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 1-9 of 9 results.

A325975 a(n) = gcd(A325977(n), A325978(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 6, 1, 1, 1, 2, 1, 4, 1, 2, 3, 1, 1, 3, 1, 2, 1, 2, 1, 12, 1, 2, 1, 4, 1, 6, 1, 1, 3, 2, 1, 1, 1, 2, 1, 2, 1, 6, 1, 4, 3, 2, 1, 4, 1, 1, 3, 2, 1, 6, 1, 8, 1, 2, 1, 12, 1, 2, 1, 1, 1, 6, 1, 2, 3, 2, 1, 3, 1, 2, 1, 4, 1, 6, 1, 2, 1, 2, 1, 4, 1, 2, 3, 4, 1, 18, 7, 4, 1, 2, 5, 12, 1, 1, 3, 1, 1, 6, 1, 2, 3
Offset: 1

Views

Author

Antti Karttunen, Jun 02 2019

Keywords

Comments

See comments in A325979 and A325981.

Crossrefs

Programs

Formula

a(n) = gcd(A325977(n), A325978(n)).
a(n) = (1/2)*gcd(A034460(n)+A325313(n), A325814(n)+A325314(n)).

A325973 Arithmetic mean of {sum of unitary divisors} and {sum of squarefree divisors}: a(n) = (1/2) * (A034448(n) + A048250(n)).

Original entry on oeis.org

1, 3, 4, 4, 6, 12, 8, 6, 7, 18, 12, 16, 14, 24, 24, 10, 18, 21, 20, 24, 32, 36, 24, 24, 16, 42, 16, 32, 30, 72, 32, 18, 48, 54, 48, 31, 38, 60, 56, 36, 42, 96, 44, 48, 42, 72, 48, 40, 29, 48, 72, 56, 54, 48, 72, 48, 80, 90, 60, 96, 62, 96, 56, 34, 84, 144, 68, 72, 96, 144, 72, 51, 74, 114, 64, 80, 96, 168, 80, 60, 43, 126
Offset: 1

Views

Author

Antti Karttunen, Jun 02 2019

Keywords

Comments

This is not multiplicative: a(4) = 4, a(9) = 7, but a(36) = 31, not 28. However, the function acts multiplicatively on certain subsequences of natural numbers, like for example when restricted to A048107, where this sequence coincides with A326043.

Examples

			For n = 36, its divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36. Of these, unitary divisors are 1, 4, 9 and 36, so A034448(36) = 1+4+9+36 = 50, while the squarefree divisors are 1, 2, 3 and 6, so A048250(36) = 1+2+3+6 = 12, thus a(36) = (50+12)/2 = 31.
For n = 495, its divisors are 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495. Of these, unitary are 1, 5, 9, 11, 45, 55, 99, 495, whose sum is A034448(495) = 720, while the squarefree divisors are 1, 3, 5, 11, 15, 33, 55, 165, and their sum is A048250(495) = 288. Thus a(495) = (720+288)/2 = 504. Also for 495, whose prime factorization is 3^2 * 5^1 * 11^1 this can be computed faster as the average of ((3^2)+1)*(5+1)*(11+1) and (3+1)*(5+1)*(11+1), thus (1/2)*(3+(3^2)+2)*(5+1)*(11+1) = 504.
		

Crossrefs

Programs

Formula

a(n) = (1/2) * (A034448(n) + A048250(n)).
a(n) = A000203(n) - A325974(n).
a(n) = n + A325977(n).
a(A048107(n)) = A326043(A048107(n)).
For n >= 1, a(2^n) = A052548(n-1) = 2^(n-1) + 2.
For n >= 1, a(3^n) = A289521(n) = (3^n + 5)/2.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(2)/zeta(3) + 1)/4 = 0.5921081944... . - Amiram Eldar, Feb 22 2024

A325981 Odd composites for which gcd(A325977(n), A325978(n)) is equal to abs(A325977(n)).

