A380375 Delete all 0's from n, then replace each remaining digit k of n by A000005(k).
1, 2, 2, 3, 2, 4, 2, 4, 3, 1, 11, 12, 12, 13, 12, 14, 12, 14, 13, 2, 21, 22, 22, 23, 22, 24, 22, 24, 23, 2, 21, 22, 22, 23, 22, 24, 22, 24, 23, 3, 31, 32, 32, 33, 32, 34, 32, 34, 33, 2, 21, 22, 22, 23, 22, 24, 22, 24, 23, 4, 41, 42, 42, 43, 42, 44, 42, 44, 43, 2, 21, 22, 22, 23, 22, 24, 22, 24, 23
Offset: 1
Examples
n = 10: delete 0, replace digit 1 by A000005(1), thus a(10) = 1. n = 24: replace digit 2 by A000005(2), replace digit 3 by A000005(3), thus a(22) = 22.
Links
- John Tyler Rascoe, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
a[n_] := FromDigits[DivisorSigma[0, DeleteCases[IntegerDigits[n], 0]]]; Array[a, 100] (* Amiram Eldar, Jan 23 2025 *)
-
Python
def A380375(n): d,y = [1,2,2,3,2,4,2,4,3],"" for i in str(n).replace("0",""): y += str(d[int(i)-1]) return int(y) # John Tyler Rascoe, Feb 27 2025
Formula
a(10^n) = 1.
Comments