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.
For n = 22 the a(22) = 2 solutions are: 22 * 24 * 33 = 132^2, and 22 * 27 * 32 * 33 = 792^2. Note that 22 * 24 * 25 * 33 = 660^2 is not a solution because the subsequence [25] has a square product.
a(14) = 4 because 14 * 15 * 16 * 18 * 20 * 21 has four distinct prime factors (2, 3, 5, and 7) and no other square product of a strictly increasing sequence starting at 14 and ending at 21 has fewer distinct prime factors.
a(48)=6 because 48*(48+2)*(48+6) is a square, but you can't square 48 with numbers from (48+1) to (48+5).
Table[k = 0; Which[IntegerQ@ Sqrt@ n, k, And[PrimeQ@ n, n > 3], k = n, True, While[Length@ Select[n Map[Times @@ # &, n + Rest@ Subsets@ Range@ k], IntegerQ@ Sqrt@ # &] == 0, k++]]; k, {n, 40}] (* Michael De Vlieger, Oct 26 2016 *)
Table[k = 0; Which[IntegerQ@Sqrt@n, k, First@Last@FactorInteger@n > Sqrt[2 n] + 1, k = First@Last@FactorInteger@n, True, While[Length@Select[n Map[Times @@ # &, n + Rest@Subsets@Range@k], IntegerQ@Sqrt@# &] == 0, k++]]; k, {n, 100}] (* Joshua Stucky, Dec 05 2022 *)
a(n) = {if(issquare(n),return(0)); if(isprime(n),if(n>3, return(n), return(n+2) )); my(l = List([n,n+1]), m=2); while(1, for(i=1, #l-2, forvec(v = vector(i, j, [2,#l-1]), if(issquare(l[1] * l[#l] * prod(k=1, #v, l[v[k]])), return(l[#l] - n)), 2)); listput(l,n+m);m++)} \\ David A. Corneth, Oct 22 2016
a(0) = 0 via 0 = 0^2 a(1) = 1 via 1 = 1^2 a(2) = 4 via 2 + 3 + 4 = 3^2 a(3) = 6 via 3 + 6 = 3^2 a(4) = 4 via 4 = 2^2 a(5) = 10 via 5 + 6 + 7 + 8 + 10 = 6^2 a(6) = 10 via 6 + 10 = 4^2
import Data.List (find) import Data.Maybe (fromJust) isSquare m = m == (integerRoot * integerRoot) where integerRoot = floor (sqrt (fromIntegral m)::Double) a277278 n | isSquare n = n | otherwise = last $ fromJust $ find (isSquare . sum) s where s = map ((n:) . map (n+)) a048793_tabf -- Peter Kagey, Oct 19 2016
a(n)=if (issquare(n), return (n)); ok = 0; d = 1; while (!ok, for (j=1, 2^d-1, b = Vecrev(binary(j)); if (issquare(n+sum(k=1,#b, b[k]*(n+k))), ok = 1; break);); if (! ok, d++);); n+d; \\ Michel Marcus, Oct 16 2016
a(0) = 0 via 0 = 0^3 a(1) = 1 via 1 = 1^3 a(2) = 4 via 2 * 4 = 2^3 a(3) = 6 via 3 * 4^2 * 6^2 = 12^3 a(4) = 9 via 4 * 6 * 9 = 6^3 a(5) = 10 via 5 * 6 * 9 * 10^2 = 30^3 a(6) = 12 via 6 * 9^2 * 12 = 18^3 a(7) = 14 via 7 * 9^2 * 12^2 * 14^2 = 252^3 a(8) = 8 via 8 = 2^3 a(9) = 16 via 9 * 12 * 16 = 12^3 a(10) = 15 via 10 * 12 * 15^2 = 30^3
For n = 19, a(19) = 28 with the sequence 19 XOR 20 XOR 27 XOR 28 = 0. A table illustrating the first eleven terms: n |a(n)| sequence ---+----+------------------- 0 | 0 | 0 1 | 3 | 1 XOR 2 XOR 3 2 | 5 | 2 XOR 3 XOR 4 XOR 5 3 | 6 | 3 XOR 5 XOR 6 4 | 7 | 4 XOR 5 XOR 6 XOR 7 5 | 10 | 5 XOR 6 XOR 9 XOR 10 6 | 9 | 6 XOR 7 XOR 8 XOR 9 7 | 12 | 7 XOR 11 XOR 12 8 | 11 | 8 XOR 9 XOR 10 XOR 11 9 | 14 | 9 XOR 10 XOR 13 XOR 14 10 | 13 | 10 XOR 11 XOR 12 XOR 13
f:= proc(n) local k,S;
S:= {n};
for k from n+1 do
