455, 187, 390, 1309, 19, 770, 663, 38, 1155, 442, 57, 910, 561, 95, 273
Offset: 0

Views

Author

Michael De Vlieger, May 12 2022

Keywords

Comments

Interpretation of A109812(n) written in binary instead as if written in "multiplicity notation", that is, as if we write 1 if divisible by prime(k+1), otherwise 0 in the k-th place. Example, decimal 12 is written in binary as 1100 = 2^2 + 2^3, and take exponents 2 and 3 and instead construe them as prime(2+1) * prime(3+1) = 5*7 = 35.
Permutation of squarefree numbers A005117.

Examples

			Table showing n, A109812(n), and b(n), the binary expansion of A109812(n) writing "." for zeros for clarity. a(n) interprets 1's in the k-th place of b(n) as prime(k+1) and thereafter takes the product. We find a(n) = A005117(j). Note that A109812(0) is not defined.
   n A109812(n) b(n)  a(n)   j
  ----------------------------
   0     -        .     1    1
   1     1        1     2    2
   2     2       1.     3    3
   3     4      1..     5    4
   4     3       11     6    5
   5     8     1...     7    6
   6     5      1.1    10    7
   7    10     1.1.    21   14
   8    16    1....    11    8
   9     6      11.    15   11
  10     9     1..1    14   10
  11    18    1..1.    33   21
  12    12     11..    35   23
  13    17    1...1    22   15
  14    14     111.   105   65
  15    32   1.....    13    9
  16     7      111    30   19
  ...
		

Crossrefs

Programs

  • Mathematica
    Clear[c, a]; nn = 60; c[_] = 0; a[0] = c[1] = j = 1; a[1] = u = 2; Do[k = u; While[Nand[c[k] == 0, BitAnd[j, k] == 0], k++]; If[k == u, While[c[u] > 0, u++]]; j = k; Set[{a[i], c[k]}, {Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, Reverse@ IntegerDigits[k, 2]], i}], {i, 2, nn}]; Array[a, nn + 1, 0]

Formula

a(0) = 1; a(n) = Product p_k where A109812(n) = Sum 2^(k-1) for n > 0.

A353955 a(n) = A019565(A353709(n)).

Original entry on oeis.org

1, 2, 3, 5, 7, 6, 11, 35, 13, 22, 15, 91, 17, 10, 21, 143, 34, 105, 19, 26, 33, 85, 14, 39, 55, 119, 78, 95, 77, 51, 65, 154, 57, 221, 70, 209, 663, 23, 110, 273, 323, 46, 165, 1547, 38, 69, 385, 442, 437, 231, 130, 391, 133, 30, 187, 247, 42, 935, 299, 114, 595
Offset: 0

Views

Author

Michael De Vlieger, May 12 2022

Keywords

Comments

Interpretation of A353709(n) written in binary instead as if written in "multiplicity notation", that is, as if we write 1 if divisible by prime(k+1), otherwise 0 in the k-th place. Example, decimal 12 is written in binary as 1100 = 2^2 + 2^3, and take exponents 2 and 3 and instead construe them as prime(2+1) * prime(3+1) = 5*7 = 35.
If A353709 is a permutation of nonnegative numbers, then this sequence is a permutation of squarefree numbers A005117.

Examples

			Table showing n, A353709(n), and b(n), the binary expansion of A353709(n) writing "." for zeros for clarity. a(n) interprets 1's in the k-th place of b(n) as prime(k+1) and thereafter takes the product. We find a(n) = A005117(j).
   n A353709(n) b(n)  a(n)   j
  ----------------------------
   1    0         .     1    1
   2    1         1     2    2
   3    2        1.     3    3
   4    4       1..     5    4
   5    8      1...     7    6
   6    3        11     6    5
   7   16     1....    11    8
   8   12      11..    35   23
   9   32    1.....    13    9
  10   17     1...1    22   15
  11    6       11.    15   11
  12   40    1.1...    91   57
  13   64   1......    17   12
  14    5       1.1    10    7
  15   10      1.1.    21   14
  16   48    11....   143   89
  ...
		

