cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

Original entry on oeis.org

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
Offset: 1

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Author

Keywords

Comments

An Ulam-type 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 n-th term (n>6) as a(n)=13+floor((n-7)/32)*S(32)+S(n-7 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).

Crossrefs

Programs

  • Haskell
    a007300 n = a007300_list !! (n-1)
    a007300_list = 2 : 5 : ulam 2 5 a007300_list
    -- Function ulam as defined in A002858.
    -- Reinhard Zumkeller, Nov 03 2011
  • Maple
    A007300:=n->if n<7 then [2, 5, 7, 9, 11, 12][n] else floor((n-7)/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(n-7,32)+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 *)

Formula

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

Extensions

More terms from Joshua Zucker, May 24 2006
More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Aug 08 2006

A100729 Period of the first difference of Ulam 1-additive sequence U(2,2n+1).

Original entry on oeis.org

32, 26, 444, 1628, 5906, 80, 126960, 380882, 2097152, 1047588, 148814, 8951040, 5406720, 242, 127842440, 11419626400, 12885001946, 160159528116, 687195466408, 6390911336402, 11728121233408, 20104735604736
Offset: 2

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Author

Ralf Stephan, Dec 03 2004

Keywords

Comments

It was proved by Akeran that a(2^k-1) = 3^(k+1) - 1.
Note that a(n)=2^(2n+1) as soon as A100730(n)=2^(2n+3)-2, that happens for n=(m-2)/2 with m>=6 being an even element of A073639.

Examples

			For k=2, we have a(3)=3^3-1=26.
		

Crossrefs

Cf. A100730 for the fundamental difference, A001857 for U(2, 3), A007300 for U(2, 5), A003668 for U(2, 7).
Cf. also A006844.

Extensions

a(3) corrected from 25 to 26 by Hugo van der Sanden and Bertram Felgenhauer (int-e(AT)gmx.de), Nov 11 2007
More terms from Balakrishnan V (balaji.iitm1(AT)gmail.com), Nov 15 2007
a(21..31) and b-file from Max Alekseyev, Dec 01 2007
Showing 1-2 of 2 results.