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 1-10 of 11 results. Next

A071734 a(n) = p(5n+4)/5 where p(k) denotes the k-th partition number.

Original entry on oeis.org

1, 6, 27, 98, 315, 913, 2462, 6237, 15035, 34705, 77231, 166364, 348326, 710869, 1417900, 2769730, 5308732, 9999185, 18533944, 33845975, 60960273, 108389248, 190410133, 330733733, 568388100, 967054374, 1629808139, 2722189979
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.
Also the coefficients in the expansion of C^5/B^6, in Watson's notation (p. 105). The connection to the partition function is in equation (3.31) with right side 5C^5/B^6 where B = x * f(-x^24), C = x^5 * f(-x^120) where f() is a Ramanujan theta function. Alternatively B = eta(q^24), C = eta(q^120). - Michael Somos, Jan 06 2015

Examples

			G.f. = 1 + 6*x + 27*x^2 + 98*x^3 + 315*x^4 + 913*x^5 + 2462*x^6 + ...
G.f. = q^19 + 6*q^43 + 27*q^67 + 98*q^91 + 315*q^115 + 913*q^139 + ...
		

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

  • Maple
    with(combinat):
    a:= n-> numbpart(5*n+4)/5:
    seq(a(n), n=0..40);  # Alois P. Heinz, Jan 07 2015
  • Mathematica
    a[ n_] := PartitionsP[ 5 n + 4] / 5; (* Michael Somos, Jan 07 2015 *)
    a[ n_] := SeriesCoefficient[ 1 / QPochhammer[ x], {x, 0, 5 n + 4}] / 5; (* Michael Somos, Jan 07 2015 *)
    nmax = 50; CoefficientList[Series[Product[(1 - x^(5*k))^5/(1 - x^k)^6, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Nov 28 2016 *)
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / eta(x + O(x^(5*n + 5))), 5*n + 4) / 5)};
    
  • PARI
    {a(n) = numbpart(5*n + 4) / 5};
    
  • PARI
    a(n)=polcoeff(prod(m=1,n,(1-x^(5*m))^5/(1-x^m +x*O(x^n))^6),n) \\ Paul D. Hanna

Formula

a(n) = (1/5)*A000041(5n+4).
G.f.: Product_{n>=1} (1 - x^(5*n))^5/(1 - x^n)^6 due to Ramanujan's identity. - Paul D. Hanna, May 22 2011
a(n) = A000041(A016897(n))/5 = A213260(n)/5. - Omar E. Pol, Jan 18 2013
Euler transform of period 5 sequence [ 6, 6, 6, 6, 1, ...]. - Michael Somos, Jan 07 2015
Expansion of q^(-19/24) * eta(q^5)^5 / eta(q)^6 in powers of q. - Michael Somos, Jan 07 2015
a(n) ~ exp(Pi*sqrt(10*n/3)) / (100*sqrt(3)*n). - Vaclav Kotesovec, Nov 28 2016

A076394 a(n) = p(11n+6)/11 where p(n) = number of partitions of n (A000041).

Original entry on oeis.org

1, 27, 338, 2835, 18566, 101955, 490253, 2121679, 8424520, 31120519, 108082568, 355805845, 1117485621, 3366123200, 9767105406, 27398618368, 74534264393, 197147918679, 508189847045, 1279140518117, 3149375120229, 7596463993261
Offset: 0

Views

Author

Jeff Burch, Nov 07 2002

Keywords

Comments

That p(11n+6) == 0 (mod 11) is one of the congruences stated by Ramanujan. - Omar E. Pol, Jan 18 2013

Crossrefs

Programs

  • Maple
    seq(combinat:-numbpart(11*n+6)/11, n=0..30); # Robert Israel, Jan 07 2015
  • Mathematica
    PartitionsP[(11*Range[0,30]+6)]/11 (* Harvey P. Dale, May 28 2015 *)
  • PARI
    a(n) = numbpart(11*n+6)/11; \\ Michel Marcus, Jan 07 2015

Formula

a(n) = A000041(A017461(n))/11 = A213256(n)/11. - 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

Views

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

A160527 Coefficients in the expansion of C^3/B^4, in Watson's notation of page 118.

Original entry on oeis.org

1, 4, 14, 40, 105, 252, 574, 1237, 2568, 5138, 9988, 18893, 34937, 63238, 112370, 196244, 337477, 572024, 956956, 1581321, 2583637, 4176495, 6684820, 10599939, 16661401, 25972485, 40171474, 61672695, 94017765, 142368024, 214211760, 320350725, 476299978
Offset: 0

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Author

N. J. A. Sloane, Nov 13 2009

Keywords

Examples

			G.f. = 1 + 4*x + 14*x^2 + 40*x^3 + 105*x^4 + 252*x^5 + 574*x^6 + ...
G.f. = q^17 + 4*q^41 + 14*q^65 + 40*q^89 + 105*q^113 + 252*q^137 + 574*q^161 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax = 50; CoefficientList[Series[Product[(1 - x^(7*k))^3 /(1 - x^k)^4, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Nov 10 2017 *)

Formula

See Maple code in A160525 for formula.
G.f.: Product_{n >= 1} (1 - x^(7*n))^3/(1 - x^n)^4. - Seiichi Manyama, Nov 06 2016
a(n) ~ exp(Pi*sqrt(50*n/21)) * 5 / (196*sqrt(3)*n). - Vaclav Kotesovec, Nov 10 2017

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

Views

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)

A220507 a(n) = spt(7n+5)/7 where spt(n) = A092269(n).

