A063427 -id:A063427 - cp's OEIS frontend

A066092 Erroneous version of A063427.

2, 6, 4, 20, 3, 42, 8, 18, 10, 110, 4, 156, 14, 10, 16, 272, 18, 342, 5, 28, 22, 506, 8, 100

Offset: 2

dead

A063647 Number of ways to write 1/n as a difference of exactly 2 unit fractions.

0, 1, 1, 2, 1, 4, 1, 3, 2, 4, 1, 7, 1, 4, 4, 4, 1, 7, 1, 7, 4, 4, 1, 10, 2, 4, 3, 7, 1, 13, 1, 5, 4, 4, 4, 12, 1, 4, 4, 10, 1, 13, 1, 7, 7, 4, 1, 13, 2, 7, 4, 7, 1, 10, 4, 10, 4, 4, 1, 22, 1, 4, 7, 6, 4, 13, 1, 7, 4, 13, 1, 17, 1, 4, 7, 7, 4, 13, 1, 13, 4, 4, 1, 22, 4, 4, 4, 10, 1, 22, 4, 7, 4, 4, 4

Offset: 1

Henry Bottomley, Jul 23 2001

Also number of ways to write 1/n as sum of exactly two distinct unit fractions. - Thomas L. York, Jan 11 2014
Also number of positive integers m such that 1/n + 1/m is a unit fraction. - Jon E. Schoenfield, Apr 17 2018
If 1/n = 1/b - 1/c then n = bc/(c-b) and 1/n = 1/(2n-b) + 1/(c+2n) (though it is also the case that 1/n = 1/(2n) + 1/(2n) equivalent to b = c = 0).
Also number of divisors of n^2 less than n. - Vladeta Jovovic, Aug 13 2001
Number of elements in the set {(x,y): x|n, y|n, xVladeta Jovovic, May 03 2002
Also number of positive integers of the form k*n/(k+n). - Benoit Cloitre, Jan 04 2002
This is similar to A062799, having the same first 29 terms. But they are different sequences.
If A001221(n) = omega(n) <= 2, then a(n) = A062799(n); if A001221(n) > 2, then a(n) > A062799(n). - Matthew Vandermast, Aug 25 2004
Number of r X s integer-sided rectangles such that r + s = 4n, r < s and (s - r) | (s * r). - Wesley Ivan Hurt, Apr 24 2020
Also number of integer-sided right triangles with 2n as a leg. Equivalent to the even indices of A046079. - Nathaniel C Beckman, May 14 2020; Jun 26 2020
a(n) is the number of positive integers k such that k+n divides k*n. - Thomas Ordowski, Dec 02 2024

			a(10) = 4 since 1/10 = 1/5 - 1/10 = 1/6 - 1/15 = 1/8 - 1/40 = 1/9 - 1/90.
a(12) = 7: the divisors of 12 are 1, 2, 3, 4, 6 and 12 and the decompositions are (1, 2), (1, 3), (1, 4), (1, 6), (1, 12), (2, 3), (3, 4).

T. D. Noe, Table of n, a(n) for n = 1..10000
Christopher J. Bradley, Solution to Problem 2175, Crux Mathematicorum, Vol. 23, No. 7, (Nov 1997), pp. 443-444.
Umberto Cerruti, Percorsi tra i numeri (in Italian), pages 3-4.
Roger B. Eggleton, Unitary Fractions: 10501, The American Mathematical Monthly, Vol. 105, No. 4 (Apr., 1998), p. 372.
Amiram Eldar, Plot of (Sum_{i=1..n} a(i))/f(n) - 1, where f(n) is an asymptotic function, for n = 2^(1..28).
Amiram Eldar, Plot of Sum_{i=1..n} a(i)/(n*log(n)*log(log(n))) for n = 2^(1..28).
Amarnath Murthy, Decomposition of the divisors of a natural number into pairwise coprime sets, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001, pp. 303-306.

Cf. A018892, A063427, A063428, A129296.
First twenty-nine terms identical to those of A062799.
Cf. A001221, A046079, A048691, A063717, A063718.
Cf. A001620, A082633, A013661, A201994, A306016.

