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%I #18 Sep 08 2022 08:46:20
%S 100,1050,3850,9500,19000,33350,53550,80600,115500,159250,212850,
%T 277300,353600,442750,545750,663600,797300,947850,1116250,1303500,
%U 1510600,1738550,1988350,2261000,2557500,2878850,3226050,3600100,4002000,4432750,4893350,5384800,5908100,6464250
%N a(n) = 25*(n + 1)*(4*n + 3)*(5*n + 4)/3.
%C 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].
%C a(n) == 0 (mod 3) if n is of the form 2*h + 3*floor(h/3 + 2/3) + 1.
%C a(n) == 0 (mod 7) if n is a member of A047278.
%D 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].
%H Colin Barker, <a href="/A300254/b300254.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F O.g.f.: 50*(2 + 13*x + 5*x^2)/(1 - x)^4 [formula 4.3 in Hirschhorn's paper].
%F E.g.f.: 25*(12 + 114*x + 111*x^2 + 20*x^3)*exp(x)/3.
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4)
%F a(n) = A014112(10*n+8) = A212964(10*n+9) = A228958(10*n+8) = A268684(5*n+4).
%t Table[25 (n + 1) (4 n + 3) (5 n + 4)/3, {n, 0, 40}]
%o (PARI) vector(40, n, n--; 25*(n+1)*(4*n+3)*(5*n+4)/3)
%o (Sage) [25*(n+1)*(4*n+3)*(5*n+4)/3 for n in (0..40)]
%o (Maxima) makelist(25*(n+1)*(4*n+3)*(5*n+4)/3, n, 0, 40);
%o (GAP) List([0..40], n -> 25*(n+1)*(4*n+3)*(5*n+4)/3);
%o (Magma) [25*(n+1)*(4*n+3)*(5*n+4)/3: n in [0..40]];
%o (Python) [25*(n+1)*(4*n+3)*(5*n+4)/3 for n in range(40)]
%o (Julia) [div(25*(n+1)*(4*n+3)*(5*n+4), 3) for n in 0:40] |> println
%o (PARI) Vec(50*(2 + 13*x + 5*x^2) / (1 - x)^4 + O(x^60)) \\ _Colin Barker_, Mar 13 2018
%Y Subsequence of A014112, A212964, A228958, A268684.
%Y Cf. A300522, A300523.
%K nonn,easy
%O 0,1
%A _Bruno Berselli_, Mar 12 2018
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