Crossrefs

Programs

  • Mathematica
    nn = 2^7; c[_] = -1; c[0] = i = 0; a[0] = c[1] = j = 1; a[1] = u = 2; Do[k = u; While[Nand[c[k] == -1, BitAnd[i, k] == 0, BitAnd[j, k] == 0], k++]; If[k == u, While[c[u] > -1, u++]]; i = j; j = k; Set[{a[n], c[k]}, {Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, Reverse@ IntegerDigits[k, 2]], n}], {n, 2, nn}]; Array[a, nn + 1, 0]

Formula

a(n) = Product p_k where A353709(n) = Sum 2^(k-1).

A354825 Dirichlet inverse of A293442, where A293442 is multiplicative with a(p^e) = A019565(e).

Original entry on oeis.org

1, -2, -2, 1, -2, 4, -2, -2, 1, 4, -2, -2, -2, 4, 4, 8, -2, -2, -2, -2, 4, 4, -2, 4, 1, 4, -2, -2, -2, -8, -2, -16, 4, 4, 4, 1, -2, 4, 4, 4, -2, -8, -2, -2, -2, 4, -2, -16, 1, -2, 4, -2, -2, 4, 4, 4, 4, 4, -2, 4, -2, 4, -2, 20, 4, -8, -2, -2, 4, -8, -2, -2, -2, 4, -2, -2, 4, -8, -2, -16, 8, 4, -2, 4, 4, 4, 4, 4
Offset: 1

Views

Author

Antti Karttunen, Jun 09 2022

Keywords

Comments

Multiplicative because A293442 is.

Crossrefs

Programs

  • PARI
    A019565(n) = { my(m=1, p=1); while(n>0, p = nextprime(1+p); if(n%2, m *= p); n >>= 1); (m); };
    A293442(n) = factorback(apply(e -> A019565(e),factor(n)[,2]));
    memoA354825 = Map();
    A354825(n) = if(1==n,1,my(v); if(mapisdefined(memoA354825,n,&v), v, v = -sumdiv(n,d,if(dA293442(n/d)*A354825(d),0)); mapput(memoA354825,n,v); (v)));

Formula

a(1) = 1, and for n > 1, a(n) = -Sum_{d|n, dA293442(n/d) * a(d).

A364919 a(0) = 1; a(n) is the smallest number m not already in the sequence such that rad(m) divides A019565(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 6, 7, 14, 21, 12, 25, 10, 15, 16, 11, 22, 27, 18, 55, 20, 33, 24, 49, 28, 63, 32, 35, 40, 45, 30, 13, 26, 39, 36, 65, 50, 75, 48, 91, 52, 81, 42, 125, 56, 105, 54, 121, 44, 99, 64, 143, 80, 117, 60, 77, 88, 147, 66, 169, 70, 135, 72, 17, 34
Offset: 0

Views

Author

Michael De Vlieger, Aug 30 2023

Keywords

Comments

Let k be a squarefree number and define R_k to be the set of numbers m such that rad(m) | k.
For n > 0, a(n) is the smallest m in R_k such that a(j) != m, j < n.
Conjecture: permutation of natural numbers.

Examples

			Let b(n) = A019565(n).
a(1) = 2 since b(1) = 2. Since 2 is prime, we find the first number in the prime power range of 2 that is not in the sequence and that is 2.
a(3) = 4 since b(3) = 6, and the smallest number m such that rad(m) | 6 that has not already appeared is 4.
a(5) = 8 since b(5) = 10. R_10 begins {1, 2, 4, 5, 8, 10, 16, ...} and the smallest number m in that list that is not already in the sequence is 8.
a(6) = 9 since b(6) = 15. R_15 begins {1, 3, 5, 9, 15, 25, ...} and the smallest m in that list not already in the sequence is 9, etc.
		

