OFFSET
1,2
COMMENTS
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
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..600
G. K. Patil, Ramanujan's Life And His Contributions In The Field Of Mathematics, International Journal of Scientific Research and Engineering Studies (IJSRES), 1(6) (2014), ISSN: 2349-8862.
Eric Weisstein's World of Mathematics, Partition Function P Congruences.
Index entries for linear recurrences with constant coefficients, signature (50,-49).
FORMULA
G.f.: x^2*(-49*x + 47)/((1 - x)*(1 - 49*x)).
a(n) = 49*a(n-1) - 2. - Vincenzo Librandi, Jul 01 2012
a(n) = 23*49^n/1176 + 1/24, n > 1. - R. J. Mathar, Oct 09 2019
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Name edited by Petros Hadjicostas, Oct 09 2019
STATUS
approved