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 15 results. Next

A336844 a(n) = A336698(A003961(n)).

Original entry on oeis.org

1, 1, 1, 7, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 61, 3, 1, 1, 7, 5, 1, 1, 1, 29, 5, 5, 5, 1, 1, 5, 23, 11, 3, 1, 101, 11, 1, 7, 3, 3, 5, 1, 23, 1, 1, 7, 91, 67, 29, 1, 59, 1, 5, 1, 1, 5, 1, 1, 5, 9, 5, 47, 547, 5, 11, 5, 33, 23, 1, 19, 39, 3, 11, 43, 5, 11, 7, 11, 61, 391, 3, 23, 59, 3, 1, 1, 9, 25, 1, 7, 49, 29, 7, 1, 137
Offset: 1

Views

Author

Antti Karttunen, Aug 06 2020

Keywords

Crossrefs

Programs

Formula

a(n) = A000265(1+A000265(A003973(n))).
a(n) = A336698(A003961(n)) = A336699(A003961(n)).

A161942 Odd part of sum of divisors of n.

Original entry on oeis.org

1, 3, 1, 7, 3, 3, 1, 15, 13, 9, 3, 7, 7, 3, 3, 31, 9, 39, 5, 21, 1, 9, 3, 15, 31, 21, 5, 7, 15, 9, 1, 63, 3, 27, 3, 91, 19, 15, 7, 45, 21, 3, 11, 21, 39, 9, 3, 31, 57, 93, 9, 49, 27, 15, 9, 15, 5, 45, 15, 21, 31, 3, 13, 127, 21, 9, 17, 63, 3, 9, 9, 195, 37, 57, 31, 35, 3, 21, 5, 93, 121, 63
Offset: 1

Views

Author

Keywords

Comments

It is conjectured that iteration of this function will always reach 1. This implies the nonexistence of odd perfect numbers. This is equivalent to the same question for A000593, which can be expressed as the sum of the divisors of the odd part of n.
Up to 20000000, there are only two odd numbers with a(n) and a(a(n)) both >= n: 81 and 18966025. See A162284.
For the nonexistence proof of odd perfect numbers, it is enough to show that this sequence has no fixed points beyond the initial one. This is equivalent to a similar condition given for A326042. - Antti Karttunen, Jun 17 2019

Crossrefs

Programs

  • Mathematica
    oddPart[n_] := n/2^IntegerExponent[n, 2]; a[n_] := oddPart[ DivisorSigma[1, n]]; Table[a[n], {n, 1, 82}] (* Jean-François Alcover, Sep 03 2012 *)
  • PARI
    oddpart(n)=n/2^valuation(n,2);
    a(n)=oddpart(sigma(n));
    
  • Python
    from sympy import divisor_sigma
    def A161942(n): return (m:=int(divisor_sigma(n)))>>(~m&m-1).bit_length() # Chai Wah Wu, Mar 17 2023
  • Scheme
    (define (A161942 n) (A000265 (A000203 n))) ;; [For the implementations of A000203 and A000265, see under the respective entries]. - Antti Karttunen, Nov 18 2017
    

Formula

Multiplicative with a(p^e) = oddpart((p^{e+1}-1)/(p-1)), where oddpart(n) = A000265(n) is the largest odd divisor of n.
a(n) = A000265(A000203(n)).
a(n) = A337194(n)-1. - Antti Karttunen, Nov 30 2024

A337194 a(n) = 1 + A000265(sigma(n)), where A000265 gives the odd part.

Original entry on oeis.org

2, 4, 2, 8, 4, 4, 2, 16, 14, 10, 4, 8, 8, 4, 4, 32, 10, 40, 6, 22, 2, 10, 4, 16, 32, 22, 6, 8, 16, 10, 2, 64, 4, 28, 4, 92, 20, 16, 8, 46, 22, 4, 12, 22, 40, 10, 4, 32, 58, 94, 10, 50, 28, 16, 10, 16, 6, 46, 16, 22, 32, 4, 14, 128, 22, 10, 18, 64, 4, 10, 10, 196, 38, 58, 32, 36, 4, 22, 6, 94, 122, 64, 22, 8, 28, 34
Offset: 1

Views

Author

Antti Karttunen, Aug 18 2020

Keywords

Crossrefs

One more than A161942.

