A019554 -id:A019554 - cp's OEIS frontend

A053143 Smallest square divisible by n.

1, 4, 9, 4, 25, 36, 49, 16, 9, 100, 121, 36, 169, 196, 225, 16, 289, 36, 361, 100, 441, 484, 529, 144, 25, 676, 81, 196, 841, 900, 961, 64, 1089, 1156, 1225, 36, 1369, 1444, 1521, 400, 1681, 1764, 1849, 484, 225, 2116, 2209, 144, 49, 100, 2601, 676, 2809

Offset: 1

Henry Bottomley, Feb 28 2000

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

Cf. A000188, A007913, A008833, A019554.

Table[i = 1; While[Mod[i^2, n] > 0, i++]; i^2, {n, 100}] (* T. D. Noe, Oct 30 2011 *)
f[p_, e_] := p^(e + Mod[e, 2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)

a(n) = n*core(n); /* Joerg Arndt, Aug 02 2012 */

a(n) = n*A007913(n) = A019554(n)^2 = n^2/A008833(n) = (n/A000188(n))^2.
Multiplicative with p^e -> p^(e + e mod 2), p prime. - Reinhard Zumkeller, Feb 09 2003
Dirichlet g.f.: zeta(2s-2)*zeta(s-2)/zeta(2s-4). - R. J. Mathar, Oct 31 2011
Sum_{k=1..n} a(k) ~ Pi^2 * n^3 / 45. - Vaclav Kotesovec, Feb 08 2019
Sum_{n>=1} 1/a(n) = 5/2. - Amiram Eldar, Jul 29 2022

A053166 Smallest positive integer for which n divides a(n)^4.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 4, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38

Offset: 1

Henry Bottomley, Feb 29 2000

According to Broughan (2002, 2003, 2006), a(n) is the "upper 4th root of n". The "lower 4th root of n" is sequence A053164. - Petros Hadjicostas, Sep 15 2019

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Cf. A000189, A000190, A008835, A013665, A015051.

f[p_, e_] := p^Ceiling[e/4]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 08 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/4)); factorback(f); \\ Michel Marcus, Jun 09 2014

a(n) = n/A000190(n) = A019554(n)/(A008835(A019554(n)^2))^(1/4).
If n is 5th-power-free (i.e., not 32, 64, 128, 243, ...) then a(n) = A007947(n).
Multiplicative with a(p^e) = p^(ceiling(e/4)). - Christian G. Bower, May 16 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(7)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3528057925... . - Amiram Eldar, Oct 27 2022

A053149 Smallest cube divisible by n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 8, 27, 1000, 1331, 216, 2197, 2744, 3375, 64, 4913, 216, 6859, 1000, 9261, 10648, 12167, 216, 125, 17576, 27, 2744, 24389, 27000, 29791, 64, 35937, 39304, 42875, 216, 50653, 54872, 59319, 1000, 68921, 74088, 79507, 10648

Offset: 1

Henry Bottomley, Feb 28 2000

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

Cf. A000188, A000189, A007913, A008834, A019554, A048798, A050985.

Cf. A002117, A013667, A330596.

a[n_] := For[k = 1, True, k++, If[ Divisible[c = k^3, n], Return[c]]]; Table[a[n], {n, 1, 44}] (* Jean-François Alcover, Sep 03 2012 *)
f[p_, e_] := p^(e + Mod[3 - e, 3]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
scdn[n_]:=Module[{c=Ceiling[Surd[n,3]]},While[!Divisible[c^3,n],c++];c^3]; Array[scdn,50] (* Harvey P. Dale, Jun 13 2020 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(f[i,2] + (3-f[i,2])%3));} \\ Amiram Eldar, Oct 27 2022

a(n) = (n/A000189(n))^3 = A008834(n)*A019554(A050985(n))^3 = n*A050985(n)^2/A000188(A050985(n))^3.
a(n) = n * A048798(n). - Franklin T. Adams-Watters, Apr 08 2009
From Amiram Eldar, Jul 29 2022: (Start)
Multiplicative with a(p^e) = p^(e + ((3-e) mod 3)).
Sum_{n>=1} 1/a(n) = Product_{p prime} ((p^3+2)/(p^3-1)) = 1.655234386560802506... . (End)
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(9)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013667*A330596/(4*A002117) = 0.1559906... . - Amiram Eldar, Oct 27 2022

A234840 Self-inverse and multiplicative permutation of integers: a(0) = 0, a(1) = 1, a(2) = 3, a(3) = 2, a(p_i) = p_{a(i+1)-1} for primes with index i > 2, and a(u * v) = a(u) * a(v) for u, v > 0.

