A048760 -id:A048760 - cp's OEIS frontend

A262690 a(n) = largest square k <= n such that A002828(n-k) = A002828(n)-1.

0, 1, 1, 1, 4, 4, 4, 4, 4, 9, 9, 9, 4, 9, 9, 9, 16, 16, 9, 9, 16, 16, 9, 9, 16, 25, 25, 25, 25, 25, 25, 25, 16, 25, 25, 25, 36, 36, 36, 36, 36, 25, 25, 25, 36, 36, 36, 36, 16, 49, 49, 49, 36, 49, 49, 49, 36, 49, 49, 49, 49, 36, 49, 49, 64, 64, 64, 49, 64, 64, 36, 49, 36, 64, 49, 49, 36, 64, 49, 49, 64, 81, 81, 81, 64, 81, 81, 81, 36, 64, 81, 81, 81, 64, 81, 81, 64, 81, 49, 81, 100

Offset: 0

Antti Karttunen, Oct 03 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..65536
A. Karttunen, Ratio a(n)/A048760(n) drawn with the help of OEIS Plot2-script

Cf. A000290, A002828, A262689, A262678.

Cf. also A048760.

(define (A262690 n) (A000290 (A262689 n)))

a(n) = A000290(A262689(n)).
Other identities. For all n >= 0:
A262678(n) = n - a(n).

A386469 The largest divisor of n whose exponents in its prime factorization are squares.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 16, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 16, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 48, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 16, 65, 66, 67, 34, 69, 70

Offset: 1

Amiram Eldar, Jul 22 2025

The largest term in A197680 that divides n.

The number of these divisors is A386470(n) and their sum is A386471(n).

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

Cf. A048760, A197680, A386470, A386471.

Similar sequences: A008833, A350390, A365683.

f[p_, e_] := p^(Floor[Sqrt[e]]^2); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

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

Multiplicative with a(p^e) = p^A048760(e).
a(n) <= n, with equality if and only if n is in A197680.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} Sum_{k>=2} (1/p^(k^2-1) - 1/p^(k^2-2)) = 0.74491327356409794092... .

A056894 If the smallest prime with a square excess of n is p then a(n)=p-n.

Original entry on oeis.org

1, 1, 4, 9, 36, 25, 16, 81, 64, 49, 36, 49, 100, 225, 64, 81, 576, 121, 144, 441, 256, 169, 144, 169, 256, 225, 196, 289, 324, 961, 400, 729, 400, 529, 324, 361, 484, 441, 400, 529, 576, 529, 576, 729, 784, 841, 900, 625, 1444, 1521, 676, 961, 900, 1225, 784

Offset: 1

Henry Bottomley, Jul 05 2000

nonn

			a(4)=9 because the smallest prime with a square excess of 4 is 13 and 13-4=9

Cf. A000040, A000196, A002496, A048760, A053186, A056892-A056898.

a(n) = A056893(n) - n = A048760(A056893(n)) = A056895(n)^2

A056895 If the smallest prime with a square excess of n is p then a(n)^2 = p - n.

Original entry on oeis.org

1, 1, 2, 3, 6, 5, 4, 9, 8, 7, 6, 7, 10, 15, 8, 9, 24, 11, 12, 21, 16, 13, 12, 13, 16, 15, 14, 17, 18, 31, 20, 27, 20, 23, 18, 19, 22, 21, 20, 23, 24, 23, 24, 27, 28, 29, 30, 25, 38, 39, 26, 31, 30, 35, 28, 45, 34, 31, 42, 31, 34, 33, 32, 33, 36, 35, 34, 75, 40, 37, 36, 41, 48, 45

Offset: 1

Henry Bottomley, Jul 05 2000

nonn

			a(4)=3 because the smallest prime with a square excess of 4 is 13 and 13 - 4 = 3^2.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A000040, A000196, A002496, A048760, A053186.

Cf. A056892, A056893, A056894, A056896, A056897, A056898.

a = {}; Do[p = 2; While[n != p - (r = Floor@Sqrt[p])^2, p = NextPrime[p]]; AppendTo[a, r], {n, 74}]; a (* Ivan Neretin, May 02 2019 *)

a(n) = {my(p=2); while(n != p-sqrtint(p)^2, p = nextprime(p+1)); sqrtint(p - n);} \\ Michel Marcus, May 05 2019

a(n) = sqrt(A056893(n)-n) = A000196(A056893(n)) = sqrt(A056894(n)).

A065731 Largest square <= n!.

Original entry on oeis.org

1, 1, 1, 4, 16, 100, 676, 4900, 40000, 362404, 3625216, 39904489, 478996996, 6226945921, 87177877081, 1307672296225, 20922784184449, 355687416544329, 6402373660047556, 121645099966283776, 2432902006216007824, 51090942169052381124, 1124000727752683686724

Offset: 0

Labos Elemer, Nov 15 2001

Harry J. Smith, Table of n, a(n) for n = 0..100

Cf. A048760, A000142, A065730-A065741.