Crossrefs

Programs

  • Mathematica
    nn = 120; rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
    f[x_] := Times @@ Prime@ Position[Reverse@ IntegerDigits[x, 2], 1][[All, 1]];
    c[] := False; c[1] = True; q[] := 1; a[0] = 1; r[_] := 1;
    Do[If[PrimeQ[#],
      While[c[Set[k, #^q[#]]], q[#]++],
      While[Or[c[r[#]], ! Divisible[#, rad[r[#]]]], r[#]++]; k = r[#] ] &[f[i]]; Set[{a[i], c[k]}, {k, True}], {i, nn}];
    Array[a, nn + 1, 0]

Formula

a(2^k) = prime(k+1).

A381500 a(n) = A019565(A187769(n)).

Original entry on oeis.org

1, 2, 3, 6, 5, 10, 15, 30, 7, 14, 21, 35, 42, 70, 105, 210, 11, 22, 33, 55, 77, 66, 110, 165, 154, 231, 385, 330, 462, 770, 1155, 2310, 13, 26, 39, 65, 91, 143, 78, 130, 195, 182, 273, 455, 286, 429, 715, 1001, 390, 546, 910, 1365, 858, 1430, 2145, 2002, 3003
Offset: 0

Views

Author

Keywords

Comments

The squarefree numbers, ordered first by largest prime factor (dividing the sequence into rows), then by number of prime factors, then lexicographically by their prime factors (written in descending order).
We index (a(n)) from offset 0, matching the choice for A019565 and similar sequences.

Examples

			Table begins:
  Row 0:  1;
  Row 1:  2;
  Row 2:  3,  6;
  Row 3:  5, 10, 15, 30;
  Row 4:  7, 14, 21, 35, 42, 70, 105, 210;
  Row 5: 11, 22, 33, 55, 77, 66, 110, 165, 154, 231, 385, 330, 462, 770, 1155, 2310;
  ...
Table of a(n) for n = 0..31, demonstrating relationship of this sequence with s = A187769:
          <-factors                    <-factors
   n  a(n)  2 3 5 7  s(n)  |   n   a(n)  2 3 5 7 11 s(n)
  -------------------------|----------------------------
   0    1   .          0   |  16    11   . . . . x   16
   1    2   x          1   |  17    22   x . . . x   17
   2    3   . x        2   |  18    33   . x . . x   18
   3    6   x x        3   |  19    55   . . x . x   20
   4    5   . . x      4   |  20    77   . . . x x   24
   5   10   x . x      5   |  21    66   x x . . x   19
   6   15   . x x      6   |  22   110   x . x . x   21
   7   30   x x x      7   |  23   165   . x x . x   22
   8    7   . . . x    8   |  24   154   x . . x x   25
   9   14   x . . x    9   |  25   231   . x . x x   26
  10   21   . x . x   10   |  26   385   . . x x x   28
  11   35   . . x x   12   |  27   330   x x x . x   23
  12   42   x x . x   11   |  28   462   x x . x x   27
  13   70   x . x x   13   |  29   770   x . x x x   29
  14  105   . x x x   14   |  30  1155   . x x x x   30
  15  210   x x x x   15   |  31  2310   x x x x x   31
  -------------------------|----------------------------
            1 2 4 8  s(n)  |             1 2 4 8 16 s(n)
             bits->                         bits->
		

Crossrefs

Programs

  • Mathematica
    a187769 = {{0}}~Join~Table[SortBy[Range[2^n, 2^(n + 1) - 1], DigitCount[#, 2, 1] &], {n, 0, 8}] // Flatten; a019565[x_] := Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[x, 2]; Map[a019565, a187769]

Formula

a(n) = A019565(A187769(n)).
As an irregular triangle T(n,k), where row 0 = {1}:
For n > 1, omega(T(n,1)) = 1, omega(T(n, 2^(n-1))) = n, thus row n is divided into n segments S such that with S, omega(T(n,k)) = m, where m = 1..n. (See A187769 for the lengths of segments associated with Pascal's triangle A007318.)
S(-1,-1) = (1).
For n >= 0:
S(n-1, n) = (); S(n, -1) = ();
for 0 <= m <= n, S(n,m) = ( A253550'(S(n-1, m)), A119416'(S(n-1, m-1)) ), where Axxx'((i_1, i_2, ..., i_j)) denotes Axxx(i_1), Axxx(i_2), ..., Axxx(i_j).
a(A163866(n)) = A098012(n).

