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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A336321 a(n) = A122111(A225546(n)).

Original entry on oeis.org

1, 2, 3, 4, 7, 5, 19, 6, 9, 11, 53, 10, 131, 23, 13, 8, 311, 15, 719, 22, 29, 59, 1619, 14, 49, 137, 21, 46, 3671, 17, 8161, 12, 61, 313, 37, 25, 17863, 727, 139, 26, 38873, 31, 84017, 118, 39, 1621, 180503, 20, 361, 77, 317, 274, 386093, 33, 71, 58, 733, 3673, 821641, 34, 1742537, 8167, 87, 18, 151, 67, 3681131, 626, 1627, 41, 7754077, 35, 16290047
Offset: 1

Views

Author

Antti Karttunen and Peter Munn, Jul 17 2020

Keywords

Comments

A122111 and A225546 are both self-inverse permutations of the positive integers based on prime factorizations, and they share further common properties. For instance, they map the prime numbers to powers of 2: A122111 maps the k-th prime to 2^k, whereas A225546 maps it to 2^2^(k-1).
In composing these permutations, this sequence maps the squarefree numbers, as listed in A019565, to the prime numbers in increasing order; and the list of powers of 2 to the "normal" numbers (A055932), as listed in A057335.

Examples

			From _Peter Munn_, Jan 04 2021: (Start)
In this set of examples we consider [a(n)] as a function a(.) with an inverse, a^-1(.).
First, a table showing mapping of the powers of 2:
  n     a^-1(2^n) =    2^n =        a(2^n) =
        A001146(n-1)   A000079(n)   A057335(n)
  0             (1)         1            1
  1               2         2            2
  2               4         4            4
  3              16         8            6
  4             256        16            8
  5           65536        32           12
  6      4294967296        64           18
  ...
Next, a table showing mapping of the squarefree numbers, as listed in A019565 (a lexicographic ordering by prime factors):
  n   a^-1(A019565(n))   A019565(n)      a(A019565(n))   a^2(A019565(n))
      Cf. {A337533}      Cf. {A005117}   = prime(n)      = A033844(n-1)
  0              1               1             (1)               (1)
  1              2               2               2                 2
  2              3               3               3                 3
  3              8               6               5                 7
  4              6               5               7                19
  5             12              10              11                53
  6             18              15              13               131
  7            128              30              17               311
  8              5               7              19               719
  9             24              14              23              1619
  ...
As sets, the above columns are A337533, A005117, A008578, {1} U A033844.
Similarly, we get bijections between sets A000290\{0} -> {1} U A070003; and {1} U A335740 -> A005408 -> A066207.
(End)
		

Crossrefs

A122111 composed with A225546.
Cf. A336322 (inverse permutation).
Other sequences used in a definition of this sequence: A000040, A000188, A019565, A248663, A253550, A253560.
Sequences used to express relationship between terms of this sequence: A003159, A003961, A297002, A334747.
Cf. A057335.
A mapping between the binary tree sequences A334866 and A253563.
Lists of sets (S_1, S_2, ... S_j) related by the bijection defined by the sequence: (A000290\{0}, {1} U A070003), ({1} U A001146, A000079, A055932), ({1} U A335740, A005408, A066207), (A337533, A005117, A008578, {1} U A033844).

Formula

a(n) = A122111(A225546(n)).
Alternative definition: (Start)
Write n = m^2 * A019565(j), where m = A000188(n), j = A248663(n).
a(1) = 1; otherwise for m = 1, a(n) = A000040(j), for m > 1, a(n) = A253550^j(A253560(a(m))).
(End)
a(A000040(m)) = A033844(m-1).
a(A001146(m)) = 2^(m+1).
a(2^n) = A057335(n).
a(n^2) = A253560(a(n)).
For n in A003159, a(2n) = b(a(n)), where b(1) = 2, b(n) = A253550(n), n >= 2.
More generally, a(A334747(n)) = b(a(n)).
a(A003961(n)) = A297002(a(n)).
a(A334866(m)) = A253563(m).

A056042 a(n) = n!/(k!)^2, where k is the largest number such that (k!)^2 divides n!.

