A000188 -id:A000188 - cp's OEIS frontend

A295657 Multiplicative with a(p^e) = p^floor((e-1)/2).

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Antti Karttunen, Nov 28 2017

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A000188, A003557, A004526, A020639, A028234, A067029, A295658, A295659.
Cf. A004709 (positions of ones), A082020, A212793.
Cf. also A008834, A053150, A061704.

Array[Apply[Times, FactorInteger[#] /. {p_, e_} /; p > 0 :> p^Floor[(e - 1)/2]] &, 105] (* Michael De Vlieger, Nov 28 2017 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^floor((f[i,2]-1)/2));} \\ Amiram Eldar, Nov 30 2022

a(1) = 1; for n > 1, a(n) = A020639(n)^A004526(A067029(n)-1) * a(A028234(n)).
a(n) = A000188(A003557(n)).
a(n) = 1 iff A212793(n) = 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 15/Pi^2 = 1.519817... (A082020). - Amiram Eldar, Nov 30 2022

A334215 T(n, k) is the greatest positive integer m such that m^k divides n; square array T(n, k), n, k > 0 read by antidiagonals downwards.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 2, 5, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2, 9, 1, 1, 1, 1, 1, 1, 1, 2, 3, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Rémy Sigrist, Apr 19 2020

			Square array starts:
  n\k|   1  2  3  4  5  6  7  8  9 10
  ---+-------------------------------
    1|   1  1  1  1  1  1  1  1  1  1
    2|   2  1  1  1  1  1  1  1  1  1
    3|   3  1  1  1  1  1  1  1  1  1
    4|   4  2  1  1  1  1  1  1  1  1
    5|   5  1  1  1  1  1  1  1  1  1
    6|   6  1  1  1  1  1  1  1  1  1
    7|   7  1  1  1  1  1  1  1  1  1
    8|   8  2  2  1  1  1  1  1  1  1
    9|   9  3  1  1  1  1  1  1  1  1
   10|  10  1  1  1  1  1  1  1  1  1
   11|  11  1  1  1  1  1  1  1  1  1
   12|  12  2  1  1  1  1  1  1  1  1
   13|  13  1  1  1  1  1  1  1  1  1
   14|  14  1  1  1  1  1  1  1  1  1
   15|  15  1  1  1  1  1  1  1  1  1
   16|  16  4  2  2  1  1  1  1  1  1

Cf. A000188, A051903, A053150, A053164, A060175, A261969, A334215, A334216.

T(n,k) = { my (f=factor(n)); prod (i=1, #f~, f[i,1]^(f[i,2]\k)) }

T(n, 1) = n.
T(n, 2) = A000188(n).
T(n, 3) = A053150(n).
T(n, 4) = A053164(n).
T(n, A051903(n)) = A261969(n).
T(n, k) = 1 for any k > A051903(n).
T(n^k, k) = n.

A375970 a(n) is the largest number k such that k^2 divides the square pyramidal number A000330(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 5, 3, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 70, 5, 3, 3, 1, 1, 1, 4, 4, 1, 1, 1, 1, 5, 1, 2, 6, 1, 1, 1, 1, 1, 1, 2, 14, 35, 5, 1, 1, 3, 3, 2, 2, 1, 1, 1, 11, 1, 5, 4, 4, 1, 1, 3, 1, 1, 1, 2, 2, 7, 5, 5, 1, 1, 1, 2, 6, 3, 1, 1, 13, 1, 1, 10, 2, 1, 1, 1, 1, 1, 3, 4, 4, 7

Offset: 1

Robert Israel, Sep 04 2024

nonn

a(n)^2 is the largest square that divides n*(n+1)*(2*n+1)/6.

			a(12) = 5 because A000330(12) = 650 = 2 * 5^2 = 13 and 5^2 is the largest square dividing 650.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000188, A000330, A119356, A375971, A375973.

g:= proc(n) local t,s,F; t:= n*(n+1)*(2*n+1)/6;
  F:= ifactors(t)[2];
  mul(s[1]^floor(s[2]/2), s=F)
end proc:
map(g, [$1..100]);

a(n) = my(m=n*(n+1)*(2*n+1)/6); sqrtint(m/core(m)); \\ Michel Marcus, Sep 06 2024

a(n) = A000188(A000330(n)).