[(NumberOfDivisors(n^2)-1)/2 : n in [1..100]]; // Vincenzo Librandi, Apr 18 2018

Table[(Length[Divisors[n^2]] - 1)/2, {n, 1, 100}]
(DivisorSigma[0,Range[100]^2]-1)/2 (* Harvey P. Dale, Apr 15 2013 *)

for(n=1,100,print1(sum(i=1,n^2,if((n*i)%(i+n),0,1)),","))

a(n)=numdiv(n^2)\2 \\ Charles R Greathouse IV, Oct 03 2016

a(n) = (tau(n^2)-1)/2.
a(n) = A018892(n)-1. If n = (p1^a1)(p2^a2)...(pt^at), a(n) = ((2*a1+1)(2*a2+1)...(2*at+1)-1)/2.
If n is prime a(n)=1. Conjecture: (1/n)*Sum_{i=1..n} a(i) = C*log(n)*log(log(n)) + o(log(n)) with C=0.7... [The conjecture is false. See the plot and the asymptotic formula below. - Amiram Eldar, Oct 03 2024]
Bisection of A046079. - Lekraj Beedassy, Jul 09 2004
a(n) = Sum_{i=1..2*n-1} (1 - ceiling(i*(4*n-i)/(4*n-2*i)) + floor(i*(4*n-i)/(4*n-2*i))). - Wesley Ivan Hurt, Apr 24 2020
Sum_{k=1..n} a(k) ~ (n/(2*zeta(2)))*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 - zeta(2) + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Oct 03 2024

A127730 Triangle read by rows: row n consists of the positive integers m where m+n divides m*n.

Original entry on oeis.org

2, 6, 4, 12, 20, 3, 6, 12, 30, 42, 8, 24, 56, 18, 72, 10, 15, 40, 90, 110, 4, 6, 12, 24, 36, 60, 132, 156, 14, 35, 84, 182, 10, 30, 60, 210, 16, 48, 112, 240, 272, 9, 18, 36, 63, 90, 144, 306, 342, 5, 20, 30, 60, 80, 180, 380, 28, 42, 126, 420, 22, 99, 220, 462

Offset: 2

Leroy Quet, Jan 26 2007

The maximum term of the n-th row, for n >= 2, is n*(n-1). The minimum term of row n is A063427(n). Row n contains A063647(n) terms (according to a comment by Benoit Cloitre). For p prime, row p^k has k terms. (Each term in row p^k is of the form p^k*(p^j -1), 1 <= j <= k.)

			Row 6 is (3,6,12,30) because 6+3 = 9 divides 6*3 = 18, 6+6 = 12 divides 6*6 = 36, 6+12 = 18 divides 6*12 = 72 and 6+30 = 36 divides 6*30 = 180.

Nathaniel Johnston, Rows n = 2..500, flattened

Cf. A063427, A063647, A127731, A191973.

for n from 2 to 20 do for m from 1 to n*(n-1) do if(m*n mod (m+n) = 0)then printf("%d, ",m): fi: od: od: # Nathaniel Johnston, Jun 22 2011

f[n_] := Select[Range[n^2], Mod[n*#, n + # ] == 0 &];Table[f[n], {n, 2, 24}] // Flatten (* Ray Chandler, Feb 13 2007 *)

arow(n)=local(d,m);d=divisors(n^2);vector(#d\2,k,m=d[ #d\2-k+1];n*(n-m)/m) \\ Franklin T. Adams-Watters, Aug 07 2009

Let d_n be the sequence of divisors of n^2 that are less than n, in reverse order. Then T(n,k) = n*(n-d_n(k))/d_n(k). - Franklin T. Adams-Watters, Aug 07 2009

Extended by Ray Chandler, Feb 13 2007

A063428 a(n) is the smallest positive integer of the form n*k/(n+k).

