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 1-6 of 6 results.

A242485 Possible values of A242480(n) in increasing order.

Original entry on oeis.org

0, 2, 3, 5, 6, 7, 8, 9, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 37, 39, 40, 41, 42, 43, 44, 45, 47, 49, 51, 52, 53, 54, 55, 56, 57, 59, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 75, 76, 77, 78, 79, 80, 81, 83
Offset: 1

Views

Author

Jaroslav Krizek, May 27 2014

Keywords

Comments

A242480(n) = (n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n = A142150(n) + A054024(n) + A229110(n) = (A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n).
Supersequence of odd numbers > 1. Complement of A242486.

Examples

			16 is in the sequence because there is a number m such that A242480(m) = 16; m = 8.

Crossrefs

Cf. A242480, A242481, A242482, A242483, A242484, A242486.

A242486 Numbers n such that A242480(x) = n has no solution.

Original entry on oeis.org

1, 4, 10, 14, 22, 26, 34, 36, 38, 46, 48, 50, 58, 60, 62, 74, 82, 84, 86, 90, 94, 98, 106, 108, 110, 118, 122, 130, 132, 134, 142, 146, 154, 156, 158, 166, 170, 178, 182, 190, 194, 202, 206, 210, 214, 218, 226, 230, 238, 242, 250, 252, 254, 262, 266, 270, 274
Offset: 1

Views

Author

Jaroslav Krizek, May 27 2014

Keywords

Comments

A242480(n) = (n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n = A142150(n) + A054024(n) + A229110(n) = (A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n).
All values of a(n) are even for n > 1. Complement of A242485.

Examples

			14 is in the sequence because there is no x whose A242480(x) = 14.

Crossrefs

Cf. A242480, A242481, A242482, A242483, A242484, A242485.

A242481 a(n) = ((n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n) / n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1
Offset: 1

Views

Author

Jaroslav Krizek, May 16 2014

Keywords

Comments

a(1) = 0. If there is no odd multiply-perfect number then a(n) = 1 or 2 for n >= 2. See A242482 = numbers m such that a(n) = 1, A242483 = numbers m such that a(n) = 2. If there are any odd multiply-perfect numbers m > 1 then a(m) = 0.

Examples

			a(8) = [(8*(8+1)/2) mod 8 + sigma(8) mod 8 + antisigma(8) mod 8] / 8 = (36 mod 8 + 15 mod 8 + 21 mod 8) / 8 = (4 + 7 + 5 ) / 8 = 2.

Links

Crossrefs

Cf. A242480, A242482, A242483, A242484, A242485, A242486.

Programs

  • Magma
    [((n*(n+1)div 2 mod n + SumOfDivisors(n) mod n + (n*(n+1)div 2-SumOfDivisors(n)) mod n))div n: n in [1..1000]]

Formula

a(n) = (A142150(n) + A054024(n) + A229110(n)) / n = ((A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n)) / n.
a(n) = A242480(n) / n.

A242482 Numbers n such that A242481(n) = ((n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n) / n = 1.

Original entry on oeis.org

2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 33, 35, 37, 39, 40, 41, 42, 43, 45, 47, 49, 51, 53, 54, 55, 56, 57, 59, 61, 63, 65, 66, 67, 69, 70, 71, 73, 75, 77, 78, 79, 80, 81, 83, 85, 87, 88, 89, 91, 93, 95, 97, 99
Offset: 1

Views

Author

Jaroslav Krizek, May 16 2014

Keywords

Comments

Numbers n such that A242480(n) = (1/2*n*(n+1)) mod n + sigma(n) mod n + antisigma(n) mod n = (A142150(n) + A054024(n) + A229110(n)) = ((A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n)) = n. Numbers n such that A242481(n) = (A142150(n) + A054024(n) + A229110(n)) / n = ((A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n)) / n = 1.
Conjecture: with number 1 complement of A242483.
Supersequence of primes (A000040).
If there is no odd multiply-perfect number, then:
(1) a(n) = union of odd numbers >= 3 and even numbers from A239719.
(2) a(n) = supersequence of odd numbers (A005408).

Examples

			6 is in sequence because [(6*(6+1)/2) mod 6 + sigma(6) mod 6 + antisigma(6) mod 6] / 6 = (21 mod 6 + 12 mod 6 + 9 mod 6) / 6 = (3 + 0 + 3 ) / 6 = 1.

Links

Crossrefs

Cf. A242480, A242481, A242483, A242484, A242485, A242486.

Programs

  • Magma
    [n: n in [1..1000] | n eq ((n*(n+1)div 2 mod n + SumOfDivisors(n) mod n + (n*(n+1)div 2-SumOfDivisors(n)) mod n))]

A242483 Numbers n such that A242481(n) = ((n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n) / n = 2.

Original entry on oeis.org

4, 8, 10, 14, 16, 22, 26, 32, 34, 36, 38, 44, 46, 48, 50, 52, 58, 60, 62, 64, 68, 72, 74, 76, 82, 84, 86, 90, 92, 94, 96, 98, 106, 108, 110, 116, 118, 122, 124, 128, 130, 132, 134, 136, 142, 144, 146, 148, 152, 154, 156, 158, 164, 166, 168, 170, 172, 178, 182
Offset: 1

Views

Author

Jaroslav Krizek, May 16 2014

Keywords

Comments

Numbers n such that A242480(n) = (n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n = (A142150(n) + A054024(n) + A229110(n)) = ((A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n)) = 2n. Numbers n such that A242481(n) = (A142150(n) + A054024(n) + A229110(n)) / n = ((A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n)) / n = 2.
Conjecture: with number 1 complement of A242482.

Examples

			8 is in sequence because [(8*(8+1)/2) mod 8 + sigma(8) mod 8 + antisigma(8) mod 8] / 8 = (36 mod 8 + 15 mod 8 + 21 mod 8) / 8 = (4 + 7 + 5 ) / 8 = 2.

Links

Crossrefs

Cf. A242480, A242481, A242482, A242484, A242485, A242486.

Programs

  • Magma
    [n: n in [1..1000] | 2 eq (((n*(n+1)div 2 mod n + SumOfDivisors(n) mod n + (n*(n+1)div 2-SumOfDivisors(n)) mod n)))div n]

A242484 Numbers n such that antisigma(n) mod n = 0, where antisigma(n) = A024816(n) = sum of numbers less than n which do not divide n.

Original entry on oeis.org

1, 2, 24, 4320, 4680, 26208, 8910720, 17428320, 20427264, 91963648, 197064960, 8583644160, 10200236032, 21857648640, 57575890944, 57629644800, 206166804480, 17116004505600, 1416963251404800, 15338300494970880
Offset: 1

Views

Author

Jaroslav Krizek, May 16 2014

Keywords

Comments

Numbers n such that antisigma(n) mod n = A229110(n) = 0.
If there are any odd multiply-perfect numbers, they are members of this sequence.
If there is no odd multiply-perfect number, then a(n) = A159907(n-1) for n >= 2.

Examples

			24 is in sequence because antisigma(24) mod 24 = 240 mod 24 = 0.

Crossrefs

Cf. A242480, A242481, A242482, A242483, A242485, A242486.

Programs

  • Magma
    [n: n in [1..1000000] | 0 eq ((n*(n+1))div 2 - SumOfDivisors(n)) mod n];
Showing 1-6 of 6 results.

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