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.

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A069782 Numbers k such that gcd(d(k^3), d(k)) = 2^w for some w.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107
Offset: 1

Views

Author

Labos Elemer, Apr 08 2002

Keywords

Comments

The first missing integer is 432 (see in A069781).

Examples

			Below 100000 only 314 integers are missing, collected in A069781.
		

Crossrefs

Programs

  • Mathematica
    f[x_] := GCD[DivisorSigma[0, x^3], DivisorSigma[0, x]]; Do[s=f[n]; If[IntegerQ[Log[2, s]], Print[{n, s}]], {n, 1, 100000}]
  • PARI
    is(n)=my(f=factor(n)[, 2], g=gcd(prod(i=1, #f, 3*f[i]+1), prod(i=1, #f, f[i]+1))); g>>valuation(g, 2)==1 \\ Charles R Greathouse IV, Oct 16 2015

A327634 Infinitary highly abundant numbers: numbers m such that isigma(m) > isigma(k) for all k < m, where isigma(k) is the sum of infinitary divisors of n (A049417).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 18, 21, 22, 24, 30, 40, 42, 54, 66, 72, 78, 88, 96, 102, 114, 120, 168, 210, 216, 264, 312, 330, 360, 378, 384, 408, 456, 480, 510, 546, 552, 600, 672, 690, 696, 744, 840, 1080, 1320, 1512, 1560, 1848, 1920, 2040, 2184, 2280, 2688
Offset: 1

Views

Author

Amiram Eldar, Sep 20 2019

Keywords

Comments

The infinitary version of A002093.

Examples

			The first 10 values of isigma(k) for k = 1 to 10 are: 1, 3, 4, 5, 6, 12, 8, 15, 10, 18. Record values are reached for all these values of k except for 7 and 9, therefore the sequence begins with 1, 2, 3, 4, 5, 6, 8, 10, ...
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(2^(-1 + Position[Reverse @ IntegerDigits[e, 2], ?(# == 1 &)])); isigma[1] = 1; isigma[n] := Times @@ (Flatten @ (f @@@ FactorInteger[n]) + 1); seq = {};sm = 0; Do[s = isigma[n]; If[s > sm, sm = s; AppendTo[seq, n]], {n, 1, 10^4}]; seq

A061234 Smallest number with prime(n)^2 divisors where prime(n) is the n-th prime.

Original entry on oeis.org

6, 36, 1296, 46656, 60466176, 2176782336, 2821109907456, 101559956668416, 131621703842267136, 6140942214464815497216, 221073919720733357899776, 10314424798490535546171949056, 13367494538843734067838845976576
Offset: 1

Views

Author

Labos Elemer, Jun 01 2001

Keywords

Examples

			1296 = 2*2*2*2*3*3*3*3 is the smallest number with 25 divisors.
		

Crossrefs

Formula

a(n) = Min_{x : d(x) = A000005(x) = p(n)^2} = 6^(p(n)-1) because x = 2^(pp-1) > 2^(p-1)3^(p-1) holds if p > 1.
a(n) = A005179(A001248(n)). - Amiram Eldar, Jun 21 2024

A069780 a(n) = gcd(d(n^3), d(n)).

Original entry on oeis.org

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

Views

Author

Labos Elemer, Apr 08 2002

Keywords

Comments

Terms are usually powers of 2. Smallest number m such that A069780(m)=2^n is A037992(n). The first n such that a(n) is not a power of 2 equals 432: a(432) = gcd(d(80621568), d(432)) = gcd(130,20) = 10.

Crossrefs

Programs

  • Mathematica
    Table[GCD[DivisorSigma[0, n^3], DivisorSigma[0, n]], {n, 1, 500}]
  • PARI
    a(n)=my(f=factor(n)[, 2]); gcd(prod(i=1, #f, 3*f[i]+1), prod(i=1, #f, f[i]+1)) \\ Charles R Greathouse IV, Oct 16 2015

Formula

a(n) = gcd(A000005(n^3), A000005(n)).

A069781 Numbers k such that gcd(d(k^3), d(k)) is not a power of 2.