[Floor(Sqrt(Factorial(n)))^2: n in [0..25]]; // Vincenzo Librandi, Dec 07 2018

Array[Floor@ Sqrt[#!]^2 &, 21, 0] (* Michael De Vlieger, Dec 06 2018 *)

a(n) = { sqrtint(n!)^2 } \\ Harry J. Smith, Oct 28 2009

a(n) = floor(sqrt(n!))^2.

a(n) = A048760(A000142(n)).

Offset changed from 1 to 0 by Harry J. Smith, Oct 28 2009

A074695 Greatest common divisor of n and floor(n^(1/2))^2.

Original entry on oeis.org

1, 1, 1, 4, 1, 2, 1, 4, 9, 1, 1, 3, 1, 1, 3, 16, 1, 2, 1, 4, 1, 2, 1, 8, 25, 1, 1, 1, 1, 5, 1, 1, 1, 1, 5, 36, 1, 2, 3, 4, 1, 6, 1, 4, 9, 2, 1, 12, 49, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 64, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 81, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 100

Offset: 1

Reinhard Zumkeller, Sep 03 2002

nonn

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000196, A000290, A048760.

Cf. A258613, A258614.

a074695 n = gcd n $ a048760 n  -- Reinhard Zumkeller, Jun 05 2015

Table[GCD[n,Floor[Sqrt[n]]^2],{n,100}] (* Harvey P. Dale, Aug 12 2018 *)

a(n) = gcd(n, sqrtint(n)^2); \\ Amiram Eldar, Nov 19 2024

a(n) = GCD(n, A048760(n)).
a(n) = n iff n is a square.
a(A258613(n)) = 1; a(A258614(n)) > 1. - Reinhard Zumkeller, Jun 05 2015

A258614 Numbers m having with the largest square <= m a common divisor > 1.

Original entry on oeis.org

4, 6, 8, 9, 12, 15, 16, 18, 20, 22, 24, 25, 30, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 56, 63, 64, 66, 68, 70, 72, 74, 76, 78, 80, 81, 84, 87, 90, 93, 96, 99, 100, 102, 104, 105, 106, 108, 110, 112, 114, 115, 116, 118, 120, 121, 132, 143, 144, 146, 147

Offset: 1

Reinhard Zumkeller, Jun 05 2015

nonn

The asymptotic density of this sequence is 1-6/Pi^2 (A229099). - Amiram Eldar, Nov 19 2024

			a(11) = 24: GCD(24,A048760(24)) = GCD(24,16) = 4 > 1.
a(12) = 25: GCD(25,A048760(25)) = GCD(25,25) = 25 > 1.
GCD(26,A048760(26)) = GCD(26,25) = 1, therefore 26 is not a term.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A074695, A048760, A229099, A258613 (complement).

a258614 n = a258614_list !! (n-1)
a258614_list = filter ((> 1) . a074695) [1..]

Select[Range[150], !CoprimeQ[#, Floor[Sqrt[#]]^2] &] (* Amiram Eldar, Nov 19 2024 *)

isok(m) = gcd(m, sqr(sqrtint(m))) > 1; \\ Michel Marcus, Jan 23 2022

A074695(a(n)) > 1.

A350674 Irregular table read by rows; the n-th row contains, in weakly decreasing order, the positive squares summing to n as obtained by the greedy algorithm.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 4, 4, 1, 4, 1, 1, 4, 1, 1, 1, 4, 4, 9, 9, 1, 9, 1, 1, 9, 1, 1, 1, 9, 4, 9, 4, 1, 9, 4, 1, 1, 16, 16, 1, 16, 1, 1, 16, 1, 1, 1, 16, 4, 16, 4, 1, 16, 4, 1, 1, 16, 4, 1, 1, 1, 16, 4, 4, 25, 25, 1, 25, 1, 1, 25, 1, 1, 1, 25, 4, 25, 4, 1, 25, 4, 1, 1

Offset: 1

Rémy Sigrist, Jan 10 2022

The n-th row has A053610(n) terms.

			The first rows are:
     1:    [1]
     2:    [1, 1]
     3:    [1, 1, 1]
     4:    [4]
     5:    [4, 1]
     6:    [4, 1, 1]
     7:    [4, 1, 1, 1]
     8:    [4, 4]
     9:    [9]
    10:    [9, 1]
    11:    [9, 1, 1]
    12:    [9, 1, 1, 1]
    13:    [9, 4]
    14:    [9, 4, 1]
    15:    [9, 4, 1, 1]
    16:    [16]

Andrew Howroyd, Table of n, a(n) for n = 1..1733 (rows 1..500)

Cf. A007961, A048760, A053610, A350698.

row(n,e=2) = { my (g=[], r); while (n, r=sqrtnint(n,e); n-=r^e; g=concat(g,[r^e])); g }

T(n, 1) = A048760(n).

T(n, A053610(n)) = A350698(n).

