A090026 Number of distinct lines through the origin in 4-dimensional cube of side length n.
0, 15, 65, 225, 529, 1185, 2065, 3745, 5841, 9105, 13025, 19105, 25521, 35361, 45825, 59905, 75425, 96865, 117841, 147505, 177041, 214961, 254401, 306321, 355249, 420929, 485489, 565265, 645377, 748081, 841841, 966881, 1086241, 1230401, 1373185, 1549825
Offset: 0
Keywords
Examples
a(2) = 65 because the 65 points with at least one coordinate=2 all make distinct lines and the remaining 15 points and the origin are on those lines.
Crossrefs
Cf. A000225, A001047, A060867, A090020, A090021, A090022, A090023, A090024 are for n dimensions with side length 1, 2, 3, 4, 5, 6, 7, 8, respectively. A049691, A090025, A090026, A090027, A090028, A090029 are this sequence for 2, 3, 4, 5, 6, 7 dimensions. A090030 is the table for n dimensions, side length k.
Programs
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Mathematica
aux[n_, k_] := If[k == 0, 0, (k + 1)^n - k^n - Sum[aux[n, Divisors[k][[i]]], {i, 1, Length[Divisors[k]] - 1}]];lines[n_, k_] := (k + 1)^n - Sum[Floor[k/i - 1]*aux[n, i], {i, 1, Floor[k/2]}] - 1;Table[lines[4, k], {k, 0, 40}]
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Python
from functools import lru_cache @lru_cache(maxsize=None) def A090026(n): if n == 0: return 0 c, j = 1, 2 k1 = n//j while k1 > 1: j2 = n//k1 + 1 c += (j2-j)*A090026(k1) j, k1 = j2, n//j2 return (n+1)**4-c+15*(j-n-1) # Chai Wah Wu, Mar 30 2021
Formula
a(n) = A090030(4, n).
a(n) = (n+1)^4 - 1 - Sum_{j=2..n+1} a(floor(n/j)). - Chai Wah Wu, Mar 30 2021
Comments