A112264 Sum of initial digits of prime factors (with multiplicity) of n.
0, 2, 3, 4, 5, 5, 7, 6, 6, 7, 1, 7, 1, 9, 8, 8, 1, 8, 1, 9, 10, 3, 2, 9, 10, 3, 9, 11, 2, 10, 3, 10, 4, 3, 12, 10, 3, 3, 4, 11, 4, 12, 4, 5, 11, 4, 4, 11, 14, 12, 4, 5, 5, 11, 6, 13, 4, 4, 5, 12, 6, 5, 13, 12, 6, 6, 6, 5, 5, 14, 7, 12, 7, 5, 13, 5, 8, 6, 7, 13, 12, 6, 8, 14, 6, 6, 5, 7, 8, 13, 8, 6, 6, 6
Offset: 1
Examples
a(4) = 4 because 4 = 2*2, so the sum of the initial digits is 2 + 2 = 4. a(11) = 1 because 11 is prime and its initial digit is 1. a(22) = 3 because 22 = 2*11, so the sum of the initial digits is 2 + 1 = 3. a(98) = 16 because 98 = 2 * 7^2, so the sum of the initial digits is 2 + 7 + 7 = 16. a(100) = 14 because 100 = 2^2 * 5^2, so the sum of the initial digits is 2 + 2 + 5 + 5 = 14. a(121) = 2 because 121 = 11^2, so the sum of the initial digits is 1 + 1 = 2. a(361) = 2 because 361 = 19^2, so the sum of the initial digits is 1 + 1 = 2.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Eric Weisstein's World of Mathematics, Prime Factor
- Eric Weisstein's World of Mathematics, Distinct Prime Factors
Programs
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Mathematica
f[1] = 0; f[n_] := Plus @@ (#[[2]] First@IntegerDigits[#[[1]]] & /@ FactorInteger[n]); Array[f, 94] (* Giovanni Resta, Jun 17 2016 *)
Formula
Extensions
a(6) and a(35) corrected by Giovanni Resta, Jun 17 2016
Comments