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.

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).

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));

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)).

A378998 Number of trailing 1-bits in the binary representation of sigma(n).

Original entry on oeis.org

1, 2, 0, 3, 0, 0, 0, 4, 1, 0, 0, 0, 0, 0, 0, 5, 0, 3, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Dec 16 2024

Keywords

Crossrefs

Cf. A000203, A007814, A088580, A028982 (positions of terms > 0), A028983 (of 0's), A072461 (of 1's), A072462 (of terms > 1), A337195, A378999 [= a(n^2)].

Programs

  • Mathematica
    IntegerExponent[DivisorSigma[1, Range[100]] + 1, 2] (* Paolo Xausa, Dec 19 2024 *)
  • PARI
    A378998(n) = valuation(sigma(n)+1,2);

Formula

a(n) = A007814(A088580(n)). [the 2-adic valuation of 1+sigma(n)]
For all n in A028982, a(n) = A337195(n).
Showing 1-4 of 4 results.