

A002858


Ulam numbers: a(1) = 1; a(2) = 2; for n>2, a(n) = least number > a(n1) which is a unique sum of two distinct earlier terms.
(Formerly M0557 N0201)


102



1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, 36, 38, 47, 48, 53, 57, 62, 69, 72, 77, 82, 87, 97, 99, 102, 106, 114, 126, 131, 138, 145, 148, 155, 175, 177, 180, 182, 189, 197, 206, 209, 219, 221, 236, 238, 241, 243, 253, 258, 260, 273, 282, 309, 316, 319, 324, 339
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OFFSET

1,2


COMMENTS

Ulam conjectured that this sequence has density 0. However, calculations up to 6.759*10^8 (Jud McCranie) indicate that the density hovers near 0.074.
A plot of the first 3 million terms shows that they lie very close to the straight line 13.51*n, so even if we cannot prove it, we believe we now know how this sequence grows (see the plots in the links below).  N. J. A. Sloane, Sep 27 2006
After a few initial terms, the sequence settles into a regular pattern of dense clumps separated by sparse gaps, with period 21.601584+. This pattern continues up to at least a(n) = 5*10^6. (This comment is just a qualitative statement about the wavelike distribution of Ulam numbers, not meant to imply that every period includes Ulam numbers.)  David W. Wilson
Don Knuth (Sep 26 2006) remarks that a(4952)=64420 and a(4953)=64682 (a gap of more than ten "dense clumps"); and there is a gap of 315 between a(18857) and a(18858).
1,2,3,47 are the only values of x < 6.759*10^8 such that x and x+1 are both Ulam numbers.  Jud McCranie, Jun 08 2001. This holds through the first 28 billion Ulam numbers  Jud McCranie, Jan 07 2016.
The integers are shown from left to right, top to bottom, with a dot where there is an Ulam number. I think his plot is 216 wide. The local density of Ulam numbers goes in waves with a period of 21.6+, so his plot shows ten cycles.
When they are arranged that way you can see the waves. The crests of the density waves don't always have Ulam numbers there but the troughs are practically void of Ulam numbers. I noticed that the ratio of that period (21.6+) to the frequency of Ulam numbers (1 in 13.52) is very close to 8/5. (End)
a(28000000000) = 378485625853  Philip Gibbs & Jud McCranie, Sep 09 2015
3 (=1+2) and 131 (=62+69) are the only two Ulam numbers in the first 28 billion Ulam numbers that are the sum of two consecutive Ulam numbers.  Jud McCranie, Jan 09 2016
Named after the PolishAmerican scientist Stanislaw Ulam (19091984).  Amiram Eldar, Jun 08 2021


REFERENCES

Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.16.2.
Richard K. Guy, Unsolved Problems in Number Theory, C4.
Donald E. Knuth, The Art of Computer Programming, Volume 4A, Section 7.1.3.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Marvin C. Wunderlich, The improbable behavior of Ulam's summation sequence, pp. 249257 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.
David Zeitlin, Ulam's sequence {U_n}, U_1=1, U_2=2, is a complete sequence, Notices Amer. Math. Soc., 22 (No. 7, 1975), Abstract 75TA267, p. A707.


LINKS

Popular Computing (Calabasas, CA), Sieves: Problem 43, Vol. 2 (No. 13, Apr 1974), pp. 67. This is Sieve #8. [Annotated and scanned copy]
David W. Wilson, Plot of initial terms, showing their quasiperiodicity as vertical bars. The image width was chosen to include approximately 10 periods. For an explanation of this picture, see Comments above.