Original entry on oeis.org

432, 576, 648, 1600, 2000, 2160, 2880, 2916, 3024, 3136, 3240, 4032, 4536, 4752, 4800, 5000, 5488, 5616, 6000, 6336, 7128, 7344, 7488, 7744, 8208, 8424, 9408, 9792, 9936, 10125, 10800, 10816, 10944, 11016, 11200, 12312, 12528, 13248, 13392
Offset: 1

Views

Author

Labos Elemer, Apr 08 2002

Keywords

Comments

The complement of this sequence in the positive integers A000027 is A069782. - M. F. Hasler, Jan 18 2015
The numbers of the form 4*3^(7*m - 1), m >= 1, are terms. - Marius A. Burtea, Oct 18 2019

Examples

			For n<100000, gcd[d(n^3),d[n]] = {5,7,10,14,20,28,40,80} which is obtained for n={20736,576,432,2880,54000,20160,2160,15120} respectively.
		

Crossrefs

Programs

  • Magma
    f:=func; [k:k in [1..14000]| not IsIntegral(Log(2,f(k)))]; // Marius A. Burtea, Oct 18 2019
  • Mathematica
    f[x_] := GCD[DivisorSigma[0, x^3], DivisorSigma[0, x]] Do[s=f[n]; If[ !IntegerQ[Log[2, s]], Print[n]], {n, 1, 100000}]
    Select[Range[14000],!IntegerQ[Log[2,GCD[DivisorSigma[0,#^3], DivisorSigma[ 0,#]]]]&] (* Harvey P. Dale, Mar 20 2018 *)
  • PARI
    is(n)=my(f=factor(n)[,2], g=gcd(prod(i=1,#f,3*f[i]+1), prod(i=1,#f,f[i]+1))); g!=1<Charles R Greathouse IV, Oct 16 2015
    

Formula

log_2(gcd(A000005(n^3), A000005(n))) is nonintegral.

A335386 Tri-unitary highly composite numbers: where the number of tri-unitary divisors (A335385) increases to a record.

Original entry on oeis.org

1, 2, 6, 24, 120, 840, 7560, 83160, 1081080, 18378360, 349188840, 8031343320, 200783583000, 5822723907000, 180504441117000, 6678664321329000, 273825237174489000, 11774485198503027000, 553400804329642269000, 27116639412152471181000, 1437181888844080972593000
Offset: 1

Views

Author

Amiram Eldar, Jun 04 2020

Keywords

Crossrefs

Analogous sequences: A002182 (highly composite), A002110 (unitary), A037992 (infinitary), A293185 (bi-unitary), A318278 (exponential), A306736 (exponential infinitary), A307845 (exponential unitary), A309141 (nonunitary), A322484 (semi-unitary).
Cf. A335385.

Programs

  • Mathematica
    f[p_, e_] := If[e == 3 || e == 6, 4, 2]; d[1] = 1; d[n_] := Times @@ (f @@@ FactorInteger[n]); dm = 0; s = {}; Do[If[(d1 = d[n]) > dm, dm = d1; AppendTo[s, n]], {n, 1, 1100000}]; s

Formula

A335385(a(n)) = 2^(n-1).

A374786 Numerator of the mean infinitary abundancy index of the infinitary divisors of n.

Original entry on oeis.org

1, 5, 7, 9, 11, 35, 15, 45, 19, 11, 23, 21, 27, 75, 77, 33, 35, 95, 39, 99, 5, 115, 47, 105, 51, 135, 133, 135, 59, 77, 63, 165, 161, 175, 33, 19, 75, 195, 63, 99, 83, 25, 87, 207, 209, 235, 95, 77, 99, 51, 245, 243, 107, 665, 23, 675, 91, 295, 119, 231, 123, 315
Offset: 1

Views

Author

Amiram Eldar, Jul 20 2024

Keywords

Comments

The infinitary abundancy index of a number k is A049417(k)/k.
The record values of a(n)/A374787(n) are attained at the terms of A037992.
The least number k such that a(k)/A374787(k) is larger than 2, 3, 4, ..., is A037992(6) = 7560, A037992(33) = 1370819010042780920891599455129161859473627856000, ... .

