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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A363521 Product of the divisors d of n such that sqrt(n) < d < n.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 4, 1, 5, 1, 24, 1, 7, 5, 8, 1, 54, 1, 50, 7, 11, 1, 576, 1, 13, 9, 98, 1, 900, 1, 128, 11, 17, 7, 1944, 1, 19, 13, 1600, 1, 2058, 1, 242, 135, 23, 1, 36864, 1, 250, 17, 338, 1, 4374, 11, 3136, 19, 29, 1, 1080000, 1, 31, 189, 512, 13, 7986, 1, 578, 23
Offset: 1

Views

Author

Wesley Ivan Hurt, Jun 07 2023

Keywords

Examples

			The divisors of 16 are {1,2,4,8,16} and the product of the divisors d of n such that sqrt(16) = 4 < d < 16 is 8, so a(16) = 8.
The divisors of 30 are {1,2,3,5,6,10,15,30} and the product of the divisors d of n such that sqrt(30) < d < 30 is 6*10*15 = 900, so a(30) = 900.

Crossrefs

Cf. A007955, A007956, A072499, A363520.

Programs

  • Mathematica
    a[n_] := Product[If[n < d^2 < n^2, d, 1], {d, Divisors[n]}]; Array[a, 100] (* Amiram Eldar, Jun 08 2023 *)
  • PARI
    a(n) = vecprod(select(x->((sqrt(n)Michel Marcus, Jun 08 2023

Formula

a(n) = Product_{d|n, sqrt(n) < d < n} d.
a(n) = A007956(n)/A072499(n).
a(n) = A007955(n)/(n*A072499(n)).

A363595 Recursive product of aliquot divisors of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 6, 1, 16, 3, 10, 1, 1728, 1, 14, 15, 2048, 1, 5832, 1, 8000, 21, 22, 1, 4586471424, 5, 26, 81, 21952, 1, 24300000, 1, 67108864, 33, 34, 35, 101559956668416, 1, 38, 39, 163840000000, 1, 130691232, 1, 85184, 91125, 46, 1, 16543163447903718821855232, 7, 125000, 51, 140608, 1, 1338925209984
Offset: 1

Views

Author

Michael De Vlieger, Jul 10 2023

Keywords

Examples

			Define S(n) to be the set of proper divisors of n.
a(2) = 1, since 2 is prime, S(2) = {1} and the product of S(2) is 1.
a(4) = 2, since S(4) = {1, 2}; S(2) = 1, hence we have (1 X 2) X 1 = 2.
a(6) = 6, since S(6) = {1, 2, 3}; 2 and 3 are primes p and both have S(p) = 1,
  hence we have (1 X 2 X 3) X 1 X 1 = 6.
a(8) = 16, since S(8) = {1, 2, 4}; a(2) = 1, a(4) = 2,
  therefore (1 X 2 X 4) X 1 X 2 = 16.
a(9) = 3, since S(9) = {1, 3}, a(3) = 1,
  therefore (1 X 3) X 1 = 3.
a(10) = 10, since S(10) = {1, 2, 5};
  a(2) = a(5) = 1, a(4) = 2,
  therefore (1 X 2 X 5) X 1 X 1 = 10.
a(12) = 1728, since S(12) = {1, 2, 3, 4, 6};
  a(2) = a(3) = 1, a(4) = 2, a(6) = 6,
  therefore (1 X 2 X 3 X 4 X 6) X 1 X 1 X 2 X 6
  = 144 X 12 = 1728.

Links

Crossrefs

Cf. A000295, A007955, A007956, analogous to A255242.

Programs

  • Mathematica
    f[x_] := f[x] = Times @@ # * Times @@ Map[f, #] &@ Most@ Divisors[x]; Table[f[n], {n, 120}]
  • PARI
    ali(n) = setminus(divisors(n), Set(n));
a(n) = my(list = List(), v = [n]); while (#v, my(w = []); for (i=1, #v, my(s=ali(v[i])); for (j=1, #s, w = concat(w, s[j]); listput(list, s[j]));); v = w;); vecprod(Vec(list)); \\ Michel Marcus, Jul 15 2023

Formula

a(n) >= A007956(n).
a(p) = 1 for prime p.
a(p^2) = p.
a(p^e) = A000295(e).
a(p*q) = p*q for primes p, q, p < q.
A007947(n) | a(n) for n with omega(n) > 2.

