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A052465
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a(n) is the smallest positive integral solution k to 24*k == 1 (mod 11^n).
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6
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6, 116, 721, 14031, 87236, 1697746, 10555551, 205427261, 1277221666, 24856698576, 154543821581, 3007660527691, 18699802411296, 363926923850606, 2262676091766811, 44035157785923321, 273783807103784126, 5328254092096721836, 33127840659557879241, 644718745143703342151, 4008468719806503388156
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OFFSET
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1,1
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COMMENTS
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Related to a Ramanujan congruence for the partition function P = A000041.
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
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LINKS
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FORMULA
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G.f.: x*(-121*x^2 + 110*x + 6)/((1 - x)*(1 - 121*x^2)). - Vincenzo Librandi, Jul 01 2012
a(n) = (1 + 23*11^n)/24, if n is even, and a(n) = (1 + 13*11^n)/24, if n is odd.
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)
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EXAMPLE
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A000041(a(3)) = A000041(721) = 161061755750279477635534762 == 0 (mod 11^3). (End)
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MATHEMATICA
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Table[PowerMod[24, -1, 11^c], {c, 20}]
CoefficientList[Series[(-121x^2+110x+6)/((1-x)(1-121*x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 01 2012 *)
LinearRecurrence[{1, 121, -121}, {6, 116, 721}, 20] (* Harvey P. Dale, Apr 27 2014 *)
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PROG
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(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
(SageMath)
def a(n): return 24.inverse_mod(11^n)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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