cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

Original entry on oeis.org

4, 24, 99, 599, 2474, 14974, 61849, 374349, 1546224, 9358724, 38655599, 233968099, 966389974, 5849202474, 24159749349, 146230061849, 603993733724, 3655751546224, 15099843343099, 91393788655599, 377496083577474
Offset: 1

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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)
		

Crossrefs

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

A052463 a(n) is the smallest nonnegative solution k to 24*k == 1 (mod 7^(2*n-2)).

Original entry on oeis.org

0, 47, 2301, 112747, 5524601, 270705447, 13264566901, 649963778147, 31848225129201, 1560563031330847, 76467588535211501, 3746911838225363547, 183598680073042813801, 8996335323579097876247
Offset: 1

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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

Crossrefs

Programs

  • 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
  • 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 *)

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

Extensions

Name edited by Petros Hadjicostas, Oct 09 2019

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

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Comments

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

Examples

			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)
		

Crossrefs

Programs

  • Magma
    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
    
  • Mathematica
    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 *)
  • PARI
    a(n) = lift(Mod(24, 11^n)^-1) \\ David A. Corneth, Jul 29 2020
    
  • SageMath
    def a(n): return 24.inverse_mod(11^n)
    print([a(n) for n in range(1, 22)]) # Peter Luschny, Jul 29 2020

Formula

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)

Extensions

More terms from David A. Corneth, Jul 29 2020

A340757 Counterexamples to a conjecture of Ramanujan about congruences related to the partition function.

Original entry on oeis.org

243, 586, 1272, 2301, 2644, 2987, 3673, 4702, 5045, 5388, 6074, 7103, 7446, 7789, 8475, 9504, 9847, 10190, 10876, 11905, 12248, 12591, 13277, 14306, 14649, 14992, 15678, 16707, 17050, 17393, 18079, 19108, 19451, 19794, 20480, 21509, 21852, 22195, 22881, 23910
Offset: 1

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Author

Washington Bomfim, Jan 19 2021

Keywords

Comments

For b in 5,7,11, and all integers n,e >= 1, Ramanujan conjectured that if (24*n-1) is divisible by b^e, the partition function p(n) = A000041(n) is also divisible by b^e.
Chowla found the first counterexample a(1) = 243. Watson showed the conjecture holds for b=5, and Atkin showed it holds for b=11. Watson showed p(n) is divisible by 7^floor((d+2)/2) when 24n-1 is divisible by 7^d, so that exceptions here are restricted to 24n-1 == 0 (mod 7^3), which is n == 243 (mod 7^3).
See A340957 for the converse, those n == 243 (mod 7^3) where the conjecture does hold.

Examples

			243 is a term because for n = 243, the condition of Ramanujan (24*n - 1) divisible by b^e is true, and p(n) is not divisible by (b^e). [We have base b=7, and exponent e=3 in this case.] Since a(1) = A182719(91), 90 numbers satisfy the conjecture before the first counterexample a(1).
		

Crossrefs

Programs

  • PARI
    seq(x) = {my( n = -100, N=0); while(N < x, n += 343; if(valuation(numbpart(n),7) < valuation(24*n-1,7), print1(n", "); N++)) };
    seq(100); \\ Gives the first 100 terms of the sequence.

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

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Author

Petros Hadjicostas, Sep 24 2019

Keywords

Comments

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.

Crossrefs

Programs

  • Mathematica
    Table[PartitionsP[49n+47]/49,{n, 0, 13}] (* Metin Sariyar, Sep 25 2019 *)
  • PARI
    a(n) = numbpart(49*n + 47)/49; \\ Michel Marcus, Sep 25 2019

Formula

a(n) = A000041(49*n + 47)/49.
Showing 1-5 of 5 results.