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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A333625 Read terms e = T(n,k) in A333624 as Product(prime(k)^e) for n in A334556.

Original entry on oeis.org

1, 8, 8, 27, 27, 216, 512, 216, 512, 648, 648, 686, 12096, 46656, 262144, 46656, 262144, 12096, 686, 192000, 139968, 192000, 139968, 1866240, 179712, 74088, 91125, 74088, 91125, 179712, 1866240, 343000, 1000000, 5832000, 4251528, 5832000, 80621568, 13824000, 1073741824
Offset: 1

Views

Author

Michael De Vlieger, May 13 2020

Keywords

Comments

Row a(n) of A067255 = row A334556(n) of A333624.
An XOR-triangle t(n) is an inverted 0-1 triangle formed by choosing a top row the binary rendition of n and having each entry in subsequent rows be the XOR of the two values above it, i.e., A038554(n) applied recursively until we reach a single bit.
Let T(n,k) address the terms in the k-th position of row n in A333624.
This sequence encodes T(n,k) via A067255 to succinctly express the number of zero-triangles in A334556(n). To decode a(n) => A333624(A334556(n)), we use A067255(a(n)).

Examples

			Relationship of this sequence to A334556 and A333624:
       n A334556(n) a(n)  Row n of A333624
       -----------------------------------
       1     1        1   0
       2    11        8   3
       3    13        8   3
       4    39       27   0, 3
       5    57       27   0, 3
       6    83      216   3, 3
       7    91      512   9
       8   101      216   3, 3
       9   109      512   9
      10   151      648   3, 4
      11   233      648   3, 4
      12   543      686   1, 0, 0, 3
      13   599    12096   6, 3, 0, 1
      14   659    46656   6, 6
      15   731   262144   18
      16   805    46656   6, 6
      ...
Let b(n) = n written in binary and let L(n) = ceiling(log_2(n)) = A070939(n). Let => be a single iteration of XOR across pairs of bits in b(n). Let t(n) be the XOR triangle initiated by b(n).
a(1) = 0, since b(1) = 1 and row 1 of A333624 is {0}. Since the XOR triangle t(1) that results from a single 1-bit merely consists of that bit and since there are no zeros in the triangle t(1), we write the single term zero in row n of A333624. thus a(n) = prime(1)^0 = 2^0 = 1.
a(2) = 8 because row A334556(2) of A333624 (i.e., the 11th row) has {3}. b(11) = 1011 => 110 => 01 => 1 (a rotationally symmetrical t(11)). We have 3 isolated zeros thus row 11 of A333624 = {3}, therefore a(2) = prime(1)^3 = 2^3 = 8.
a(4) = 27 because row A334556(4) of A333624 (i.e., the 39th row) has {0, 3}. b(39) = 100111 => 10100 => 1110 => 001 => 01 => 1 (a rotationally symmetrical t(39)). We have 3 isolated triangles of zeros with edge length 2, thus row 39 of A333624 = {0, 3}, therefore a(4) = prime(1)^0 * prime(2)^3 = 2^0 * 3^3 = 27.
		

Crossrefs

Programs

  • Mathematica
    With[{s = Rest[Import["https://oeis.org/A334556/b334556.txt", "Data"][[All, -1]] ]}, Map[With[{w = NestWhileList[Map[BitXor @@ # &, Partition[#, 2, 1]] &, IntegerDigits[#, 2], Length@ # > 1 &]}, If[Length@ # == 0, 1, Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, #] &@ ReplacePart[ConstantArray[0, Max@ #[[All, 1]]], Map[#1 -> #2 & @@ # &, #]]] &@ Tally@ Flatten@ Array[If[# == 1, Map[If[First@ # == 1, Nothing, Length@ #] &, Split@ w[[#]] ], Map[If[First@ # == -1, Length@ #, Nothing] &, Split[w[[#]] - Most@ w[[# - 1]] ] ]] &, Length@ w]] /. -Infinity -> 0 &, s[[1 ;; 36]]]]

A334896 Read terms e = T(n,k) in A333624 as Product(prime(k)^e) for n in A334769.

