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 1-8 of 8 results.

A326062 a(1) = gcd((sigma(n)-A032742(n))-n, n-A032742(n)), where A032742 gives the largest proper divisor of n.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 6, 1, 1, 1, 10, 2, 12, 1, 2, 1, 16, 3, 18, 2, 2, 1, 22, 12, 1, 1, 2, 14, 28, 3, 30, 1, 2, 1, 2, 1, 36, 1, 2, 10, 40, 3, 42, 2, 6, 1, 46, 4, 1, 1, 2, 2, 52, 3, 2, 4, 2, 1, 58, 6, 60, 1, 2, 1, 2, 3, 66, 2, 2, 1, 70, 3, 72, 1, 2, 2, 2, 3, 78, 2, 1, 1, 82, 14, 2, 1, 2, 4, 88, 9, 2, 2, 2, 1, 2, 12, 96, 1, 6, 1, 100, 3, 102, 2, 2
Offset: 1

Views

Author

Antti Karttunen, Jun 06 2019

Keywords

Comments

See comments in A326063 and A326064.

Crossrefs

Programs

Formula

a(1) = 1; for n > 1, a(n) = gcd(A060681(n), A318505(n)).
a(n) = gcd((A000203(n)-A032742(n))-n, n-A032742(n)).

A318508 a(n) = A032742(n) AND A001065(n)-A032742(n), where AND is bitwise-and (A004198) and A001065 = sum of proper divisors and A032742 = the largest proper divisor of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 3, 4, 0, 0, 8, 0, 8, 4, 3, 0, 8, 1, 1, 0, 14, 0, 11, 0, 0, 0, 1, 6, 0, 0, 3, 4, 20, 0, 1, 0, 18, 2, 3, 0, 16, 1, 16, 0, 16, 0, 3, 2, 4, 0, 1, 0, 14, 0, 3, 20, 0, 4, 33, 0, 0, 4, 35, 0, 4, 0, 1, 24, 2, 8, 35, 0, 0, 9, 1, 0, 34, 0, 3, 4, 32, 0, 33, 8, 14, 4, 3, 2, 32, 0, 16, 0, 2, 0, 51, 0, 52, 32
Offset: 1

Views

Author

Antti Karttunen, Aug 28 2018

Keywords

Crossrefs

Programs

Formula

a(n) = A004198(A032742(n), A318505(n)).
For n > 1, a(n) = A001065(n) - A318506(n) = (A001065(n) - A318507(n))/2.

A318506 a(n) = A032742(n) OR A001065(n)-A032742(n), where OR is bitwise-or (A003986) and A001065 = sum of proper divisors and A032742 = the largest proper divisor of n.

Original entry on oeis.org

1, 1, 1, 3, 1, 3, 1, 7, 3, 7, 1, 14, 1, 7, 5, 15, 1, 13, 1, 14, 7, 11, 1, 28, 5, 15, 13, 14, 1, 31, 1, 31, 15, 19, 7, 55, 1, 19, 13, 30, 1, 53, 1, 22, 31, 23, 1, 60, 7, 27, 21, 30, 1, 63, 15, 60, 23, 31, 1, 94, 1, 31, 21, 63, 15, 45, 1, 58, 23, 39, 1, 119, 1, 39, 25, 62, 11, 55, 1, 106, 31, 43, 1, 106, 23, 43, 29, 60, 1, 111, 13, 62, 31
Offset: 1

Views

Author

Antti Karttunen, Aug 28 2018

Keywords

Crossrefs

Programs

Formula

a(n) = A003986(A032742(n), A318505(n)).
For n > 1, a(n) = A001065(n) - A318508(n).

A318507 a(n) = A032742(n) XOR A001065(n)-A032742(n), where XOR is bitwise-or (A003987) and A001065 = sum of proper divisors and A032742 = the largest proper divisor of n.