A055491 Smallest square divisible by n divided by largest square which divides n.

Original entry on oeis.org

1, 4, 9, 1, 25, 36, 49, 4, 1, 100, 121, 9, 169, 196, 225, 1, 289, 4, 361, 25, 441, 484, 529, 36, 1, 676, 9, 49, 841, 900, 961, 4, 1089, 1156, 1225, 1, 1369, 1444, 1521, 100, 1681, 1764, 1849, 121, 25, 2116, 2209, 9, 1, 4, 2601, 169, 2809, 36, 3025, 196, 3249, 3364

Offset: 1

Henry Bottomley, Jun 28 2000

			a(12) = 36/4 = 9.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A056551, A056552.
Cf. A000188, A000290, A007913, A008833, A019554, A053143.
Cf. A013661, A013664.

a055491 = (^ 2) . a007913  -- Reinhard Zumkeller, Jul 23 2014

With[{sqs=Range[100]^2},Table[SelectFirst[sqs,Divisible[#,n]&]/ SelectFirst[ Reverse[sqs],Divisible[n,#]&],{n,60}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Feb 18 2018 *)
f[p_, e_] := p^(2 * Mod[e, 2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(2*(f[i,2]%2)));} \\ Amiram Eldar, Oct 27 2022

If n is written as Product(Pj^Ej) then a(n) = Product(Pj^(2*(Ej mod 2))).
a(n) = A053143(n)/A008833(n) = A007913(n)^2 = (A019554(n)/A000188(n))^2 = A000290(n)/A008833(n)^2.
Sum_{k=1..n} a(k) ~ c * n^3, where c = (zeta(6)/(3*zeta(2))) = 2*Pi^4/945 = 0.206156... . - Amiram Eldar, Oct 27 2022
Dirichlet g.f.: zeta(s-2) * zeta(2*s) / zeta(2*s-4). - Amiram Eldar, Sep 16 2023

A056043 Let k be largest number such that k^2 divides n!; a(n) = k/floor(n/2)!.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 3, 6, 6, 2, 2, 2, 6, 3, 3, 2, 2, 2, 2, 2, 2, 2, 10, 10, 30, 30, 30, 12, 12, 3, 3, 6, 30, 10, 10, 10, 30, 6, 6, 2, 2, 2, 30, 60, 60, 30, 210, 42, 42, 42, 42, 28, 28, 2, 2, 4, 4, 4, 4, 4, 84, 21, 21, 14, 14, 14, 42, 6, 6, 2, 2, 2, 10, 10, 70, 140, 140, 14, 126, 126

Offset: 1

Labos Elemer, Jul 25 2000

nonn

			For n = 7, 7! = 5040 = 144*35, so 12 is its largest square-root-divisor, A000188(5040), and it is divisible by 6 = 3!, so a(7) = 12/3! = 2.

Cf. A000142, A000188, A001405, A004526, A055772, A081123.

f[p_, e_] := p^Floor[e/2]; b[1] = 1; a[n_] := (Times @@ f @@@ FactorInteger[n!]) / Floor[n/2]!; Array[a, 100] (* Amiram Eldar, May 24 2024 *)

a(n) = A000188(n!)/floor(n/2)! = A055772(n)/A000142(A004526(n)) = A055772(n)/A081123(n). [Corrected by Amiram Eldar, May 24 2024]

A056194 Characteristic cube divisor of n!: a(n) = A056191(n!).

Original entry on oeis.org

1, 1, 1, 8, 8, 1, 1, 8, 8, 1, 1, 27, 27, 216, 1000, 1000, 1000, 125, 125, 1, 9261, 74088, 74088, 343, 343, 2744, 74088, 216, 216, 125, 125, 1000, 35937000, 4492125, 12326391, 12326391, 12326391, 98611128, 8024024008, 125375375125

Offset: 1

Labos Elemer, Aug 02 2000

nonn

Amiram Eldar, Table of n, a(n) for n = 1..4054

Cf. A000142, A055229, A000188, A008833, A007913, A055231, A055230, A055772, A055071, A055204, A055773 A056552, A056195.

f[p_, e_] := If[OddQ[e] && e > 1, p^3, 1]; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* Amiram Eldar, Sep 06 2020 *)

a(n) = A056191(A000142(n)). - Amiram Eldar, Sep 06 2020

A056195 a(n) = n! divided by its characteristic cube divisor A056194.