Original entry on oeis.org

1, 2, 2, 4, 2, 6, 4, 6, 5, 10, 3, 12, 7, 6, 8, 16, 6, 18, 4, 12, 11, 22, 6, 20, 13, 18, 12, 28, 5, 30, 16, 22, 17, 10, 9, 36, 19, 26, 8, 40, 6, 42, 22, 18, 23, 46, 12, 42, 25, 34, 26, 52, 18, 30, 7, 38, 29, 58, 10, 60, 31, 14, 32, 40, 22, 66, 34, 46, 20, 70, 8, 72, 37, 30, 38, 28, 26

Offset: 2

Henry Bottomley, Jul 19 2001

nonn

Or, smallest b such that 1/n + 1/c = 1/b has integer solutions.
Largest b is (n-1) since 1/n + 1/(n(n-1)) = 1/(n-1).
a(n) = smallest k such that k*n/(k-n) is an integer. - Derek Orr, May 29 2014

			a(6) = 2 because 6*3/(6+3) = 2 is the smallest integer of the form 6*k/(6+k).
a(10) = 5 since 1/10 + 1/10 = 1/5, 1/10 + 1/15 = 1/6, 1/10 + 1/40 = 1/8, 1/10 + 1/90 = 1/9 and so the first sum provides the value.

Harry J. Smith, Table of n, a(n) for n = 2..1000

Cf. A018892, A063427, A127730, A063647, A063648, A063649, A066092.

spi[n_]:=Module[{k=1},While[!IntegerQ[(n*k)/(n+k)],k++];(n*k)/(n+k)]; Array[ spi,80,2] (* Harvey P. Dale, May 05 2022 *)

a(n)={my(k=1); if(n>1, while (n*k%(n + k), k++); n*k/(n + k))} \\ Harry J. Smith, Aug 20 2009

a(n) = n*A063427(n)/(n + A063427(n)) = 2n - A063649(n).

If n is prime a(n) = n - 1. - Benoit Cloitre, Dec 31 2001

New description from Benoit Cloitre, Dec 31 2001
Entry revised by N. J. A. Sloane, Feb 13 2007
Definition revised by Franklin T. Adams-Watters, Aug 07 2009

A339144 a(n) is the smallest positive integer such that n*a(n) contains n+a(n) as a substring. If no such number exists then a(n) = -1.

Original entry on oeis.org

-1, 2, -1, 68, -1, -1, -1, 44, -1, 890, 110, 60, 44, 35, 30, 27, 25, 23, 22, 20, 929, 19, 18, 88, 17, -1, 16, 16, 68, 15, 15, 60, 58, 56, 14, 14, 14, 371, 48, 360, 336, 562, 9104, 8, 13, 13, 283, 39, 269, 450, 37, 452, 245, 18, 679, 34, 225, 33, 2053, 12, 12, 12, 12, 12, 30, 369, 889, 4, 16961

Offset: 1

Scott R. Shannon, Nov 25 2020

For n = 1, 3, 5, 6, 7, 9, 26 no value has been found for which n*a(n) contains n + a(n) as a substring (obviously true for n = 1) for a(n) up to 5x10^10. It is likely, although unproven, that this is the complete list of values for which a(n) = -1.

The sequence values display erratic behavior. Most of the term values appear random but there are ranges of n values with the same value. The largest such range for the first one million terms is a(501000) to a(501499), 500 terms, all of which have a(n) = 1002. Patterns also appear for n value corresponding to multiples of powers-of-ten. For example if n=10^k then a(n) = 89*10^k. The largest value in the first one million terms is a(554635) = 879948670.

			a(2) = 2 as 2*2 = 4 which contains 2 + 2 = 4 as a substring.
a(4) = 68 as 4*68 = 272 which contains 4+68 = 72 as a substring.
a(69) = 16961 as 69*16961 = 1170309 which contains 69+16961 = 17030 as a substring.
a(501000) = 1002 as 501000*1002 = 502002000 which contains 501000+1002 = 502002 as a substring. This is the first of 500 consecutive terms with a(n) = 1002.
a(554635) = 879948670 as 554635*879948670 = 488050330585450 which contain 554635+879948670 = 880503305 as a substring. This is the largest value of a(n) for the first one million terms.

Scott R. Shannon, Table of n, a(n) for n = 1..20000.

