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%I #38 Feb 12 2022 03:38:22
%S 1,3,1,8,1
%N Maximum length of arithmetic progression with difference n such that each term k has tau(k) = n.
%C a(n) = 1 for all odd n.
%C a(10) >= 5, as witnessed by 43920665884407841463671+10*j, for j=0..4. - _Giovanni Resta_, Jul 28 2013
%C 9 <= a(6) <= 13; a(8) = 17; 7 <= a(10) <= 8; 10 <= a(12) <= 103. - _Hugo van der Sanden_, Nov 29 2016
%C From _Vladimir Letsko_, Nov 12 2017: (Start)
%C a(10) >= 7 since tau(n+10*j) = 10 for j = 0..6, where n = 14050704001368114927829875896053677879533000445528831984807324824611106055054250255923644575057624816866566109483087319903747624457245979936363.
%C 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.
%C 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.
%C 5 <= a(22) <= 8 since tau(n+22*j) = 22 for j = 0..4, where n = 1779938154456103755564381033732365223441543154296787.
%C 5 <= a(26) <= 8 since tau(n+26*j) = 26 for j = 0..4, where n = 1245939824866421006701844954340329129368712284774265869140521.
%C 5 <= a(34) <= 8 since tau(n+34*j) = 34 for j = 0..4, where n = 6938810980364904492570379879375548344505899946381175576077064821014404296739.
%C 5 <= a(46) <= 8 since tau(n+46*j) = 46 for j = 0..4, where n = 238126419116398949002327631305276551305298646600638655722184434490882642040819831057336330413818359191.
%C 5 <= a(58) <= 8 since tau(n+58*j) = 58 for j = 0..4, where n = 7704824180751624694570876810656527362795506581322310649236093848083558127895667820824262841185358447776614390313625335693359143.
%C 5 <= a(62) <= 8 since tau(n+62*j) = 62 for j = 0..4, where n = 1815849596256775454944849876198119384742848046265852804696322354947434625695582072075372178480692269137534575212736614048480987548827877.
%C (End)
%e 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.
%Y Cf. A064491, A165497, A165499, A165500.
%K hard,nonn,more
%O 1,2
%A _Hugo van der Sanden_, Sep 21 2009
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