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 A052466 a(n) is the smallest positive solution k to 24*k == 1 (mod 13^n). 6

%I

%S 6,162,1007,27371,170176,4625692,28759737,781741941,4860395546,

%T 132114388022,821406847267,22327331575711,138817757188116,

%U 3773319036295152,23460200964791597,637690917133880681,3964773963049779886,107769764995625835082,670046799755412800727

%N a(n) is the smallest positive solution k to 24*k == 1 (mod 13^n).

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

%C Atkin and O'Brien (1967) proved that for all integral n >= 1, there is an integral constant K(n) not divisible by 13 s.t. P(169*m - 7) == K(n)*P(m) (mod 13^n) for all integral m >= 1 that satisfy 24*m == 1 (mod 13^n). In particular, P(169*a(n) - 7) == K(n)*P(a(n)) (mod 13^n) for all n >= 1. Unfortunately, the calculation of the integral constants K(n) depends on several recursions found in the paper. (For each n, there are infinitely many such K(n)'s, but one may choose the smallest one that satisfies the above property.) See Theorem 2, p. 444, in their paper, even though their P is different that the P = A000041 here. - _Petros Hadjicostas_, Jul 29 2020

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

%C Assume n = 2*m, where m >= 1, and 24*k == 1 (mod 13^(2*m)), where k >= 1. Then there is integer x = x(k) s.t. 24*k - 1 = 169^m*x. Then 1 = 24*k - 169^m*x == 0 - 1^m*x == -x (mod 24). With x = x(k) = 23, we find a(2*m), the smallest value of k >= 1 that satisfies 24*k == 1 (mod 13^(2*m)). Thus, a(2*m) = (1 + 23*13^(2*m))/24.

%C Assume now n = 2*m + 1, where m >= 0, and 24*k == 1 (mod 13^(2*m+1)), where k >= 1. Then there is integer x = x(k) s.t. 24*k - 1 = 13*169^m*x. Then 1 = 24*k - 13*169^m*x == 0 - 13*1^m*x == -13*x (mod 24). With x = x(k) = 11, we find a(2*m+1), the smallest value of k >= 1 that satisfies 24*k == 1 (mod 13^(2*m)). Thus, a(2*m+1) = (1 + 11*13^(2*m+1))/24. (End)

%H Vincenzo Librandi, <a href="/A052466/b052466.txt">Table of n, a(n) for n = 1..900</a>

%H A. O. L. Atkin and J. N. O'Brien, <a href="https://doi.org/10.1090/S0002-9947-1967-0214540-7">Some Properties of p(n) and c(n) Modulo Powers of 13</a>, Trans. Amer. Math. Soc. 126 (1967), 442-459.

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

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,169,-169).

%F G.f.: x*(-169*x^2 + 156*x + 6)/((1 - x)*(1 - 13*x)*(1 + 13*x)). - _Vincenzo Librandi_, Jul 01 2012

%F a(n) = a(n-1) + 169*a(n-2) - 169*a(n-3). - _Vincenzo Librandi_, Jul 01 2012

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

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

%F a(n) - a(n-1) = 12*13^(n-1) for n even >= 2, and 5*13^(n-1) for n odd >= 3. (End)

%e From _Petros Hadjicostas_, Jul 29 2020: (Start)

%e The only value of the constant K(n) that appears explicitly in Atkin and O'Brien (1967) is K(2) = 45 (see p. 453). We then have

%e P(169*a(2) - 7) - K(2)*P(a(2)) = P(169*162 - 7) - 45*P(162) = A000041(27371) - 45*A000041(162) = A000041(27371) - 5846125708665 == 0 (mod 13^2).

%e Thus, we must have A000041(27371) == 99 (mod 169). (End)

%t Table[PowerMod[24, -1, 13^d], {d, 20}]

%t CoefficientList[Series[(-169x^2+156x+6)/((1-x)(1-13x)(1+13x)),{x,0,40}],x] (* _Vincenzo Librandi_, Jul 01 2012 *)

%t LinearRecurrence[{1,169,-169},{6,162,1007},30] (* _Harvey P. Dale_, Mar 15 2015 *)

%o (MAGMA) I:=[6, 162, 1007]; [n le 3 select I[n] else Self(n-1)+169*Self(n-2)-169*Self(n-3): n in [1..20]]; // _Vincenzo Librandi_, Jul 01 2012

%o (PARI) a(n) = lift(Mod(24, 13^n)^-1) \\ _Petros Hadjicostas_, Jul 29 2020

%o (SageMath)

%o def a(n): return 24.inverse_mod(13^n)

%o print([a(n) for n in range(1, 20)]) # _Peter Luschny_, Jul 30 2020

%Y Cf. A000041, A052462, A052463, A052465.

%K nonn,easy

%O 1,1

%A _Eric W. Weisstein_

%E Name edited by _Petros Hadjicostas_, Jul 29 2020

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