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a(n) = floor(n/5)*floor((n+1)/5)*floor((n+2)/5)*floor((n+3)/5)*floor((n+4)/5).
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%I #43 Jan 10 2023 07:59:38

%S 0,0,0,0,0,1,2,4,8,16,32,48,72,108,162,243,324,432,576,768,1024,1280,

%T 1600,2000,2500,3125,3750,4500,5400,6480,7776,9072,10584,12348,14406,

%U 16807,19208,21952,25088,28672,32768,36864,41472,46656,52488,59049,65610,72900,81000

%N a(n) = floor(n/5)*floor((n+1)/5)*floor((n+2)/5)*floor((n+3)/5)*floor((n+4)/5).

%C For n >= 5, a(n) is the maximal product of 5 positive integers with sum n. - _Wesley Ivan Hurt_, Jun 29 2022

%H <a href="/index/Rec#order_22">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,4,-8,4,0,0,-6,12,-6,0,0,4,-8,4,0,0,-1,2,-1).

%F From _R. J. Mathar_, May 08 2013: (Start)

%F a(n) = +2*a(n-1) -a(n-2) +4*a(n-5) -8*a(n-6) +4*a(n-7) -6*a(n-10) +12*a(n-11) -6*a(n-12) +4*a(n-15) -8*a(n-16) +4*a(n-17) -a(n-20) +2*a(n-21) -a(n-22).

%F G.f.: x^5 *(x^10 -2*x^9 +4*x^8 -4*x^7 +8*x^6 -8*x^5 +8*x^4 -4*x^3 +4*x^2 -2*x+1) *(1+x)^2 / ( (x^4+x^3+x^2+x+1)^4 *(x-1)^6 ). (End)

%F a(5*m) = m^5 (A000584). - _Bernard Schott_, Sep 21 2022

%F Sum_{n>=5} 1/a(n) = 1 + zeta(5). - _Amiram Eldar_, Jan 10 2023

%t CoefficientList[Series[x^5*(x^10 - 2*x^9 + 4*x^8 - 4*x^7 + 8*x^6 - 8*x^5 + 8*x^4 - 4*x^3 + 4*x^2 - 2*x + 1)*(1 + x)^2/((x^4 + x^3 + x^2 + x + 1)^4*(x - 1)^6), {x, 0, 60}], x] (* _Wesley Ivan Hurt_, Jun 29 2022 *)

%o (Maxima) A008382(n):=floor(n/5)*floor((n+1)/5)*floor((n+2)/5)*floor((n+3)/5)*floor((n+4)/5)$

%o makelist(A008382(n),n,0,30); /* _Martin Ettl_, Oct 26 2012 */

%Y Maximal product of k positive integers with sum n, for k = 2..10: A002620 (k=2), A006501 (k=3), A008233 (k=4), this sequence (k=5), A008881 (k=6), A009641 (k=7), A009694 (k=8), A009714 (k=9), A354600 (k=10).

%Y Cf. A000584, A013663.

%K nonn

%O 0,7

%A _N. J. A. Sloane_