

A007300


a(1)=2, a(2)=5; for n >= 3, a(n) is smallest number which is uniquely of the form a(j) + a(k) with 1 <= j < k < n.
(Formerly M1328)


18



2, 5, 7, 9, 11, 12, 13, 15, 19, 23, 27, 29, 35, 37, 41, 43, 45, 49, 51, 55, 61, 67, 69, 71, 79, 83, 85, 87, 89, 95, 99, 107, 109, 119, 131, 133, 135, 137, 139, 141, 145, 149, 153, 155, 161, 163, 167, 169, 171, 175, 177, 181, 187, 193, 195, 197, 205, 209, 211, 213, 215
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OFFSET

1,1


COMMENTS

An Ulamtype sequence  see A002858 for many further references, comments, etc.
I have a note saying that this is periodic mod 126. Is that correct?  N. J. A. Sloane, Apr 29 2006
Comments from Joshua Zucker, May 24 2006: "Concerning the conjecture about periodicity mod 126. Out of the first 300 terms, only the 2 and 12 are even. But if you neglect those first 6 terms, mod 2 they're all odd, mod 9 it goes: 0 4 6 1 7 4 6 8 7 2 4 6 8 5 0 8 1 2 5 7 0 2 4 6 1 5 0 2 8 1 5 7 which appears to repeat indefinitely and mod 7 it goes: 0 2 6 1 3 0 2 6 5 4 6 1 2 6 1 3 5 4 1 2 4 0 5 0 2 4 6 1 5 2 6 1 which also appears to repeat indefinitely.
"So it seems as though neglecting the first few terms, it is indeed periodic mod 126 with period 32. In fact it appears that after the first few terms, a(n+32) = a(n) + 126. But this is only based on the first few hundred terms and is not proved!
"The Mathworld link cites a proof that sequences of this type (2,n) have only two even terms and another proof that sequences with only finitely many even terms must eventually have periodic first differences. So I think the period 32 difference of 126 conjecture may be proved in those references."
Given that the sequence of first differences is periodic with period 32 after the first 6 terms (3,2,2,2,1,1), the repeating digits being p=(2,4,4,4,2,6,2,4,2,2,4,2,4,6,6,2,2,8,4,2,2,2,6,4,8,2,10,12,2,2,2,2), one can calculate the nth term (n>6) as a(n)=13+floor((n7)/32)*S(32)+S(n7 mod 32) where S(k)=sum(p(i),i=1..k): (S(k);k=0..32)=(0, 2, 6, 10, 14, 16, 22, 24, 28, 30, 32, 36, 38, 42, 48, 54, 56, 58, 66, 70, 72, 74, 76, 82, 86, 94, 96, 106, 118, 120, 122, 124, 126).  M. F. Hasler, Nov 25 2007


REFERENCES

R. K. Guy, Unsolved Problems in Number Theory, Section C4.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1).


FORMULA

For n > 6, a(n+32) = a(n) + 126.  T. D. Noe, Jan 21 2008


MAPLE

A007300:=n>if n<7 then [2, 5, 7, 9, 11, 12][n] else floor((n7)/32)*126+[13, 15, 19, 23, 27, 29, 35, 37, 41, 43, 45, 49, 51, 55, 61, 67, 69, 71, 79, 83, 85, 87, 89, 95, 99, 107, 109, 119, 131, 133, 135, 137][modp(n7, 32)+1] fi; # M. F. Hasler, Nov 25 2007


MATHEMATICA

theList = {2, 5}; Print[2]; Print[5]; For[i=1, i <= 500, i++, count=0; For[j=1, j <= Length[theList]1, j++, For[k=j+1, k <= Length[theList], k++, If[theList[[j]]+theList[[k]] == i, count++ ]; ]; ]; If[count == 1, Print[i]; theList = Append[theList, i]; ]; ]; (* Sam Handler (sam_5_5_5_0(AT)yahoo.com), Aug 08 2006 *)
Nest[Append[#, SelectFirst[Union@ Select[Tally@ Map[Total, Select[Permutations[#, {2}], #1 < #2 & @@ # &]], Last@ # == 1 &][[All, 1]], Function[k, FreeQ[#, k]]]] &, {2, 5}, 59] (* Michael De Vlieger, Nov 16 2017 *)


PROG

(Haskell)
a007300 n = a007300_list !! (n1)
a007300_list = 2 : 5 : ulam 2 5 a007300_list
 Function ulam as defined in A002858.


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS

More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Aug 08 2006


STATUS

approved