Original entry on oeis.org

648, 648, 686, 12096, 12096, 686, 192000, 139968, 192000, 139968, 1866240, 179712, 179712, 1866240, 814968, 2101248, 102036672, 331776000, 102036672, 331776000, 2101248, 814968, 179712000, 4423680000, 1866240000, 131010048, 179712000, 4423680000, 1866240000, 131010048
Offset: 1

Views

Author

Michael De Vlieger, May 23 2020

Keywords

Comments

Row a(n) of A067255 = row A334769(n) of A333624.
An XOR-triangle t(n) is an inverted 0-1 triangle formed by choosing a top row the binary rendition of n and having each entry in subsequent rows be the XOR of the two values above it, i.e., A038554(n) applied recursively until we reach a single bit.
Let T(n,k) address the terms in the k-th position of row n in A333624.
This sequence encodes T(n,k) via A067255 to succinctly express the number of zero-triangles in A334769(n). To decode a(n) => A333624(A334769(n)), we use A067255(a(n)).

Examples

			a(1) = 648, since b(A334769(1)) = b(151) = 10010111, which generates T(151) as shown below, replacing 1 with "@" and 0 with ".":
  @ . . @ . @ @ @
   @ . @ @ @ . .
    @ @ . . @ .
     . @ . @ @
      @ @ @ .
       . . @
        . @
         @
In this figure we see 3 zero-triangles of side length k = 1, and 4 of side length k = 2, therefore, T(1,1) = 3 and T(1,2) = 4. This becomes 2^3 * 3^4 = 8 * 81 = 648.
Relationship of this sequence to A334556 and A333624:
        n  A334769(n)  a(n)  Row n of A333624
      --------------------------------------
       1    151       648    3, 4
       2    233       648    3, 4
       3    543       686    1, 0, 0, 3
       4    599     12096    6, 3, 0, 1
       5    937     12096    6, 3, 0, 1
       6    993       686    1, 0, 0, 3
       7   1379    192000    9, 1, 3
       8   1483    139968    6, 7
       9   1589    192000    9, 1, 3
      10   1693    139968    6, 7
      11   2359   1866240    9, 6, 1
      12   2391    179712    9, 3, 0, 0, 0, 1
      13   3753    179712    9, 3, 0, 0, 0, 1
      14   3785   1866240    9, 6, 1
      15   8607    814968    3, 3, 0, 3, 1
      16   9559   2101248   12, 3, 0, 0, 0, 0, 0, 1
      ...
		

Crossrefs

Programs

  • Mathematica
    With[{s = Rest[Import["https://oeis.org/A334769/b334769.txt", "Data"][[All, -1]]]}, Map[With[{w = NestWhileList[Map[BitXor @@ # &, Partition[#, 2, 1]] &, IntegerDigits[#, 2], Length@ # > 1 &]}, If[Length@ # == 0, 1, Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, #] &@ ReplacePart[ConstantArray[0, Max@ #[[All, 1]]], Map[#1 -> #2 & @@ # &, #]]] &@ Tally@ Flatten@ Array[If[# == 1, Map[If[First@ # == 1, Nothing, Length@ #] &, Split@ w[[#]]], Map[If[First@ # == -1, Length@ #, Nothing] &, Split[w[[#]] - Most@ w[[# - 1]]]]] &, Length@ w]] /. -Infinity -> 0 &, s[[1 ;; 29]]]]

A336510 a(n) = Sum_{p | A055204(n)} 2^(pi(p) - 1).

Original entry on oeis.org

0, 1, 3, 3, 7, 4, 12, 13, 13, 8, 24, 26, 58, 51, 53, 53, 117, 116, 244, 240, 250, 235, 491, 488, 488, 457, 459, 451, 963, 964, 1988, 1989, 2007, 1942, 1946, 1946, 3994, 3867, 3897, 3900, 7996, 7991, 16183, 16167, 16163, 15906, 32290, 32288, 32288, 32289, 32355
Offset: 1

Views

Author

Michael De Vlieger, Sep 18 2020

Keywords

Comments

All terms of A055204 are squarefree by definition, therefore we can compress the terms of A055204 by interpreting the terms of reverse(A067255(A055204(n))) as a binary number and converted to decimal.

