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

%I A300523 #27 Sep 08 2022 08:46:20
%S A300523 35,220,680,1540,2925,4960,7770,11480,16215,22100,29260,37820,47905,
%T A300523 59640,73150,88560,105995,125580,147440,171700,198485,227920,260130,
%U A300523 295240,333375,374660,419220,467180,518665,573800,632710,695520,762355,833340,908600,988260,1072445
%N A300523 a(n) = (5*n + 5)*(5*n + 6)*(5*n + 7)/6.
%C A300523 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].
%C A300523 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)"].
%D A300523 Ali H. Al-Saedi, Using Periodicity to Obtain Partition Congruences, Journal of Number Theory, Vol. 178, 2017, pages 158-178.
%D A300523 Michael D. Hirschhorn, Congruences modulo 5 for partitions into at most four parts, The Fibonacci Quarterly, Vol. 56, Number 1, 2018, pages 34-37.
%H A300523 Colin Barker, <a href="/A300523/b300523.txt">Table of n, a(n) for n = 0..1000</a>
%H A300523 Ali H. Al-Saedi, <a href="https://arxiv.org/abs/1609.03633">Using Periodicity to Obtain Partition Congruences</a>, arXiv:1609.03633 [math.NT], 2017, pages 12-13.
%H A300523 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A300523 O.g.f.: 5*(7 + 16*x + 2*x^2)/(1 - x)^4 [formula 4.2 in Hirschhorn's paper].
%F A300523 E.g.f.: 5*(42 + 222*x + 165*x^2 + 25*x^3)*exp(x)/6.
%F A300523 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F A300523 a(-n) = -A300522(n-2).
%t A300523 Table[(5 n + 5) (5 n + 6) (5 n + 7)/6, {n, 0, 40}]
%t A300523 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 *)
%o A300523 (PARI) vector(40, n, n--; (5*n+5)*(5*n+6)*(5*n+7)/6)
%o A300523 (Sage) [(5*n+5)*(5*n+6)*(5*n+7)/6 for n in (0..40)]
%o A300523 (Maxima) makelist((5*n+5)*(5*n+6)*(5*n+7)/6, n, 0, 40);
%o A300523 (GAP) List([0..40], n -> (5*n+5)*(5*n+6)*(5*n+7)/6);
%o A300523 (Magma) [(5*n+5)*(5*n+6)*(5*n+7)/6: n in [0..40]];
%o A300523 (Python) [(5*n+5)*(5*n+6)*(5*n+7)/6 for n in range(40)]
%o A300523 (Julia) [div((5*n+5)*(5*n+6)*(5*n+7), 6) for n in 0:40] |> println
%Y A300523 Subsequence of A160790.
%Y A300523 Cf. A000292, A300522.
%K A300523 nonn,easy
%O A300523 0,1
%A A300523 _Bruno Berselli_, Mar 08 2018