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

A327965 "Tamed variant" of arithmetic derivative: a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 5, 1, 3, 6, 7, 1, 1, 1, 9, 2, 2, 1, 21, 1, 6, 10, 13, 1, 11, 10, 15, 1, 2, 1, 31, 1, 5, 14, 19, 3, 15, 1, 21, 1, 17, 1, 41, 1, 3, 39, 25, 1, 7, 14, 45, 5, 14, 1, 3, 1, 23, 22, 31, 1, 23, 1, 33, 51, 3, 18, 61, 1, 18, 26, 59, 1, 39, 1, 39, 55, 5, 18, 71, 1, 11, 1, 43, 1, 31, 22, 45, 2, 35, 1, 123, 5, 6
Offset: 0

Views

Author

Antti Karttunen, Oct 01 2019

Keywords

Comments

Applying A327938 to the result of A003415(n) ensures that all terms stay in A048103, and that all iteration paths will (hopefully) terminate in zero. See A327966.

Crossrefs

Programs

  • PARI
    A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
    A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
    A327965(n) = if(n<=1,0,A327938(A003415(n)));

Formula

a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).

A085731 Greatest common divisor of n and its arithmetic derivative.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 4, 3, 1, 1, 4, 1, 1, 1, 16, 1, 3, 1, 4, 1, 1, 1, 4, 5, 1, 27, 4, 1, 1, 1, 16, 1, 1, 1, 12, 1, 1, 1, 4, 1, 1, 1, 4, 3, 1, 1, 16, 7, 5, 1, 4, 1, 27, 1, 4, 1, 1, 1, 4, 1, 1, 3, 64, 1, 1, 1, 4, 1, 1, 1, 12, 1, 1, 5, 4, 1, 1, 1, 16, 27, 1, 1, 4, 1, 1, 1, 4, 1, 3, 1, 4, 1, 1, 1, 16
Offset: 1

Views

Author

Reinhard Zumkeller, Jul 20 2003

Keywords

Comments

a(n) = 1 iff n is squarefree (A005117), cf. A068328.
This sequence is very probably multiplicative. - Mitch Harris, Apr 19 2005

Crossrefs

Programs

  • Haskell
    a085731 n = gcd n $ a003415 n -- Reinhard Zumkeller, May 10 2011
    
  • Mathematica
    d[0] = d[1] = 0; d[n_] := d[n] = n*Total[Apply[#2/#1 &, FactorInteger[n], {1}]]; a[n_] := GCD[n, d[n]]; Table[a[n], {n, 1, 96}] (* Jean-François Alcover, Feb 21 2014 *)
    f[p_, e_] := p^If[Divisible[e, p], e, e - 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 31 2023 *)
  • PARI
    a(n) = {my(f = factor(n)); for (i=1, #f~, if (f[i,2] % f[i,1], f[i,2]--);); factorback(f);} \\ Michel Marcus, Feb 14 2016

Formula

a(n) = GCD(n, A003415(n)).
Multiplicative with a(p^e) = p^e if p divides e; a(p^e) = p^(e-1) otherwise. - Eric M. Schmidt, Oct 22 2013
From Antti Karttunen, Feb 28 2021: (Start)
Thus a(A276086(n)) = A328572(n), by the above formula and the fact that A276086 is a permutation of A048103.
a(n) = n / A083346(n) = A190116(n) / A086130(n). (End)

A327936 Multiplicative with a(p^e) = p if e >= p, otherwise 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2
Offset: 1

Views

Author

Antti Karttunen, Oct 01 2019

Keywords

Examples

			For n = 12 = 2^2 * 3^1, only prime factor p = 2 satisfies p^p | 12, thus a(12) = 2.
For n = 108 = 2^2 * 3^3, both prime factors p = 2 and p = 3 satisfy p^p | 108, thus a(108) = 2*3 = 6.
		

Crossrefs

Differs from A129252 for the first time at n=108.

