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.

Showing 1-10 of 10 results.

A084301 a(n) = sigma(n) mod 6.

Original entry on oeis.org

1, 3, 4, 1, 0, 0, 2, 3, 1, 0, 0, 4, 2, 0, 0, 1, 0, 3, 2, 0, 2, 0, 0, 0, 1, 0, 4, 2, 0, 0, 2, 3, 0, 0, 0, 1, 2, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 4, 3, 3, 0, 2, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 2, 1, 0, 0, 2, 0, 0, 0, 0, 3, 2, 0, 4, 2, 0, 0, 2, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 4, 0, 2, 0, 0, 0, 2, 3, 0, 1, 0, 0, 2, 0, 0
Offset: 1

Views

Author

Labos Elemer, Jun 02 2003

Keywords

Crossrefs

Sequences sigma(n) mod k: A053866 (k=2), A074941 (k=3), A105824 (k=4), A105825 (k=5), A084301 (k=6), A105826 (k=7), A105827 (k=8).
Cf. A074627 (locations of 0), A074628 (locations of 2), A067051 (locations of 3), A074630 (locations of 4), A074384 (locations of 5).

Programs

Formula

a(n) = A010875(A000203(n)). - Antti Karttunen, Nov 07 2017

A067051 The smallest k>1 such that k divides sigma(k*n) is equal to 3.

Original entry on oeis.org

2, 8, 18, 32, 49, 50, 72, 98, 128, 162, 169, 196, 200, 242, 288, 338, 361, 392, 441, 450, 512, 578, 648, 676, 722, 784, 800, 882, 961, 968, 1058, 1152, 1225, 1250, 1352, 1369, 1444, 1458, 1521, 1568, 1682, 1764, 1800, 1849, 1922, 2048, 2178, 2312, 2450, 2592
Offset: 1

Views

Author

Benoit Cloitre, Jul 26 2002

Keywords

Comments

The smallest m>1 such that m divides sigma(m*n) is 2, 3 or 6.
Appears to be the same sequence as A074629. - Ralf Stephan, Aug 18 2004. [Proof: Mathar link]
Square terms are in A074216. Nonsquare terms appear to be A001105 except {0}. - Michel Marcus, Dec 26 2013

Crossrefs

Subsequence of A087943.

Programs

  • Magma
    [n: n in [1..3*10^3] | (SumOfDivisors(n) mod 6) eq 3]; // Vincenzo Librandi, Dec 11 2015
  • Maple
    select(t -> numtheory:-sigma(t) mod 6 = 3, [$1..10000]); # Robert Israel, Dec 11 2015
  • Mathematica
    Select[Range@ 2600, Mod[DivisorSigma[1, #], 6] == 3 &] (* Michael De Vlieger, Dec 10 2015 *)
  • PARI
    isok(n) = (sigma(2*n) % 2) && !(sigma(3*n) % 3); \\ Michel Marcus, Dec 26 2013
    

Formula

{n: A000203(n) mod 6 = 3.} (Old definition of A074629) - Labos Elemer, Aug 26 2002
In the prime factorization of n, no odd prime has odd exponent, and 2 has odd exponent or at least one prime == 1 (mod 6) has exponent == 2 (mod 6). - Robert Israel, Dec 11 2015
{n: A049605(n) = 3}. - R. J. Mathar, May 19 2020
{n: A084301(n) = 3 }. - R. J. Mathar, May 19 2020
A087943 INTERSECT A028982. - R. J. Mathar, May 30 2020

A074627 Numbers n such that sigma(n) is divisible by 6.

Original entry on oeis.org

5, 6, 10, 11, 14, 15, 17, 20, 22, 23, 24, 26, 29, 30, 33, 34, 35, 38, 40, 41, 42, 44, 45, 46, 47, 51, 53, 54, 55, 56, 58, 59, 60, 62, 65, 66, 68, 69, 70, 71, 74, 77, 78, 80, 82, 83, 85, 86, 87, 88, 89, 90, 92, 94, 95, 96, 99, 101, 102, 104, 105, 106, 107, 110, 113, 114, 115
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Comments

n=10: sigma(10) = 1+2+5+10 = 18 = 3*6.

Crossrefs

Programs

  • Mathematica
    Select[Range@ 120, Divisible[DivisorSigma[1, #], 6] &] (* Michael De Vlieger, Feb 25 2017 *)
  • PARI
    isok(n) = !(sigma(n) % 6); \\ Michel Marcus, Dec 17 2013

Formula

A000203(n) modulo 6 = 0.
{n: A084301(n) = 0 }. - R. J. Mathar, May 19 2020
A087943 INTERSECT A028983. - R. J. Mathar, May 19 2020

A074628 Numbers k such that sigma(k) == 2 mod 6.

