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 A052465 a(n) is the smallest positive integral solution k to 24*k == 1 (mod 11^n). 6

%I

%S 6,116,721,14031,87236,1697746,10555551,205427261,1277221666,

%T 24856698576,154543821581,3007660527691,18699802411296,

%U 363926923850606,2262676091766811,44035157785923321,273783807103784126,5328254092096721836,33127840659557879241,644718745143703342151,4008468719806503388156

%N a(n) is the smallest positive integral solution k to 24*k == 1 (mod 11^n).

%C Related to a Ramanujan congruence for the partition function P = A000041.

%C Extending work of Ramanujan, Atkin (1967) proved that P(m) == 0 (mod 11^n) when 24*m == 1 (mod 11^n). In particular, P(a(n)) == 0 (mod 11^n). - _Petros Hadjicostas_, Jul 29 2020

%H Vincenzo Librandi, <a href="/A052465/b052465.txt">Table of n, a(n) for n = 1..900</a>

%H A. O. L. Atkin, <a href="https://doi.org/10.1017/S0017089500000045">Proof of a Conjecture of Ramanujan</a>, Glasgow Math. J. 8 (1967), 14-32.

%H G. K. Patil, <a href="https://web.archive.org/web/20150911053452/http://www.ijsres.com/2014/vol-1_issue-6/paper_8.pdf">Ramanujan's Life And His Contributions In The Field Of Mathematics</a>, International Journal of Scientific Research and Engineering Studies (IJSRES), 1(6) (2014), ISSN: 2349-8862.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PartitionFunctionPCongruences.html">Partition Function Congruences</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,121,-121).

%F G.f.: x*(-121*x^2 + 110*x + 6)/((1 - x)*(1 - 121*x^2)). - _Vincenzo Librandi_, Jul 01 2012

%F a(n) = a(n-1) + 121*a(n-2) - 121*a(n-3). - _Vincenzo Librandi_, Jul 01 2012

%F A000041(a(n)) == 0 (mod 11^n). - _Petros Hadjicostas_, Jul 29 2020

%F From _Petros Hadjicostas_, Aug 02 2020: (Start)

%F a(n) = (1 + 23*11^n)/24, if n is even, and a(n) = (1 + 13*11^n)/24, if n is odd.

%F a(n) - a(n-1) = 10*11^(n-1), if n is even >= 2, and 5*11^(n-1), if n is odd >= 3. (End)

%e From _Petros Hadjicostas_, Jul 29 2020:

%e A000041(a(1)) = A000041(6) = 11 == 0 (mod 11^1).

%e A000041(a(2)) = A000041(116) = 1188908248 == 0 (mod 11^2).

%e A000041(a(3)) = A000041(721) = 161061755750279477635534762 == 0 (mod 11^3). (End)

%t Table[PowerMod[24, -1, 11^c], {c, 20}]

%t CoefficientList[Series[(-121x^2+110x+6)/((1-x)(1-121*x^2)),{x,0,30}],x] (* _Vincenzo Librandi_, Jul 01 2012 *)

%t LinearRecurrence[{1,121,-121},{6,116,721},20] (* _Harvey P. Dale_, Apr 27 2014 *)

%o (MAGMA) I:=[6, 116, 721]; [n le 3 select I[n] else Self(n-1)+121*Self(n-2)-121*Self(n-3): n in [1..20]]; // _Vincenzo Librandi_, Jul 01 2012

%o (PARI) a(n) = lift(Mod(24, 11^n)^-1) \\ _David A. Corneth_, Jul 29 2020

%o (SageMath)

%o def a(n): return 24.inverse_mod(11^n)

%o print([a(n) for n in range(1, 22)]) # _Peter Luschny_, Jul 29 2020

%Y Cf. A000041, A052462, A052463, A052466.

%K nonn,easy

%O 1,1

%A _Eric W. Weisstein_

%E More terms from _David A. Corneth_, Jul 29 2020

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Last modified April 13 12:29 EDT 2021. Contains 342936 sequences. (Running on oeis4.)