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A052466 a(n) is the smallest positive solution k to 24*k == 1 (mod 13^n). 6
6, 162, 1007, 27371, 170176, 4625692, 28759737, 781741941, 4860395546, 132114388022, 821406847267, 22327331575711, 138817757188116, 3773319036295152, 23460200964791597, 637690917133880681, 3964773963049779886, 107769764995625835082, 670046799755412800727 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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 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 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)

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).

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)

EXAMPLE

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)

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 *)

PROG

(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

(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

CROSSREFS

Cf. A000041, A052462, A052463, A052465.

Sequence in context: A193370 A332116 A015086 * A280477 A078535 A177781

Adjacent sequences:  A052463 A052464 A052465 * A052467 A052468 A052469

KEYWORD

nonn,easy

AUTHOR

Eric W. Weisstein

EXTENSIONS

Name edited by Petros Hadjicostas, Jul 29 2020

STATUS

approved

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Last modified March 3 06:51 EST 2021. Contains 341759 sequences. (Running on oeis4.)