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 1-4 of 4 results.

A349169 Numbers k such that k * gcd(sigma(k), A003961(k)) is equal to the odd part of {sigma(k) * gcd(k, A003961(k))}, where A003961 shifts the prime factorization one step towards larger primes, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 15, 105, 3003, 3465, 13923, 45045, 264537, 459459, 745875, 1541475, 5221125, 8729721, 10790325, 14171625, 29288025, 34563375, 57034575, 71430975, 99201375, 109643625, 144729585, 205016175, 255835125, 295708875, 356080725, 399242025, 419159475, 449323875, 928602675, 939495375, 1083656925, 1941623775, 1962350685, 2083228875
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2021

Keywords

Comments

Numbers k such that A348990(k) [= k/gcd(k, A003961(k))] is equal to A348992(k), which is the odd part of A349162(k), thus all terms must be odd, as A348990 preserves the parity of its argument.
Equally, numbers k for which gcd(A064987(k), A191002(k)) is equal to A000265(gcd(A064987(k), A341529(k))).
Also odd numbers k for which A348993(k) = A319627(k).
Odd terms of A336702 are given by the intersection of this sequence and A349174.
Conjectures:
(1) After 1, all terms are multiples of 3. (Why?)
(2) After 1, all terms are in A104210, in other words, for all n > 1, gcd(a(n), A003961(a(n))) > 1. Note that if we encountered a term k with gcd(k, A003961(k)) = 1, then we would have discovered an odd multiperfect number.
(3) Apart from 1, 15, 105, 3003, 13923, 264537, all other terms are abundant. [These apparently are also the only terms that are not Zumkeller, A083207. Note added Dec 05 2024]
(4) After 1, all terms are in A248150. (Cf. also A386430).
(5) After 1, all terms are in A348748.
(6) Apart from 1, there are no common terms with A349753.
Note: If any of the last four conjectures could be proved, it would refute the existence of odd perfect numbers at once. Note that it seems that gcd(sigma(k), A003961(k)) < k, for all k except these four: 1, 2, 20, 160.
Questions:
(1) For any term x here, can 2*x be in A349745? (Partial answer: at least x should be in A191218 and should not be a multiple of 3). Would this then imply that x is an odd perfect number? (Which could explain the points (1) and (4) in above, assuming the nonexistence of opn's).

Crossrefs

Programs

  • Mathematica
    Select[Range[10^6], #1/GCD[#1, #3] == #2/(2^IntegerExponent[#2, 2]*GCD[#2, #3]) & @@ {#, DivisorSigma[1, #], Times @@ Map[NextPrime[#1]^#2 & @@ # &, FactorInteger[#]]} &] (* Michael De Vlieger, Nov 11 2021 *)
  • PARI
    A000265(n) = (n >> valuation(n, 2));
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    isA349169(n) = { my(s=sigma(n),u=A003961(n)); (n*gcd(s,u) == A000265(s)*gcd(n,u)); }; \\ (Program simplified Nov 30 2021)

Formula

For all n >= 1, A007949(A000203(a(n))) = A007949(a(n)). [sigma preserves the 3-adic valuation of the terms of this sequence] - Antti Karttunen, Nov 29 2021

Extensions

Name changed and comment section rewritten by Antti Karttunen, Nov 29 2021

A319627 Primorial deflation of n (denominator): Let f be the completely multiplicative function over the positive rational numbers defined by f(p) = A034386(p) for any prime number p; f constitutes a permutation of the positive rational numbers; let g be the inverse of f; for any n > 0, a(n) is the denominator of g(n).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 5, 1, 4, 3, 7, 1, 11, 5, 2, 1, 13, 2, 17, 3, 10, 7, 19, 1, 9, 11, 8, 5, 23, 1, 29, 1, 14, 13, 3, 1, 31, 17, 22, 3, 37, 5, 41, 7, 4, 19, 43, 1, 25, 9, 26, 11, 47, 4, 21, 5, 34, 23, 53, 1, 59, 29, 20, 1, 33, 7, 61, 13, 38, 3, 67, 1, 71, 31, 6
Offset: 1

Views

Author

Rémy Sigrist, Sep 25 2018

Keywords

Comments

See A319626 for the corresponding numerators and additional comments.

Examples

			f(21/5) = (2*3) * (2*3*5*7) / (2*3*5) = 42, hence g(42) = 21/5 and a(42) = 5.
		

