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

6, 162, 1007, 27371, 170176, 4625692, 28759737, 781741941, 4860395546, 132114388022, 821406847267, 22327331575711, 138817757188116, 3773319036295152, 23460200964791597, 637690917133880681, 3964773963049779886, 107769764995625835082, 670046799755412800727

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

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

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

From Petros Hadjicostas, Aug 02 2020: (Start)

Assume n = 2*m, where m >= 1, and 24*k == 1 (mod 13^(2*m)), where k >= 1. Then there is an 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.

Assume now n = 2*m + 1, where m >= 0, and 24*k == 1 (mod 13^(2*m+1)), where k >= 1. Then there is an 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)

Examples

			From _Petros Hadjicostas_, Jul 29 2020: (Start)
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
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).
Thus, we must have A000041(27371) == 99 (mod 169). (End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..900
• A. O. L. Atkin and J. N. O'Brien, Some Properties of p(n) and c(n) Modulo Powers of 13, Trans. Amer. Math. Soc. 126 (1967), 442-459.
• Eric Weisstein's World of Mathematics, Partition Function P Congruences.
• Index entries for linear recurrences with constant coefficients, signature (1,169,-169).

Crossrefs

Cf. A000041, A052462, A052463, A052465.

Programs

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

Mathematica

Table[PowerMod[24, -1, 13^d], {d, 20}]
CoefficientList[Series[(-169x^2+156x+6)/((1-x)(1-13x)(1+13x)),{x,0,40}],x] (* Vincenzo Librandi, Jul 01 2012 *)
LinearRecurrence[{1,169,-169},{6,162,1007},30] (* Harvey P. Dale, Mar 15 2015 *)

PARI

a(n) = lift(Mod(24, 13^n)^-1) \\ Petros Hadjicostas, Jul 29 2020

SageMath

def a(n): return 24.inverse_mod(13^n)
print([a(n) for n in range(1, 20)]) # Peter Luschny, Jul 30 2020

Formula

G.f.: x*(-169*x^2 + 156*x + 6)/((1 - x)*(1 - 13*x)*(1 + 13*x)). - Vincenzo Librandi, Jul 01 2012

a(n) = a(n-1) + 169*a(n-2) - 169*a(n-3). - Vincenzo Librandi, Jul 01 2012

From Petros Hadjicostas, Aug 02 2020: (Start)

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

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

Extensions

Name edited by Petros Hadjicostas, Jul 29 2020