Programs

  • Mathematica
    Array[Apply[Times, FactorInteger[#] /. {p_, e_} /; IntegerQ@ p :> If[e >= p, p, 1]] &, 120] (* Michael De Vlieger, Oct 01 2019 *)
  • PARI
    A327936(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]>=f[k,1])); factorback(f); };

Formula

Multiplicative with a(p^e) = p if e >= p, otherwise 1.
A001221(a(n)) = A129251(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + (p-1)/p^p) = 1.3443209052633459342... . - Amiram Eldar, Nov 07 2022

A351458 Numbers k for which k * gcd(sigma(k), A276086(k)) is equal to sigma(k) * gcd(k, A276086(k)), where A276086 is the primorial base exp-function, and sigma gives the sum of divisors of its argument.

Original entry on oeis.org

1, 10, 56, 9196, 9504, 56160, 121176, 239096, 354892, 411264, 555520, 716040, 804384, 904704, 1063348, 1387386, 1444352, 1454112, 1884800, 2708640, 3317248, 3548920, 4009824, 4634784, 6179712, 6795360, 7285248, 14511744, 16328466, 28377216, 29855232, 31940280, 37444736, 42711552, 49762944, 52815744
Offset: 1

Views

Author

Antti Karttunen, Feb 13 2022

Keywords

Comments

Numbers k such that k * A324644(k) = A000203(k) * A324198(k).
Numbers k such that gcd(A064987(k), A324580(k)) = gcd(A064987(k), A351252(k)).
Numbers k such that their abundancy index [sigma(k)/k] is equal to A324644(k)/A324198(k). See A364286.
A324644 gives odd values for even numbers and for the odd squares. A324198 is odd on all arguments, therefore on odd squares the above equation reduces to odd * odd = odd * odd, and on odd nonsquares as odd * even = even * odd. It is an open question whether there are any odd terms after the initial a(1)=1.
If k is even, but not a multiple of 3, then A276086(k) is a multiple of 3, but not even (i.e., is an odd multiple of 3). If for such k also sigma(k) = 3*k, then A007949(A324644(k)) = min(A007949(sigma(k)), A007949(A276086(k))) = 1, while A007949(A324198(k)) = min(A007949(k), A007949(A276086(k))) = 0, therefore all such k's do occur in this sequence, for example, the two known terms of A005820 (3-perfect numbers) that are not multiples of three: 459818240, 51001180160, but also any hypothetical term of A005820 of the form 4u+2, where 2u+1 is not multiple of 3, and which by necessity is then also an odd perfect number.
Similarly, of the 65 known 5-multiperfect numbers (A046060), those 20 that are not multiples of five are included in this sequence. Note that all 65 are multiples of six.
It is conjectured that the intersection of this sequence with the multiperfect numbers (A007691) gives A323653, see comments in the latter.
For all even terms k of this sequence, A007814(A000203(k)) = A007814(k), sigma preserves the 2-adic valuation, and A007949(A000203(k)) >= A007949(k), i.e., does not decrease the 3-adic valuation. The condition is equivalence (=) when k is a multiple of 6. With odd terms, any hypothetical odd perfect number x would yield a one greater 2-adic valuation for sigma(x) than for x, but would satisfy the main condition of this sequence. - Corrected Feb 17 2022
If k is a nonsquare positive odd number (in A088828), then it must be a term of A191218. - Antti Karttunen, Mar 10 2024

Crossrefs

Cf. also A351549.