MAPLE

UlamList := proc(len) local isUlam, nextUlam, behead; behead := u > u[2..numelems(u)]; isUlam := proc(n, h, u, r) local hu, tu, hr, tr; hu := u[1]; hr := r[1]; if h = 2 then return false fi; if hr <= hu then return evalb(h = 1) fi; if (hr + hu) = n then tu := behead(u); tr := behead(r); return isUlam(n, h+1, tu, tr) fi; if (hr + hu) < n then tu := behead(u): return isUlam(n, h, tu, r) fi; tr := behead(r); isUlam(n, h, u, tr) end: nextUlam := proc(n, u, r) if isUlam(n, 0, u, r) then if nops(u) = len1 then return [op(u), n] fi; nextUlam(n+1, [op(u), n], [n, op(r)]) else nextUlam(n+1, u, r) fi end: nextUlam(3, [1, 2], [2, 1]) end:


MATHEMATICA

Ulam4Compiled = Compile[{{nmax, _Integer}, {init, _Integer, 1}, {s, _Integer}}, Module[{ulamhash = Table[0, {nmax}], ulam = init}, ulamhash[[ulam]] = 1; Do[ If[Quotient[Plus @@ ulamhash[[i  ulam]], 2] == s, AppendTo[ulam, i]; ulamhash[[i]] = 1], {i, Last[init] + 1, nmax}]; ulam]]; ulams = Ulam4Compiled[355, {1, 2}, 1]
(* Second program: *)
ulams = {1, 2}; Do[AppendTo[ulams, n = Last[ulams]; While[n++; Length[DeleteCases[Intersection[ulams, n  ulams], n/2, 1, 1]] != 2]; n], {100}]; ulams (* JeanFrançois Alcover, Sep 08 2011 *)
findUlams[s_List, j_Integer] := Block[{k = s[[1]] + 1, ss = Plus @@@ Subsets[s, {j}]}, While[ Count[ss, k] != 1, k++]; Append[s, k]]; ulams = Nest[findUlams[#, 2] &, {1, 2}, 70] (* Robert G. Wilson v, Jul 05 2014 *)


PROG

(Haskell)
a002858 n = a002858_list !! (n1)
a002858_list = 1 : 2 : ulam 2 2 a002858_list
ulam :: Int > Integer > [Integer] > [Integer]
ulam n u us = u' : ulam (n + 1) u' us where
u' = f 0 (u+1) us'
f 2 z _ = f 0 (z + 1) us'
f e z (v:vs)  z  v <= v = if e == 1 then z else f 0 (z + 1) us'
 z  v `elem` us' = f (e + 1) z vs
 otherwise = f e z vs
us' = take n us
(Python)
def isUlam(n, h, u, r):
if h == 2: return False
hu = u[0]; hr = r[0]
if hr <= hu: return h == 1
if hr + hu > n: r = r[1:]
elif hr + hu < n: u = u[1:]
else: h += 1; r = r[1:]; u = u[1:]
return isUlam(n, h, u, r)
def UlamList(length):
u = [1, 2]; r = [2, 1]; n = 2
while len(u) < length:
n += 1
if isUlam(n, 0, u[:], r[:]):
u.append(n); r.insert(0, n)
return u
(Julia)
function isUlam(u, n, h, i, r)
h == 2 && return false
ur = u[r]; ui = u[i]
ur <= ui && return h == 1
if ur + ui > n
r = 1
elseif ur + ui < n
i += 1
else
h += 1; i += 1; r = 1
end
isUlam(u, n, h, i, r)
end
function UlamList(len)
u = Array{Int, 1}(undef, len)
u[1] = 1; u[2] = 2; i = 2; n = 2
while i < len
n += 1
if isUlam(u, n, 0, 1, i)
i += 1
u[i] = n
end
end
return u
end
(PARI) aupto(N)= my(seen=vector(N), U=[]); seen[1]=seen[2]=1; for(i=1, N, if(1==seen[i], for(j=1, #U, my(sum=i+U[j]); if(sum>N, break); seen[sum]++); U=concat(U, i))); U \\ Ruud H.G. van Tol, Dec 29 2022


CROSSREFS



KEYWORD

nonn,nice


AUTHOR



EXTENSIONS



STATUS

approved