S:= S union map(Bits:-Xor,S,k);
if member(0,S) then return k fi;
od;
end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Jan 12 2023
f[n_] := Module[{k, S}, S = {n}; For[k = n+1, True, k++, S = S ~Union~ BitXor[S, k]; If[MemberQ[S, 0], Return[k]]]];
f[0] = 0;
f /@ Range[0, 100] (* Jean-François Alcover, Jan 22 2023, after Robert Israel *)
a(n)= if (n==0, return (0), my (x=[n],y); for (m=n+1, oo, if (vecmin(y=[bitxor(v,m) | v<-x])==0, return (m), x=setunion(x,Set(y))))) \\ Rémy Sigrist, Jan 12 2023
a245530 = product . a245499_row
Table[k = 0; While[Length@ # == 0 &@ Set[f, Select[Rest@ Subsets@ Range@ k, IntegerQ@ Sqrt[n (Times @@ # &[n + #])] &]], k++]; If[IntegerQ@ Sqrt@ n, k = {n}, k = n + Prepend[First@ f, 0]]; Times @@ k, {n, 22}] (* Michael De Vlieger, Oct 26 2016 *)
For n = 1, a(1) = 1 with sequence 1 = 1^4. For n = 2, a(2) = 4 with sequence 2^2 * 4 = 2^4. For n = 3, a(3) = 6 with sequence 3^2 * 4 * 6^2 = 6^4. For n = 4, a(4) = 4 with sequence 4^2 = 2^4. For n = 5, a(5) = 10 with sequence 5 * 8^3 * 10^3 = 40^4. For n = 6, a(6) = 9 with sequence 6^2 * 8^2 * 9 = 12^4. For n = 7, a(7) = 14 with sequence 7^2 * 8^2 * 14^2 = 28^4.
a(1) = 1 via [1]
a(2) = 6 via [2, 3, 6]
a(3) = 6 via [2, 3, 6]
a(4) = 12 via [2, 4, 6, 12]
a(5) = 20 via [2, 4, 5, 20]
a(6) = 6 via [2, 3, 6]
a(7) = 28 via [2, 4, 7, 14, 28]
a(8) = 24 via [2, 3, 8, 24]
a(9) = 18 via [2, 3, 9, 18]
a(10) = 15 via [2, 3, 10, 15]
a(11) > 30
a(12) = 12 via [2, 4, 6, 12]
a(13) > 30
a(14) = 28 via [2, 4, 7, 14, 28]
a(15) = 15 via [2, 3, 10, 15]
a(16) > 30
a(17) > 30
a(18) = 18 via [2, 3, 9, 18]
From _Jon E. Schoenfield_, Feb 15 2020: (Start)
For n=31, the largest prime or prime power divisor of n is P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6} has no subset sum that includes 1/(n/P) = 1/1 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7} does have such a subset sum, namely, 1/1 + 1/3 + 1/7 = 31/21, so a(31) >= 7*31 = 217. In fact, the numbers 1*31=31, 3*31=93, and 7*31=217 are elements of many sets of integers that include n=31, include no element > 217, and have a reciprocal sum of 1 (one such set is {2,3,12,28,31,93,217}), so a(31)=217.
For n=62, the largest prime or prime power divisor of n is again P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4} has no subset sum that includes 1/(n/P) = 1/2 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5} does have such a subset sum, namely, 1/2 + 1/3 + 1/5 = 31/30, so a(62) >= 5*31 = 155. In fact, the numbers 2*31=62, 3*31=93, and 5*31=155 are elements of many sets of integers that include n=62, include no element > 155, and have a reciprocal sum of 1 (one such set is {2,3,12,20,62,93,155}), so a(62)=155.
(End)
Table[SelectFirst[Range@ 20, MemberQ[Map[Total, 1/DeleteCases[Rest@ Subsets[Range@ #, #], w_ /; FreeQ[w, n]]], 1] &] /. k_ /; MissingQ@ k -> 0, {n, 12}] (* Michael De Vlieger, Aug 18 2016, Version 10.2, values of a(n) > 20 appear as 0 *)
Comments