A300522 a(n) = (5*n + 3)*(5*n + 4)*(5*n + 5)/6.

10, 120, 455, 1140, 2300, 4060, 6545, 9880, 14190, 19600, 26235, 34220, 43680, 54740, 67525, 82160, 98770, 117480, 138415, 161700, 187460, 215820, 246905, 280840, 317750, 357760, 400995, 447580, 497640, 551300, 608685, 669920, 735130, 804440, 877975, 955860, 1038220, 1125180

Offset: 0

Author

Bruno Berselli, Mar 08 2018

Keywords

• nonn
• easy

Comments

Al-Saedi has discovered that p(10*n+2,4) + p(10*n+3,4) + p(10*n+4,4) == 0 (mod 5), where p(m,k) denote the number of partitions of m into at most k parts [see Theorem 3.6, formula 24, in Links and References sections].

Hirschhorn showed that p(10*n+2,4) + p(10*n+3,4) + p(10*n+4,4) = (5*n+3)*(5*n+4)*(5*n+5)/6 [see References section: paragraph 3, "Proofs of (1.5)-(1.8)"].

The sequence binomial(5*m,3), m>=0, begins 0, 0, 0, 10, 120, 455, 1140, 2300, 4060, ... - N. J. A. Sloane, Jun 13 2020

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, A007318, A300523.

Programs

GAP

List([0..40], n -> (5*n+3)*(5*n+4)*(5*n+5)/6);

Julia

[div((5*n+3)*(5*n+4)*(5*n+5), 6) for n in 0:40] |> println

Magma

[(5*n+3)*(5*n+4)*(5*n+5)/6: n in [0..40]];

Mathematica

Table[(5 n + 3) (5 n + 4) (5 n + 5)/6, {n, 0, 40}]

Maxima

makelist((5*n+3)*(5*n+4)*(5*n+5)/6, n, 0, 40);

PARI

vector(40, n, n--; (5*n+3)*(5*n+4)*(5*n+5)/6)

Python

[(5*n+3)*(5*n+4)*(5*n+5)/6 for n in range(40)]

Sage

[(5*n+3)*(5*n+4)*(5*n+5)/6 for n in (0..40)]

Formula

O.g.f.: 5*(2 + 16*x + 7*x^2)/(1 - x)^4 [formula 4.1 in Hirschhorn's paper].

E.g.f.: 5*(12 + 132*x + 135*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) = -A300523(n-2).

Sum_{n>=0} 1/a(n) = 3*sqrt(5+2/sqrt(5))*Pi/10 - 9*sqrt(5)*log(phi)/10 - 3*log(5)/4. - Amiram Eldar, Jan 04 2022