Original entry on oeis.org

1, 1, 1, 3, 1, 0, 1, 7, 2, 6, 1, 12, 1, 4, 1, 15, 1, 5, 1, 6, 3, 8, 1, 20, 4, 14, 13, 0, 1, 20, 1, 31, 15, 18, 1, 55, 1, 16, 9, 10, 1, 52, 1, 4, 29, 20, 1, 44, 6, 11, 21, 14, 1, 60, 13, 56, 23, 30, 1, 80, 1, 28, 1, 63, 11, 12, 1, 58, 19, 4, 1, 115, 1, 38, 1, 60, 3, 20, 1, 106, 22, 42, 1, 72, 23, 40, 25, 28, 1, 78, 5, 48, 27, 44, 21, 92, 1
Offset: 1

Views

Author

Antti Karttunen, Aug 28 2018

Keywords

Comments

Note that here zeros occur only on even perfect numbers (even terms of A000396), in contrast to A318457, which would be zero also for any hypothetical odd perfect number. - Antti Karttunen, Aug 29 2018

Crossrefs

Programs

Formula

a(n) = A003987(A032742(n), A318505(n)).
For n > 1, a(n) = A001065(n) - 2*A318508(n).

A326064 Odd composite numbers n, not squares of primes, such that (A001065(n) - A032742(n)) divides (n - A032742(n)), where A032742 gives the largest proper divisor, and A001065 is the sum of proper divisors.

Original entry on oeis.org

117, 775, 10309, 56347, 88723, 2896363, 9597529, 12326221, 12654079, 25774633, 29817121, 63455131, 105100903, 203822581, 261019543, 296765173, 422857021, 573332713, 782481673, 900952687, 1129152721, 3350861677, 3703086229, 7395290407, 9347001661, 9350506057
Offset: 1

Views

Author

Antti Karttunen, Jun 06 2019

Keywords

Comments

Nineteen initial terms factored:
n a(n) factorization A060681(a(n))/A318505(a(n))
1: 117 = 3^2 * 13, (3)
2: 775 = 5^2 * 31, (10)
3: 10309 = 13^2 * 61, (39)
4: 56347 = 29^2 * 67, (58)
5: 88723 = 17^2 * 307, (136)
6: 2896363 = 41^2 * 1723, (820)
7: 9597529 = 73^2 * 1801, (1314)
8: 12326221 = 59^2 * 3541, (1711)
9: 12654079 = 113^2 * 991, (904)
10: 25774633 = 71^2 * 5113, (2485)
11: 29817121 = 97^2 * 3169, (2328)
12: 63455131 = 89^2 * 8011, (3916)
13: 105100903 = 101^2 * 10303, (5050)
14: 203822581 = 157^2 * 8269, (6123)
15: 261019543 = 349^2 * 2143, (2094)
16: 296765173 = 131^2 * 17293, (8515)
17: 422857021 = 233^2 * 7789, (6757)
18: 573332713 = 331^2 * 5233, (4965)
19: 782481673 = 167^2 * 28057, (13861).
Note how the quotient (in the rightmost column) seems always to be a multiple of non-unitary prime factor and less than the unitary prime factor.
For p, q prime, if p^2+p+1 = kq and k+1|p-1, then p^2*q is in this sequence. - Charlie Neder, Jun 09 2019

Crossrefs

Subsequence of A326063.

Programs

  • Mathematica
    Select[Range[15, 10^6 + 1, 2], And[! PrimePowerQ@ #1, Mod[#1 - #2, #2 - #3] == 0] & @@ {#, DivisorSigma[1, #] - #, Divisors[#][[-2]]} &] (* Michael De Vlieger, Jun 22 2019 *)
  • PARI
    A032742(n) = if(1==n,n,n/vecmin(factor(n)[,1]));
    A060681(n) = (n-A032742(n));
    A318505(n) = if(1==n,0,(sigma(n)-A032742(n))-n);
    isA326064(n) = if((n%2)&&(2!=isprimepower(n)), my(s=A032742(n), t=sigma(n)-s); (gcd(t-n, n-A032742(n)) == t-n), 0);

Extensions

More terms from Amiram Eldar, Dec 24 2020

A326061 Sum of all other divisors of n except the largest proper divisor. a(1) = 0 by convention.

Original entry on oeis.org

0, 2, 3, 5, 5, 9, 7, 11, 10, 13, 11, 22, 13, 17, 19, 23, 17, 30, 19, 32, 25, 25, 23, 48, 26, 29, 31, 42, 29, 57, 31, 47, 37, 37, 41, 73, 37, 41, 43, 70, 41, 75, 43, 62, 63, 49, 47, 100, 50, 68, 55, 72, 53, 93, 61, 92, 61, 61, 59, 138, 61, 65, 83, 95, 71, 111, 67, 92, 73, 109, 71, 159, 73, 77, 99, 102, 85, 129, 79, 146, 94, 85, 83
Offset: 1

Views

Author

Antti Karttunen, Jun 06 2019

Keywords

Crossrefs

Programs

Formula

a(n) = A000203(n) - A032742(n).
For n > 1, a(n) = n + A318505(n).

A326063 Composite numbers n such that (A001065(n) - A032742(n)) divides (n - A032742(n)), where A032742 gives the largest proper divisor, and A001065 is the sum of proper divisors.

Original entry on oeis.org

4, 6, 9, 25, 28, 49, 117, 121, 169, 289, 361, 496, 529, 775, 841, 961, 1369, 1681, 1849, 2209, 2809, 3481, 3721, 4489, 5041, 5329, 6241, 6889, 7921, 8128, 9409, 10201, 10309, 10609, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 22201, 22801, 24649, 26569, 27889, 29929, 32041, 32761, 36481, 37249, 38809, 39601, 44521, 49729, 51529, 52441
Offset: 1

Views

Author

Antti Karttunen, Jun 06 2019

Keywords

Comments

Composite numbers n such that A318505(n) [sum of divisors of n excluding n itself and the second largest of them, A032742(n)] divides A060681(n) [the largest difference between consecutive divisors of n, = n - A032742(n)].
Numbers k such that A326062(k) = A318505(k).
Question: Is it possible that this sequence could contain a term with more than one non-unitary prime factor? If not, then there are no odd perfect numbers. (See e.g., A326137).

Examples

			For n = 9 = 3*3, its divisors are [1, 3, 9], thus A318505(9) = 1 and A060681(9) = 9-3 = 6, and 1 divides 6, so 9 is included, like all squares of primes.
For n = 117 = 3^2 * 13,its divisors are [1, 3, 9, 13, 39, 117], thus A318505(117) = 1+3+9+13 = 26 and A060681(117) = (117-39) = 78, which is a multiple of 26, thus 117 is included in the sequence.
		

Crossrefs

Subsequences: A000396, A001248, A326064 (odd terms that are not squares of primes).

Programs

A318504 SumXOR of divisors of n, up to, but not including the second largest of them A032742(n); a(1) = 0 by convention.

Original entry on oeis.org

0, 0, 0, 1, 0, 3, 0, 3, 1, 3, 0, 4, 0, 3, 2, 7, 0, 6, 0, 2, 2, 3, 0, 10, 1, 3, 2, 0, 0, 9, 0, 15, 2, 3, 4, 7, 0, 3, 2, 0, 0, 15, 0, 12, 14, 3, 0, 22, 1, 12, 2, 10, 0, 29, 4, 6, 2, 3, 0, 26, 0, 3, 12, 31, 4, 27, 0, 22, 2, 5, 0, 5, 0, 3, 8, 20, 6, 17, 0, 4, 11, 3, 0, 14, 4, 3, 2, 18, 0, 3, 6, 16, 2, 3, 4, 46, 0, 10, 0, 5, 0, 53, 0, 24, 26
Offset: 1

Views

Author

Antti Karttunen, Aug 28 2018

Keywords

Crossrefs

Programs

  • PARI
    A032742(n) = if(1==n,n,n/vecmin(factor(n)[,1]));
    A318504(n) = { my(v=0); fordiv(n,d,if(d<A032742(n), v = bitxor(v,d))); (v); };

Formula

a(n) = A032742(n) XOR A227320(n).
For n > 1, a(n) = A106409(n) XOR A178910(n).
Showing 1-8 of 8 results.