A064491 -id:A064491 - cp's OEIS frontend

A096335 Number of iterations of n -> n + tau(n) needed for the trajectory of n to join the trajectory of A064491, or -1 if the two trajectories never merge.

0, 0, 2, 0, 1, 3, 0, 1, 0, 2, 8, 0, 7, 1, 6, 5, 6, 0, 5, 3, 4, 3, 4, 0, 3, 2, 13, 2, 13, 1, 12, 0, 11, 1, 10, 8, 10, 0, 9, 7, 9, 0, 8, 1, 7, 1, 8, 6, 7, 0, 6, 6, 6, 5, 5, 0, 4, 5, 4, 26, 3, 4, 2, 0, 2, 3, 2, 3, 1, 2, 0, 25, 0, 2, 0, 2, 1, 1, 1, 1, 0, 1, 39, 24, 38

Offset: 1

Jason Earls, Jun 28 2004

nonn

Conjecture: For any positive integer starting value n, iterations of n -> n + tau(n) will eventually join A064491 (verified for all n up to 50000).

The graph looks like a forest of stalks. The tops of the stalks form A036434. - N. J. A. Sloane, Jan 17 2013

			a(6)=3 because the trajectory for 1 (sequence A064491) starts
1->2->4->7->9->12->18->24->32->38->42...
and the trajectory for 6 starts
6->10->14->18->24->32->38->42->50->56...
so the sequence beginning with 6 joins A064491 after 3 steps.

Claudia Spiro, Problem proposed at West Coast Number Theory Meeting, 1977. - From N. J. A. Sloane, Jan 11 2013

T. D. Noe, Table of n, a(n) for n = 1..11000
T. D. Noe, Logarithmic plot of 10^6 terms

Cf. A000005, A036434, A064491, A096335.

s = 1; t = Join[{s}, Table[s = s + DivisorSigma[0, s], {n, 2, 1000}]]; mx = Max[t]; Table[r = n; gen = 0; While[r < mx && ! MemberQ[t, r], gen++; r = r + DivisorSigma[0, r]]; If[r >= mx, gen = -1]; gen, {n, 100}] (* T. D. Noe, Jan 13 2013 *)

Escape clause added to definition by N. J. A. Sloane, Nov 09 2020

A113866 Primes in the sequence A064491.

Original entry on oeis.org

2, 7, 71, 73, 811, 821, 823, 883, 937, 947, 953, 977, 983, 997, 1031, 1033, 1051, 1063, 1093, 1103, 1153, 1171, 1181, 1193, 1231, 1277, 1279, 1289, 1291, 1297, 1303, 1321, 1381, 1427, 1429, 1439, 1451, 1453, 1471, 1481, 1483, 29599, 29641, 29683, 29717

Offset: 1

John L. Drost, Jan 25 2006

if the sequence of iterates hits an odd square, then the succeeding f(n) is even. The iterates will remain even until f(n) is an even square, so the sequence grows slowly for a long time, then has big 'jumps'. Is the sequence infinite?

			If f(n) = n + tau(n), then the functional iterates of 7 are 7,9,12,18,24,32,38,42,50,56,64 and 71 which is prime so a(2)=7, a(3)=71

Robert Israel, Table of n, a(n) for n = 1..4453

Cf. A064491.

a:= 1: R:= NULL: count:= 0:
while count < 100 do
  a:= a+numtheory:-tau(a);
  if isprime(a) then count:= count+1; R:= R, a;  fi
od:
R; # Robert Israel, Nov 14 2020

Select[NestList[#+DivisorSigma[0, #]&, 1,2535] ,PrimeQ] (* James C. McMahon, Jul 07 2024 *)

A140483 Squares in A064491.

Original entry on oeis.org

1, 4, 9, 64, 81, 784, 1521, 29584, 34225, 132496, 136161, 4999696, 5076009, 15492096, 15547249, 299982400, 300086329, 1849172004, 1849430025

Offset: 1

Franklin T. Adams-Watters, Jun 25 2008

nonn

These are the places where A064491 changes parity.

Cf. A064491, A140484.

a(16)-a(19) from Donovan Johnson, Oct 06 2009

A062249 a(n) = n + d(n), where d(n) = number of divisors of n, cf. A000005.

Original entry on oeis.org

2, 4, 5, 7, 7, 10, 9, 12, 12, 14, 13, 18, 15, 18, 19, 21, 19, 24, 21, 26, 25, 26, 25, 32, 28, 30, 31, 34, 31, 38, 33, 38, 37, 38, 39, 45, 39, 42, 43, 48, 43, 50, 45, 50, 51, 50, 49, 58, 52, 56, 55, 58, 55, 62, 59, 64, 61, 62, 61, 72, 63, 66, 69, 71, 69, 74, 69, 74, 73, 78, 73

Offset: 1

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 01 2001

nonn

Number of cyclic subgroups of dihedral group with 2n elements.

a(n) is the n-th smallest number not a divisor of n. - J. Lowell, Apr 06 2008

Harry J. Smith, Table of n, a(n) for n=1..1000

Cf. A007503, A060710, A000005, A049820.

