A052463 -id:A052463 - cp's OEIS frontend

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

Eric W. Weisstein

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

			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)

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).

Cf. A000041, A052463, A052465, A052466.

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

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 *)

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

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

Name edited by Petros Hadjicostas, Jul 29 2020

A052465 a(n) is the smallest positive integral solution k to 24*k == 1 (mod 11^n).

Original entry on oeis.org

6, 116, 721, 14031, 87236, 1697746, 10555551, 205427261, 1277221666, 24856698576, 154543821581, 3007660527691, 18699802411296, 363926923850606, 2262676091766811, 44035157785923321, 273783807103784126, 5328254092096721836, 33127840659557879241, 644718745143703342151, 4008468719806503388156

Offset: 1

Eric W. Weisstein

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

Extending work of Ramanujan, Atkin (1967) proved that P(m) == 0 (mod 11^n) when 24*m == 1 (mod 11^n). In particular, P(a(n)) == 0 (mod 11^n). - Petros Hadjicostas, Jul 29 2020

			From _Petros Hadjicostas_, Jul 29 2020: (Start)
A000041(a(1)) = A000041(6) = 11 == 0 (mod 11^1).
A000041(a(2)) = A000041(116) = 1188908248 == 0 (mod 11^2).
A000041(a(3)) = A000041(721) = 161061755750279477635534762 == 0 (mod 11^3). (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..900
A. O. L. Atkin, Proof of a Conjecture of Ramanujan, Glasgow Math. J. 8 (1967), 14-32.
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 (1,121,-121).

Cf. A000041, A052462, A052463, A052466.

I:=[6, 116, 721]; [n le 3 select I[n] else Self(n-1)+121*Self(n-2)-121*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jul 01 2012

Table[PowerMod[24, -1, 11^c], {c, 20}]
CoefficientList[Series[(-121x^2+110x+6)/((1-x)(1-121*x^2)),{x,0,30}],x] (* Vincenzo Librandi, Jul 01 2012 *)
LinearRecurrence[{1,121,-121},{6,116,721},20] (* Harvey P. Dale, Apr 27 2014 *)

a(n) = lift(Mod(24, 11^n)^-1) \\ David A. Corneth, Jul 29 2020

def a(n): return 24.inverse_mod(11^n)
print([a(n) for n in range(1, 22)]) # Peter Luschny, Jul 29 2020

G.f.: x*(-121*x^2 + 110*x + 6)/((1 - x)*(1 - 121*x^2)). - Vincenzo Librandi, Jul 01 2012
a(n) = a(n-1) + 121*a(n-2) - 121*a(n-3). - Vincenzo Librandi, Jul 01 2012
A000041(a(n)) == 0 (mod 11^n). - Petros Hadjicostas, Jul 29 2020
From Petros Hadjicostas, Aug 02 2020: (Start)
a(n) = (1 + 23*11^n)/24, if n is even, and a(n) = (1 + 13*11^n)/24, if n is odd.
a(n) - a(n-1) = 10*11^(n-1), if n is even >= 2, and 5*11^(n-1), if n is odd >= 3. (End)

More terms from David A. Corneth, Jul 29 2020

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

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein

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)

			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)

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).

Cf. A000041, A052462, A052463, A052465.

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

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 *)

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

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

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)

Name edited by Petros Hadjicostas, Jul 29 2020

A327770 a(n) = (23 * 7^(2*n) + 1)/24. Sequence related to the properties of the partition function A000041 modulo a power of 7.