Original entry on oeis.org

45, 495, 585, 765, 855, 1305, 18837, 21525, 31635, 38295, 45315, 50445, 51255, 60435, 63495, 68085, 77265, 96615, 1403115, 2446353, 3411975, 3999465, 4091745, 4233537, 4287255, 4631319, 10813425, 10967085, 11490345, 15578199, 16143309, 16329645, 16633071, 17179515, 17311203, 17355915, 21159075, 21933975, 22579725
Offset: 1

Views

Author

Antti Karttunen, Jun 02 2019

Keywords

Comments

Provided that A325977 and A325978 are never zero on same n, these are odd composite numbers n such that A325977(n) is not zero and divides A325978(n).
Based on the first 147 terms it seems that this sequence is a subsequence of A072357, that is each term has exactly one prime factor with exponent 2, with one or more other prime factors that are all unitary (i.e., each term satisfies A001222(x) - A001221(x) = 1). On the other hand, it has been proved that no odd perfect number, if such numbers exist at all, can have such a factorization (see A326137 and a link to P. P. Nielsen's paper there).
Nineteen initial terms factorize as:
45 = 3^2 * 5^1,
495 = 3^2 * 5^1 * 11^1,
585 = 3^2 * 5^1 * 13^1,
765 = 3^2 * 5^1 * 17^1,
855 = 3^2 * 5^1 * 19^1,
1305 = 3^2 * 5^1 * 29^1,
18837 = 3^2 * 7^1 * 13^1 * 23^1,
21525 = 3^1 * 5^2 * 7^1 * 41^1,
31635 = 3^2 * 5^1 * 19^1 * 37^1,
38295 = 3^2 * 5^1 * 23^1 * 37^1,
45315 = 3^2 * 5^1 * 19^1 * 53^1,
50445 = 3^2 * 5^1 * 19^1 * 59^1,
51255 = 3^2 * 5^1 * 17^1 * 67^1,
60435 = 3^2 * 5^1 * 17^1 * 79^1,
63495 = 3^2 * 5^1 * 17^1 * 83^1,
68085 = 3^2 * 5^1 * 17^1 * 89^1,
77265 = 3^2 * 5^1 * 17^1 * 101^1,
96615 = 3^2 * 5^1 * 19^1 * 113^1,
1403115 = 3^1 * 5^1 * 7^2 * 23^1 * 83^1,
and the 62nd term as a(62) = 2919199437 = 3^2 * 7^1 * 11^1 * 43^1 * 163^1 * 601^1.
If we select a subsequence of terms for which the quotient A325978(n)/A325977(n) is positive, then we are left with the following numbers: 495, 585, 31635, 38295, 45315, 51255, 60435, 63495, 1403115, 3999465, etc. which is a subsequence of A326070.

Crossrefs

Programs

A325977 a(n) = (1/2)*(A034460(n) + A325313(n)).

Original entry on oeis.org

0, 1, 1, 0, 1, 6, 1, -2, -2, 8, 1, 4, 1, 10, 9, -6, 1, 3, 1, 4, 11, 14, 1, 0, -9, 16, -11, 4, 1, 42, 1, -14, 15, 20, 13, -5, 1, 22, 17, -4, 1, 54, 1, 4, -3, 26, 1, -8, -20, -2, 21, 4, 1, -6, 17, -8, 23, 32, 1, 36, 1, 34, -7, -30, 19, 78, 1, 4, 27, 74, 1, -21, 1, 40, -11, 4, 19, 90, 1, -20, -38, 44, 1, 44, 23, 46, 33, -16, 1, 36, 21, 4
Offset: 1

Views

Author

Antti Karttunen, Jun 02 2019

Keywords

Comments

Question: Are n = 1, 4, 24, 240, 349440 (A325963) the only positions of zeros in this sequence?

Crossrefs

Programs

Formula

a(n) = (1/2)*(A034460(n) + A325313(n)).
a(n) = A325973(n) - n.
a(n) = A325978(n) - A033879(n).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(2)/zeta(3) - 1)/4 = 0.0921081944... . - Amiram Eldar, Feb 22 2024

A325978 a(n) = (1/2)*(A325314(n) + A325814(n)).