Examples

			For n = 4, 4 has 2 infinitary divisors, 1 and 4. Their infinitary abundancy indices are isigma(1)/1 = 1 and isigma(4)/4 = 5/4, and their mean infinitary abundancy index is (1 + 5/4)/2 = 9/8. Therefore a(4) = numerator(9/8) = 9.
		

Crossrefs

Similar sequences: A374777/A374778, A374783/A374784.

Programs

  • Mathematica
    f[p_, e_] := p^(2^(-1 + Position[Reverse@IntegerDigits[e, 2], ?(# == 1 &)])); a[1] = 1; a[n] := Numerator[Times @@ (1 + 1/(2*Flatten@ (f @@@ FactorInteger[n])))]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), b); numerator(prod(i = 1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], 1 + 1/(2*f[i, 1]^(2^(#b-k))), 1))));}

Formula

Let f(n) = a(n)/A374787(n). Then:
f(n) = (Sum_{d infinitary divisor of n} isigma(d)/d) / id(n), where isigma(n) is the sum of infinitary divisors of n (A049417), and id(n) is their number (A037445).
f(n) is multiplicative with f(p^e) = Product{k>=1, e_k=1} (1 + 1/(2*p^(2^(k+1)))), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.
f(n) = (Sum_{d infinitary divisor of n} d*id(d)) / (n*id(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} f(k) = Product_{P} (1 + 1/(2*P*(P+1))) = 1.21407233718434377029..., where P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376). For comparison, the asymptotic mean of the infinitary abundancy index over all the positive integers is 1.461436... = 2 * A327574.
Lim sup_{n->oo} f(n) = oo (i.e., f(n) is unbounded).
f(n) <= A374777(n)/A374778(n) with equality if and only if n is squarefree (A005117).
f(n) >= A374783(n)/A374784(n) with equality if and only if n is in A138302.

A333931 Recursive highly composite numbers: numbers with a record number of recursive divisors (A282446).

Original entry on oeis.org

1, 2, 4, 6, 12, 30, 36, 60, 180, 420, 900, 1260, 4620, 6300, 13860, 44100, 55440, 69300, 180180, 485100, 720720, 900900, 2882880, 3063060, 6306300, 12252240, 15315300, 49008960, 58198140, 107207100, 232792560, 290990700, 931170240, 1163962800, 2036934900, 4655851200
Offset: 1

Views

Author

Amiram Eldar, Apr 10 2020

Keywords

Comments

The corresponding record values are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 48, 54, ...

Crossrefs

Subsequence of A025487.
Cf. A282446.
Analogous sequences: A002182 (highly composite), A002110 (unitary), A037992 (infinitary), A293185 (bi-unitary), A309141 (nonunitary), A318278 (exponential).

Programs

  • Mathematica
    recDivNum[1] = 1; recDivNum[n_] := recDivNum[n] = Times @@ (1 + recDivNum/@ (Last /@ FactorInteger[n])); rm = 0; s = {}; Do[r = recDivNum[n]; If[r > rm, rm = r; AppendTo[s, n]], {n, 1, 10^4}]; s

A343020 a(n) is the smallest prime p such that tau(p+1) = 2^n.

Original entry on oeis.org

2, 5, 23, 167, 839, 7559, 128519, 1081079, 20540519, 397837439, 8031343319, 188972783999, 3212537327999, 125568306863999, 2888071057871999, 190487121512687999, 4381203794791823999, 215961289494494543999, 13283916764437951631999, 540119185025730854543999, 26465840066260811872655999, 1356699703068812438127791999
Offset: 1

Views

Author

Jaroslav Krizek, Apr 02 2021

Keywords

Comments

tau(m) = the number of divisors of m (A000005).
Sequences of primes p such that tau(p+1) = 2^n for 2 <= n <= 5:
n = 2: 5, 7, 13, 37, 61, 73, 157, 193, 277, 313, 397, 421, ...
n = 3: 23, 29, 41, 53, 101, 103, 109, 113, 127, 137, 151, ...
n = 4: 167, 263, 269, 311, 383, 389, 439, 461, 509, 569, ...
n = 5: 839, 1319, 1511, 1559, 1847, 1889, 2039, 2309, 2687, ...
Conjecture: a(n) is also the smallest number m such that tau(m+1) = tau(m)^n.