Cf. A339107, A063427, A063428, A328095, A333923, A332580.

isok(n, k) = #strsplit(Str(n*k), Str(n+k)) > 1;
a(n) = {if (vecsearch([1, 3, 5, 6, 7, 9, 26], n), return (-1)); my(k=1); while (! isok(k, n), k++); k;} \\ Michel Marcus, Dec 02 2020 and Jan 23 2021

A339403 a(n) is the smallest positive integer such that n*a(n) contains the string n+a(n) in reverse as a substring. If no such number exists then a(n) = -1.

Original entry on oeis.org

0, -1, 2, 24, 37, 26, 34, 35, 57, 9, -1, 12, 11, 45, 193, 228, 28, 51, 23, 44, 841, 11, 27, 18, 3, 626, 5, 22, 16, 46716, 56, 41, 33, 32, 6, 7, 21, 4, 3, 24, 592, 31, 7, 619, 19, 13, 38, 2, 117, 5, 463, 17, 34, 308, 33, 36, 30, 8, 31, 4, 23, 21, 648, 124, 921, 903, 386, 395, 4, 334, 755, 31, 563

Offset: 0

Scott R. Shannon, Dec 03 2020

This is a variation of A339144 where, instead of the n*a(n) containing n+a(n) as a substring, it contains the reverse of the string n+a(n), including any leading zeros.
Based on a search limit of 5x10^9 up to n = 100000 the values of n for which no a(n) is found are n = 10^k, with k>=0, and 17500. A test of 175000 and 1750000 also found no a(n) indicating that all values of the form 17500*10^k may have no term for a(n).
It is found that when n = 200*10^k, with k>=0, the corresponding value for a(n) is significantly larger than neighboring terms. As an example a(20000) = 666843331, which is the largest term up to n = 100000.
Unlike A339144, which contains multiple consecutive terms with the same value of a(n), in this sequence the largest consecutive run of the same a(n) in the first 100000 terms is only two. The first term of these pairs occurs at n = 110, 121, 2717, 4368, 7916, 10100, 11211, 13231, 17271, 44573, 63529.

			a(3) = 24 as 3*24 = 72 which contains reverse(3+24) = reverse(27) = 72 as a substring.
a(6) = 34 as 6*34 = 204 which contains reverse(6+34) = reverse(40) = 04 as a substring. Note the leading zero is included.
a(29) = 46716 as 29*46716 = 1354764 which contains reverse(29+4671) = reverse(46745) = 54764 as a substring.
a(110) = 11 as 110*11 = 1210 which contains reverse(110+11) = reverse(121) = 121 as a substring. This is the first of two consecutive terms with a(n) = 11.
a(20000) = 666843331 as 20000*666843331 = 13336866620000 which contains reverse(20000+666843331) = reverse(666863331) = 133368666 as a substring. This is the largest value in the first 100000 terms.

Scott R. Shannon, Table of n, a(n) for n = 0..10000

Cf. A339144, A339107, A063427, A063428, A328095, A333923, A332580.

isok(n, k) = #strsplit(Str(n*k), concat(Vecrev(Str(n+k)))) > 1;
ispt(n) = my(t); ispower(n,,&t) && (t==10);
a(n) = {if ((n==1) || (n==10) || ispt(n), return (-1)); my(k=0); while (! isok(n, k), k++); k;} \\ Michel Marcus, Jan 22 2021

A063649 Largest b such that 1/n=1/c+1/b has integer solutions with c>b.

Original entry on oeis.org

3, 4, 6, 6, 10, 8, 12, 12, 15, 12, 21, 14, 21, 24, 24, 18, 30, 20, 36, 30, 33, 24, 42, 30, 39, 36, 44, 30, 55, 32, 48, 44, 51, 60, 63, 38, 57, 52, 72, 42, 78, 44, 66, 72, 69, 48, 84, 56, 75, 68, 78, 54, 90, 80, 105, 76, 87, 60, 110, 62, 93, 112, 96, 90, 110, 68, 102, 92, 120

Offset: 2

Henry Bottomley, Jul 23 2001

Smallest b is (n+1) since 1/n = 1/(n(n+1))+1/(n+1).

			a(10)=15 since 1/10=1/20+1/20=1/30+1/15=1/35+1/14=1/60+1/12=1/110+1/11, but the first sum does not have c>b, leaving the second sum to provide the value.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A018892, A063427, A063428, A063647, A063648.

f:= proc(n) local b;
  for b from 2*n-1 by -1 do
     if n*b mod (b-n) = 0 then return b fi
od
end proc:
map(f, [$2..100]); # Robert Israel, Dec 01 2019

a[n_] := n + SelectFirst[Divisors[n^2] // Reverse, #Jean-François Alcover, Jun 07 2020 *)

a(n) = n*A063648(n)/(A063648(n)-n) = 2n-A063428(n).
From Robert Israel, Dec 01 2019: (Start)
a(n) = n + A063717(n).
a(n) = n + 1 if and only if n is prime. (End)

A339107 a(1) = 3; for n>1, a(n) = the smallest positive number not occurring earlier such that a(n-1)*a(n) is divisible by a(n-1)+a(n), where a(n) is not a prime or 4.

Original entry on oeis.org

3, 6, 12, 24, 8, 56, 42, 21, 28, 70, 30, 15, 10, 40, 60, 20, 80, 48, 16, 112, 84, 14, 35, 140, 105, 120, 72, 9, 18, 36, 45, 90, 135, 108, 54, 27, 216, 270, 180, 144, 240, 160, 96, 32, 224, 168, 126, 63, 378, 189, 252, 315, 210, 231, 132, 44, 77, 462, 22, 99, 198, 165, 110, 374, 204, 68, 221, 2652

Offset: 1

Scott R. Shannon, Nov 23 2020

nonn

Given a(n-1) the candidates for a(n) are k*a(n-1)/(a(n-1)-k), where 1<=kA063647(n). The values of a(n-1)*a(n)/(a(n-1)+a(n)) are given by the companion sequence A339133.
The first term is 3 as one can easily show 1 and 2 cannot occur in the sequence; if a(n-1)=1 then 1*a(n)/(1+a(n)) has no integer solution while if a(n-1)=2 then 2*a(n)/(2+a(n)) has the one integer solution a(n)=2, but that is a(n-1).
One can also show that a(n) can never be a prime. The only way a(n)=p can be produced is if a(n-1) = p*(p-1). However the only candidate for a(n+1) if a(n)=p is the number p*(p-1), but that is a(n-1). Thus allowing a(n) to be a prime will halt the sequence; if a number of the form p*(p-1) occurs in the sequence the smallest candidate other than p must be chosen for the next term.
Likewise to avoid halting the sequence the number 4 cannot be chosen; if a(n-1)=4 the only candidates for a(n) would be 4 and 12, but a(3)=12 and 4 cannot occur again, thus a(n-1)=4 would halt the sequence.
It is likely all numbers other than 1,2,4 and the primes appear in the sequence, although this is unknown. The smallest composite not to have appeared after 100 thousand terms is 794. The one fixed point in the first 100 thousand terms is a(20572) = 20572.

			a(2) = 6. The only candidate for a(2) that satisfies 3*a(2) being divisible by 3+a(2) is a(2) = 6.
a(3) = 12. The candidates for a(3) given a(2) = 6 are 3,6,12,30, all of which satisfy 6*a(3) being divisible by 6+a(3). 3 and 6 have already appeared so the next smallest candidate is chosen, being 12.
a(4) = 24. The candidates for a(4) given a(3) = 12 are 4,6,12,24,36,60,132, all of which satisfy 12*a(4) being divisible by 12+a(4). 4 is not allowed as there would be no candidates for a(5), and 6 and 12 have already appeared, so the next smallest candidate is chosen, being 24.
a(5) = 8. There are 10 candidates for a(5) given a(4) = 24, the smallest that has not appeared is 8.

Robert Israel, Table of n, a(n) for n = 1..10000
Scott R. Shannon, Line graph of the first 100000 terms.

Cf. A339133, A063647, A063427, A027750, A000040, A098550, A336957, A064413.

