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A052463
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a(n) is the smallest nonnegative solution k to 24*k == 1 (mod 7^(2*n-2)).
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6
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0, 47, 2301, 112747, 5524601, 270705447, 13264566901, 649963778147, 31848225129201, 1560563031330847, 76467588535211501, 3746911838225363547, 183598680073042813801, 8996335323579097876247
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OFFSET
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1,2
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COMMENTS
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Related to a Ramanujan congruence for the partition function P = A000041.
In other words, a(n) = k such that 24*k (mod 7^(2*n-2) ) == 1. - N. J. A. Sloane, Oct 08 2019
If b(n) = a(n) + 7^(2*n-2)*r, where r is a nonnegative integer, then there is an integer s >= 0 such that 24*b(n) = 24*a(n) + 24*7^(2*n-2)*r = 7^(2*n-2)*s + 1 + 24*7^(2*n-2)*r = 7^(2*n-2)*(24*r+s) + 1 == 1 (mod 7^(2*n-2)). Thus, we insist that a(n) is the smallest k >= 0 such that 24*k == 1 (mod 7^(2*n-2)). - Petros Hadjicostas, Oct 09 2019
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LINKS
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FORMULA
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G.f.: x^2*(-49*x + 47)/((1 - x)*(1 - 49*x)).
a(n) = 23*49^n/1176 + 1/24, n > 1. - R. J. Mathar, Oct 09 2019
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MATHEMATICA
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Table[PowerMod[24, -1, 7^(2b-2)], {b, 20}]
CoefficientList[Series[(-49x^2+47x)/((1-x)(1-49x)), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 01 2012 *)
LinearRecurrence[{50, -49}, {0, 47, 2301}, 20] (* Harvey P. Dale, Aug 23 2021 *)
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PROG
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(Magma) I:=[0, 47]; [n le 2 select I[n] else 49*Self(n-1)-2: n in [1..20]]; // Vincenzo Librandi, Jul 01 2012
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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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