A052462 a(n) is the minimal positive integral solution k to 24*k == 1 (mod 5^n).

4, 24, 99, 599, 2474, 14974, 61849, 374349, 1546224, 9358724, 38655599, 233968099, 966389974, 5849202474, 24159749349, 146230061849, 603993733724, 3655751546224, 15099843343099, 91393788655599, 377496083577474

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

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

Extending work of Ramanujan, Watson (1938) proved that P(m) == 0 (mod 5^n) if 24*m == 1 (mod 5^n). In particular, P(a(n)) == 0 (mod 5^n). - Petros Hadjicostas, Jul 29 2020

Examples

			From _Petros Hadjicostas_, Jul 29 2020: (Start)
A000041(a(1)) = A000041(4) = 5 == 0 (mod 5).
A000041(a(2)) = A000041(24) = 1575 == 0 (mod 5^2).
A000041(a(3)) = A000041(99) = 169229875 == 0 (mod 5^3).
A000041(a(4)) = A000041(599) = 435350207840317348270000 == 0 (mod 5^4). (End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• G. N. Watson, Ramanujans Vermutung über Zerfällungsanzahlen, J. Reine Angew. Math. (Crelle), 179 (1938), 97-128.
• Eric Weisstein's World of Mathematics, Partition Function P Congruences.
• Index entries for linear recurrences with constant coefficients, signature (1,25,-25).

Crossrefs

Cf. A000041, A052463, A052465, A052466.

Programs

Magma

I:=[4, 24, 99]; [n le 3 select I[n] else Self(n-1)+25*Self(n-2)-25*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jul 01 2012

Mathematica

Table[PowerMod[24, -1, 5^a], {a, 21}]
CoefficientList[Series[(-25x^2+20x+4)/((1-x)(1-5x)(1+5x)),{x,0,30}],x] (* Vincenzo Librandi, Jul 01 2012 *)

PARI

a(n) = lift(Mod(24, 5^n)^-1) \\ David A. Corneth and Petros Hadjicostas, Jul 29 2020

Formula

G.f.: x*(-25*x^2 + 20*x + 4)/((1 - x)*(1 - 5*x)*(1 + 5*x)).

a(n) = (1 + (21 + 2*(-1)^n)*5^n)/24. - Bruno Berselli, Apr 04 2011

a(n) = a(n-1) + 25*a(n-2) - 25*a(n-3). - Vincenzo Librandi, Jul 01 2012

A000041(a(n)) == 0 (mod 5^n). - Petros Hadjicostas, Jul 29 2020

Extensions

Name edited by Petros Hadjicostas, Jul 29 2020