Original entry on oeis.org

0, 1, 3, 2, 9, 19, 6, 61, 27, 4, 57, 11, 18, 281, 183, 38, 81, 101, 12, 5, 171, 122, 33, 263, 54, 361, 843, 8, 549, 29, 114, 59, 243, 22, 303, 1159, 36, 1811, 15, 562, 513, 1091, 366, 157, 99, 76, 789, 409, 162, 3721, 1083, 202, 2529, 541, 24, 209, 1647, 10, 87, 31

Offset: 0

Antti Karttunen, Dec 31 2013

The permutation satisfies A008578(a(n)) = a(A008578(n)) for all n, and is self-inverse.
The sequence of fixed points begins as 0, 1, 6, 11, 29, 36, 66, 95, 107, 121, 149, 174, 216, 313, 319, 396, 427, ... and is itself multiplicative in a sense that if a and b are fixed points, then also a*b is a fixed point.
The records are 0, 1, 3, 9, 19, 61, 281, 361, 843, 1159, 1811, 3721, 5339, 5433, 17141, 78961, 110471, 236883, 325679, ...
and they occur at positions 0, 1, 2, 4, 5, 7, 13, 25, 26, 35, 37, 49, 65, 74, 91, 169, 259, 338, 455, ...
(Note how the permutations map squares to squares, and in general keep the prime signature the same.)
Composition with similarly constructed A235199 gives the permutations A234743 & A234744 with more open cycle-structure.
The result of applying a permutation of the prime numbers to the prime factors of n. - Peter Munn, Dec 15 2019

			a(4) = a(2 * 2) = a(2)*a(2) = 3*3 = 9.
a(5) = a(p_3) = p_{a(3+1)-1} = p_{9-1} = p_8 = 19.
a(11) = a(p_5) = p_{a(5+1)-1} = p_{a(6)-1} = p_5 = 11.

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for sequences that are permutations of the natural numbers

List below gives similarly constructed permutations, which all force a swap of two small numbers, with (the rest of) primes permuted with the sequence itself and the new positions of composite numbers defined by the multiplicative property. Apart from the first one, all satisfy A000040(a(n)) = a(A000040(n)) except for a finite number of cases (with A235200, substitute A065091 for A000040):
A235200 (swaps 3 & 5).
A235199 (swaps 5 & 7).
A235201 (swaps 3 & 4).
A235487 (swaps 7 & 8).
A235489 (swaps 8 & 9).
Cf. A064614, A234743/A234744, A235485/A235486, A235493/A235494, also A317930.
Properties preserved by the sequence as a function: A000005, A001221, A001222, A051903, A101296.
A007913, A007947, A008578, A019554, A055231, A059895, A059896, A059897 are used to express relationships between terms of this sequence.

a[n_] := a[n] = Switch[n, 0, 0, 1, 1, 2, 3, 3, 2, _, Product[{p, e} = pe; Prime[a[PrimePi[p] + 1] - 1]^e, {pe, FactorInteger[n]}]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 21 2021 *)

A234840(n) = if(n<=1,n,my(f = factor(n)); for(i=1, #f~, if(2==f[i,1], f[i,1]++, if(3==f[i,1], f[i,1]--, f[i,1] = prime(-1+A234840(1+primepi(f[i,1])))))); factorback(f)); \\ Antti Karttunen, Aug 23 2018

a(0) = 0, a(1) = 1, a(2) = 3, a(3) = 2, a(p_i) = p_{a(i+1)-1} for primes with index i > 2, and a(u * v) = a(u) * a(v) for u, v > 0.
From Peter Munn, Dec 14 2019. These identities would hold also if a(n) applied any other permutation of the prime numbers to the prime factors of n: (Start)
A000005(a(n)) = A000005(n).
A001221(a(n)) = A001221(n).
A001222(a(n)) = A001222(n).
A051903(a(n)) = A051903(n).
A101296(a(n)) = A101296(n).
a(A007913(n)) = A007913(a(n)).
a(A007947(n)) = A007947(a(n)).
a(A019554(n)) = A019554(a(n)).
a(A055231(n)) = A055231(a(n)).
a(A059895(n,k)) = A059895(a(n), a(k)).
a(A059896(n,k)) = A059896(a(n), a(k)).
a(A059897(n,k)) = A059897(a(n), a(k)).
(End)

A102631 a(n) = n^2 / (squarefree kernel of n).