Original entry on oeis.org

1, 2, 6, 6, 30, 20, 140, 70, 630, 7, 77, 924, 12012, 3432, 51480, 12870, 218790, 48620, 923780, 184756, 3879876, 705432, 16224936, 2704156, 67603900, 10400600, 280816200, 178296, 5170584, 155117520, 4808643120, 601080390, 19835652870
Offset: 1

Views

Author

Labos Elemer, Jul 25 2000

Keywords

Comments

Least integer of the form n!/{(n-k)!}^2.
Similar to but different from A001405.

Examples

			E.g. for n=9, 10, 11, 12, a(n)=630, 7, 77, 924 while the corresponding central binomial coefficients are 126, 252, 462, 924 respectively.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Min[ Select[ Table[ n!/(n - k)!^2, {k, n}], IntegerQ[ # ] &]]; Table[ f[n], {n, 33}] (Robert G. Wilson v)

A059591 Squarefree part of n^2+1.

Original entry on oeis.org

1, 2, 5, 10, 17, 26, 37, 2, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 13, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 41, 1090, 1157, 1226, 1297, 1370, 5, 1522, 1601, 2, 1765, 74, 1937, 2026, 2117, 2210, 2305, 2402, 2501, 2602
Offset: 0

Views

Author

Marc LeBrun, Jan 25 2001

Keywords

Comments

Related to period-1 continued fractions [z,z,z,...].

Examples

			a(7)=2 since 7^2+1 = 50 = 25*2 = (5^2)*2.
		

Crossrefs

Programs

  • Maple
    A:= proc(n)
    local F;
    F:= select(t -> t[2]::odd, ifactors(n^2+1)[2]);
    mul(t[1],t=F)
    end proc:
    map(A, [$0..100]); # Robert Israel, Jun 18 2015
  • Mathematica
    Table[Times @@ Flatten[Table[#1, {#2}] & @@@ Select[FactorInteger[n^2 + 1], OddQ@ Last@ # &]], {n, 120}] (* Michael De Vlieger, Jun 19 2015 *)
  • PARI
    a(n) = core(n^2+1); \\ Michel Marcus, Jun 18 2015

Formula

a(n) = A007913(n^2+1).
a(n)*A059592(n)^2 = n^2+1.

A160696 Largest k such that k^2 divides prime(n)+1.

Original entry on oeis.org

1, 2, 1, 2, 2, 1, 3, 2, 2, 1, 4, 1, 1, 2, 4, 3, 2, 1, 2, 6, 1, 4, 2, 3, 7, 1, 2, 6, 1, 1, 8, 2, 1, 2, 5, 2, 1, 2, 2, 1, 6, 1, 8, 1, 3, 10, 2, 4, 2, 1, 3, 4, 11, 6, 1, 2, 3, 4, 1, 1, 2, 7, 2, 2, 1, 1, 2, 13, 2, 5, 1, 6, 4, 1, 2, 8, 1, 1, 1, 1, 2, 1, 12, 1, 2, 2, 15, 1, 1, 4, 6, 4, 2, 2, 10, 6, 1, 3, 2, 1, 2, 3, 2
Offset: 1

Views

Author

Reinhard Zumkeller, May 24 2009

Keywords

Comments

A160697 and A160698 give record values and where they occur.

Crossrefs

Programs

  • PARI
    a(n) = core(prime(n)+1, 1)[2]; \\ Michel Marcus, Nov 06 2022

Formula

a(A049097(n)) = 1; a(A049098(n)) > 1;
a(n) = A000188(A008864(n)).

A283484 Odd bisection of A283983; square root of the largest square dividing A277324.

Original entry on oeis.org

1, 1, 3, 1, 3, 3, 15, 1, 3, 15, 45, 15, 15, 15, 105, 1, 3, 105, 225, 525, 1575, 1125, 1575, 105, 105, 525, 1575, 525, 105, 105, 1155, 1, 3, 1155, 1575, 3675, 7875, 275625, 55125, 5775, 17325, 275625, 4134375, 55125, 55125, 275625, 121275, 1155, 1155, 40425, 385875, 202125, 606375, 1929375, 606375, 5775, 8085, 40425, 121275, 40425, 1155, 1155, 15015, 1, 3
Offset: 0

Views

Author

Antti Karttunen, Mar 25 2017

Keywords

Crossrefs

Programs

Formula

a(n) = A283983((2*n)+1).
a(n) = A000188(A277324(n)).
A001222(a(n)) = A284265(n).

