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

%I A300522 #33 Sep 08 2022 08:46:20
%S A300522 10,120,455,1140,2300,4060,6545,9880,14190,19600,26235,34220,43680,
%T A300522 54740,67525,82160,98770,117480,138415,161700,187460,215820,246905,
%U A300522 280840,317750,357760,400995,447580,497640,551300,608685,669920,735130,804440,877975,955860,1038220,1125180
%N A300522 a(n) = (5*n + 3)*(5*n + 4)*(5*n + 5)/6.
%C A300522 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].
%C A300522 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)"].
%C A300522 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
%D A300522 Ali H. Al-Saedi, Using Periodicity to Obtain Partition Congruences, Journal of Number Theory, Vol. 178, 2017, pages 158-178.
%D A300522 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 A300522 Colin Barker, <a href="/A300522/b300522.txt">Table of n, a(n) for n = 0..1000</a>
%H A300522 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 A300522 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A300522 O.g.f.: 5*(2 + 16*x + 7*x^2)/(1 - x)^4 [formula 4.1 in Hirschhorn's paper].
%F A300522 E.g.f.: 5*(12 + 132*x + 135*x^2 + 25*x^3)*exp(x)/6.
%F A300522 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F A300522 a(-n) = -A300523(n-2).
%F A300522 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
%t A300522 Table[(5 n + 3) (5 n + 4) (5 n + 5)/6, {n, 0, 40}]
%o A300522 (PARI) vector(40, n, n--; (5*n+3)*(5*n+4)*(5*n+5)/6)
%o A300522 (Sage) [(5*n+3)*(5*n+4)*(5*n+5)/6 for n in (0..40)]
%o A300522 (Maxima) makelist((5*n+3)*(5*n+4)*(5*n+5)/6, n, 0, 40);
%o A300522 (GAP) List([0..40], n -> (5*n+3)*(5*n+4)*(5*n+5)/6);
%o A300522 (Magma) [(5*n+3)*(5*n+4)*(5*n+5)/6: n in [0..40]];
%o A300522 (Python) [(5*n+3)*(5*n+4)*(5*n+5)/6 for n in range(40)]
%o A300522 (Julia) [div((5*n+3)*(5*n+4)*(5*n+5), 6) for n in 0:40] |> println
%Y A300522 Subsequence of A160790.
%Y A300522 Cf. A000292, A007318, A300523.
%K A300522 nonn,easy
%O A300522 0,1
%A A300522 _Bruno Berselli_, Mar 08 2018