Crossrefs

Cf. A025487 (positions of 1's), A064989, A329900, A358217 [= bigomega(a(n))].
Cf. A319626 (numerators, see comments there).
Cf. also A307035, A337377, A348990 [= a(A003961(n))], A349169 (odd numbers k such that A348993(k) = a(k)), A354365/A354366.

Programs

  • Mathematica
    Array[#2/GCD[#1, #2] & @@ {#, Apply[Times, Map[If[#1 <= 2, 1, NextPrime[#1, -1]]^#2 & @@ # &, FactorInteger[#]]]} &, 120] (* Michael De Vlieger, Aug 27 2020 *)
  • PARI
    a(n) = my (f=factor(n)); denominator(prod(i=1, #f~, my (p=f[i,1]); (p/if (p>2, precprime(p-1), 1))^f[i,2]))

Formula

a(n) = A064989(n) / gcd(n, A064989(n)).
a(n) = 1 iff n belongs to A025487.

Extensions

"Primorial deflation" prefixed to the name by Antti Karttunen, Apr 29 2022

A348992 a(n) = A000265(sigma(n)) / gcd(sigma(n), A003961(n)), where A003961(n) is fully multiplicative with a(prime(k)) = prime(k+1), and sigma is the sum of divisors function.

Original entry on oeis.org

1, 1, 1, 7, 3, 1, 1, 5, 13, 3, 3, 7, 7, 1, 3, 31, 9, 13, 5, 1, 1, 3, 3, 1, 31, 7, 1, 7, 15, 3, 1, 7, 3, 9, 3, 91, 19, 5, 7, 5, 21, 1, 11, 7, 39, 3, 3, 31, 57, 31, 9, 49, 27, 1, 9, 5, 1, 15, 15, 1, 31, 1, 13, 127, 3, 3, 17, 7, 3, 3, 9, 13, 37, 19, 31, 35, 3, 7, 5, 31, 121, 21, 21, 7, 27, 11, 3, 5, 45, 39, 7, 7, 1, 3
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2021

Keywords

Comments

Denominator of ratio A003961(n) / A161942(n).

Crossrefs

Odd part of A349162.
Cf. A349161 (numerators).

Programs

  • Mathematica
    Array[#1/(2^IntegerExponent[#1, 2]*GCD[##]) & @@ {DivisorSigma[1, #], Times @@ Map[NextPrime[#1]^#2 & @@ # &, FactorInteger[#]]} &, 94] (* Michael De Vlieger, Nov 11 2021 *)
  • PARI
    A000265(n) = (n >> valuation(n, 2));
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A348992(n) = { my(s=sigma(n)); (A000265(s)/gcd(s,A003961(n))); };

Formula

a(n) = A161942(n) / A342671(n) = A000265(A349162(n)).
a(n) = A003961(A348993(n)).

A348994 a(n) = A003961(n) / gcd(n, A003961(n)), where A003961(n) is fully multiplicative with a(prime(k)) = prime(k+1).

Original entry on oeis.org

1, 3, 5, 9, 7, 5, 11, 27, 25, 21, 13, 15, 17, 33, 7, 81, 19, 25, 23, 63, 55, 39, 29, 45, 49, 51, 125, 99, 31, 7, 37, 243, 65, 57, 11, 25, 41, 69, 85, 189, 43, 55, 47, 117, 35, 87, 53, 135, 121, 147, 95, 153, 59, 125, 91, 297, 115, 93, 61, 21, 67, 111, 275, 729, 119, 65, 71, 171, 145, 33, 73, 75, 79, 123, 49, 207
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2021

Keywords

Comments

Numerator of ratio A003961(n) / n. This ratio is fully multiplicative, and a(n) / A348990(n) = A319626(A003961(n)) / A319627(A003961(n)) gives it in its lowest terms.

Crossrefs

Cf. A003961, A319626, A319627, A348990 (denominators).

Programs

  • Mathematica
    Array[#2/GCD[##] & @@ {#, If[# == 1, 1, Times @@ Map[NextPrime[#1]^#2 & @@ # &, FactorInteger[#]]]} &, 76] (* Michael De Vlieger, Nov 11 2021 *)
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A348994(n) = (A003961(n) / gcd(n, A003961(n)));

Formula

a(n) = A003961(n) / gcd(n, A003961(n)).
a(n) = A319626(A003961(n)).
Showing 1-4 of 4 results.