A383180 Irregular table T(n,k) = A010846(A019565(2^n + k)).

Original entry on oeis.org

1, 2, 2, 5, 2, 6, 5, 18, 2, 6, 5, 19, 5, 20, 16, 68, 2, 7, 6, 22, 5, 21, 18, 77, 5, 22, 17, 79, 16, 74, 60, 283, 2, 7, 6, 23, 5, 23, 18, 80, 5, 22, 18, 82, 16, 78, 62, 295, 5, 24, 19, 87, 16, 82, 64, 315, 15, 80, 62, 316, 55, 290, 226, 1161
Offset: 0

Views

Author

Michael De Vlieger, May 09 2025

Keywords

Examples

			Triangle begins:
  0: 1;
  1: 2;
  2: 2, 5;
  3: 2, 6, 5, 18;
  4: 2, 6, 5, 19, 5, 20, 16, 68;
  5: 2, 7, 6, 22, 5, 21, 18, 77, 5, 22, 17, 79, 16, 74, 60, 283;
   ...
		

Crossrefs

Programs

  • Mathematica
    (* Load the "theta" program at the Mathematica link in A369609, then: *)
    f[x_] := Times @@ Prime@ Position[Reverse@ IntegerDigits[x, 2], 1][[All, 1]]; Table[theta[f[2^n + k] ], {n, 0, 7}, {k, 0, 2^n - 1}]

Formula

T(0,0) = 1.
T(n,0) = 2.
T(n,2^(n-1)-1) = A363061(n).

A103788 a(n) = number of ks that make primorial P(n)/A019565(k)-A019565(k) prime.

Original entry on oeis.org

0, 1, 3, 6, 13, 28, 39, 78, 138, 207, 437, 865, 1423, 2750, 4904, 8861, 16201, 33346, 58534, 111878, 208914, 397522
Offset: 1

Views

Author

Lei Zhou, Feb 16 2005

Keywords

Examples

			P(2)/A(0)-A(0)=6-1=5 is prime, so a(2)=1;
P(4)/A(k)-A(k): 210/2-2=103; 210/3-3=67; 210/6-6=29; 210/5-5=37; 210/10-10=11; 210/7-7=23; so a(4)=6;
		

Crossrefs

Programs

  • Mathematica
    npd = 1; Do[npd = npd*Prime[n]; tn = 0; tt = 1; cp = npd/tt - tt; ct = 0; While[IntegerQ[cp], If[(cp > 0) && PrimeQ[cp], ct = ct + 1]; tn = tn + 1; tt = 1; k1 = tn; o = 1; While[k1 > 0, k2 = Mod[k1, 2]; If[k2 == 1, tt = tt*Prime[o]]; k1 = ( k1 - k2)/2; o = o + 1]; cp = npd/tt - tt]; Print[ct], {n, 1, 22}]

A103792 Index k of the first occurrence of A019565(2n-1) as the smallest term that makes prime(k)-A019565(2n-1) prime.

Original entry on oeis.org

3, 5, 13, 25, 67, 79, 140, 127, 345, 129, 222, 206, 479, 1008, 1577, 766, 2583, 869, 1406, 3427, 5367, 4215, 4141, 9716, 23067, 5030, 13586, 7502, 17340, 19211, 14991, 30961, 27008, 82915, 84387, 91387, 92294, 32886, 30890, 70886, 271430, 131908
Offset: 1

Views

Author

Lei Zhou, Feb 28 2005

Keywords

Examples

			n=1: A019565(2n-1)=2; Prime(3)-2=3 is prime, so a(1)=3;
Prime(4)-A019565(1)=5 is prime, not counted;
n=2: A019565(2n-1)=6; Prime(5)-A019565(1)=9 is not prime; ... Prime(5)-6=5 is prime, so a(2)=5;
Prime(6)-A019565(1)=11 is prime, not counted;
...
Prime(12)-A019565(3)=31 is prime, not counted;
n=3; A019565(2n-1)=10; Prime(13)-2=39, Prime(13)-6=35; Prime(13)-10=31 is prime, so a(3)=13.
		

