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a(n) = Product_{i=0..8} floor((n+i)/9).
12

%I #42 Aug 21 2024 12:26:49

%S 0,0,0,0,0,0,0,0,0,1,2,4,8,16,32,64,128,256,512,768,1152,1728,2592,

%T 3888,5832,8748,13122,19683,26244,34992,46656,62208,82944,110592,

%U 147456,196608,262144,327680,409600,512000,640000,800000,1000000,1250000,1562500

%N a(n) = Product_{i=0..8} floor((n+i)/9).

%C For n >= 9, a(n) is the maximal product of 9 positive integers with sum n. - _Wesley Ivan Hurt_, Jul 08 2022

%H Paolo Xausa, <a href="/A009714/b009714.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_74">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 0, 0, 0, 0, 0, 0, 8, -16, 8, 0, 0, 0, 0, 0, 0, -28, 56, -28, 0, 0, 0, 0, 0, 0, 56, -112, 56, 0, 0, 0, 0, 0, 0, -70, 140, -70, 0, 0, 0, 0, 0, 0, 56, -112, 56, 0, 0, 0, 0, 0, 0, -28, 56, -28, 0, 0, 0, 0, 0, 0, 8, -16, 8, 0, 0, 0, 0, 0, 0, -1, 2, -1).

%F a(9*n) = n^9. - _Bernard Schott_, Nov 20 2022

%F a(9*n+j) = n^(9-j)*(n+1)^j for 0 <= j <= 8. - _Robert Israel_, Nov 21 2022

%F Sum_{n>=9} 1/a(n) = 1 + zeta(9). - _Amiram Eldar_, Jan 10 2023

%t A009714[n_] := Product[Floor[(n + i)/9], {i, 0, 8}];

%t Array[A009714, 50, 0] (* _Paolo Xausa_, Aug 21 2024 *)

%o (PARI) a(n) = prod(k=0, 8, floor((n+k)/9)); \\ _Georg Fischer_, Nov 07 2019

%Y Maximal product of k positive integers with sum n, for k = 2..10: A002620 (k=2), A006501 (k=3), A008233 (k=4), A008382 (k=5), A008881 (k=6), A009641 (k=7), A009694 (k=8), this sequence (k=9), A354600 (k=10).

%Y Cf. A001017 (n^9, a subsequence), A013667.

%K nonn,easy

%O 0,11

%A _N. J. A. Sloane_