Programs

Formula

a(n) = 1+A000265(A000203(n)).
a(n) = A336698(n) * 2^A337195(n).
A007949(a(n)) = A337196(n).

A336700 Numbers k such that the odd part of (1+k) divides (1 + odd part of sigma(k)).

Original entry on oeis.org

1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 2431, 2943, 3775, 4095, 8191, 13311, 14335, 16383, 17407, 21951, 22527, 32767, 34335, 44031, 57855, 65535, 85375, 131071, 204799, 262143, 376831, 524287, 923647, 1048575, 1562623, 1632255, 2056191, 2097151, 2744319, 4194303, 6815743, 8388607, 8781823, 10059775, 16777215
Offset: 1

Views

Author

Antti Karttunen, Aug 02 2020

Keywords

Comments

Numbers k for which A337194(k) = 1+A161942(k) is a multiple of A000265(1+k).
Conjecture: After 1, all terms are of the form 4u+3 (in A004767). If this could be proved, then it would refute at once the existence of both the odd perfect numbers and the quasiperfect numbers. Concentrating on the latter is probably easier, as it is known that all quasiperfect numbers must be odd squares, thus k is of the form 4u+1, in which case the condition given in A336701 that A000265(1+A000265(sigma(k))) must be equal to A000265(1+k) reduces to a simpler form, A000265(1+sigma(k)) = (1+k)/2, and as k = s^2, with s odd, so (s^2 + 1)/2 should divide 1+sigma(s^2). Does that condition allow any other solutions than s=1 ? See A337339.

Crossrefs

Subsequences: A000225, A336701 (terms where the quotient is a power of 2).

Programs

  • Mathematica
    Block[{f}, f[n_] := n/2^IntegerExponent[n, 2]; Select[Range[2^20], Mod[f[1 + f[DivisorSigma[1, #]]], f[1 + #]] == 0 &] ] (* Michael De Vlieger, Aug 22 2020 *)
  • PARI
    A000265(n) = (n>>valuation(n,2));
    isA336700(n) = !((1+A000265(sigma(n)))%A000265(1+n));

A336701 Numbers k for which A000265(1+A000265(sigma(k))) is equal to A000265(1+k).

Original entry on oeis.org

1, 3, 7, 15, 31, 127, 1023, 8191, 34335, 57855, 131071, 524287, 2147483647
Offset: 1

Views

Author

Antti Karttunen, Aug 02 2020

Keywords

Comments

Numbers k such that A336698(k) [= A000265(1+A161942(k))] is equal to A000265(1+k).
Numbers k such that A337194(k) = 2^e * A000265(1+k), for some e >= 1, where that e = A337195(k).
Any odd perfect number would trivially satisfy this condition.
Also, all hypothetical quasiperfect numbers, numbers k that satisfy sigma(k) = 2k+1, would be members.
Question: Is A066175 a subsequence of this sequence?
From Antti Karttunen, Aug 23 2020: (Start)
Numbers k such that (1+k) = 2^e * A336698(k), for some e >= 0.
Thus numbers k such that for some e >= 0, (1+k) = 2^(e-A337195(k)) * A337194(k), or equally, that A337194(k) = 2^(A337195(k)-e) * (1+k).
Conjecture: There are no even terms. This is equivalent to claim that there are no k such that A336698(k) = 1+k: If we assume that k is even, then in above equations we set e=0, and the requirement will then become that A337194(k) = 2^A337195(k)*(1+k), thus 1+k = A336698(k) = A000265(1+A000265(sigma(k))).
(End)

Crossrefs

Subsequence of A336700.
Cf. A000668 (a subsequence).
See also comments in A326042, A332223.