A382286 a(n) is the least k such that floor(sqrt(nk/d(nk))) = floor(sqrt(d(n*k))), where d(k) is the largest divisor of k which is <= sqrt(k).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 4, 2, 1, 2, 9, 2, 9, 2, 2, 1, 16, 2, 16, 1, 2, 5, 16, 1, 1, 5, 3, 1, 25, 1, 25, 1, 3, 8, 1, 1, 36, 8, 3, 1, 36, 1, 36, 3, 2, 8, 36, 1, 1, 2, 6, 3, 49, 2, 2, 1, 6, 13, 49, 2, 49, 13, 2, 1, 2, 2, 64, 4, 6, 2, 64, 2, 64, 18, 2, 4, 2, 2, 64, 4, 1, 18, 81, 2, 4, 18, 9, 4, 81

Offset: 1

Hassan Baloui, Mar 20 2025

nonn

In the case of semiprime numbers n=p*q, the sequence a(n) chooses the first two multiples of the prime factors p and q such that these multiples (say k*p, m*q) are between two consecutive squares, then a(n)=m*k.
In the case of any composite number n, for each pair of conjugate divisors (d_i, n/d_i) we choose the first two multiples say (k_i*d_i, m_i*n/d_i) which are between two consecutive squares. Then a(n) is the smallest product k_i*m_i where i runs over half the number of divisors of n (since only pairs are considered).
a(n) can also factor numbers as follow:
gcd(n,A000196(a(n)*n)+1-sqrt((A000196(a(n)*n)+1)^2-a(n)*n))
gcd(n,A000196(a(n)*n)+1+sqrt((A000196(a(n)*n)+1)^2-a(n)*n)) are proper divisors of n if sqrt(a(n)*n)-A000196(a(n)*n) < 1/2
Or
gcd(n,(2*A000196(a(n)*n)+1-sqrt((2*A000196(a(n)*n)+1)^2-4*a(n)*n))/2)
gcd(n,(2*A000196(a(n)*n)+1+sqrt((2*A000196(a(n)*n)+1)^2-4*a(n)*n))/2) are proper divisors of n if sqrt(a(n)*n)-A000196(a(n)*n) > 1/2

			Let C(k)=floor(sqrt(k/d(k)))-floor(sqrt(d(k))), where d(k) is the largest divisor of k which is <= sqrt(k):
a(1)=a(2)=a(3)=1 since  C(n)=0, for n=1,2,3.
a(4)=1 (and generally a(n^2)=1).
a(5)=4, since C(k*5) > 0 for k=1 to 3 and C(4*k)=0.
a(6)=1 since C(6)=0.
a(7)=4 since C(k*7) > 0 for k=1 to 3 and C(4*7)=0.
a(8)=2 since C(8)=1 and C(2*8)=0.
a(9)=1
a(11)=9 since C(k*11) > 0 for k=1 to 8 and C(9*11)=0.

Cf. A000196, A033676, A033677, A048760.

d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
a(n) = my(k=1); while (sqrtint(n*k/d(n*k)) != sqrtint(d(n*k)), k++); k; \\ Michel Marcus, Mar 21 2025

a(n) < A048760(n) iff n is a composite number.
a(n) = A048760(n) iff n is a prime number.
a(p^(2*m+1)) = p*a(p)/A048760(p) for any pair (p,m) of a prime number p and a natural number m (except the case (2,2)).
a(n^2) = 1.
If we combines the first two properties we get floor(a(n)/A048760(n))=A010051(n),
And partial sum of floor(a(n)/A048760(n)) from 2 to N = A000720(N).

A386470 The number of divisors of n whose exponents in their prime factorization are squares.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 3, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 3, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4

Offset: 1

Amiram Eldar, Jul 22 2025

First differs from A365171 and A369310 at n = 32.
First differs from A365488 at n = 128.
The number of terms in A197680 that divide n.
The sum of these divisors is A386471(n) and the largest of them is A386469(n).

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

Cf. A000005, A005117, A197680, A365171, A365488, A369310, A386469, A386471.

f[p_, e_] := Floor[Sqrt[e]] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = vecprod(apply(x -> sqrtint(x) + 1, factor(n)[, 2]));

Multiplicative with a(p^e) = A048760(e) + 1.

a(n) <= A000005(n), with equality if and only if n is squarefree (A005117).

cp's OEIS Frontend

A262690 a(n) = largest square k <= n such that A002828(n-k) = A002828(n)-1.

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A386469 The largest divisor of n whose exponents in its prime factorization are squares.

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A056894 If the smallest prime with a square excess of n is p then a(n)=p-n.

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A056895 If the smallest prime with a square excess of n is p then a(n)^2 = p - n.

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Mathematica

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A065731 Largest square <= n!.

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Magma

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Extensions

A074695 Greatest common divisor of n and floor(n^(1/2))^2.

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A258614 Numbers m having with the largest square <= m a common divisor > 1.

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Haskell

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A350674 Irregular table read by rows; the n-th row contains, in weakly decreasing order, the positive squares summing to n as obtained by the greedy algorithm.

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A382286 a(n) is the least k such that floor(sqrt(nk/d(nk))) = floor(sqrt(d(n*k))), where d(k) is the largest divisor of k which is <= sqrt(k).