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.

A323244 a(1) = 0; and for n > 1, a(n) = A033879(A156552(n)).

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 1, 6, 0, 5, 1, 10, 1, 16, 2, 6, 1, 12, 1, 18, -3, 18, 1, 22, -4, 46, 4, 22, 1, 10, 1, 30, 14, 82, -2, 14, 1, 256, -12, 22, 1, 36, 1, 66, 8, 226, 1, 46, -12, 19, 8, 130, 1, 28, -19, 70, -12, 748, 1, 42, 1, 1362, 16, 22, 10, 42, 1, 214, 254, 40, 1, 38, 1, 3838, 10, 406, -10, 106, 1, 78, -12, 5458, 1, 26, -72, 12250, -348, 30, 1, 12
Offset: 1

Views

Author

Antti Karttunen, Jan 10 2019

Keywords

Comments

After a(1) = 0, the other zeros occur for k >= 1, at A005940(1+A000396(k)), which, provided no odd perfect numbers exist, is equal to A324201(k) = A062457(A000043(k)): 9, 125, 161051, 410338673, ..., etc.
There are 2321 negative terms among the first 10000 terms.

Crossrefs

Cf. A324201 (positions of zeros, conjectured), A324551 (of negative terms), A324720 (of nonnegative terms), A324721 (of positive terms), A324731, A324732.
Cf. A329644 (Möbius transform).
Cf. A323174, A324055, A324185, A324546 for other permutations of deficiency, and also A324574, A324575, A324654.

Programs

  • Mathematica
    Array[2 # - If[# == 0, 0, DivisorSigma[1, #]] &@ Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ #]] &, 90] (* Michael De Vlieger, Apr 21 2019 *)
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n))));
    A323244(n) = if(1==n, 0, my(k=A156552(n)); (2*k)-sigma(k));
    
  • Python
    from sympy import divisor_sigma, primepi, factorint
    def A323244(n): return (lambda n: (n<<1)-divisor_sigma(n))(sum((1< 1 else 0 # Chai Wah Wu, Mar 10 2023

Formula

a(n) = 2*A156552(n) - A323243(n).
a(1) = 0; and for n > 1, a(n) = A033879(A156552(n)).
a(n) = A323248(n) + A001222(n) = (A323247(n) - A323243(n)) + A001222(n).
From Antti Karttunen, Mar 12 2019 & Nov 23 2019: (Start)
a(n) = Sum_{d|n} (2*A297112(d) - A324543(d)) = Sum_{d|n} A329644(d).
A002487(a(n)) = A324115(n).
a(n) = A329638(n) - A329639(n).
a(n) = A329645(n) - A329646(n).
(End)

A329644 Möbius transform of A323244, the deficiency of A156552(n).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 4, -1, 3, 1, 5, 1, 14, 0, 0, 1, 9, 1, 12, -5, 16, 1, 8, -5, 44, 4, 5, 1, 2, 1, 24, 12, 80, -4, -4, 1, 254, -14, 0, 1, 22, 1, 47, 7, 224, 1, 24, -13, 19, 6, 83, 1, 12, -21, 44, -14, 746, 1, 14, 1, 1360, 20, -8, 8, 9, 1, 131, 252, 24, 1, 12, 1, 3836, 13, 149, -12, 71, 1, 56, -16, 5456, 1, -21, -74, 12248, -350, -40, 1
Offset: 1

Views

Author

Antti Karttunen, Nov 21 2019

Keywords

Comments

The first eleven zeros occur at n = 1, 15, 16, 40, 96, 119, 120, 160, 893, 2464, 6731. There are 3091 negative terms among the first 10000 terms.
Applying this function to the divisors of the first four terms of A324201 reveals the following pattern:
------------------------------------------------------------------------------------
A324201(n) divisors a(n) applied Sum of positive
to each: terms, A329610
9: [1, 3, 9] -> [0, 1, -1] 1
125: [1, 5, 25, 125] -> [0, 1, -5, 4] 5
161051: [1, 11, 121, 1331, 14641, 161051] -> [0, 1, -29, 4, -240, 264] 269
410338673: [1, 17, 289, 4913, 83521, 1419857, 24137569, 410338673]
-> [0, 1, -125, 4, -1008, 1032, -5048, 5144] 6181
The positive and negative terms seem to alternate, and the fourth term (from case n=125 onward) is always 4. See also array A329637.

Crossrefs

Programs

  • PARI
    A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552
    A323244(n) = if(1==n, 0, my(k=A156552(n)); (2*k)-sigma(k));
    A329644(n) = sumdiv(n,d,moebius(n/d)*A323244(d));

Formula

a(n) = Sum_{d|n} A008683(n/d) * A323244(d).
a(n) = Sum_{d|n} A008683(n/d) * (2*A156552(d) - A323243(d)).
a(1) = 0; for n > 1, a(n) = 2*A297112(n) - A324543(n) = 2^A297113(n) - A324543(n).
a(n) = A329642(n) - A329643(n).
For all n >= 1, a(A000040(n)^2) = A323244(A000040(n)^2)-1 = -A036563(n).
For all primes p, a(p^3) = A323244(p^3) - A323244(p^2) = 4.

A329645 Inverse Möbius transform of A329642.

Original entry on oeis.org

0, 1, 1, 3, 1, 6, 1, 9, 1, 7, 1, 19, 1, 18, 4, 15, 1, 19, 1, 28, -1, 20, 1, 49, -3, 48, 5, 43, 1, 27, 1, 45, 16, 84, 0, 50, 1, 258, -10, 58, 1, 59, 1, 89, 13, 228, 1, 107, -11, 23, 10, 181, 1, 51, -17, 121, -10, 750, 1, 112, 1, 1364, 16, 67, 12, 84, 1, 301, 256, 65, 1, 128, 1, 3840, 11, 667, -8, 150, 1, 148, -7, 5460, 1, 143, -70, 12252
Offset: 1

Views

Author

Antti Karttunen, Nov 21 2019

Keywords

Crossrefs

Programs

Formula

a(n) = Sum_{d|n} A329642(d).
a(n) = A323244(n) + A329646(n).

A329643 a(n) = Sum_{d|n} [-1 == A008683(n/d)] * A323244(d), where A323244(x) gives the deficiency of A156552(x).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 2, 1, 2, 0, 6, 0, 2, 2, 6, 0, 4, 0, 7, 2, 2, 0, 16, 1, 2, 0, 18, 0, 11, 0, 6, 2, 2, 2, 22, 0, 2, 2, 24, 0, 17, 0, 20, 2, 2, 0, 28, 1, 1, 2, 48, 0, 16, 2, 28, 2, 2, 0, 39, 0, 2, -3, 30, 2, 36, 0, 84, 2, 19, 0, 36, 0, 2, -2, 258, 2, 38, 0, 28, 4, 2, 0, 69, 2, 2, 2, 72, 0, 31, 2, 228, 2, 2, 2, 76, 0, 4, 14, 37, 0, 94, 0, 136, -3
Offset: 1

Views

Author

Antti Karttunen, Nov 21 2019

Keywords

Crossrefs

Cf. A329646 (inverse Möbius transform).

Programs

  • PARI
    A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552
    A323243(n) = if(1==n,0,sigma(A156552(n)));
    A329643(n) = sumdiv(n,d,(-1==moebius(n/d))*((2*A156552(d))-A323243(d)));

Formula

a(n) = Sum_{d|n} [-1 == A008683(n/d)] * (2*A156552(d) - A323243(d)).
a(n) = A329642(n) - A329644(n).
For all n, a(A000040(n)) = 0, a(A006881(n)) = 2.
Showing 1-4 of 4 results.