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, -1, 3, 10, 11, 0, 13, 14, 15, -5, 17, 0, 19, 2, 21, 22, 23, -12, 10, 26, 3, 4, 29, 30, 31, -13, 33, 34, 35, -24, 37, 38, 39, -14, 41, 42, 43, 8, 9, 46, 47, -36, 21, 5, 51, 10, 53, -18, 55, -16, 57, 58, 59, -12, 61, 62, 15, -29, 65, 66, 67, 14, 69, 70, 71, -72, 73, 74, 15, 16, 77, 78, 79, -46, 3, 82, 83, -12, 85
Offset: 1

Views

Author

Antti Karttunen, Jun 02 2019

Keywords

Comments

Question: Are a(12) = 0 and a(18) = 0 the only zeros in this sequence?

Crossrefs

Programs

Formula

a(n) = (1/2)*(A325314(n) + A325814(n)).
a(n) = n - A325974(n).
a(n) = A033879(n) + A325977(n).
Sum_{k=1..n} a(k) ~ c * n^2, where c = 3/4 - zeta(2)*(1/2 - 1/(4*zeta(3))) = 0.2696411609... . - Amiram Eldar, Feb 22 2024

A326070 Numbers k such that A325977(k) has the same sign as A325978(k).

Original entry on oeis.org

2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 80, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 100
Offset: 1

Views

Author

Antti Karttunen, Jun 07 2019

Keywords

Comments

Here A325977(k) = A325973(k) - k and A325978(k) = k - A325974(k), where A325973(k) is the average of {sum of unitary divisors} and {sum of squarefree divisors} = (1/2) * (A034448(k) + A048250(k)) while A325974(k) is the average of {sum of non-unitary divisors} and {sum of nonsquarefree divisors} = (1/2)*(A048146(k) + A162296(k)). Only if signs of A325977(k) and A325978(k) are equal can their difference A325978(k) - A325977(k) = (k - A325974(k)) - (A325973(k) - k) = 2k - (A325973(k) + A325974(k)) = 2k - A000203(k) = A033879(k) be zero, which happens when k is a perfect number (in A000396).

Crossrefs

Cf. A326071 (complement), A326072, A000396 (a subsequence).

Programs

A348984 a(n) = gcd(sigma(n), A325973(n)), where A325973 is the arithmetic mean of {sum of squarefree divisors} and {sum of unitary divisors}.

Original entry on oeis.org

1, 3, 4, 1, 6, 12, 8, 3, 1, 18, 12, 4, 14, 24, 24, 1, 18, 3, 20, 6, 32, 36, 24, 12, 1, 42, 8, 8, 30, 72, 32, 9, 48, 54, 48, 1, 38, 60, 56, 18, 42, 96, 44, 12, 6, 72, 48, 4, 1, 3, 72, 14, 54, 24, 72, 24, 80, 90, 60, 24, 62, 96, 8, 1, 84, 144, 68, 18, 96, 144, 72, 3, 74, 114, 4, 20, 96, 168, 80, 6, 1, 126, 84, 32, 108
Offset: 1

Views

Author

Antti Karttunen, Nov 06 2021

Keywords

Comments

This is not multiplicative. The first point where a(m*n) = a(m)*a(n) does not hold for coprime m and n is 108 = 4*27, where a(108) = 4, although a(4) = 1 and a(27) = 8.

Crossrefs

Differs from A348047 for the first time at n=108, where a(108) = 4, while A348047(108) = 8.
Cf. also A348733, A348946.

Programs

  • Mathematica
    f1[p_, e_] := p + 1; f2[p_, e_] := p^e + 1; s[1] = 1; s[n_] := (Times @@ f1 @@@ (f = FactorInteger[n]) + Times @@ f2 @@@ f)/2; a[n_] := GCD[s[n], DivisorSigma[1, n]]; Array[a, 100] (* Amiram Eldar, Nov 06 2021 *)
  • PARI
    A325973(n) = (1/2)*sumdiv(n, d, d*(issquarefree(d) + (1==gcd(d, n/d))));
    A348984(n) = gcd(sigma(n), A325973(n));

Formula

a(n) = gcd(A000203(n), A325973(n)).
a(n) = gcd(A000203(n), A325974(n)) = gcd(A325973(n), A325974(n)).
a(n) = A000203(n) / A348985(n) = A325973(n) / A348986(n).