Original entry on oeis.org

1, 2, 6, 3, 15, 720, 5040, 5040, 45360, 3628800, 39916800, 17740800, 230630400, 403603200, 1307674368, 20922789888, 355687428096, 51218989645824, 973160803270656, 2432902008176640000, 5516784599040000

Offset: 1

Labos Elemer, Aug 02 2000

nonn

			n = 10, a(10) = 10! because g(10!) = 1.
n = 9, a(9) = 45320 because 9! = 2*2*2*2*2*2*2*3*3*5*7 and g(9!) = 2, so a(9) = 9!/8.

Amiram Eldar, Table of n, a(n) for n = 1..491

Cf. A055229, A055230, A055071, A055772, A055773.

f[p_, e_] := If[OddQ[e] && e > 1, p^3, 1]; a[n_] := n! / (Times @@ f @@@ FactorInteger[n!]); Array[a, 20] (* Amiram Eldar, Sep 06 2020 *)

The CCD of n! is the cube of g = A055229(n!) = A055230(n). So a(n) = n!/ggg = L*L*f where L = A000188(n!)/A055229(n!) = A055772(n)/A055230(n) and f = A055231(n!) = A055773(n).

A064368 Number of 2 X 2 symmetric singular matrices with entries from {0,...,n}.

Original entry on oeis.org

1, 4, 7, 10, 15, 18, 21, 24, 29, 36, 39, 42, 47, 50, 53, 56, 65, 68, 75, 78, 83, 86, 89, 92, 97, 108, 111, 118, 123, 126, 129, 132, 141, 144, 147, 150, 163, 166, 169, 172, 177, 180, 183, 186, 191, 198, 201, 204, 213, 228, 239, 242, 247, 250, 257, 260, 265, 268

Offset: 0

Vladeta Jovovic, Sep 27 2001

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..10000

Cf. A000010, A000188, A064276, A059306, A062801, A059976, A039623.

Cf. A001620, A306016.

f[p_, e_] := p^Floor[e/2]; a[0] = 0; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; With[{max = 100}, 1 + Range[0, max] + 2 * Accumulate[Array[a, max + 1, 0]]] (* Amiram Eldar, Nov 07 2024 *)

a(n) = n + 1 + 2*sum(k=1, n, sumdiv(k, d, issquare(d)*eulerphi(sqrtint(d)))) \\ Michel Marcus, Jun 17 2013

a(n) = n + 1 + 2*Sum_{k=1..n} Sum_{d^2|k} phi(d), where phi = Euler totient function A000010.

a(n) ~ (n/zeta(2)) * (log(n) + 3*gamma - 1 + zeta(2) - 2*zeta'(2)/zeta(2)), where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 07 2024

A087050 Square root of the largest square >1 dividing the n-th nonsquarefree number.

Original entry on oeis.org

2, 2, 3, 2, 4, 3, 2, 2, 5, 3, 2, 4, 6, 2, 2, 3, 4, 7, 5, 2, 3, 2, 2, 3, 8, 2, 6, 5, 2, 4, 9, 2, 2, 3, 2, 4, 7, 3, 10, 2, 6, 4, 2, 3, 2, 11, 2, 5, 3, 8, 2, 3, 2, 2, 12, 7, 2, 5, 2, 3, 2, 4, 9, 2, 2, 13, 3, 2, 5, 4, 6, 2, 2, 3, 8, 14, 3, 10, 2, 3, 4, 2, 6, 2, 4, 15, 2, 2, 3, 2, 4, 11, 9, 2, 7, 2, 5, 6, 16, 2, 3

Offset: 1

Wolfdieter Lang, Sep 08 2003

			n=10, A013929(10) = 27, a(10)^2 = 3^2 = 9. 27 = 9*3.
n=39, A013929(39) = 100, a(39)^2 = 10^2 = 100.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A087049, A013929, A000188, A000290.