Original entry on oeis.org

2, 34, 260, 1498, 6956, 28024, 100953, 333680, 1026540, 2976024, 8197962, 21608760, 54788100, 134217717, 318816426, 736549424, 1659169712, 3652248590, 7870890952, 16633964444, 34522173765, 70450341042, 141526909340, 280158178412
Offset: 0

Views

Author

Omar E. Pol, Jan 18 2013

Keywords

Comments

That spt(7n+5) == 0 (mod 7) is one of the congruences stated by George E. Andrews. See theorem 2 in the Andrews' paper. See also A220505 and A220513.

Crossrefs

Programs

  • Mathematica
    b[n_, i_] := b[n, i] = If[n==0 || i==1, n, {q, r} = QuotientRemainder[n, i]; If[r == 0, q, 0] + Sum[b[n - i*j, i - 1], {j, 0, n/i}]];
    spt[n_] := b[n, n];
    a[n_] := spt[7 n + 5]/7;
    Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jan 30 2019, after Alois P. Heinz in A092269 *)

Formula

a(n) = A092269(A017041(n))/7 = A220502(n)/7.

A277958 Expansion of Product_{n>=1} (1 - x^(7*n))^7/(1 - x^n)^8 in powers of x.

Original entry on oeis.org

1, 8, 44, 192, 726, 2464, 7704, 22521, 62281, 164252, 415796, 1015334, 2401462, 5519640, 12363062, 27047913, 57917068, 121588588, 250638216, 507974950, 1013409244, 1992161935, 3862461694, 7392045512, 13975011909, 26116935550, 48277368020, 88320521108, 159993054081
Offset: 0

Views

Author

Seiichi Manyama, Nov 06 2016

Keywords

Examples

			G.f.: 1 + 8*x + 44*x^2 + 192*x^3 + 726*x^4 + 2464*x^5 + 7704*x^6 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Product[(1 - x^(7*k))^7 /(1 - x^k)^8 , {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Nov 10 2017 *)

Formula

G.f.: Product_{n>=1} (1 - x^(7*n))^7/(1 - x^n)^8.
A213261(n) = 7*A160527(n) + 49*a(n - 1) for n > 0 due to Ramanujan's congruences.
a(n) ~ exp(Pi*sqrt(98*n/21)) / (1372*sqrt(3)*n). - Vaclav Kotesovec, Nov 10 2017

A333435 Partition numbers A000041(k*x_n + y_n) are known to be divisible by prime(n); sequence gives the list of y_n.

Original entry on oeis.org

4, 5, 6, 237, 2623, 815655
Offset: 3

Views

Author

Frank Ellermann, Mar 21 2020

Keywords

Comments

Grime notes that Ramanujan's pattern for a(3), a(4), a(5) or prime(3), prime(4), prime(5) cannot be directly extended to prime(6) = 13, and shows solutions for 13, 17, 19.

Examples

			All {partition( 5k+4)} are divisible by prime(3) = 5, so a(3) = 4.
All {partition( 7k+5)} are divisible by prime(4) = 7, so a(4) = 5.
All {partition(11k+6)} are divisible by prime(5) = 11, so a(5) = 6.
		

Crossrefs

Cf. A333436 (y_n), A000040 (primes), A000041 (partitions).
Cf. A071734 (p(5k+4)/5), A071746 (p(7k+5)/7), A076394 (p(11k+6)/11).
Cf. A213260 (p(5k+4)).

A333436 Partition numbers A000041(k*x_n + y_n) are known to be divisible by prime(n); sequence gives the list of x_n.

Original entry on oeis.org

5, 7, 11, 17303, 206839, 1977147619
Offset: 3

Views

Author

Frank Ellermann, Mar 21 2020

Keywords

Comments

Grime notes that Ramanujan's pattern for a(3), a(4), a(5) or prime(3), prime(4), prime(5) cannot be directly extended to prime(6) = 13, and shows solutions for 13, 17, 19.

Examples

			All {partition( 5k+4)} are divisible by prime(3) = 5, so a(3) = 5.
All {partition( 7k+5)} are divisible by prime(4) = 7, so a(4) = 7.
All {partition(11k+6)} are divisible by prime(5) = 11, so a(5) = 11.
		

Crossrefs

Cf. A333435 (x_n), A000040 (primes), A000041 (partitions).
Cf. A071734 (p(5k+4)/5), A071746 (p(7k+5)/7), A076394 (p(11k+6)/11).
Cf. A213260 (p(5k+4)).
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