Examples

			A055204(1) = 1, the empty product; by convention a(1) = 0.
5! = 120 = 2^3 * 3 * 5, therefore 2 * 3 * 5 = 30 is the squarefree part, which we write "111", a 1 in the first three places to signify a product of the first three primes. Interpreting "111" as a binary number yields 8. Thus a(5) = 8.
13! = 6227020800 = 2^10 * 3^5 * 5^2 * 7 * 11 * 13; its squarefree part is 3 * 7 * 11 * 13 = 3003, a product of the 2nd, 4th, 5th, and 6th primes. Therefore we write "111010", which, interpreted as a binary number and converted to decimal, is 58. Thus a(13) = 58.
Table illustrating the first terms of this sequence, with b(n) = A055204(n):
               Multiplicities of p|b(n)
   n      b(n)   2  3  5  7 11 13 17 -> Binary   a(n)
  --------------------------------------------------
   1        1    .  .  .  .  .  .  .         0     0
   2        2    1  .  .  .  .  .  .         1     1
   3        6    1  1  .  .  .  .  .        11     3
   4        6    1  1  .  .  .  .  .        11     3
   5       30    1  1  1  .  .  .  .       111     7
   6        5    .  .  1  .  .  .  .       100     4
   7       35    .  .  1  1  .  .  .      1100    12
   8       70    1  .  1  1  .  .  .      1101    13
   9       70    1  .  1  1  .  .  .      1101    13
  10        7    .  .  .  1  .  .  .      1000     8
  11       77    .  .  .  1  1  .  .     11000    24
  12      231    .  1  .  1  1  .  .     11010    26
  13     3003    .  1  .  1  1  1  .    111010    58
  14      858    1  1  .  .  1  1  .    110011    51
  15     1430    1  .  1  .  1  1  .    110101    53
  16     1430    1  .  1  .  1  1  .    110101    53
  17    24310    1  .  1  .  1  1  1   1110101   117
  18    12155    .  .  1  .  1  1  1   1110100   116
  ...
		

Crossrefs

Programs

  • Mathematica
    Block[{nn = 51, k, p}, k = PrimePi@ nn; Array[Set[p[Prime@ #], 0] &, k]; {0}~Join~Reap[Do[Map[Set[p[#1], Mod[p[#1] + Mod[#2, 2], 2]] & @@ # &, FactorInteger@ i]; Sow[FromDigits[Array[p[Prime[k - # + 1]] &, k], 2]], {i, 2, nn}]][[-1, 1]]] (* or *)
    Block[{nn = 51, k = 1}, Reap[Do[Map[If[Mod[k, #] == 0, k /= #, k *= #] &, Flatten[ConstantArray[#1, #2] & @@@ FactorInteger@ i]]; Sow[If[k == 1, 0, Total@ Map[2^(PrimePi[#] - 1) &, FactorInteger[k][[All, 1]] ] ] ], {i, nn}]][[-1, 1]]]

A343033 Array T(n, k), n, k > 0, read by antidiagonals; a variant of lunar multiplication (A087062) based on prime exponents of numbers (see Comments section for precise definition).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 2, 5, 2, 1, 1, 5, 3, 3, 5, 1, 1, 6, 7, 4, 7, 6, 1, 1, 7, 15, 5, 5, 15, 7, 1, 1, 2, 11, 6, 11, 6, 11, 2, 1, 1, 3, 3, 7, 35, 35, 7, 3, 3, 1, 1, 10, 5, 4, 13, 30, 13, 4, 5, 10, 1, 1, 11, 21, 9, 5, 77, 77, 5, 9, 21, 11, 1
Offset: 1

Views

Author

Rémy Sigrist and N. J. A. Sloane, Apr 03 2021

Keywords

Comments

To compute T(n, k):
- write the prime exponents of n and of k on two lines, right aligned (these lines correspond to rows of A067255 in reversed order),
- to "multiply" two prime numbers: take the smallest,
- to "add" two prime numbers: take the largest,
- for example, for T(12, 14):
(11 7 5 3 2)
12 --> 1 2
14 --> x 1 0 0 1
---------
1 1
0 0
0 0
+ 1 1
-----------
1 1 0 1 1 --> 462 = T(12, 14)
This sequence is closely related to lunar multiplication (A087062):
- for any b > 1, let S_b be the set of nonnegative integers m such that A051903(m)< b,
- there is a natural bijection f from S_b to the set of nonnegative integers:
f(Product_{k >= 0} prime(k)^d(k)) = Sum_{k >= 0} d(k) * b^k,
- let g be the inverse of f,
- then for any numbers n and k in S_b, we have:
T(n, k) = g(f(n) "*" f(k)) where "*" denotes lunar product in base b,
- the corresponding addition table is A003990.

Examples

			Array T(n, k) begins:
  n\k|  1   2   3   4   5    6    7   8   9   10   11   12   13   14
  ----  -  --  --  --  --  ---  ---  --  --  ---  ---  ---  ---  ---
    1|  1   1   1   1   1    1    1   1   1    1    1    1    1    1
    2|  1   2   3   2   5    6    7   2   3   10   11    6   13   14  --> A007947
    3|  1   3   5   3   7   15   11   3   5   21   13   15   17   33  --> A328915
    4|  1   2   3   4   5    6    7   4   9   10   11   12   13   14  --> A007948
    5|  1   5   7   5  11   35   13   5   7   55   17   35   19   65
    6|  1   6  15   6  35   30   77   6  15  210  143   30  221  462
    7|  1   7  11   7  13   77   17   7  11   91   19   77   23  119
    8|  1   2   3   4   5    6    7   8   9   10   11   12   13   14
    9|  1   3   5   9   7   15   11   9  25   21   13   45   17   33
   10|  1  10  21  10  55  210   91  10  21  110  187  210  247  910
   11|  1  11  13  11  17  143   19  11  13  187   23  143   29  209
   12|  1   6  15  12  35   30   77  12  45  210  143   60  221  462
   13|  1  13  17  13  19  221   23  13  17  247   29  221   31  299
   14|  1  14  33  14  65  462  119  14  33  910  209  462  299  238
		

Crossrefs

Programs

  • PARI
    T(n,k) = { my (r=1, pp=factor(n)[,1]~, qq=factor(k)[,1]~); for (i=1, #pp, for (j=1, #qq, my (p=prime(primepi(pp[i])+primepi(qq[j])-1), v=valuation(r, p), w=min(valuation(n, pp[i]), valuation(k, qq[j]))); if (w>v, r*=p^(w-v)))); r }

Formula

T(n, k) = T(k, n).
T(n, 1) = 1.
T(n, 2) = A007947(n).
T(n, 3) = A328915(n).
T(n, 4) = A007948(n).
T(n, n) = A343035(n).
A051903(T(n, k)) = min(A051903(n), A051903(k)).

A363250 Numbers in A363063 arranged in lexicographic order according to ordered prime signature (i.e., multiplicities of prime power factors p^k, written in order of p).

Original entry on oeis.org

1, 2, 4, 12, 8, 24, 16, 48, 144, 720, 32, 96, 288, 1440, 864, 4320, 21600, 151200, 64, 192, 576, 2880, 1728, 8640, 43200, 302400, 128, 384, 1152, 5760, 3456, 17280, 86400, 604800, 10368, 51840, 259200, 1814400, 256, 768, 2304, 11520, 6912, 34560, 172800, 1209600, 20736, 103680, 518400, 3628800, 62208
Offset: 0

Views

Author

Michael De Vlieger, May 23 2023

Keywords

Comments

The sequence is also readable as an irregular triangle by rows in which row n lists the terms divisible by 2^k but not by 2^(k+1).
Numbers m in A363063 are products of prime powers p(j)^S(j), j = 1..N, where p(j) is the j-th prime, such that p(j+1)^S(j+1) < p(j)^S(j). As consequence of definition of A363063, S(j) > S(j+1), hence multiplicities S(j) are distinct. Consequently, A363063 is a subset of A025487; m is a product of primorials. A025487 in turn is a subset of A055932.
These qualities enable us to write an algorithm that increments S(j) or drops the last term in S until we can increment S(j) to attain a solution. This algorithm generates terms in lexicographic order as described in the Name. The same qualities enable expression of m = Product p(j)^S(j) instead as Sum 2^(S(j)-1), a strictly increasing sequence.

Examples

			Table of n, a(n), and multiplicities S(j) written such that Product p(j)^S(j) = a(n). a(n) = A000079(i) is shown in the penultimate column, while a(n) = A347284(k) appears in the last column.
   n      a(n) multiplicities  i    k
  -----------------------------------
   0:       1                  0    0
   1:       2           1      1    1
   2:       4         2        2
   3:      12         2 1           2
   4:       8       3          3
   5:      24       3   1           3
   6:      16     4            4
   7:      48     4     1
   8:     144     4   2
   9:     720     4   2 1           4
  10:      32   5              5
  11:      96   5       1
  12:     288   5     2
  13:    1440   5     2 1
  14:     864   5   3
  15:    4320   5   3   1
  16:   21600   5   3 2
  17:  151200   5   3 2 1           5
  ...
Sequence read as an irregular triangle T(n, k):
  n\k   1    2    3     4     5     6      7       8
  ---------------------------------------------------
  0:    1
  1:    2
  2:    4   12
  3:    8   24
  4:   16   48  144   720
  5:   32   96  288  1440   864  4320  21600  151200
  6:   64  192  576  2880  1728  8640  43200  302400
  ...
		