Cf. A001610, A013661, A306016.

f[p_, e_] := p^Floor[e/2]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; s /@ Select[Range[300], !SquareFreeQ[#] &] (* Amiram Eldar, Feb 11 2021 *)

from math import isqrt, prod
from sympy import mobius, factorint
def A087050(n):
    def f(x): return n+sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return prod(p**(e>>1) for p, e in factorint(m).items() if e>1) # Chai Wah Wu, Jul 22 2024

a(n)^2 is the largest square factor (from A000290) of the nonsquarefree number A013929(n), n>=1.
a(n) = A000188(A013929(n)). - Amiram Eldar, Feb 11 2021
Sum_{k=1..n} a(k) ~ (n/(2*(zeta(2)-1))) * (log(n) + 3*gamma - 3 - 2*zeta'(2)/zeta(2) - log(1-1/zeta(2))), where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 14 2024

A087726 Number of elements X in the matrix ring M_2(Z_n) such that X^2 == 0 mod n.

Original entry on oeis.org

1, 4, 9, 28, 25, 36, 49, 112, 153, 100, 121, 252, 169, 196, 225, 640, 289, 612, 361, 700, 441, 484, 529, 1008, 1225, 676, 1377, 1372, 841, 900, 961, 2560, 1089, 1156, 1225, 4284, 1369, 1444, 1521, 2800, 1681, 1764, 1849, 3388, 3825, 2116, 2209, 5760, 4753, 4900, 2601, 4732

Offset: 1

Yuval Dekel (dekelyuval(AT)hotmail.com), Sep 28 2003

Conjecture: a(n)=n^2 if and only if n is squarefree. [Ben Branman, Mar 22 2013]
Preceding conjecture is true in the case where n is squarefree. - Eric M. Schmidt, Mar 23 2013
It appears that a(p^k) = (1+3*p^2 + 2*k*(p^2-1) + (-1)^k*(p^2-1))*p^(2*k-2)/4 for primes p.  Since the sequence is multiplicative, this would imply the conjecture. - Robert Israel, Jun 10 2015
A proof of the formula for k=1 can be done easily (see pdf). - Manfred Scheucher, Jun 10 2015

Manfred Scheucher, Table of n, a(n) for n = 1..1000
Manfred Scheucher, A proof of the formula for k=1

Cf. A066907, A000188.

#include
#include
int main(int argc,char** argv)
{
  long ct = 0;
  int n = atoi(argv[1]);
  int a,b,c,d;
  for(a=0;aManfred Scheucher, Jun 09 2015 */

f:= proc(n)
  local tot, S, a, mult, sa, d, ad, g, cands;
  tot:= 0;
  S:= ListTools:-Classify(t -> t^2 mod n, [$0..n-1]);
  for a in numtheory:-divisors(n) do
    mult:= numtheory:-phi(n/a);
    sa:= a^2 mod n;
    for d in S[sa] do
       g:= igcd(a+d,n);
       cands:= [seq(i*n/g, i=0..g-1)];
       tot:= tot + mult * numboccur(sa,[seq(seq(s*t,s=cands),t=cands)] mod n);
    od
  od;
  tot
end proc:
map(f, [$1..100]); # Robert Israel, Jun 09 2015

a[m_] := Count[Table[Mod[MatrixPower[Partition[IntegerDigits[n, m, 4], 2], 2], m] == {{0, 0}, {0, 0}}, {n, 0, m^4 - 1}], True]; Table[a[n], {n,2,30}] (* Ben Branman, Mar 22 2013 *)

More terms from Ben Branman, Mar 22 2013

More terms from Manfred Scheucher, Jun 09 2015

cp's OEIS Frontend

A295657 Multiplicative with a(p^e) = p^floor((e-1)/2).

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A334215 T(n, k) is the greatest positive integer m such that m^k divides n; square array T(n, k), n, k > 0 read by antidiagonals downwards.

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A375970 a(n) is the largest number k such that k^2 divides the square pyramidal number A000330(n).

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A055491 Smallest square divisible by n divided by largest square which divides n.

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A056043 Let k be largest number such that k^2 divides n!; a(n) = k/floor(n/2)!.

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A056194 Characteristic cube divisor of n!: a(n) = A056191(n!).

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A056195 a(n) = n! divided by its characteristic cube divisor A056194.

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A064368 Number of 2 X 2 symmetric singular matrices with entries from {0,...,n}.

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