A036288 a(n) = 1 + integer log of n: if the prime factorization of n is n = Product (p_j^k_j) then a(n) = 1 + Sum (p_j * k_j) (cf. A001414).

1, 3, 4, 5, 6, 6, 8, 7, 7, 8, 12, 8, 14, 10, 9, 9, 18, 9, 20, 10, 11, 14, 24, 10, 11, 16, 10, 12, 30, 11, 32, 11, 15, 20, 13, 11, 38, 22, 17, 12, 42, 13, 44, 16, 12, 26, 48, 12, 15, 13, 21, 18, 54, 12, 17, 14, 23, 32, 60, 13, 62, 34, 14, 13, 19, 17, 68, 22

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn

Comments

If this function is iterated then, starting at any number n >= 7, we will always reach an 8 - see A212813, A212814, A212815. - N. J. A. Sloane, May 30 2012

a(n) = 1 + Sum_{k=1..A001221(n)} A027748(k) * A124010(k). - Reinhard Zumkeller, May 30 2012

Examples

			12 = 2^2 * 3 so a(12) = 1 + 2^2 + 3 = 8.
		

References

• Bellamy, O. S.; Cadogan, C. C. Subsets of positive integers: their cardinality and maximality properties. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 167--178, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561043 (82b:10006) - From N. J. A. Sloane, May 30 2012
• R. Honsberger, Problem 89, Another Curious Sequence, Mathematical Morsels, MAA, 1978, pp. 223-227.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• J. B. Roberts, Problem E2356, Amer. Math. Monthly, 79 (1972); solution by H. Kappus, loc. cit., 80 (1973), p. 810.

Crossrefs

Equals A001414 + 1.

Cf. A212813, A212814, A212815, A212816, A212908, A212909.

Programs

Haskell

a036288 n = 1 + sum (zipWith (*)
            (a027748_row n) (map fromIntegral $ a124010_row n))
-- Reinhard Zumkeller, May 30 2012

Maple

f:=proc(n) local i,t1; t1:=ifactors(n)[2]; 1+add( t1[i][1]*t1[i][2], i=1..nops(t1)); end; # N. J. A. Sloane, May 30 2012

Mathematica

f[1]=1;f[n_]:=Total[Apply[Times,FactorInteger[n],1]]+1;f/@Range@68 (* Ivan N. Ianakiev, Apr 18 2016 *)

PARI

A036288(n)=1+(n=factor(n))[,1]~*n[,2]  \\ M. F. Hasler, May 30 2012

Extensions

Edited by N. J. A. Sloane, Jun 01 2012