A229971 Palindromes n whose product of proper divisors is a palindrome > 1 and not equal to n.

Original entry on oeis.org

4, 9, 121, 212, 1001, 10201, 110011, 1100011, 10100101, 11000011, 101000101, 110000011, 1010000101, 1100000011, 10000000001, 10100000101, 1000000000001, 10000000000001, 10011100000111001, 10022212521222001, 10100101110100101, 10101100100110101
Offset: 1

Views

Author

Derek Orr, Oct 04 2013

Keywords

Comments

Palindromes in the sequence A229970.

Examples

			The product of the proper divisors of 4 is 2 (also a palindrome, different from 4). So, 4 is a member of this sequence.
The proper divisors of 1001 are 1, 7, 11, 13, 77, 91, and 143. 1*7*11*13*77*91*143 = 1001^3 = 1003003001 (also a palindrome, different from 1001). So, 1001 is a member of this sequence.

Crossrefs

Cf. A229970, A007956.

Programs

  • Mathematica
    palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; fQ[n_] := Block[{s = Times @@ Most@ Divisors@ n}, And[palQ@ s, s > 1, s != n]]; Select[Select[Range@ 1000000, palQ], fQ] (* Michael De Vlieger, Apr 06 2015 *)
ppdpQ[n_]:=Module[{pp=Times@@Most[Divisors[n]]},AllTrue[{n,pp},PalindromeQ]&&pp>1&&pp!=n]; Select[Range[115*10^4],ppdpQ] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Sep 18 2022 *)
  • PARI
    pal(n)=d=digits(n);Vecrev(d)==d
for(n=1,10^6,D=divisors(n);p=prod(i=1,#D-1,D[i]);if(pal(n)&&pal(p)&&p-1&&p-n,print1(n,", "))) \\ Derek Orr, Apr 05 2015
  • Python
    from sympy import divisors
def PD(n):
  p = 1
  for i in divisors(n):
    if i != n:
      p *= i
  return p
def pal(n):
  r = ''
  for i in str(n):
    r = i + r
  return r == str(n)
{print(n,end=', ') for n in range(1,10**6) if pal(n) and pal(PD(n)) and (PD(n)-1) and PD(n)-n}
# Simplified by Derek Orr, Apr 05 2015
# Syntax error fixed by Robert C. Lyons, Mar 17 2023

Formula

A229970 INTERSECT A002113.

Extensions

a(7)-a(22) from Giovanni Resta, Oct 06 2013
Name edited by Derek Orr, Apr 05 2015

A278474 Numbers n such that the product of proper divisors of n ends with n and n is not a multiplicatively perfect number (A007422).

Original entry on oeis.org

24, 36, 76, 375, 376, 432, 624, 625, 693, 875, 999, 2499, 4557, 8307, 9375, 9376, 9999, 34375, 40625, 47943, 50001, 59375, 81249, 90624, 90625, 99999, 109376, 186432, 218751, 586432, 609375, 690624, 718751, 781249, 890625, 954193, 968751, 999999, 2109375, 2890624, 2890625
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 23 2016

Keywords

Comments

Numbers n such that A007956(n) == n (mod A011557(A055642(n))) and A000005(n) <> 4.

Examples

			24 is in the sequence because 24 has 7 proper divisors {1,2,3,4,6,8,12} and 1*2*3*4*6*8*12 = 13824;
36 is in the sequence because 36 has 8 proper divisors {1,2,3,4,6,9,12,18} and 1*2*3*4*6*9*12*18 = 279936;
76 is in the sequence because 76 has 5 proper divisors {1,2,4,19,38} and 1*2*4*19*38 = 5776, etc.

Links

Crossrefs

Cf. A000005, A007422, A007956, A011557, A055642, A210994

Programs

  • Mathematica
    Select[Range[3000000], Mod[Sqrt[#1]^DivisorSigma[0, #1]/#1, 10^IntegerLength[#1]] == #1 && Sqrt[#1]^DivisorSigma[0, #1] != #1^2 & ]
Previous Showing 41-44 of 44 results.

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