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 1-7 of 7 results.

A071746 a(n) = p(7n+5)/7 where p(k) denotes the k-th partition number.

Original entry on oeis.org

1, 11, 70, 348, 1449, 5334, 17822, 55165, 160215, 441105, 1159752, 2929465, 7142275, 16873472, 38749850, 86737678, 189672868, 405991500, 852077072, 1756048833, 3558408287, 7098041203, 13951818365, 27047831797, 51760979985
Offset: 0

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Author

Benoit Cloitre, Jun 24 2002

Keywords

Comments

One of the congruences related to the partition numbers stated by Ramanujan.

References

  • Berndt and Rankin, "Ramanujan: letters and commentaries", AMS-LMS, History of mathematics, vol. 9, pp. 192-193.
  • G. H. Hardy, Ramanujan, Cambridge Univ. Press, 1940. - From N. J. A. Sloane, Jun 07 2012

Crossrefs

Programs

  • Magma
    a:= func< n | NumberOfPartitions((7*n+5)) div 7 >; [ a(n) : n in [0..30]]; // Vincenzo Librandi, Nov 30 2015
  • Mathematica
    Table[PartitionsP[7n+5]/7, {n, 0, 24}] (* Jean-François Alcover, Nov 30 2015 *)
  • PARI
    a(n)=if(n<0, 0, n=7*n+5; polcoeff(1/eta(x+x*O(x^n)),n)/7)
    
  • PARI
    {a(n)=local(A,B); if(n<0, 0, A=x*O(x^n); B=eta(x^7+A); A=eta(x+A); polcoeff( B^3/A^4 +x*7*B^7/A^8, n))} /* Michael Somos, Jan 01 2006 */
    
  • PARI
    a(n) = numbpart(7*n+5)/7; \\ Michel Marcus, Nov 30 2015
    

Formula

a(n) = (1/7)*A000041(7n+5).
a(n) = A000041(A017041(n))/7 = A213261(n)/7. - Omar E. Pol, Jan 18 2013

A213261 a(n) = p(7*n + 5), where p(k) = number of partitions of k = A000041(k).

Original entry on oeis.org

7, 77, 490, 2436, 10143, 37338, 124754, 386155, 1121505, 3087735, 8118264, 20506255, 49995925, 118114304, 271248950, 607163746, 1327710076, 2841940500, 5964539504, 12292341831, 24908858009, 49686288421, 97662728555, 189334822579, 362326859895, 684957390936, 1280011042268, 2366022741845, 4328363658647, 7840656226137
Offset: 0

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Author

N. J. A. Sloane, Jun 07 2012

Keywords

Comments

It is known that a(n) is divisible by 7 (see A071746).

Crossrefs

Programs

Formula

a(n) = A000041(A017041(n)). - Omar E. Pol, Jan 18 2013
a(n) = 7 * A071746(n). - Joerg Arndt, Nov 06 2016

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

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Author

Petros Hadjicostas, Sep 24 2019

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    a(n) = (23 * 7^(2*n) + 1)/24; \\ Michel Marcus, Sep 25 2019
    
  • PARI
    Vec((1 - 3*x) / ((1 - x)*(1 - 49*x)) + O(x^20)) \\ Colin Barker, Sep 25 2019

Formula

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)

A327582 a(n) = (17 * 7^(2*n+1) + 1)/24. Sequence related to the properties of the partition function A000041 modulo a power of 7.

Original entry on oeis.org

5, 243, 11905, 583343, 28583805, 1400606443, 68629715705, 3362856069543, 164779947407605, 8074217422972643, 395636653725659505, 19386196032557315743, 949923605595308471405, 46546256674170115098843, 2280766577034335639843305, 111757562274682446352321943
Offset: 0

Views

Author

Petros Hadjicostas, Sep 23 2019

Keywords

Comments

If p(n) = A000041(n) is the partition function, Watson (1938) proved that p(7^(2*m+1)*n + a(m)) == 0 mod 7^(m+1) for n >= 0 and m >= 1.
It is well-known that this result is true even for m = 0 (cf. A071746 and the references there).

Examples

			For m=1 and n=0, p(7^(2*1+1)*0 + a(1)) = p(243) = 133978259344888 = 7^2 * 2734250190712.
For m=1 and n=1, p(7^(2*1+1)*1 + a(1)) = p(586) = 224282898599046831034631 = 7^2 * 4577202012225445531319.
		

Crossrefs

Programs

  • PARI
    a(n) = (17 * 7^(2*n+1) + 1)/24; \\ Michel Marcus, Sep 25 2019
    
  • PARI
    Vec((5 - 7*x) / ((1 - x)*(1 - 49*x)) + O(x^15)) \\ Colin Barker, Sep 27 2019

Formula

From Colin Barker, Sep 27 2019: (Start)
G.f.: (5 - 7*x) / ((1 - x)*(1 - 49*x)).
a(n) = 50*a(n-1) - 49*a(n-2) for n>1.
(End)

A327713 Exceptional class of numbers k such that p(25*k + 24) == 0 (mod 125), where p() = A000041().

Original entry on oeis.org

6, 26, 60, 65, 70, 81, 96, 126, 135, 141, 175, 176, 196, 205, 206, 226, 305, 310, 330, 340, 346, 371, 380, 435, 436, 440, 460, 480, 481, 516, 595, 611, 646, 650, 665, 666, 685, 696, 700, 710, 716, 725, 730, 736, 745, 751, 760, 765, 775, 780, 811, 826, 841, 860, 871
Offset: 1

Views

Author

Petros Hadjicostas, Sep 23 2019

Keywords

Comments

The unexceptional class consists of the numbers k == (2, 3, or 4) (mod 5). Watson (1938, p. 111) proved that such numbers k satisfy p(25*k + 24) == 0 (mod 125).
(p(25*a(m) + 24)/125: m >= 1) = (3177000598, 140513239982045202108972, 23104937422373952975695974907848646058, ...).

Examples

			p(25*6 + 24) = p(174) = 397125074750 = 3177000598 * 125 (the only example in Watson (1938)).
		

Crossrefs

Programs

  • PARI
    is(n) = n % 5 < 2 && numbpart(25*n+24)%125==0 \\ David A. Corneth, Sep 23 2019

Extensions

More terms from David A. Corneth, Sep 23 2019

A160524 Exceptional class of numbers k such that p(5k+4) == 0 (mod 25), where p() = A000041().

Original entry on oeis.org

8, 15, 17, 37, 41, 46, 51, 53, 55, 65, 75, 77, 102, 106, 110, 116, 130, 131, 138, 140, 147, 157, 158, 165, 166, 167, 178, 180, 183, 192, 197, 217, 222, 225, 233, 235, 251, 258, 285, 287, 302, 310, 315, 321, 325, 328, 333, 336, 340, 355, 368, 371, 377, 380, 393, 416, 418, 420, 430, 432, 441, 447
Offset: 1

Views

Author

N. J. A. Sloane, Nov 13 2009

Keywords

Comments

The unexceptional class consists of the numbers k == 4 (mod 5).
(p(5*a(m) + 4)/25: m >= 1) = (3007, 553946, 1999837, 61090943985, 341143252095, 2634063438811, 18381830017947, 38993374797785, 81633034103003, ...) - Petros Hadjicostas, Sep 23 2019

Crossrefs

Programs

Extensions

More terms from Petros Hadjicostas, Sep 23 2019

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

Views

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 1-7 of 7 results.