A284264 a(n) = A001222(A283983(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 2, 0, 0, 0, 1, 1, 2, 1, 3, 1, 2, 0, 2, 2, 2, 0, 3, 0, 0, 0, 1, 1, 3, 1, 4, 2, 4, 1, 5, 3, 5, 1, 5, 2, 3, 0, 3, 2, 4, 2, 5, 2, 4, 0, 3, 3, 3, 0, 4, 0, 0, 0, 1, 1, 4, 1, 5, 3, 5, 1, 6, 4, 8, 2, 7, 4, 5, 1, 6, 5, 8, 3, 10, 5, 7, 1, 7, 5, 8, 2, 7, 3, 4, 0, 4, 3, 6, 2, 8, 4, 7, 2, 8, 5, 9, 2, 8, 4, 5, 0, 5, 3, 6, 3, 7, 3, 6, 0
Offset: 0

Views

Author

Antti Karttunen, Mar 25 2017

Keywords

Comments

a(n) = Sum_{c} floor(c/2), where c ranges over each coefficient of terms c * x^k in the Stern polynomial B(n,x), thus sum of the halved terms (for odd terms floored down) on row n of table A125184.

Crossrefs

Cf. A023758 (gives the positions of zeros).

Programs

Formula

a(n) = A001222(A283983(n)).
Other identities and observations. For all n >= 0:
a(2n) = a(n).
a(n) = (1/2) * (A002487(n) - A277700(n)).
2*a(n) <= A284272(n).

A365549 The number of exponentially odd divisors of the square root of the largest square dividing n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Sep 08 2023

Keywords

Comments

First differs from A278908, A307848, A323308 and A358260 at n = 64.
The number of exponentially odd divisors of the largest square dividing n is the same as the number of squares dividing n, A046951(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := 2 + Floor[(e-2)/4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> 2 + (x-2)\4, factor(n)[, 2]));

Formula

a(n) = A322483(A000188(n)).
a(n) >= 1 with equality if and only if n is squarefree (A005117).
Multiplicative with a(p^e) = 2 + floor((e-2)/4).
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 1/p^(2*s) - 1/p^(4*s)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 + 1/p^2 - 1/p^4) = 1.54211628314015874165... .

A381522 Sequence where k is appended after every k^2 occurrences of 1, with multiple values following a 1 listed in order.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 2, 4, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 5, 1, 1, 3, 1, 2, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 2, 3, 6, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 4, 1, 7, 1, 5, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 4, 8, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 6
Offset: 1

Views

Author

Jwalin Bhatt, Feb 26 2025

Keywords

Comments

The frequencies of the terms follow the zeta distribution with parameter value 2.
The geometric mean approaches exp(-zeta'(2)/zeta(2)) A381456 in the limit. In general, if the sequence was formed by every k^s occurrences, it would approach e^(-zeta'(s)/zeta(s)).
Considered as an irregular triangle, the n-th row lists the divisors of the square root of the largest square dividing n.

Examples

			After every 4 ones we see a 2, after every 9 ones we see a 3 and so on.
		

Crossrefs

Programs

  • PARI
    lista(n)={my(L=List()); for(n=1, n, fordiv(sqrtint(n/core(n)), d, listput(L,d))); Vec(L[1..n])} \\ Andrew Howroyd, Feb 26 2025
  • Python
    from itertools import islice
    def zeta_distribution_generator():
        num_ones, num_reached = 0, 1
        while num_ones := num_ones+1:
            yield 1
            for num in range(2, num_reached+2):
                if num_ones % (num*num) == 0:
                    yield num
                    num_reached += num == num_reached+1
    A381522 = list(islice(zeta_distribution_generator(), 120))
    

A229297 Number of solutions to x^2 == n (mod 2*n) for 0 <= x < 2*n.

Original entry on oeis.org

1, 0, 1, 2, 1, 0, 1, 0, 3, 0, 1, 2, 1, 0, 1, 4, 1, 0, 1, 2, 1, 0, 1, 0, 5, 0, 3, 2, 1, 0, 1, 0, 1, 0, 1, 6, 1, 0, 1, 0, 1, 0, 1, 2, 3, 0, 1, 4, 7, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 3, 8, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 5, 2, 1, 0, 1, 4, 9, 0, 1, 2, 1, 0
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    A[n_] := Sum[If[Mod[a^2, 2*n] == n, 1, 0], {a, 0, 2*n - 1}]; Array[A, 100]
    f[p_, e_] := If[OddQ[e], p^((e - 1)/2), p^(e/2)]; f[2, e_] := If[OddQ[e], 0, 2^(e/2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jan 01 2023 *)
  • PARI
    a(n)={sum(i=0, 2*n-1, i^2%(2*n)==n)} \\ Andrew Howroyd, Aug 06 2018
    
  • PARI
    a(n)={if(valuation(n,2)%2==1, 0, core(n, 1)[2])} \\ Andrew Howroyd, Aug 07 2018

Formula

From Andrew Howroyd, Aug 07 2018: (Start)
Multiplicative with a(2^e) = 0 for odd e and 2^floor(e/2) for even e, and a(p^e) = p^floor(e/2) for p>=3. [corrected by Georg Fischer, Aug 01 2022].
a(n) = A000188(n) for odd n, a(2^k) = 1 + (-1)^k for k > 0. (End)
From Amiram Eldar, Jan 01 2023: (Start)
Dirichlet g.f.: zeta(2*s-1)*zeta(s)/(zeta(2*s)*(1+1/2^s)).
Sum_{k=1..n} a(k) ~ (n*log(n) + (3*gamma + log(2)/3 - 2*zeta'(2)/zeta(2) - 1)*n)*2/Pi^2, where gamma is Euler's constant (A001620). (End)

A254732 a(n) is the least k > n such that n divides k^2.

Original entry on oeis.org

2, 4, 6, 6, 10, 12, 14, 12, 12, 20, 22, 18, 26, 28, 30, 20, 34, 24, 38, 30, 42, 44, 46, 36, 30, 52, 36, 42, 58, 60, 62, 40, 66, 68, 70, 42, 74, 76, 78, 60, 82, 84, 86, 66, 60, 92, 94, 60, 56, 60, 102, 78, 106, 72, 110, 84, 114, 116, 118, 90, 122, 124, 84, 72
Offset: 1

Views

Author

Peter Kagey, Feb 06 2015

Keywords

Comments

A073353(n) <= a(n) <= 2*n. Any prime that divides n must also divide a(n), and because n divides (2*n)^2.
Are all terms even? -Harvey P. Dale, Aug 07 2025

Examples

			a(12) = 18 because 12 divides 18^2, but 12 does not divide 13^2, 14^2, 15^2, 16^2, or 17^2.
		

Crossrefs

Cf. A254733 (similar, with k^3), A254734 (similar, with k^4), A073353 (similar, with limit m->infinity of k^m).
Cf. A253905.

Programs

  • Haskell
    a254732 n = head [k | k <- [n + 1 ..], mod (k ^ 2) n == 0]
    -- Reinhard Zumkeller, Feb 07 2015
    
  • Mathematica
    lk[n_]:=Module[{k=n+1},While[!Divisible[k^2,n],k++];k]; Array[lk,70] (* Harvey P. Dale, Nov 05 2017 *)
    Table[Module[{k=n+1},While[PowerMod[k,2,n]!=0,k++];k],{n,70}] (* Harvey P. Dale, Aug 07 2025 *)
  • PARI
    a(n)=for(k=n+1,2*n,if(k^2%n==0,return(k)))
    vector(100,n,a(n)) \\ Derek Orr, Feb 06 2015
    
  • PARI
    a(n)=my(t=factorback(factor(n)[,1])); forstep(k=n+t,2*n,t,if(k^2%n==0, return(k))) \\ Charles R Greathouse IV, Feb 07 2015
    
  • Python
    def A254732(n):
        k = n + 1
        while pow(k,2,n):
            k += 1
        return k # Chai Wah Wu, Feb 15 2015
  • Ruby
    def a(n)
      (n+1..2*n).find { |k| k**2 % n == 0 }
    end
    

Formula

a(n) = sqrt(n*A072905(n)).
a(n) = A019554(n)*(A000188(n)+1).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + zeta(3)/zeta(2) = 1 + A253905 = 1.73076296940143849872... . - Amiram Eldar, Feb 17 2024
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