Original entry on oeis.org

7, 13, 19, 21, 28, 31, 37, 39, 43, 52, 57, 61, 63, 67, 73, 76, 79, 84, 93, 97, 103, 109, 111, 112, 117, 124, 127, 129, 139, 148, 151, 156, 157, 163, 171, 172, 175, 181, 183, 189, 193, 199, 201, 208, 211, 219, 223, 228, 229, 237, 241, 244, 252
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Examples

			For k=39: sigma(39) = 1+3+13+39 = 56 = 6*9 + 2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[300],Mod[DivisorSigma[1,#],6]==2&] (* Harvey P. Dale, Nov 14 2014 *)
  • PARI
    isok(n) = ((sigma(n) % 6) == 2); \\ Michel Marcus, Dec 19 2013

Formula

A000203(n) mod 6 = 2.
{n: A084301(n) = 2}. - R. J. Mathar, May 19 2020

A072862 The smallest k>1 such that k divides sigma(k*n) is equal to 6.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 64, 81, 100, 121, 144, 225, 256, 289, 324, 400, 484, 529, 576, 625, 729, 841, 900, 1024, 1089, 1156, 1296, 1600, 1681, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2809, 2916, 3025, 3364, 3481, 3600, 4096, 4356, 4624, 4761, 5041
Offset: 1

Views

Author

Benoit Cloitre, Jul 26 2002

Keywords

Comments

The smallest k>1 such that k divides sigma(k*n) is always 2, 3 or 6. Sequence generates squares but not all the squares (49, 169, 196, 361, 441, 676, 784, ... are the first squares not in the sequence).
This is not the sequence of solutions to sigma(n) == 1 (mod 6), because 2401 is in this sequence but sigma(2401) == 5 (mod 6). See A074384 for more examples. - R. J. Mathar, May 19 2020

Formula

{n: A049605(n) = 6}. - R. J. Mathar, May 19 2020

A074630 Numbers k such that sigma(k) == 4 mod 6.

Original entry on oeis.org

3, 12, 27, 48, 75, 91, 108, 133, 192, 217, 243, 247, 259, 273, 300, 301, 343, 363, 364, 399, 403, 427, 432, 469, 481, 511, 532, 553, 559, 589, 651, 675, 679, 703, 721, 741, 763, 768, 777, 793, 817, 819, 867, 868, 871, 889, 903, 949, 972, 973, 988, 1027, 1029
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Examples

			k=48 is a term because sigma(48) = 1+2+3+4+6+8+12+16+24+48 = 124 = 6*20 + 4. [corrected by _Harvey P. Dale_, Jan 17 2013]
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1100],Mod[DivisorSigma[1,#],6]==4&] (* Harvey P. Dale, Jan 17 2013 *)

Formula

Mod(A000203(n), 6) = 4.
{n: A084301(n) = 4}. - R. J. Mathar, May 19 2020

A074625 Triangular array T(n,k) (n >= 1, 1 <= k <= n) read by rows, where T(n,k) = smallest number x such that Mod[sigma[x],n]=k.

Original entry on oeis.org

1, 1, 3, 1, 7, 2, 1, 5, 2, 3, 1, 4, 2, 3, 8, 1, 7, 2, 3, 2401, 5, 1, 29, 2, 3, 6, 5, 4, 1, 10, 2, 3, 9, 5, 4, 7, 1, 19, 2, 3, 13, 5, 4, 7, 10, 1, 6, 2, 3, 8, 5, 4, 7, 18, 19, 1, 9, 2, 3, 24, 5, 4, 7, 16, 21, 43, 1, 13, 2, 3, 2401, 5, 4, 7, 49, 31213, 9604, 6, 1, 8, 2, 3, 10, 5, 4, 7, 33, 22
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Comments

In the table output, one can observe constant diagonals (or lines in the square output). The indices of these are: 1, 3, 4, 6, 7, 8, 12, 13, ... (see A002191). And the corresponding values are: 1, 2, 3, 5, 4, 7, 6, 9, ... (see A002192). - Michel Marcus, Dec 19 2013

Examples

			Triangle begins
1;
1,3;
1,7,2;
1,5,2,3;
1,4,2,3,8; ...
		

Crossrefs

Programs

  • Mathematica
    {k=0, s=0, fl=1}; Table[Print["#"]; Table[fl=1; Print[{r, m}]; Do[s=Mod[DivisorSigma[1, n], m]; If[(s==r)&&(fl==1), Print[n]; fl=0], {n, 1, 150000}], {r, 0, m-1}], {m, 1, 25}]

Formula

Min{x; Mod[sigma[x], n]=r}, r=1..n, n=1, ...