Cf. A064491 (iteration, start=1).

a062249 n = a000005 n + n  -- Reinhard Zumkeller, Mar 29 2014

with(numtheory):seq(n+tau(n), n=1..71) ; # Zerinvary Lajos, Jun 04 2008

Table[n + DivisorSigma[0, n], {n, 100}] (* Indranil Ghosh, Apr 12 2017 *)

a(n) = n + numdiv(n) \\ Harry J. Smith, Aug 03 2009

from sympy.ntheory import divisor_count
[n + divisor_count(n) for n in range(101)] # Indranil Ghosh, Apr 12 2017

a(n) = n + A000005(n). - Omar E. Pol, Dec 12 2008
From Ilya Gutkovskiy, Apr 12 2017: (Start)
G.f.: x/(1 - x)^2 + Sum_{k>=1} x^k/(1 - x^k).
Dirichlet g.f.: zeta(s)^2 + zeta(s-1). (End)

Formula and more terms from Vladeta Jovovic, Jul 03 2001

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

Hugo van der Sanden, Sep 21 2009

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)

			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.

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

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

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.

			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.

Cf. A064491, A165497, A165498, A165501.

A165495 First result not divisible by 4 when iterating k -> k+tau(k) from 2(2n-1)^2.

Original entry on oeis.org

7, 38, 71, 122, 178, 265, 415, 486, 602, 799, 927, 1230, 1321, 1486, 1810, 1951, 2214, 2474, 2830, 3194, 3386, 3738, 4206, 4639, 4874, 5358, 5838, 6222, 6534, 7101, 7482, 8145, 8570, 9010, 9722, 10922, 10922, 11434, 11882, 12571, 13274

Offset: 1

Hugo van der Sanden, Sep 21 2009

This sequence and A165494 explore a critical aspect of the behavior of A064491.

			For a(2) we start at 2.3^2=18; 18+tau(18)=24; 24+tau(24)=32; 32+tau(32)=38, which is not divisible by 4, so a(2)=38.

Cf. A064491, A165494, A165496

A165497 a(n) starts arithmetic progression of n terms separated by tau(a(n)), each term having the same number of divisors.

Original entry on oeis.org

1, 3, 3, 60, 60, 201, 481, 5989, 3122037, 4434429, 13576837, 183894465, 187925171, 209072257, 1498642520, 12239200420, 20220712468, 20220712468, 875023683404, 992997544772, 2721798771116, 9770941874212, 9770941874212

Offset: 1

Hugo van der Sanden, Sep 21 2009

a(15) > 10^9.
a(19) > 10^11. - Donovan Johnson, Sep 24 2009
a(24) > 10^13. - Giovanni Resta, Aug 02 2013

			tau(60) = tau(72) = tau(84) = tau(96) = tau(108) = 12. This is the first such progression of length greater than 3, so a(4) and a(5) are both 60.

Cf. A064491, A165498, A165499.

has(n)=my(t=numdiv(n),s=1); while(numdiv(n+=t)==t, s++); s
a(n)=my(k); while(has(k++)Charles R Greathouse IV, Apr 24 2015

a(15)-a(18) from Donovan Johnson, Sep 24 2009

a(19)-a(23) from Giovanni Resta, Aug 02 2013

A085864 a(1) = 2, a(n+1) = a(n)*{tau(a(n))}.

Original entry on oeis.org

2, 4, 12, 72, 864, 20736, 933120, 117573120, 42326323200, 36315985305600, 146426052752179200, 1475974611741966336000, 37784950060594338201600000, 2031318915257551621718016000000, 212719716805770805826310635520000000

Offset: 1

Amarnath Murthy, Jul 06 2003

nonn

a(1) = 1, a(n+1) = a(n) + tau(a(n)) gives A064491.

			a(4) = 72 and tau(72)= 12, hence a(5) = 72*12= 864.

Cf. A064491.

NestList[# DivisorSigma[0,#]&,2,20] (* Harvey P. Dale, Apr 23 2023 *)

for(n=1,16,if(n==1,a=2,a*=numdiv(a)); print(a); )

More terms from Ray Chandler, Jul 16 2003

A165494 Iterations of k -> k+tau(k) from 2(2n-1)^2 until result not divisible by 4.

Original entry on oeis.org

2, 3, 3, 3, 2, 3, 7, 2, 3, 6, 2, 9, 3, 2, 10, 3, 2, 2, 5, 7, 3, 4, 7, 10, 5, 5, 10, 8, 2, 7, 3, 8, 7, 4, 9, 36, 13, 9, 2, 6, 7, 5, 2, 2, 9, 4, 30, 2, 16, 4, 6, 6, 4, 11, 24, 16, 11, 9, 3, 11, 15, 2, 9, 10, 2, 26, 2, 10, 3, 3, 4, 13, 11, 5, 12, 3, 7, 17, 21, 17, 17, 12, 5, 4, 3, 8, 19, 9, 2

Offset: 1

Hugo van der Sanden, Sep 21 2009

This sequence and A165495 explore a critical aspect of the behavior of A064491.

			For a(2) we start at 2.3^2=18; 18+tau(18)=24; 24+tau(24)=32; 32+tau(32)=38, which is not divisible by 4, so a(2)=3 for the 3 iterations needed.

Cf. A064491, A165495

cp's OEIS Frontend

A096335 Number of iterations of n -> n + tau(n) needed for the trajectory of n to join the trajectory of A064491, or -1 if the two trajectories never merge.

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A113866 Primes in the sequence A064491.

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A140483 Squares in A064491.

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A062249 a(n) = n + d(n), where d(n) = number of divisors of n, cf. A000005.

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

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

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A165495 First result not divisible by 4 when iterating k -> k+tau(k) from 2(2n-1)^2.

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A165497 a(n) starts arithmetic progression of n terms separated by tau(a(n)), each term having the same number of divisors.

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A085864 a(1) = 2, a(n+1) = a(n)*{tau(a(n))}.

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