A014112 -id:A014112 - cp's OEIS frontend

100, 1050, 3850, 9500, 19000, 33350, 53550, 80600, 115500, 159250, 212850, 277300, 353600, 442750, 545750, 663600, 797300, 947850, 1116250, 1303500, 1510600, 1738550, 1988350, 2261000, 2557500, 2878850, 3226050, 3600100, 4002000, 4432750, 4893350, 5384800, 5908100, 6464250

Offset: 0

Bruno Berselli, Mar 12 2018

Hirschhorn has discovered that p(20*n+11,4) + p(20*n+12,4) + p(20*n+13,4) = 25*(n + 1)*(4*n + 3)*(5*n + 4)/3, where p(m,k) denote the number of partitions of m into at most k parts. Therefore, p(20*n+11,4) + p(20*n+12,4) + p(20*n+13,4) == 0 (mod 50) [see Hirschhorn's paper in References section].
a(n) == 0 (mod 3) if n is of the form 2*h + 3*floor(h/3 + 2/3) + 1.
a(n) == 0 (mod 7) if n is a member of A047278.

Michael D. Hirschhorn, Congruences modulo 5 for partitions into at most four parts, The Fibonacci Quarterly, Vol. 56, Number 1, 2018, pages 32-37 [the equation 1.7 contains a typo].

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Subsequence of A014112, A212964, A228958, A268684.

Cf. A300522, A300523.

List([0..40], n -> 25*(n+1)*(4*n+3)*(5*n+4)/3);

[div(25*(n+1)*(4*n+3)*(5*n+4), 3) for n in 0:40] |> println

[25*(n+1)*(4*n+3)*(5*n+4)/3: n in [0..40]];

Table[25 (n + 1) (4 n + 3) (5 n + 4)/3, {n, 0, 40}]

makelist(25*(n+1)*(4*n+3)*(5*n+4)/3, n, 0, 40);

vector(40, n, n--; 25*(n+1)*(4*n+3)*(5*n+4)/3)

Vec(50*(2 + 13*x + 5*x^2) / (1 - x)^4 + O(x^60)) \\ Colin Barker, Mar 13 2018

[25*(n+1)*(4*n+3)*(5*n+4)/3 for n in range(40)]

[25*(n+1)*(4*n+3)*(5*n+4)/3 for n in (0..40)]

O.g.f.: 50*(2 + 13*x + 5*x^2)/(1 - x)^4 [formula 4.3 in Hirschhorn's paper].
E.g.f.: 25*(12 + 114*x + 111*x^2 + 20*x^3)*exp(x)/3.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4)
a(n) = A014112(10*n+8) = A212964(10*n+9) = A228958(10*n+8) = A268684(5*n+4).

cp's OEIS Frontend

A300254 a(n) = 25(n + 1)(4n + 3)(5*n + 4)/3.

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cp's OEIS Frontend

A300254 a(n) = 25*(n + 1)*(4*n + 3)*(5*n + 4)/3.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Julia

Magma

Mathematica

Maxima

PARI

PARI

Python

Sage

Formula

A300254 a(n) = 25(n + 1)(4n + 3)(5*n + 4)/3.