A348985 Numerator of ratio sigma(n) / A325973(n), where A325973 is the arithmetic mean of {sum of squarefree divisors} and {sum of unitary divisors}.

Original entry on oeis.org

1, 1, 1, 7, 1, 1, 1, 5, 13, 1, 1, 7, 1, 1, 1, 31, 1, 13, 1, 7, 1, 1, 1, 5, 31, 1, 5, 7, 1, 1, 1, 7, 1, 1, 1, 91, 1, 1, 1, 5, 1, 1, 1, 7, 13, 1, 1, 31, 57, 31, 1, 7, 1, 5, 1, 5, 1, 1, 1, 7, 1, 1, 13, 127, 1, 1, 1, 7, 1, 1, 1, 65, 1, 1, 31, 7, 1, 1, 1, 31, 121, 1, 1, 7, 1, 1, 1, 5, 1, 13, 1, 7, 1, 1, 1, 7, 1, 57, 13
Offset: 1

Views

Author

Antti Karttunen, Nov 06 2021

Keywords

Comments

This is not multiplicative. The first point where a(m*n) = a(m)*a(n) does not hold for coprime m and n is 108 = 4*27, where a(108) = 70 <> 35 = 7*5 = a(4)*(27).

Crossrefs

Differs from A348048 for the first time at n=108, where a(108) = 70, while A348048(108) = 35.
Cf. also A348948.

Programs

  • Mathematica
    f1[p_, e_] := p + 1; f2[p_, e_] := p^e + 1; s[1] = 1; s[n_] := (Times @@ f1 @@@ (f = FactorInteger[n]) + Times @@ f2 @@@ f)/2; a[n_] := Numerator[DivisorSigma[1, n]/s[n]]; Array[a, 100] (* Amiram Eldar, Nov 06 2021 *)
  • PARI
    A325973(n) = (1/2)*sumdiv(n, d, d*(issquarefree(d) + (1==gcd(d, n/d))));
    A348985(n) = { my(s=sigma(n)); (s/gcd(s, A325973(n))); };

Formula

a(n) = A000203(n) / A348984(n) = sigma(n) / gcd(sigma(n), A325973(n)).

A348986 Denominator of ratio sigma(n) / A325973(n), where A325973 is the arithmetic mean of {sum of squarefree divisors} and {sum of unitary divisors}.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 2, 7, 1, 1, 4, 1, 1, 1, 10, 1, 7, 1, 4, 1, 1, 1, 2, 16, 1, 2, 4, 1, 1, 1, 2, 1, 1, 1, 31, 1, 1, 1, 2, 1, 1, 1, 4, 7, 1, 1, 10, 29, 16, 1, 4, 1, 2, 1, 2, 1, 1, 1, 4, 1, 1, 7, 34, 1, 1, 1, 4, 1, 1, 1, 17, 1, 1, 16, 4, 1, 1, 1, 10, 43, 1, 1, 4, 1, 1, 1, 2, 1, 7, 1, 4, 1, 1, 1, 2, 1, 29, 7, 74, 1, 1, 1, 2, 1
Offset: 1

Views

Author

Antti Karttunen, Nov 06 2021

Keywords

Comments

This is not multiplicative: a(4) = 4 and a(9) = 7, but a(36) = 31, not 28.

Crossrefs

Cf. also A348947.

Programs

  • Mathematica
    f1[p_, e_] := p + 1; f2[p_, e_] := p^e + 1; s[1] = 1; s[n_] := (Times @@ f1 @@@ (f = FactorInteger[n]) + Times @@ f2 @@@ f)/2; a[n_] := Denominator[DivisorSigma[1, n]/s[n]]; Array[a, 100] (* Amiram Eldar, Nov 06 2021 *)
  • PARI
    A325973(n) = (1/2)*sumdiv(n, d, d*(issquarefree(d) + (1==gcd(d, n/d))));
    A348986(n) = { my(am=A325973(n)); (am/gcd(sigma(n),am)); };

Formula

a(n) = A325973(n) / A348984(n) = A325973(n) / gcd(A000203(n), A325973(n)).
Showing 1-9 of 9 results.