A362006 a(n) is the minimum integer m such that floor(e^n) = floor(Sum_{k=0..m} (n^k)/(k!)).

0, 1, 4, 8, 9, 12, 15, 17, 19, 24, 25, 29, 30, 34, 37, 39, 41, 44, 48, 49, 52, 55, 59, 61, 62, 66, 68, 70, 74, 79, 79, 82, 84, 89, 89, 92, 96, 98, 102, 103, 106, 110, 112, 114, 116, 122, 124, 126, 128, 132, 133, 137, 138, 141, 144, 147, 151, 152, 154, 158, 161

Offset: 0

Author

Luca Onnis, Apr 03 2023

Keywords

• nonn

Comments

Conjecture: a(n) ~ e*n as n->infinity.

Conjecture: a(n) <= 3n for all n.

The second one would imply: A000149(n) = floor(Sum_{k=0..3n} (n^k)/(k!)).

Examples

			a(3) = 8 since floor(e^3) = 20, floor(Sum_{k=0..8} (n^k)/(k!)) = 20 and "8" is the minimum because floor(Sum_{k=0..7} (n^k)/(k!)) = 19.
		

Crossrefs

Cf. A000149.

Programs

Mathematica

f[n_, m_] := Floor[Sum[(n^k)/(k!), {k, 0, m}]] - Floor[E^n];
a[n_] := Min[Flatten[Position[Table[f[n, m], {m, 0, 150}], 0]]] - 1;
Table[a[n], {n, 1, 50}]

PARI

a(n) = my(m=0, x=floor(exp(n)), y=1); while(floor(y) != x, m++; y += n^m/m!); m; \\ Michel Marcus, Apr 14 2023