Examples

			a(4) = 167 because 167 is the smallest prime p such that tau(p+1) = 16 = 2^4.
		

Crossrefs

Programs

  • Magma
    Ax:=func; [Ax(n): n in [1..7]]
    
  • Mathematica
    Do[p = 1; While[DivisorSigma[0, Prime[p] + 1] != 2^n, p++]; Print[n, " ", Prime[p]], {n, 1, 9}] (* Vaclav Kotesovec, Apr 03 2021 *)
  • PARI
    a(n) = my(t=2^n); forprime(p=2, oo, if(numdiv(p+1)==t, return(p))); \\ Jinyuan Wang, Apr 02 2021
    
  • Python
    from sympy import isprime,nextprime
    primes=[2]
    def solve(v,k,i,j):
        global record,stack,primes
        if k==0:
            if isprime(v-1):
                record=v
            return True
        sizeok=False
        cnt=True
        while cnt:
            if i>=len(primes):
                primes.append(nextprime(primes[-1]))
            if jBert Dobbelaere, Apr 11 2021

Extensions

a(11) from Jinyuan Wang, Apr 02 2021
More terms from David A. Corneth, Apr 09 2021

A036538 Number of integers m <= 2^n such that d(m) = 2^k for some k = 0, 1, 2, 3, ...

Original entry on oeis.org

2, 3, 7, 12, 23, 45, 89, 178, 356, 707, 1409, 2822, 5639, 11273, 22546, 45088, 90165, 180315, 360637, 721258, 1442491, 2884973, 5769941, 11539858, 23079721, 46159395, 92318705, 184637321, 369274467, 738548867, 1477097749, 2954195452, 5908390605, 11816780739
Offset: 1

Views

Author

Keywords

Comments

a(n+1)/a(n) is very close to 2; a(n)/2^n is near 0.7.
As n goes to infinity, lim a(n)/2^n = 0.687827... (A327839; see comments in A036537). - Vladimir Shevelev, Feb 28 2017

Examples

			Of the numbers 1 .. 2^4 = 16, only 4, 9, 12 and 16 are not in A036537, so a(4) = 16 - 4 = 12.
		

Crossrefs

Programs

  • Maple
    IversonBrackets := expr -> subs(true=1, false=0, expr):
    A := proc(n) option remember; {seq(2^k, k=0..n)} end:
    h := proc(n) option remember; add(evalb(numtheory:-tau(j) in A(n)), j=2^(n-1) + 1..2^n); IversonBrackets(%) end:
    a := n -> 1 + add(h(k), k=1..n); seq(a(n), n=1..17); # Peter Luschny, May 14 2018
  • Mathematica
    Table[Count[#, 1] &@ Table[1 - Sign[# - Floor@ #] &@Log[2, #] &@ DivisorSigma[0, x], {x, 1, 2^m}], {m, 1, 20}] (* original program edited by Michael De Vlieger, Mar 01 2017, or *)
    1 + Accumulate@ Table[Count[Range[2^(n - 1) + 1, 2^n], k_ /; IntegerQ@ Log2@ DivisorSigma[0, k]], {n, 20}] (* Michael De Vlieger, Feb 28 2017 *)
  • PARI
    a(n) = sum(k=1, 2^n, d = numdiv(k); (d<=2) || (ispower(d,,&p) && (p==2))); \\ Michel Marcus, May 14 2018
    
  • Python
    from sympy import factorint
    def A036538(n): return sum(1 for m in range(1,(1<Chai Wah Wu, Jun 22 2023

Formula

a(n) = number of 1s in f(tau(k)) mapped across k = 1..2^n, with f(x):= 1-sign(log_2 x - floor( log_2 x )). - Michael De Vlieger, Mar 01 2017

Extensions

a(20)-a(26) from Michael De Vlieger, Feb 28 2017
a(27)-a(34) from Giovanni Resta, May 14 2018
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