Original entry on oeis.org

1, 47, 2301, 112747, 5524601, 270705447, 13264566901, 649963778147, 31848225129201, 1560563031330847, 76467588535211501, 3746911838225363547, 183598680073042813801, 8996335323579097876247, 440820430855375795936101, 21600201111913414000868947

Offset: 0

Petros Hadjicostas, Sep 24 2019

If p(n) = A000041(n) is the partition function, Watson (1938) proved that p(7^(2*m)*n + a(m)) == 0 mod 7^(m+1) for n >= 0 and m >= 1. (Obviously, this is not always true for m = 0).
For m=1 and n=0, p(7^(2*1)*0 + a(1)) = p(47) = 7^(1+1) * 2546.
For m=1 and n=1, p(7^(2*1)*1 + a(1)) = p(96) = 7^(1+1) * 2410496.
For m=1 and n=2, p(7^(2*1)*2 + a(1)) = p(145) = 7^(1+1) * 508344041.
For m=2 and n=0, p(7^(2*2)*0 + a(2)) = p(2301) = 7^(2+1) * 49629361905981812695622866669844910256876089360.
Essentially the same as A052463. - R. J. Mathar, Oct 08 2019

Colin Barker, Table of n, a(n) for n = 0..500
G. N. Watson, Ramanujans Vermutung über Zerfällungsanzahlen, J. Reine Angew. Math. (Crelle), 179 (1938), 97-128; see pp. 118 and 124.
Eric Weisstein's World of Mathematics, Partition Function P Congruences.
Wikipedia, G. N. Watson.
Index entries for linear recurrences with constant coefficients, signature (50,-49).

Cf. A052463, A071746, A213261, A327714, A327582.

CoefficientList[Series[(1 - 3 x)/((1 - x) (1 - 49 x)), {x, 0, 15}], x] (* Michael De Vlieger, Sep 27 2019 *)
LinearRecurrence[{50,-49},{1,47},20] (* Harvey P. Dale, Mar 09 2023 *)

a(n) = (23 * 7^(2*n) + 1)/24; \\ Michel Marcus, Sep 25 2019

Vec((1 - 3*x) / ((1 - x)*(1 - 49*x)) + O(x^20)) \\ Colin Barker, Sep 25 2019

From Colin Barker, Sep 25 2019: (Start)
G.f.: (1 - 3*x) / ((1 - x)*(1 - 49*x)).
a(n) = 50*a(n-1) - 49*a(n-2) for n>1.
(End)

A327771 a(n) = p(49*n + 47)/49, where p(k) denotes the k-th partition number (i.e., A000041).

Original entry on oeis.org

2546, 2410496, 508344041, 48286178405, 2734250190712, 106823899382728, 3143746885297470, 73830872731991927, 1440681502991063990, 24058683492974200054, 351628923073820626951, 4577202012225445531319, 53811955397591074514675, 577896157936323089053580

Offset: 0

Petros Hadjicostas, Sep 24 2019

nonn

Watson (1938), p. 120, proved that p(7*n + 5) == 0 (mod 7) and p(49*n + 47) == 0 (mod 49) for n >= 0, where p() = A000041(). For more general congruence results modulo a power of 7 by George Neville Watson regarding the partition function, see A327582 and A327770.

G. N. Watson, Ramanujans Vermutung über Zerfällungsanzahlen, J. Reine Angew. Math. (Crelle), 179 (1938), 97-128; see p. 120.
Eric Weisstein's World of Mathematics, Partition Function P Congruences.
Wikipedia, G. N. Watson.

Cf. A000041, A052462, A052463, A052465, A052466, A071746, A213261, A327714, A327582, A327770.

Table[PartitionsP[49n+47]/49,{n, 0, 13}] (* Metin Sariyar, Sep 25 2019 *)

a(n) = numbpart(49*n + 47)/49; \\ Michel Marcus, Sep 25 2019

a(n) = A000041(49*n + 47)/49.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A052465 a(n) is the smallest positive integral solution k to 24*k == 1 (mod 11^n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A327770 a(n) = (23 * 7^(2*n) + 1)/24. Sequence related to the properties of the partition function A000041 modulo a power of 7.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A327771 a(n) = p(49*n + 47)/49, where p(k) denotes the k-th partition number (i.e., A000041).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula