A257843 -id:A257843 - cp's OEIS frontend

A073101 Number of integer solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

0, 0, 1, 1, 2, 5, 5, 6, 4, 9, 7, 15, 4, 14, 33, 22, 4, 21, 9, 30, 25, 22, 19, 45, 10, 17, 25, 36, 7, 72, 17, 62, 27, 22, 59, 69, 9, 29, 67, 84, 7, 77, 12, 56, 87, 39, 32, 142, 16, 48, 46, 53, 13, 82, 92, 124, 37, 30, 25, 178, 11, 34, 147, 118, 49, 94, 15, 67, 51, 176, 38, 191, 7

Offset: 1

Robert G. Wilson v, Aug 18 2002

nonn

In 1948 Erdős and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y + 1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z). All of the solutions can be printed by removing the comment symbols from the Mathematica program. For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247). See A075248 for Sierpiński's conjecture for 5/n.

See (A257839, A257840, A257841) for the lexicographically smallest solutions, and A257843 for the differences between these and those with largest z-value. - M. F. Hasler, May 16 2015

			a(5)=2 because there are two solutions: 4/5 = 1/2 + 1/4 + 1/20 and 4/5 = 1/2 + 1/5 + 1/10.

T. D. Noe, Table of n, a(n) for n = 1..1000, (corrected by _Peter Luschny_, Jan 19 2019)
Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227.
David Eppstein, Algorithms for Egyptian Fractions
Paul Erdős, Az 1/z_1 + 1/z_2 + ... + 1/z_n = a/b egyenlet egész számú megoldásairól, (On a Diophantine equation), Mat. Lapok, 1:192-210, 1050. Math. Rev. 13:208b.
Ron Knott, Egyptian Fractions
Eric Weisstein's World of Mathematics, Egyptian Fraction
Wikipedia, Erdős-Straus conjecture

Cf. A075245, A075246, A075247, A075248.

Cf. A192787 (# distinct solutions with x <= y <= z).

import Data.Ratio ((%), numerator, denominator)
a073101 n = length [(x,y) |
   x <- [n `div` 4 + 1 .. 3 * n `div` 4],   let y' = recip $ 4%n - 1%x,
   y <- [floor y' + 1 .. floor (2*y') + 1], let z' = recip $ 4%n - 1%x - 1%y,
   denominator z' == 1 && numerator z' > y && y > x]
-- Reinhard Zumkeller, Jan 03 2011

A:= proc(n)
   local x,t, p,q,ds,zs,ys,js, tot,j;
tot:= 0;
for x from 1+floor(n/4) to ceil(3*n/4)-1 do
    t:= 4/n - 1/x;
    p:= numer(t);
    q:= denom(t);
    ds:= convert(select(d -> (d < q) and d + q mod p = 0,
          numtheory:-divisors(q^2)),list);
    ys:= map(d -> (d+q)/p, ds);
    zs:= map(d -> (q^2/d+q)/p, ds);
    js:= select(j -> ys[j] > x,[$1..nops(ds)]);
    tot:= tot + nops(js);
od;
tot;
end proc:
seq(A(n),n=2..100); # Robert Israel, Aug 22 2014

(* download Egypt.m from D. Eppstein's site and put it into MyOwn directory underneath Mathematica\AddOns\StandardPackages *) Needs["MyOwn`Egypt`"]; Table[ Length[ EgyptianFraction[4/n, Method -> Lexicographic, MaxTerms -> 3, MinTerms -> 3, Duplicates -> Disallow, OutputFormat -> Plain]], {n, 5, 80}]
m = 4; For[lst = {}; n = 2, n <= 100, n++, cnt = 0; xr = n/m; If[IntegerQ[xr], xMin = xr + 1, xMin = Ceiling[xr]]; If[IntegerQ[3xr], xMax = 3xr - 1, xMax = Floor[3xr]]; For[x = xMin, x <= xMax, x++, yr = 1/(m/n - 1/x); If[IntegerQ[yr], yMin = yr + 1, yMin = Ceiling[yr]]; If[IntegerQ[2yr], yMax = 2yr + 1, yMax = Ceiling[2yr]]; For[y = yMin, y <= yMax, y++, zr = 1/(m/n - 1/x - 1/y); If[y > x && zr > y && IntegerQ[zr], z = zr; cnt++; (*Print[n, " ", x, " ", y, " ", z]*)]]]; AppendTo[lst, cnt]]; lst
f[n_] := Length@ Solve[4/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 72, 2] (* Robert G. Wilson v, Jul 17 2013 *)

A073101(n)=sum(c=n\4+1,n*3\4,sum(b=c+1,ceil(2/(t=4/n-1/c))-1,numerator(t-1/b)==1)) \\ M. F. Hasler, May 15 2015

Edited by T. D. Noe, Sep 10 2002

Extended to offset 1 with a(1) = 0 by M. F. Hasler, May 16 2015

A257850 a(n) = floor(n/10) * (n mod 10).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 0, 8

Offset: 0

M. F. Hasler, May 10 2015

Equivalently, write n in base 10, multiply the last digit by the number with its last digit removed.
See A142150(n-1) for the base 2 analog and A257843 - A257849 for the base 3 - base 9 variants.
The first 100 terms coincide with those of A035930 (maximal product of any two numbers whose concatenation is n), A171765 (product of digits of n, or 0 for n<10), A257297 ((initial digit of n)*(n with initial digit removed)), but the sequence is of course different from each of these three.
The terms a(10) - a(100) also coincide with those of A007954 (product of decimal digits of n).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,-1).

Cf. A142150 (the base 2 analog), A115273, A257844 - A257849.

Cf. also A007954, A035930, A171765, A257297.

[Floor(n/10)*(n mod 10): n in [0..100]]; // Vincenzo Librandi, May 11 2015

Table[Floor[n/10] Mod[n, 10], {n, 100}] (* Vincenzo Librandi, May 11 2015 *)

PARI
```
a(n,b=10)=(n=divrem(n,b))[1]*n[2]
```

def A257850(n): return n//10*(n%10) # M. F. Hasler, Sep 01 2021

a(n) = 2*a(n-10)-a(n-20). - Colin Barker, May 11 2015

G.f.: x^11*(9*x^8+8*x^7+7*x^6+6*x^5+5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^4-x^3+x^2-x+1)^2*(x^4+x^3+x^2+x+1)^2). - Colin Barker, May 11 2015

cp's OEIS Frontend

A073101 Number of integer solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

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