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.

Previous Showing 11-17 of 17 results.

A160521 Coefficients in the expansion of C^7/B^8, in Watson's notation of page 106.

Original entry on oeis.org

1, 8, 44, 192, 726, 2457, 7648, 22220, 60993, 159478, 399906, 966600, 2261630, 5139897, 11378988, 24598683, 52033372, 107890610, 219630050, 439535138, 865784403, 1680352500, 3216454360, 6077280123, 11343018559, 20928404349, 38194869384, 68989715838
Offset: 0

Views

Author

N. J. A. Sloane, Nov 13 2009

Keywords

Examples

			x^27+8*x^51+44*x^75+192*x^99+726*x^123+2457*x^147+7648*x^171+...
		

Crossrefs

Cf. Product_{n>=1} (1 - x^(5*n))^k/(1 - x^n)^(k + 1): A160461 (k=1), A160462 (k=2), A160463 (k=3), A160506 (k=4), A071734 (k=5), A160460 (k=6), this sequence (k=7), A278555 (k=12), A278556 (k=18), A278557 (k=24), A278558 (k=30).

Programs

  • Mathematica
    nmax = 50; CoefficientList[Series[Product[(1 - x^(5*k))^7/(1 - x^k)^8, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Nov 28 2016 *)

Formula

See Maple code in A160458 for formula.
a(n) ~ sqrt(11) * exp(Pi*sqrt(22*n/5)) / (2500*n). - Vaclav Kotesovec, Nov 28 2016

A213260 p(5n+4) where p(k) = number of partitions of k = A000041(k).

Original entry on oeis.org

5, 30, 135, 490, 1575, 4565, 12310, 31185, 75175, 173525, 386155, 831820, 1741630, 3554345, 7089500, 13848650, 26543660, 49995925, 92669720, 169229875, 304801365, 541946240, 952050665, 1653668665, 2841940500, 4835271870, 8149040695, 13610949895, 22540654445, 37027355200, 60356673280, 97662728555, 156919475295
Offset: 0

Views

Author

N. J. A. Sloane, Jun 07 2012

Keywords

Comments

It is known that a(n) is divisible by 5 (see A071734).

Crossrefs

Programs

  • Mathematica
    Table[PartitionsP[5n+4],{n,0,40}] (* Harvey P. Dale, May 30 2013 *)
  • PARI
    a(n) = numbpart(5*n+4); \\ Michel Marcus, Jan 07 2015
    
  • Python
    from sympy.functions import partition
    def a(n): return partition(5*n+4)
    print([a(n) for n in range(33)]) # Michael S. Branicky, May 30 2021

Formula

a(n) = A000041(A016897(n)). - Omar E. Pol, Jan 18 2013

A220505 a(n) = spt(5n+4)/5 where spt(n) = A092269(n).

Original entry on oeis.org

2, 16, 88, 364, 1309, 4126, 11992, 32368, 82590, 200487, 467152, 1049224, 2283364, 4829302, 9959035, 20069790, 39612612, 76703340, 145945332, 273224940, 503888206, 916373028, 1644925432, 2916814954, 5113148026, 8866911378, 15220453704
Offset: 0

Views

Author

Omar E. Pol, Jan 18 2013

Keywords

Comments

That spt(5n+4) == 0 (mod 5) is one of the congruences stated by George E. Andrews. See theorem 2 in the Andrews' paper. See also A220507 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[5n+4]/5;
    Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Jan 30 2019, after Alois P. Heinz in A092269 *)

Formula

a(n) = A092269(A016897(n))/5 = A220485(n)/5.

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

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)).
Previous Showing 11-17 of 17 results.