Crossrefs

Programs

  • Mathematica
    A019565 = Function[tn, k1 = tn; o = 1; tt = 1; While[k1 > 0, k2 = Mod[k1, 2]; If[k2 == 1, tt = tt*Prime[o]]; k1 = (k1 - k2)/2; o = o + 1]; tt]; Array[fa, {1, 500}]; Do[fa[n] = 0, {n, 1, 500}]; n = 2; npd = Prime[n]; ct = 1; wt = 1; While[wt < 200, cr = (ct + 1)/2; If[fa[cr] == 0, fa[cr] = n; While[fa[wt] > 0, Print[fa[wt]]; wt = wt + 1]]; n = n + 1; npd = Prime[n]; ct = 1; tt = ct; cp = npd - A019565[tt]; While[ ! (PrimeQ[cp]), ct = ct + 1; tt = ct; cp = npd - A019565[tt]]]

A103801 Indices n such that A019565(n)+/-2 are both primes.

Original entry on oeis.org

4, 6, 10, 14, 22, 26, 34, 38, 62, 78, 106, 138, 142, 162, 190, 194, 230, 258, 274, 278, 302, 402, 430, 458, 494, 498, 534, 550, 618, 670, 674, 690, 694, 742, 770, 814, 822, 902, 934, 1014, 1030, 1038, 1062, 1090, 1102, 1106, 1142, 1146, 1182, 1222, 1234
Offset: 1

Views

Author

Lei Zhou, Feb 22 2005

Keywords

Examples

			A019565(4)=5, 3 and 7 are both primes, so a(1)=4;
A019565(6)=15, 13 and 17 are both primes, so a(2)=6;
		

Crossrefs

Programs

  • Mathematica
    A019565 = Function[tn, k1 = tn; o = 1; tt = 1; While[k1 > 0, k2 = Mod[k1, 2]; If[k2 == 1, tt = tt*Prime[o]]; k1 = (k1 - k2)/2; o = o + 1]; tt]; Do[cp1 = A019565[n] - 2; cp2 = cp1 + 4; If[PrimeQ[cp1] && PrimeQ[cp2], Print[n]], {n, 0, 20000}]

A109163 a(n) = A019565(n-th prime).

Original entry on oeis.org

3, 6, 10, 30, 42, 70, 22, 66, 330, 770, 2310, 130, 182, 546, 2730, 1430, 6006, 10010, 102, 510, 238, 3570, 1122, 2618, 442, 2210, 6630, 9282, 15470, 4862, 510510, 114, 266, 798, 2090, 6270, 14630, 1482, 7410, 17290, 16302, 27170, 570570, 646, 3230
Offset: 1

Views

Author

Leroy Quet, Aug 18 2005

Keywords

Examples

			The 5th prime is 11 (decimal), which is 1011 in binary. So a(5) is the product of the primes corresponding to the 1's of 1011, 2*3*7 = 42.
		

Crossrefs

Cf. A019565.

Programs

  • Mathematica
    Table[m = 1; o = 1; k1 = Prime[k]; While[k1 > 0, k2 = Mod[k1, 2]; If[k2 == 1, m = m*Prime[o]]; k1 = (k1 - k2)/2; o = o + 1]; m, {k, 1, 55}] (* Stefan Steinerberger, Mar 19 2006 *)

Extensions

More terms from Stefan Steinerberger, Mar 19 2006
Previous Showing 101-110 of 340 results. Next