A084303 Smallest x such that sigma(x) mod 6 = n.

Original entry on oeis.org

5, 1, 7, 2, 3, 2401
Offset: 0

Views

Author

Labos Elemer, Jun 02 2003

Keywords

Comments

Row 6 of A074625 (apart from different ordering). - Michel Marcus, Dec 19 2013

Examples

			n=5: sigma(2401) = 1+7+49+343+2401 = 2801 = 6*466+5, hence a(5)=2401.
		

Crossrefs

Programs

  • Mathematica
    Table[k = 1; While[Mod[DivisorSigma[1, k], 6] != n, k++]; k, {n, 0, 5}] (* Michael De Vlieger, Mar 25 2017 *)
  • PARI
    a(n) = {my(x = 1); while((sigma(x) % 6) != n, x++); x;} \\ Michel Marcus, Dec 18 2013

A074629 Duplicate of A067051.

Original entry on oeis.org

2, 8, 18, 32, 49, 50, 72, 98, 128, 162, 169, 196, 200, 242, 288, 338, 361, 392, 441, 450, 512, 578, 648, 676, 722, 784, 800, 882, 961, 968, 1058, 1152, 1225, 1250, 1352, 1369, 1444, 1458, 1521, 1568, 1682, 1764, 1800, 1849, 1922, 2048, 2178, 2312, 2450, 2592
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Comments

Square terms are in A074216. Nonsquare terms appear to be A001105 except {0}. - Michel Marcus, Dec 26 2013
In the prime factorization of n, no odd prime has odd exponent, and 2 has odd exponent or at least one prime == 1 (mod 6) has exponent == 2 (mod 6). - Robert Israel, Dec 11 2015

Examples

			n=32: sigma(32) = 63 = 6*10 + 3.
		

Crossrefs

Appears to be the same sequence as A067051. - Ralf Stephan, Aug 18 2004

Programs

  • Magma
    [n: n in [1..3*10^3] | (SumOfDivisors(n) mod 6) eq 3]; // Vincenzo Librandi, Dec 11 2015
  • Maple
    select(t -> numtheory:-sigma(t) mod 6 = 3, [$1..10000]); # Robert Israel, Dec 11 2015
  • Mathematica
    Select[Range@ 2600, Mod[DivisorSigma[1, #], 6] == 3 &] (* Michael De Vlieger, Dec 10 2015 *)
  • PARI
    isok(n) = (sigma(n) % 6) == 3; \\ Michel Marcus, Dec 26 2013
    

Formula

A000203(n) mod 6 = 3.
{n: A084301(n) = 3 }. - R. J. Mathar, May 19 2020

A331121 a(n) is the smallest positive integer k for which tau(k) does not divide sigma(n).

Original entry on oeis.org

2, 2, 4, 2, 6, 16, 4, 2, 2, 6, 16, 4, 4, 16, 16, 2, 6, 2, 4, 6, 4, 16, 16, 24, 2, 6, 4, 4, 6, 16, 4, 2, 16, 6, 16, 2, 4, 24, 4, 6, 6, 16, 4, 16, 6, 16, 16, 4, 2, 2, 16, 4, 6, 36, 16, 36, 4, 6, 24, 16, 4, 16, 4, 2, 16, 16
Offset: 1

Views

Author

Lechoslaw Ratajczak, Jan 10 2020

Keywords

Comments

Consecutive t satisfying the equation a(t) = 2 are consecutive elements of A028982 (squares and twice squares).
Conjecture: consecutive u satisfying the equation a(u) = 4 are consecutive elements of a sequence defined as follows: (A024614 \ A088535) \ A074384. The conjecture was checked for 10^6 consecutive integers.

Examples

			a(10) = 6 because sigma(10) = 18 is divisible by (tau(1) = 1), (tau(2) = 2), (tau(3) = 2), (tau(4) = 3), (tau(5) = 2), and is not divisible by (tau(6) = 4).
		

Crossrefs

Programs

  • Mathematica
    Array[Block[{k = 1}, While[Mod[DivisorSigma[1, #], DivisorSigma[0, k]] == 0, k++]; k] &, 66] (* Michael De Vlieger, Jan 31 2020 *)
  • Maxima
    a(n):=(for k:1 while mod(divsum(n), length(divisors(k))) = 0 do z:k, z+1) $ makelist(a(n), n, 1, 100, 1);
    
  • PARI
    a(n) = my(k=1, sn=sigma(n)); while ((sn % numdiv(k)) == 0, k++); k; \\ Michel Marcus, Jan 10 2020
Showing 1-10 of 10 results.