Original entry on oeis.org

1, 2, 3, 8, 5, 6, 7, 32, 27, 10, 11, 24, 13, 14, 15, 128, 17, 54, 19, 40, 21, 22, 23, 96, 125, 26, 243, 56, 29, 30, 31, 512, 33, 34, 35, 216, 37, 38, 39, 160, 41, 42, 43, 88, 135, 46, 47, 384, 343, 250, 51, 104, 53, 486, 55, 224, 57, 58, 59, 120, 61, 62, 189, 2048, 65, 66, 67

Offset: 1

Reinhard Zumkeller, Feb 25 2005

Index of first occurrence of n in A019554. - Franklin T. Adams-Watters, Nov 17 2006

Michel Marcus, Table of n, a(n) for n = 1..5000

Cf. A000290, A007947, A003557, A005117, A019554, A350390.

a[n_] := n^2/Times @@ FactorInteger[n][[All, 1]];
Array[a, 70] (* Jean-François Alcover, Jun 11 2019 *)

a(n) = my(f=factor(n)); for (k=1, #f~, f[k,2] = 2*f[k,2]-1); factorback(f); \\ Michel Marcus, Aug 20 2017

def A102631(n) :
    p = n
    for a in factor(n) :
        if a[1] > 1 :
            p = p * a[0]^(a[1]-1)
    return p
[A102631(n) for n in (1..67)] # Peter Luschny, Feb 07 2012

a(n) = A000290(n)/A007947(n) = n*A003557(n);
a(n) = n iff n is squarefree: a(A005117(n)) = A005117(n).
Multiplicative with a(p^e) = p^{2e-1}. - Franklin T. Adams-Watters, Nov 17 2006
Dirichlet g.f.: Product_{p prime} (1 - p/(p^2 - p^s)). - Amiram Eldar, Aug 28 2023
a(n) = A350390(n^2). - Amiram Eldar, Nov 30 2023

A015052 a(n) is the smallest positive integer m such that m^5 is divisible by n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 4, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38, 77, 78

Offset: 1

R. Muller

A multiplicative companion function n/a(n) = 1,1,1,2,1,1,1,4,3,1,1,2,1,1,1,8,1,... can be defined using the 5th instead of the 4th power in A000190, which differs from A000190 and also from A003557. - R. J. Mathar, Jul 14 2012

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Henry Ibstedt, Surfing on the Ocean of Numbers, Erhus Univ. Press, Vail, 1997.
Florentin Smarandache, Collected Papers, Vol. II, Tempus Publ. Hse, Bucharest, 1996.
Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015053 (outer 6th root).

Cf. A013667.

f[p_, e_] := p^Ceiling[e/5]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/5)); factorback(f); \\ Michel Marcus, Feb 15 2015

Multiplicative with a(p^e) = p^(ceiling(e/5)). - Christian G. Bower, May 16 2005

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(9)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6 + 1/p^7 - 1/p^8) = 0.3523622369... . - Amiram Eldar, Oct 27 2022

Corrected by David W. Wilson, Jun 04 2002

Name reworded by Jon E. Schoenfield, Oct 28 2022

A015053 Smallest positive integer for which n divides a(n)^6.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 2, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38, 77, 78

Offset: 1

R. Muller (Research37(AT)aol.com)

Differs from A007947 as follows: A007947(128)=2, a(128)=4; A007947(256)=2, a(256)=4; A007947(384)=6, a(384)=12; A007947(512)=2, a(512)=4; A007947(640)=10, a(640)=20, etc. - R. J. Mathar, Oct 28 2008

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Henry Ibstedt, Surfing on the Ocean of Numbers, Erhus Univ. Press, Vail, 1997.
Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (5th outer root).

Cf. A013669.

spi[n_]:=Module[{k=1},While[PowerMod[k,6,n]!=0,k++];k]; Array[spi,80] (* Harvey P. Dale, Feb 29 2020 *)
f[p_, e_] := p^Ceiling[e/6]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/6)); factorback(f); \\ Michel Marcus, Feb 15 2015

Multiplicative with a(p^e) = p^ceiling(e/6). - Christian G. Bower, May 16 2005

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(11)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6 + 1/p^7 - 1/p^8 + 1/p^9 - 1/p^10) = 0.3522558764... . - Amiram Eldar, Oct 27 2022

Corrected by David W. Wilson, Jun 04 2002

A061503 a(n) = Sum_{k=1..n} tau(k^2), where tau is the number of divisors function A000005.