Programs

  • PARI
    A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
    isA351458(n) = { my(s=sigma(n), z=A276086(n)); (n*gcd(s,z))==(s*gcd(n,z)); };
    
  • PARI
    A006530(n) = if(1==n, n, my(f=factor(n)); f[#f~, 1]); \\ Works OK with rationals also!
    isA351458(n) = { my(orgn=n, s=sigma(n), abi=s/n, p=2, q=A006530(abi), d, e1, e2); while((1!=abi)&&(p<=q), d = n%p; e1 = min(d, valuation(s, p)); e2 = min(d, valuation(orgn, p)); d = e1-e2; if(valuation(abi,p)!=d, return(0), abi /= (p^d)); n = n\p; p = nextprime(1+p)); (abi==1); }; \\ (This implementation does not require the construction of largish intermediate numbers, A276086, but might still be slower and return a few false positives on the long run, so please check the results with the above program). - Antti Karttunen, Feb 19 2022

A327939 Multiplicative with a(p^e) = p^(e-(e mod p)).

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 27, 4, 1, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16, 1, 1, 1, 4, 1, 27, 1, 4, 1, 1, 1, 4, 1, 1, 1, 64, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16, 27, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 108, 1, 1, 1, 16
Offset: 1

Views

Author

Antti Karttunen, Oct 01 2019

Keywords

Comments

Fixed points of the map x -> gcd(x, A003415(x)), i.e., if we start iterating with A085731 from any x = n (>= 1), we will eventually reach a(n), after which the result does not change anymore. This was found by LODA miner (see C. Krause link), and is easily seen to be true by Eric M. Schmidt's multiplicative formula for A085731. Note also that this sequence is idempotent, meaning a(a(n)) = a(n) for all n. - Antti Karttunen, Apr 05 2021
The largest divisor of n that is a term of A072873. - Amiram Eldar, Sep 14 2023

Crossrefs

Differs from A234957 for the first time at n=27.

Programs

  • Mathematica
    f[p_, e_] := p^(e - Mod[e, p]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 14 2023 *)
  • PARI
    A327939(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]-(f[k,2]%f[k,1]))); factorback(f); };

Formula

Multiplicative with a(p^e) = p^(e-(e mod p)).
a(n) = n / A327938(n).

A327966 Number of iterations of "tamed variant of arithmetic derivative", A327965 needed to reach 0 from n, or -1 if zero is never reached.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Oct 01 2019

Keywords

Comments

Conjecture: from all n, zero is eventually reached.

Crossrefs

Cf. A003415, A256750, A327938, A327965, A327967 (indices of the records).

Programs

  • PARI
    A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
    A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
    A327965(n) = if(n<=1,0,A327938(A003415(n)));
    A327966(n) = { my(k=0); while(n>0, k++; n = A327965(n)); (k); };
    \\ Or alternatively, as a recurrence:
    A327966(n) = if(!n,0,1+A327966(A327965(n)));

Formula

a(0) = 0; for n > 0, a(n) = 1 + a(A327965(n)).
a(p) = 2 for all primes p.

A327968 a(0) = a(1) = 0, a(prime) = 1, and for all other numbers, a(n) = the first noncomposite reached when iterating A327965, or -1 if no noncomposite is ever reached.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 5, 1, 3, 5, 7, 1, 1, 1, 5, 2, 2, 1, 7, 1, 5, 7, 13, 1, 11, 7, 2, 1, 2, 1, 31, 1, 5, 5, 19, 3, 2, 1, 7, 1, 17, 1, 41, 1, 3, 1, 7, 1, 7, 5, 1, 5, 5, 1, 3, 1, 23, 13, 31, 1, 23, 1, 5, 5, 3, 7, 61, 1, 7, 2, 59, 1, 1, 1, 1, 1, 5, 7, 71, 1, 11, 1, 43, 1, 31, 13, 1, 2, 3, 1, 11, 5, 5, 19, 5, 5, 17, 1, 7, 1, 3, 1, 5, 1, 41, 71
Offset: 0

Views

Author

Antti Karttunen, Oct 02 2019

Keywords

Comments

Of the prime terms, 5 seems to be the most common (8886 occurrences among the first 100001 terms). See also A327975.