R:= 3: a:= 3: S:= {4,3}:
for i from 2 to 100 do
  Cands:= remove (t -> t < 1 or isprime(t), map(`-`,numtheory:-divisors(a^2),a) minus S);
  a:= min(Cands); R:= R, a; S:= S union {a};
od:
R; # Robert Israel, Mar 23 2023

A063648 Smallest c such that 1/n=1/c+1/b has integer solutions with c>b.

Original entry on oeis.org

6, 12, 12, 30, 15, 56, 24, 36, 30, 132, 28, 182, 42, 40, 48, 306, 45, 380, 45, 70, 66, 552, 56, 150, 78, 108, 77, 870, 66, 992, 96, 132, 102, 84, 84, 1406, 114, 156, 90, 1722, 91, 1892, 132, 120, 138, 2256, 112, 392, 150, 204, 156, 2862, 135, 176, 120, 228, 174

Offset: 2

Henry Bottomley, Jul 23 2001

nonn

Largest c is n(n+1) since 1/n=1/(n(n+1))+1/(n+1)

			a(10)=30 since 1/10=1/20+1/20=1/30+1/15=1/35+1/14=1/60+1/12=1/110+1/11, but the first sum does not have c>b, leaving the second sum to provide the value.

Cf. A018892, A063427, A063428, A063647, A063649.

a(n) =n*A063649(n)/(A063649(n)-n) =A063427(n)+2n.

A063718 a(n) is the smallest divisor of n^2 that is greater than n.

Original entry on oeis.org

4, 9, 8, 25, 9, 49, 16, 27, 20, 121, 16, 169, 28, 25, 32, 289, 27, 361, 25, 49, 44, 529, 32, 125, 52, 81, 49, 841, 36, 961, 64, 99, 68, 49, 48, 1369, 76, 117, 50, 1681, 49, 1849, 88, 75, 92, 2209, 64, 343, 100, 153, 104, 2809, 81, 121, 64, 171, 116, 3481, 72, 3721, 124

Offset: 2

Vladeta Jovovic, Aug 12 2001

nonn

Larger of two distinct numbers with minimum sum whose geometric mean is n. E.g., a(12) = 16 as 12^2 = 144 = 1*144 = 2*72 = 3*48 = 4*36 = 6*24 = 8*18 = 9*16, etc. - Amarnath Murthy, Feb 15 2003

			a(45)=75 because divisors of 45^2 are {1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 135, 225, 405, 675, 2025} and the smallest element greater than 45 is 75.

Harry J. Smith, Table of n, a(n) for n = 2..1000

A063648(n) = n + a(n), A063717(n) = n^2/A063718(n), A063427(n) = n - a(n).

Cf. A063717.

with(numtheory): for n from 2 to 200 do a := divisors(n^2): b := a[(tau(n^2)-1)/2]: printf(`%d,`,n^2/b); od:

sdgn[n_]:=Select[Divisors[n^2],#>n&,1]; Flatten[Array[sdgn,70]] (* Harvey P. Dale, Jun 18 2012 *)

{ for (n=2, 1000, d=divisors(n^2); write("b063718.txt", n, " ", d[2 + length(d)\2]) ) } \\ Harry J. Smith, Aug 28 2009

cp's OEIS Frontend

A066092 Erroneous version of A063427.

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A063647 Number of ways to write 1/n as a difference of exactly 2 unit fractions.

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A127730 Triangle read by rows: row n consists of the positive integers m where m+n divides m*n.

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A063428 a(n) is the smallest positive integer of the form n*k/(n+k).

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A339144 a(n) is the smallest positive integer such that n*a(n) contains n+a(n) as a substring. If no such number exists then a(n) = -1.

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A339403 a(n) is the smallest positive integer such that n*a(n) contains the string n+a(n) in reverse as a substring. If no such number exists then a(n) = -1.

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A063649 Largest b such that 1/n=1/c+1/b has integer solutions with c>b.

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A339107 a(1) = 3; for n>1, a(n) = the smallest positive number not occurring earlier such that a(n-1)*a(n) is divisible by a(n-1)+a(n), where a(n) is not a prime or 4.

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