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A112264
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Sum of initial digits of prime factors (with multiplicity) of n.
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3
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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
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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).
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EXAMPLE
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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.
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MATHEMATICA
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f[1] = 0; f[n_] := Plus @@ (#[[2]] First@IntegerDigits[#[[1]]] & /@ FactorInteger[n]); Array[f, 94] (* Giovanni Resta, Jun 17 2016 *)
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CROSSREFS
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KEYWORD
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base,easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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