A300522 -id:A300522 - cp's OEIS frontend

A300523 a(n) = (5n + 5)(5n + 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

Bruno Berselli, Mar 08 2018

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)"].

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.

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).

Subsequence of A160790.

Cf. A000292, A300522.

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

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

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

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 *)

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

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

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

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

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).

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

Original entry on oeis.org

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

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

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

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cp's OEIS Frontend

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Julia

Magma

Mathematica

Maxima

PARI

Python

Sage

Formula

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

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

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