Original entry on oeis.org

1, 4, 7, 12, 15, 24, 27, 34, 39, 48, 51, 66, 69, 78, 87, 96, 99, 114, 117, 132, 141, 150, 153, 174, 179, 188, 195, 210, 213, 240, 243, 254, 263, 272, 281, 306, 309, 318, 327, 348, 351, 378, 381, 396, 411, 420, 423, 450, 455, 470, 479, 494, 497

Offset: 1

N. J. A. Sloane, Jun 14 2001

nonn

a(n) is the number of pairs of positive integers <= n with their LCM <= n. - Andrew Howroyd, Sep 01 2019

Mentioned by Steven Finch in a posting to the Number Theory List (NMBRTHRY(AT)LISTSERV.NODAK.EDU), Jun 13 2001.

Harry J. Smith, Table of n, a(n) for n = 1..1024
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, vol. 101, (2002), pp. 105-114.
Steven R. Finch, Errata and Addenda to Mathematical Constants, p. 19.
Vaclav Kotesovec, Graph - the asymptotic ratio
Eric Weisstein's World of Mathematics, Stieltjes Constants

Cf. A000005, A061502. Partial sums of A048691.

List([1..60],n->Sum([1..n],k->Tau(k^2))); # Muniru A Asiru, Mar 09 2019

with(numtheory): a:=n->add(tau(k^2),k=1..n): seq(a(n),n=1..60); # Muniru A Asiru, Mar 09 2019

DivisorSigma[0, Range[60]^2] // Accumulate (* Jean-François Alcover, Nov 25 2013 *)

for (n=1, 1024, write("b061503.txt", n, " ", sum(k=1, n, numdiv(k^2)))) \\ Harry J. Smith, Jul 23 2009

t=0;v=vector(60,n,t+=numdiv(n^2)) \\ Charles R Greathouse IV, Nov 08 2012

from math import prod
from sympy import factorint
def A061503(n): return sum(prod(2*e+1 for e in factorint(k).values()) for k in range(1,n+1)) # Chai Wah Wu, May 10 2022

def A061503(n) :
    tau = sloane.A000005
    return add(tau(k^2) for k in (1..n))
[ A061503(i) for i in (1..19)] # Peter Luschny, Sep 15 2012

a(n) = Sum_{j=1..n^2} floor(n/A019554(j)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jul 20 2002
a(n) = Sum_{i=1..n} 2^omega(i) * floor(n/i). - Enrique Pérez Herrero, Sep 15 2012
a(n) ~ 3/Pi^2 * n log^2 n. - Charles R Greathouse IV, Nov 08 2012
a(n) ~ 3*n/Pi^2 * (log(n)^2 + log(n)*(-2 + 6*g - 24*z/Pi^2) + 2 - 6*g + 6*g^2 - 6*sg1 + 288*z^2/Pi^4 - 24*(-z + 3*g*z + z2)/ Pi^2), where g is the Euler-Mascheroni constant A001620, sg1 is the first Stieltjes constant (see A082633), z = Zeta'(2) (see A073002), z2 = Zeta''(2) = A201994. - Vaclav Kotesovec, Jan 30 2019
a(n) = Sum_{k=1..n} A064608(floor(n/k)). - Daniel Suteu, Mar 09 2019

Name corrected by Peter Luschny, Sep 15 2012

A186646 Every fourth term of the sequence of natural numbers 1,2,3,4,... is halved.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 4, 9, 10, 11, 6, 13, 14, 15, 8, 17, 18, 19, 10, 21, 22, 23, 12, 25, 26, 27, 14, 29, 30, 31, 16, 33, 34, 35, 18, 37, 38, 39, 20, 41, 42, 43, 22, 45, 46, 47, 24, 49, 50, 51, 26, 53, 54, 55, 28, 57, 58, 59, 30, 61, 62, 63, 32, 65, 66, 67, 34, 69, 70, 71, 36, 73, 74, 75, 38, 77, 78, 79, 40, 81, 82, 83, 42, 85, 86, 87, 44, 89, 90, 91, 46, 93, 94, 95, 48, 97, 98, 99

