A300523 a(n) = (5*n + 5)*(5*n + 6)*(5*n + 7)/6.

35, 220, 680, 1540, 2925, 4960, 7770, 11480, 16215, 22100, 29260, 37820, 47905, 59640, 73150, 88560, 105995, 125580, 147440, 171700, 198485, 227920, 260130, 295240, 333375, 374660, 419220, 467180, 518665, 573800, 632710, 695520, 762355, 833340, 908600, 988260, 1072445

Offset: 0

Author

Bruno Berselli, Mar 08 2018

Keywords

• nonn
• easy

Comments

Al-Saedi has discovered that p(10*n+6,4) + p(10*n+7,4) + p(10*n+8,4) == 0 (mod 5), where p(m,k) denote the number of partitions of m into at most k parts [see Theorem 3.6, formula 23, in Links and References sections].

Hirschhorn showed that p(10*n+6,4) + p(10*n+7,4) + p(10*n+8,4) = (5*n+5)*(5*n+6)*(5*n+7)/6 [see References section: paragraph 3, "Proofs of (1.5)-(1.8)"].

References

• Ali H. Al-Saedi, Using Periodicity to Obtain Partition Congruences, Journal of Number Theory, Vol. 178, 2017, pages 158-178.
• Michael D. Hirschhorn, Congruences modulo 5 for partitions into at most four parts, The Fibonacci Quarterly, Vol. 56, Number 1, 2018, pages 34-37.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Ali H. Al-Saedi, Using Periodicity to Obtain Partition Congruences, arXiv:1609.03633 [math.NT], 2017, pages 12-13.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Subsequence of A160790.

Cf. A000292, A300522.

Programs

GAP

List([0..40], n -> (5*n+5)*(5*n+6)*(5*n+7)/6);

Julia

[div((5*n+5)*(5*n+6)*(5*n+7), 6) for n in 0:40] |> println

Magma

[(5*n+5)*(5*n+6)*(5*n+7)/6: n in [0..40]];

Mathematica

Table[(5 n + 5) (5 n + 6) (5 n + 7)/6, {n, 0, 40}]
Table[Times@@(5n+{5,6,7})/6,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{35,220,680,1540},40] (* Harvey P. Dale, Oct 22 2019 *)

Maxima

makelist((5*n+5)*(5*n+6)*(5*n+7)/6, n, 0, 40);

PARI

vector(40, n, n--; (5*n+5)*(5*n+6)*(5*n+7)/6)

Python

[(5*n+5)*(5*n+6)*(5*n+7)/6 for n in range(40)]

Sage

[(5*n+5)*(5*n+6)*(5*n+7)/6 for n in (0..40)]

Formula

O.g.f.: 5*(7 + 16*x + 2*x^2)/(1 - x)^4 [formula 4.2 in Hirschhorn's paper].

E.g.f.: 5*(42 + 222*x + 165*x^2 + 25*x^3)*exp(x)/6.

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

a(-n) = -A300522(n-2).