Crossrefs

Programs

  • Mathematica
    nn = 12;
     f[x_] := Times @@ MapIndexed[Prime[First[#2]]^#1 &, x];
     {1}~Join~Reap[Do[s = {i}; Sow[2^i]; Set[k, 1];
         Do[
          If[Prime[k]^s[[-1]] > Prime[k + 1],
           AppendTo[s, 1]; k++; Sow[f[s]],
           If[Length[s] == 1, Break[],
            If[Prime[k - 1]^(s[[-2]]) > Prime[k]^(s[[-1]] + 1),
             s[[-1]]++; Sow[f[s]],
             While[And[k > 1,
               Prime[k - 1]^(s[[-2]]) < Prime[k]^(s[[-1]] + 1)], k--;
              s = s[[1 ;; k]]]; If[k == 1, Break[], s[[-1]]++; Sow[f[s]] ]
              ] ] ], {j, Infinity}], {i, nn}]][[-1, -1]]
  • Python
    from sympy import nextprime,oo
    from itertools import islice
    primes = [2] # global list of first primes
    def f(pi, ppmax):
        # Generate numbers with nonincreasing prime-powers <= ppmax, starting at the (pi+1)-st prime.
        if len(primes) <= pi: primes.append(nextprime(primes[-1]))
        p0 = primes[pi]
        if ppmax < p0:
            yield 1
            return
        pp = 1
        while pp <= ppmax:
            for x in f(pi+1, pp):
                yield pp*x
            pp *= p0
    def A363250_list(nterms):
        return list(islice(f(0,oo),nterms)) # Pontus von Brömssen, May 25 2023

Formula

Seen as an irregular triangle, the first term in row i is 2^i, and the last term in row i is A347284(i).

Extensions

Edited by Michael De Vlieger/_Peter Munn_, May 27 2025

A363596 a(n) = (Product_{k=1..pi(n+1)} prime(k)^floor(n/(prime(k)-1) ) )/(n+1)!.

Original entry on oeis.org

1, 1, 2, 1, 6, 2, 12, 3, 10, 2, 12, 2, 420, 60, 24, 3, 90, 10, 420, 42, 660, 60, 360, 30, 3276, 252, 56, 4, 120, 8, 3696, 231, 3570, 210, 36, 2, 103740, 5460, 840, 42, 13860, 660, 27720, 1260, 19320, 840, 5040, 210, 198900, 7956, 10296, 396, 11880, 440, 6384, 228
Offset: 0

Views

Author

Michael De Vlieger, Aug 03 2023

Keywords

Comments

Motivated by Proposition 3.2, p. 10 of the Bedhouche-Farhi paper.
Observations regarding prime power decomposition of terms in a(0..20737):
For n > 300, most terms are in A361098 but not in A286708. A303606 is a subset of A286708, which is a subset of A361098, which in turn is a subset of A126706, numbers that are neither prime powers nor squarefree.
a(34) = 36 is the only term in A286708 (more specifically, in A303606).
a(35) = 2 is the last prime term.
a(29) = 8 is the only composite prime power.
a(190) = 221760 is the last term in A002182, but a(61) = a(102) = 720720 is the largest.
a(191) = 2310 is the last primorial term.
a(1055) = 2207550414530882190 is the last squarefree term. If there are further squarefree terms a(n), n is likely to belong to -1 (mod 24).
a(7055) = 1733187515208453605856007490304335826298500960 is the last term that is not in A361098. a(n) not in A361098 is likely to belong to -1 (mod 24).

Examples

			The table below relates b(n) = A091137(n) to a(n), with (n+1)!*a(n) = k!*m = b(n), where k! is the largest factorial that divides b(n).
 n  A067255(b(n)) (n+1)!*a(n)   k! * m
---------------------------------------
 0  0                1! * 1     1! * 1
 1  1                2! * 1     2! * 1
 2  2.1              3! * 2     3! * 2
 3  3.1              4! * 1     4! * 1
 4  4.2.1            5! * 6     6! * 1
 5  5.2.1            6! * 2     6! * 2
 6  6.3.1.1          7! * 12    7! * 12
 7  7.3.1.1          8! * 3     8! * 3
 8  8.4.2.1          9! * 10   10! * 1
 9  9.4.2.1         10! * 2    10! * 2
10  10.5.2.1.1      11! * 12   12! * 1
11  11.5.2.1.1      12! * 2    12! * 2
12  12.6.3.2.1.1    13! * 420  15! * 2
13  13.6.3.2.1.1    14! * 60   15! * 4
14  14.7.3.2.1.1    15! * 24   15! * 24
15  15.7.3.2.1.1    16! * 3    16! * 3
16  16.8.4.2.1.1.1  17! * 90   18! * 5
...
		

