

A096436


a(n) = the number of squared primes and 1's needed to sum to n.


3



1, 2, 3, 1, 2, 3, 4, 2, 1, 2, 3, 3, 2, 3, 4, 4, 3, 2, 3, 4, 4, 3, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 4, 3, 4, 5, 5, 4, 3, 4, 5, 5, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 4, 3, 4, 5, 5, 4, 3, 4, 5, 5, 4, 5, 6, 2, 3, 4, 5, 3, 4, 5, 6, 4, 3, 4, 5, 5, 4, 5, 6, 6, 5, 4, 5, 6, 6, 5, 6, 2, 3, 4, 5, 3, 4, 5, 6
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OFFSET

1,2


COMMENTS

a(n) has a new maximum at n=1,2,3,7,24,73,266,795.
I suspect that a(n) <= 9 for all n.  Robert G. Wilson v, Sep 18 2004


LINKS

Table of n, a(n) for n=1..105.


EXAMPLE

a(5) = 2 because 5=4+1.
a(17) = 3 because 17=9+4+4.
A number may have many such sums: 27=25+1+1=9+9+9, 50=25+25=49+1.


MATHEMATICA

f[n_] := Block[{d = n, k = PrimePi[ Sqrt[n]], sp = {}}, While[d > 3, While[p = Prime[k]; d >= p^2, AppendTo[sp, p]; d = d  p^2]; k ]; While[d != 0, AppendTo[sp, 1]; d = d  1]; If[Position[sp, 3] != {} && sp[[ 3]] == 1, sp = Delete[Drop[sp, 3], Position[sp, 3][[1]]]; AppendTo[sp, {2, 2, 2}]]; Reverse[ Sort[ Flatten[ sp]]]]; Table[ Length[ f[n]], {n, 105}] (* Robert G. Wilson v, Sep 20 2004 *)


CROSSREFS

Cf. A001248, A002828, A045698, A051034, A063274.
Sequence in context: A002828 A191091 A098066 * A053610 A264031 A104246
Adjacent sequences: A096433 A096434 A096435 * A096437 A096438 A096439


KEYWORD

nonn,easy


AUTHOR

Tom Raes (tommy1729(AT)hotmail.com), Aug 10 2004


EXTENSIONS

Edited and extended by Robert G. Wilson v, Sep 18 2004
Edited by Don Reble, Apr 23 2006


STATUS

approved



