A112264 - OEIS

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

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OFFSET

1,2

COMMENTS

For primes p, elements of A000040, a(p) = A000030(p). The cumulative sum of this sequence is A112265. Primes in the cumulative sum are A112266. This is a base 10 sequence, the base 1 equivalent is A001222(n) = BigOmega(n) = e_1 + e_2 + ... + e_k, the number of prime factors (with multiplicity), where k = A001221(n) = SmallOmega(n). The base 2 equivalent is equal to the base 1 equivalent.

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

FORMULA

a(1) = 0 and given the prime factorization n = (p_1)^(e_1) * (p_2)^(e_2) * ... * (p_k)^(e_k) then a(n) = (e_1)*A000030(p_1) + (e_2)*A000030(p_2) + ... + (e_k)*A000030(p_l).

EXAMPLE

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.

MATHEMATICA

f[1] = 0; f[n_] := Plus @@ (#[[2]] First@IntegerDigits[#[[1]]] & /@ FactorInteger[n]); Array[f, 94] (* Giovanni Resta, Jun 17 2016 *)

CROSSREFS

Cf. A000030, A000040, A001221, A001222, A112266.