Crossrefs

Programs

  • PARI
    A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
    A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
    A327965(n) = if(n<=1,0,A327938(A003415(n)));
    A327968(n) = if(n<=1,0,if(isprime(n),1, while((n>1)&&!isprime(n), n = A327965(n)); (n)));

A342007 Multiplicative with a(p^e) = p^floor(e/p).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1
Offset: 1

Views

Author

Antti Karttunen, Mar 04 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Array[Times @@ Map[#1^Floor[#2/#1] & @@ # &, FactorInteger[#]] &, 105] (* Michael De Vlieger, Mar 12 2021 *)
  • PARI
    A342007(n) = { my(f = factor(n)); for(k=1, #f~, f[k, 2] = floor(f[k, 2]/f[k, 1])); factorback(f); };

Formula

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + (p-1)/(p^p-p)) = 1.6270951877598772517... . - Amiram Eldar, Nov 07 2022

A328618 Multiplicative with a(p^e) = p^e if p = 2 or e is a multiple of p, otherwise a(p^e) = p^((p*floor(e/p)) + (2e mod p)).

Original entry on oeis.org

1, 2, 9, 4, 25, 18, 49, 8, 3, 50, 121, 36, 169, 98, 225, 16, 289, 6, 361, 100, 441, 242, 529, 72, 625, 338, 27, 196, 841, 450, 961, 32, 1089, 578, 1225, 12, 1369, 722, 1521, 200, 1681, 882, 1849, 484, 75, 1058, 2209, 144, 2401, 1250, 2601, 676, 2809, 54, 3025, 392, 3249, 1682, 3481, 900, 3721, 1922, 147, 64, 4225, 2178, 4489, 1156, 4761, 2450, 5041, 24
Offset: 1

Views

Author

Antti Karttunen, Oct 23 2019

Keywords

Crossrefs

Cf. A328619 (inverse permutation).

Programs

  • Mathematica
    a[n_] := Product[{p, e} = pe; If[p == 2 || Divisible[e, p], p^e, p^((p*Floor[e/p]) + Mod[2e, p])], {pe, FactorInteger[n]}];
    Array[a, 100] (* Jean-François Alcover, Nov 21 2021 *)
  • PARI
    A328618(n) = { my(f = factor(n), m, q); for(k=1, #f~, q = (f[k, 2]\f[k, 1]); m = (f[k, 2]%f[k, 1]); if(m&&(f[k,1]!=2), f[k, 2] = q*f[k, 1] + ((2*f[k, 2])%f[k, 1]))); factorback(f); };

Formula

For all n >= 0, A276085(a(A276086(n))) = A328622(n).

A327937 Multiplicative with a(p^e) = p^(p-1) if e >= p, otherwise a(p^e) = p^e.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 9, 10, 11, 6, 13, 14, 15, 2, 17, 18, 19, 10, 21, 22, 23, 6, 25, 26, 9, 14, 29, 30, 31, 2, 33, 34, 35, 18, 37, 38, 39, 10, 41, 42, 43, 22, 45, 46, 47, 6, 49, 50, 51, 26, 53, 18, 55, 14, 57, 58, 59, 30, 61, 62, 63, 2, 65, 66, 67, 34, 69, 70, 71, 18, 73, 74, 75, 38, 77, 78, 79, 10, 9, 82, 83, 42, 85, 86, 87, 22
Offset: 1

Views

Author

Antti Karttunen, Oct 01 2019

Keywords

Comments

All terms are in A048103.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e >= p, p^(p - 1), p^e]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 07 2022 *)
  • PARI
    A327937(n) = { my(f = factor(n)); for(k=1, #f~, if(f[k,2]>=f[k,1], f[k,2] = f[k,1]-1)); factorback(f); };

Formula

Multiplicative with a(p^e) = p^(p-1) if e >= p, otherwise a(p^e) = p^e.
For all n, A129251(a(n)) = 0, A327936(a(n)) = 1.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} (1 - 1/((p+1)*p^(p-1))) = 0.40498413761092854213... . - Amiram Eldar, Nov 07 2022
Showing 1-10 of 15 results. Next