Crossrefs

Programs

  • Mathematica
    Table[j = 1; ( Times @@ Reap[While[Sow[#^Floor[n/(# - 1)]] &[Prime[j]] > 1, j++]][[-1, 1]] )/Factorial[n + 1], {n, 0, 60}]
  • Python
    from math import prod, factorial
    from sympy import sieve
    def A363596(n: int) -> int:
        numer = prod(p ** (n // (p - 1)) for p in sieve.primerange(2, n + 2))
        return numer // factorial(n + 1)
    print([A363596(n) for n in range(56)])  # Peter Luschny, Aug 17 2025

Formula

a(n) = A091137(n)/(n+1)!.

A371630 Numbers k that set records in A372720.

Original entry on oeis.org

1, 2, 6, 12, 30, 60, 120, 210, 420, 840, 1260, 1680, 2520, 4620, 9240, 13860, 18480, 27720, 32760, 55440, 65520, 102960, 110880, 120120, 180180, 240240, 360360, 556920, 720720, 1081080, 1441440, 1884960, 2162160, 2827440, 2882880, 3063060, 3603600, 4084080, 6126120
Offset: 1

Views

Author

Michael De Vlieger, Jun 04 2024

Keywords

Comments

In other words, numbers k that set records for d(k) - f(k), where d = A000005 and f = A008479.
Largest primorial in this sequence is A002110(4) = 210.
The primorials A002110(0..4) are the only squarefree numbers in this sequence.
{a(n)} \ A002110(0..4) is contained in A126706.
Not a subset of A060735; a(13) = 2520 is not in A060735. Though common for small n, the set of a(n) in A060735 is likely finite; the restriction is connected with the finite number of primorials in the sequence.
Not a subset of A025487 or A055932; a(19) = 32760 is the smallest term without a primorial kernel.
The prime shape of a(n) appears to feature exponents m of prime power factors p^m | a(n) that are nonincreasing as pi(p) increases.

Examples

			Table of a(n) and A371634(n) = b(n) for n = 1..20. Asterisks in the a(n) column denote squarefree terms while "+" denotes numbers not in A055932 (i.e., in A080259).
   n     a(n)  A067255(a(n))            d(n)-f(n) = b(n)
  ------------------------------------------------------
   1       1*  1                          1 -  1 =   0
   2       2*  2                          2 -  1 =   1
   3       6*  2 * 3                      4 -  1 =   3
   4      12   2^2 * 3                    6 -  2 =   4
   5      30*  2 * 3 * 5                  8 -  1 =   7
   6      60   2^2 * 3 * 5               12 -  2 =  10
   7     120   2^3 * 3 * 5               16 -  4 =  12
   8     210*  2 * 3 * 5 * 7             16 -  1 =  15
   9     420   2^2 * 3 * 5 * 7           24 -  2 =  22
  10     840   2^3 * 3 * 5 * 7           32 -  4 =  28
  11    1260   2^2 * 3^2 * 5 * 7         36 -  6 =  30
  12    1680   2^4 * 3 * 5 * 7           40 -  8 =  32
  13    2520   2^3 * 3^2 * 5 * 7         48 - 11 =  37
  14    4620   2^2 * 3 * 5 * 7 * 11      48 -  2 =  46
  15    9240   2^3 * 3 * 5 * 7 * 11      64 -  4 =  60
  16   13860   2^2 * 3^2 * 5 * 7 * 11    72 -  6 =  66
  17   18480   2^4 * 3 * 5 * 7 * 11      80 -  8 =  72
  18   27720   2^3 * 3^2 * 5 * 7 * 11    96 - 12 =  84
  19   32760+  2^3 * 3^2 * 5 * 7 * 13    96 - 11 =  85
  20   55440   2^4 * 3^2 * 5 * 7 * 11   120 - 20 = 100
		

Crossrefs

A381240 Indices of records in A381096.

