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 1-3 of 3 results.

A064491 a(1) = 1, a(n+1) = a(n) + tau(a(n)), where tau(n) (A000005) is the number of divisors of n.

Original entry on oeis.org

1, 2, 4, 7, 9, 12, 18, 24, 32, 38, 42, 50, 56, 64, 71, 73, 75, 81, 86, 90, 102, 110, 118, 122, 126, 138, 146, 150, 162, 172, 178, 182, 190, 198, 210, 226, 230, 238, 246, 254, 258, 266, 274, 278, 282, 290, 298, 302, 306, 318, 326, 330, 346, 350, 362, 366, 374
Offset: 1

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Oct 04 2001

Keywords

Comments

a(n) = partial sums of A165930(n). [Jaroslav Krizek, Sep 30 2009]

References

  • Claudia Spiro, Problem proposed at West Coast Number Theory Meeting, 1977. [If you change the starting term, does the resulting sequence always join this one? Does the parity of terms change infinitely often?] - From N. J. A. Sloane, Jan 11 2013

Links

Crossrefs

Cf. A000005, A165930, A165494, A165495, A165496, A165497, A165498, A165499.

Programs

  • Haskell
    a064491 n = a064491_list !! (n-1)
a064491_list = iterate a062249 1  -- Reinhard Zumkeller, Mar 29 2014
  • Mathematica
    a[n_] := a[n] = a[n - 1] + DivisorSigma[0, a[n - 1]]; a[1] = 1; Table[a[n], {n, 1, 57}] (* Jean-François Alcover, Oct 11 2012 *)
NestList[#+DivisorSigma[0,#]&,1,60] (* Harvey P. Dale, Feb 05 2017 *)
  • PARI
    { for (n=1, 1000, if (n>1, a+=numdiv(a), a=1); write("b064491.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 16 2009
  • Python
    from itertools import islice
from sympy import divisor_count
def A064491gen(): # generator of terms
    n = 1
    yield n
    while True:
        n += divisor_count(n)
        yield n
A064491_list = list(islice(A064491gen(),20)) # Chai Wah Wu, Dec 13 2021

Formula

It seems likely that there exist constants c_1 and c_2 such that c_1*n*log(n) < a(n) < c_2*n*log(n) for all sufficiently large n. - Franklin T. Adams-Watters, Jun 25 2008
a(n+1) = A062249(a(n)). - Reinhard Zumkeller, Mar 29 2014

Extensions

Beginning of sequence corrected by T. D. Noe, Sep 13 2007

A165498 Maximum length of arithmetic progression with difference n such that each term k has tau(k) = n.

Original entry on oeis.org

1, 3, 1, 8, 1
Offset: 1

Views

Author

Hugo van der Sanden, Sep 21 2009

Keywords

Comments

a(n) = 1 for all odd n.
a(10) >= 5, as witnessed by 43920665884407841463671+10*j, for j=0..4. - Giovanni Resta, Jul 28 2013
9 <= a(6) <= 13; a(8) = 17; 7 <= a(10) <= 8; 10 <= a(12) <= 103. - Hugo van der Sanden, Nov 29 2016
From Vladimir Letsko, Nov 12 2017: (Start)
a(10) >= 7 since tau(n+10*j) = 10 for j = 0..6, where n = 14050704001368114927829875896053677879533000445528831984807324824611106055054250255923644575057624816866566109483087319903747624457245979936363.
Congruences x^6 == 14 (mod 3^5) and x^6 == -14 (mod 3^5) have no solutions. On other hand, tau(n+14*j) = 14 for j = 0..4, where n = 1330022329820905436281789742546819. Therefore a(14) = 5.
Similarly, congruences x^18 == 38 (mod 3^17) and x^18 == -38 (mod 3^17) have no solutions. On other hand, tau(n+38*j) = 38 for j = 0..4, where n = 4362267871759873721878756446028394403250067166871580001529317952259003864288330077973. Therefore a(38) = 5.
5 <= a(22) <= 8 since tau(n+22*j) = 22 for j = 0..4, where n = 1779938154456103755564381033732365223441543154296787.
5 <= a(26) <= 8 since tau(n+26*j) = 26 for j = 0..4, where n = 1245939824866421006701844954340329129368712284774265869140521.
5 <= a(34) <= 8 since tau(n+34*j) = 34 for j = 0..4, where n = 6938810980364904492570379879375548344505899946381175576077064821014404296739.
5 <= a(46) <= 8 since tau(n+46*j) = 46 for j = 0..4, where n = 238126419116398949002327631305276551305298646600638655722184434490882642040819831057336330413818359191.
5 <= a(58) <= 8 since tau(n+58*j) = 58 for j = 0..4, where n = 7704824180751624694570876810656527362795506581322310649236093848083558127895667820824262841185358447776614390313625335693359143.
5 <= a(62) <= 8 since tau(n+62*j) = 62 for j = 0..4, where n = 1815849596256775454944849876198119384742848046265852804696322354947434625695582072075372178480692269137534575212736614048480987548827877.
(End)

Examples

			When tau(k) = 4, k cannot be divisible by 9 unless k = 27. An arithmetic progression of 9 terms with difference 4 must have a term divisible by 9, and k=27 is not part of a progression of 9 terms with tau(k)=4, so a(4) must be less than 9. Since a progression of 8 terms is achievable (e.g. starting at 5989), a(4) = 8 is proved.

Crossrefs

Cf. A064491, A165497, A165499, A165500.

A165499 First term of maximal arithmetic progression with difference n, such that each term k has tau(k) = n.

Original entry on oeis.org

1, 3, 4, 5989, 16
Offset: 1

Views

Author

Hugo van der Sanden, Sep 21 2009; updated Nov 29 2016

Keywords

Comments

a(6) <= 161804009483982959337354063701 if A165498(6) = 9, and at least 1e14.
a(8) = 380017309607.
a(10) <= 43920665884407841463671 if A165498(10) = 5 (found by Giovanni Resta), and at least 1e12.
a(12) <= 11673662470957217427690002629075 if A165498(12) = 10, and at least 1e10.
a(16) = 2a(8).
A165498(n) = 1 for odd n, so a(7) = 64; a(9) = 36; a(11) = 1024; a(13) = 4096; a(15) = 144; etc.

Examples

			A165498(4) = 8, and exhaustive search finds tau(5989) = tau(5993) = tau(5997) = tau(6001) = tau(6005) = tau(6009) = tau(6013) = tau(6017) = 4 is the first example of an 8-term progression, so a(4) = 5989.

Crossrefs

Cf. A064491, A165497, A165498, A165501.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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