A009714 a(n) = Product_{i=0..8} floor((n+i)/9).
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 768, 1152, 1728, 2592, 3888, 5832, 8748, 13122, 19683, 26244, 34992, 46656, 62208, 82944, 110592, 147456, 196608, 262144, 327680, 409600, 512000, 640000, 800000, 1000000, 1250000, 1562500
Offset: 0
Links
- Paolo Xausa, Table of n, a(n) for n = 0..10000
- Index entries for linear recurrences with constant coefficients, 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).
Crossrefs
Programs
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Mathematica
A009714[n_] := Product[Floor[(n + i)/9], {i, 0, 8}]; Array[A009714, 50, 0] (* Paolo Xausa, Aug 21 2024 *)
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PARI
a(n) = prod(k=0, 8, floor((n+k)/9)); \\ Georg Fischer, Nov 07 2019
Formula
a(9*n) = n^9. - Bernard Schott, Nov 20 2022
a(9*n+j) = n^(9-j)*(n+1)^j for 0 <= j <= 8. - Robert Israel, Nov 21 2022
Sum_{n>=9} 1/a(n) = 1 + zeta(9). - Amiram Eldar, Jan 10 2023
Comments