Original entry on oeis.org

1, 6, 10, 12, 14, 18, 22, 24, 28, 30, 42, 54, 60, 66, 78, 84, 90, 102, 114, 120, 126, 132, 138, 150, 168, 180, 198, 204, 210, 240, 252, 264, 270, 294, 300, 330, 360, 378, 390, 420, 450, 462, 480, 504, 510, 540, 546, 570, 600, 630, 660, 690, 714, 750, 780, 810, 840
Offset: 1

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Author

Michael De Vlieger, Feb 18 2025

Keywords

Comments

Let f(k) = A381096(k) = k - phi(k) - tau(k/rad(k)) = k - A000010(k) - A005361(k), where phi = A000010, tau = A000005, and rad = A007947. This sequence contains k such that f(k) > f(j) for j < k as k increases.
Apart from a(1) = 1, terms are in A024619.
Conjecture 1: For i > 1, A002110(i) is in this sequence.
Conjecture 2: Intersection with A001694 (i.e., in A286708) is {900, 1800}.

Examples

			Let g(n) = A067255(n) be the exponents of prime power factors p^m | n, writing "." for m = 0 and ending at the pi(gpf(n))-th term. Example: for n = 84, g(84) = {2, 1, 0, 1}, therefore we write "21.1" for concision in the table below.
Table of first 12 terms.
   n  a(n)  g(a(n)) f(a(n))
  --------------------------
   1    1   .           0
   2    6   11          3
   3   10   1.1         5
   4   12   21          6
   5   14   1..1        7
   6   18   12         10
   7   22   1...1      11
   8   24   31         13
   9   28   2..1       14
  10   30   111        21
  11   42   11.1       29
  12   54   13         33
		

Crossrefs

Programs

  • Mathematica
    r = 0; nn = 2^20; f[x_] := x - EulerPhi[x] - DivisorSigma[0, x/Apply[Times, FactorInteger[x][[All, 1]] ] ]; {1}~Join~Reap[Monitor[Do[If[# > r, r = #; Sow[n]] &[f[n] ], {n, nn}], n]]

A070829 Array showing which primes divide n >= 2.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, May 17 2002

Keywords

Comments

In the Kac reference this array is called rho_{p}(n) := 1 if p divides n else 0.
The row length sequence is A061395(n),n>=2: [1,2,1,3,2,4,1,2,3,5,2,6,4,3,...] (the index of the largest prime dividing n). All row entries beyond these numbers are 0, hence they are not shown. The n=1 row would have 0 for all entries.
The column sequences (without leading zeros) give for m>=1 periodic sequences with the period: 1 followed by p(m)-1 zeros. They start with n=p(m) := A000040(m).

Examples

			{1}, {0, 1}, {1}, {0, 0, 1}, {1, 1}, {0, 0, 0, 1}, {1}, {0, 1}, {1, 0, 1}...
Row n=10: {1,0,1} because p(1)=2 and p(3)= 5 divides 10.
		

References

  • Mark Kac, A Personal History of the Scottish Book, pp. 17-27, in R. D. Mauldin (ed.), The Scottish Book, Birkhäuser, Boston, Basel, 1981.

Crossrefs

Cf. A067255 (array with multiplicities).

Formula

a(n, m)=1 if p(m), m>=1, divides n>=2, with the prime p(m) := A000040(m), else 0.

A134363 Irregular triangle read by rows where n-th row (of A061395(n) terms, for n>=2) is such that n = Product_{j=1..A061395(n)} prime(j)^(Sum_{k=1..j} T(n,k)). Row 1 is {0}.

Original entry on oeis.org

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

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Author

Leroy Quet, Oct 22 2007

Keywords

Comments

The rows of this triangle also give all the ordered ways that a finite number of integers can be arranged so that their partial sums, from left to right, are all nonnegative and their total sum is positive.

Examples

			Triangle begins:
  0;
  1;
  0, 1;
  2;
  0, 0, 1;
  1, 0;
  0, 0, 0, 1;
  3;
  ...
Row 20 is {2, -2, 1}. So 20 = prime(1)^T(20,1) * prime(2)^(T(20,1) + T(20,2)) * prime(3)^(T(20,1) + T(20,2) + T(20,3)) = 2^2 * 3^(2 - 2) * 5^(2 - 2 + 1) = 2^2 * 3^0 * 5^1.
		

Crossrefs

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