Offset: 1

R. J. Mathar, Feb 25 2011

a(n) is the length of the period of the sequence k^2 mod n, k=1,2,3,4,..., i.e., the length of the period of A000035 (n=2), A011655 (n=3), A000035 (n=4), A070430 (n=5), A070431 (n=6), A053879 (n=7), A070432 (n=8), A070433 (n=9), A008959 (n=10), A070434 (n=11), A070435 (n=12) etc.
From Franklin T. Adams-Watters, Feb 24 2011: (Start)
Clearly if gcd(n,m) = 1, a(nm) = lcm(a(n),a(m)), so it suffices to establish this for prime powers.
If p is a prime, the period must divide p, but k^2 mod p is not constant, so a(p) = p.
a(p^e), e > 1, must be divisible by a(p^(e-1)), and must divide p^e. If p != 2, (p^(e-1)+1)^2 =  p^(2e-2)+2p^(e-1)+1 == 2p^(e-1)+1 (mod p^2), so a(p^e) != p^(e-1); it must then be e.
By inspection, a(4) = 2 and a(8) = 4.
This leaves a(2^e), e > 3. But then (2^(e-2)+1)^2 = 2^(2e-4)+2^(e-1)+1 == 2^(e-1)+1 (mod 2^e), so a(n) > 2^(e-2). On the other hand, (2^(e-1)+c)^2 = 2^(2e-2)+c2^e+c^2 == c^2 (mod 2^e). Hence the period is 2^(e-1). (End)

Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A000224 (size of the set of moduli of k^2 mod n), A019554, A060819, A061037, A090129, A142705, A164115, A283971.

A186646 := proc(n) if n mod 4 = 0 then n/2 ; else n ; end if; end proc ;

Flatten[Table[{n,n+1,n+2,(n+3)/2},{n,1,101,4}]] (* or *) LinearRecurrence[ {0,0,0,2,0,0,0,-1},{1,2,3,2,5,6,7,4},100] (* Harvey P. Dale, May 30 2014 *)
Table[n (7 - (-1)^n - 2 Cos[n Pi/2])/8, {n, 100}] (* Federico Provvedi , Jan 02 2018 *)

a(n)=if(n%4,n,n/2) \\ Charles R Greathouse IV, Oct 16 2015

def A186646(n): return n if n&3 else n>>1 # Chai Wah Wu, Jan 10 2023

a(n) = 2*a(n-4) - a(n-8).
a(4n) = 2n; a(4n+1) = 4n+1; a(4n+2) = 4n+2; a(4n+3) = 4n+3.
a(n) = n/A164115(n).
G.f.: x*(1 + 2*x + 3*x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6) / ( (x-1)^2*(1+x)^2*(x^2+1)^2 ).
Dirichlet g.f.: (1-2/4^s)*zeta(s-1).
A019554(n) | a(n). - Charles R Greathouse IV, Feb 24 2011
a(n) = n*(7 - (-1)^n - (-i)^n - i^n)/8, with i=sqrt(-1). - Bruno Berselli, Feb 25 2011
Multiplicative with a(p^e)=2^e if p=2 and e<=1; a(p^e)=2^(e-1) if p=2 and e>=2; a(p^e)=p^e otherwise. - David W. Wilson, Feb 26 2011
a(n) * A060819(n+2) = A142705(n+1) = A061037(2n+2). - Paul Curtz, Mar 02 2011
a(n) = n - (n/2)*floor(((n-1) mod 4)/3). - Gary Detlefs, Apr 14 2013
a(2^n) = A090129(n+1). - R. J. Mathar, Oct 09 2014
a(n) = n*(7 - (-1)^n - 2*cos(n*Pi/2))/8. - Federico Provvedi, Jan 02 2018
E.g.f.: (1/4)*x*(4*cosh(x) + sin(x) + 3*sinh(x)). - Stefano Spezia, Jan 26 2020
Sum_{k=1..n} a(k) ~ (7/16) * n^2. - Amiram Eldar, Nov 28 2022

A346602 Decimal expansion of 3*zeta(3)/Pi^2.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Jul 30 2021

			0.365381484700719249363018365653857319764...

This constant describes the growth of A019554 and A346597.

Cf. A002117.

RealDigits[3*Zeta[3]/Pi^2, 10, 100][[1]] (* Amiram Eldar, Jul 30 2021 *)

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