Programs

  • Mathematica
    Block[{f}, f[n_] := n/2^IntegerExponent[n, 2]; Select[Range[2^20], f[1 + f[DivisorSigma[1, #]]] == f[1 + #] &] ] (* Michael De Vlieger, Aug 22 2020 *)
  • PARI
    A000265(n)  = (n>>valuation(n,2));
    isA336701(n) = (A000265(1+A000265(sigma(n))) == A000265(1+n));

A336699 a(n) = A000265(1+A000265(sigma(A000265(n)))), where A000265(k) gives the odd part of k, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 5, 7, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 5, 1, 7, 5, 3, 1, 1, 11, 1, 3, 1, 5, 1, 1, 1, 29, 1, 5, 1, 7, 3, 5, 1, 3, 1, 1, 1, 1, 1, 7, 1, 11, 1, 9, 5, 1, 1, 5, 7, 19, 5, 1, 3, 1, 1, 3, 1, 61, 11, 11, 1, 7, 3, 1, 1, 23, 5, 1, 1, 1, 1, 1, 1, 25, 29, 5, 1, 13, 5, 7, 1, 1
Offset: 1

Views

Author

Antti Karttunen, Aug 02 2020

Keywords

Comments

See the "lacunae" in the scatter plot. - Antti Karttunen, Mar 27 2022

Crossrefs

Programs

Formula

a(n) = A000265(1+A000265(A000593(n))) = A000265(1+A161942(A000265(n))).
a(n) = A336698(A000265(n)).
From Antti Karttunen, Mar 27 2022: (Start)
a(n) = A351565(A000593(n)).
[The following formulas were discovered by Sequence Machine]:
a(n) = A351565(A002131(n)) = A000265(1+A000265(A002131(n))).
a(n) = A336698(1+A322250(n)).
a(n) = A171435(A000593(n)+A082903(n)).
(End)

A337195 The 2-adic valuation of 1+A000265(sigma(n)), where A000265 gives the odd part.

Original entry on oeis.org

1, 2, 1, 3, 2, 2, 1, 4, 1, 1, 2, 3, 3, 2, 2, 5, 1, 3, 1, 1, 1, 1, 2, 4, 5, 1, 1, 3, 4, 1, 1, 6, 2, 2, 2, 2, 2, 4, 3, 1, 1, 2, 2, 1, 3, 1, 2, 5, 1, 1, 1, 1, 2, 4, 1, 4, 1, 1, 4, 1, 5, 2, 1, 7, 1, 1, 1, 6, 2, 1, 1, 2, 1, 1, 5, 2, 2, 1, 1, 1, 1, 6, 1, 3, 2, 1, 4, 1, 1, 1, 3, 1, 1, 1, 4, 6, 1, 2, 3, 1, 2, 2, 1, 1, 2
Offset: 1

Views

Author

Antti Karttunen, Aug 18 2020

Keywords

Crossrefs

Programs

Formula

a(n) = A007814(A337194(n)) = A007814(1+A000265(A000203(n))).

A351565 Odd part of Kimberling's paraphrases: a(n) = A000265(A003602(n)).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 5, 3, 3, 1, 7, 1, 1, 1, 9, 5, 5, 3, 11, 3, 3, 1, 13, 7, 7, 1, 15, 1, 1, 1, 17, 9, 9, 5, 19, 5, 5, 3, 21, 11, 11, 3, 23, 3, 3, 1, 25, 13, 13, 7, 27, 7, 7, 1, 29, 15, 15, 1, 31, 1, 1, 1, 33, 17, 17, 9, 35, 9, 9, 5, 37, 19, 19, 5, 39, 5, 5, 3, 41, 21, 21, 11, 43, 11, 11, 3, 45, 23, 23, 3, 47
Offset: 1

Views

Author

Antti Karttunen, Mar 27 2022

Keywords

Crossrefs

Cf. A000265, A003602, A023758 (gives the positions of 1's after its initial zero-term).
Cf. also A336698, A336699.

Programs

Formula

a(n) = A000265(A003602(n)) = A000265(1+A000265(n)).

A337196 The 3-adic valuation of 1+A000265(sigma(n)), where A000265 gives the odd part.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 18 2020

Keywords

Crossrefs

Cf. A337197 (the first occurrence of each n).

Programs

Formula

a(n) = A007949(A337194(n)) = A007949(1+A000265(A000203(n))).
a(n) = A007949(A336698(n)).

A337198 Number of distinct prime factors in A337194(n) = 1+A000265(sigma(n)), where A000265(k) gives the odd part of k.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 20 2020

Keywords

Comments

The first 3 occurs at a(137).

Crossrefs

Programs

Formula

a(n) = A001221(A337194(n)) = 1 + A001221(A336698(n)).
Showing 1-10 of 15 results. Next