A016945 -id:A016945 - cp's OEIS frontend

A249827 Row 3 of A246278: replace in 2n each prime factor p(k) with prime p(k+2).

5, 25, 35, 125, 55, 175, 65, 625, 245, 275, 85, 875, 95, 325, 385, 3125, 115, 1225, 145, 1375, 455, 425, 155, 4375, 605, 475, 1715, 1625, 185, 1925, 205, 15625, 595, 575, 715, 6125, 215, 725, 665, 6875, 235, 2275, 265, 2125, 2695, 775, 295, 21875, 845, 3025, 805, 2375, 305, 8575, 935, 8125, 1015, 925, 335, 9625, 355, 1025, 3185, 78125, 1045, 2975, 365

Offset: 1

Antti Karttunen, Nov 23 2014

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Row 3 of A246278.

Cf. A003961, A016945, A048673, A084967, A249824, A249734.

(define (A249827 n) (A003961 (A249734 n)))
(define (A249827 n) (A246278bi 3 n)) ;; Alternative implementation. Code for A246278bi given in A246278.

a(n) = A003961(A249734(n)).
a(n) = A003961(A016945(A048673(n)-1)).
a(n) = A084967(A249824(n)). [Permutation of A084967.]

A262997 a(n+3) = a(n) + 24*n + 40, a(0)=0, a(1)=5, a(2)=19.

Original entry on oeis.org

0, 5, 19, 40, 69, 107, 152, 205, 267, 336, 413, 499, 592, 693, 803, 920, 1045, 1179, 1320, 1469, 1627, 1792, 1965, 2147, 2336, 2533, 2739, 2952, 3173, 3403, 3640, 3885, 4139, 4400, 4669, 4947, 5232, 5525, 5827, 6136, 6453, 6779, 7112, 7453, 7803, 8160, 8525

Offset: 0

Paul Curtz, Oct 07 2015

The hexasections of A262397(n) are
0,  1,  4,  9, 16,  25,  36, ...  = A000290(n)
0,  5, 19, 40, 69, 107, 152, ...  = a(n)
0,  1,  5, 11, 18,  28,  40, ...  = A240438(n+1)
1,  9, 25, 49, 81, 121, 169, ...  = A016754(n)
0,  2,  7, 13, 21,  32,  44, ...  = A262523(n)
3, 13, 32, 59, 93, 136, 187, ...  = e(n+1).
The five-step recurrence in FORMULA is valuable for the six sequences.
Consider a(n) extended from right to left with their first two differences:
...,  59,  32,  13,  3, 0,  5, 19, 40, 69, ...
..., -27, -19, -10, -3, 5, 14, 21, 29, 38, ...
...,   8,   9,   7,  8, 9,  7,  8,  9,  7, ... .
From 0,the first row is
1) from right to left: e(n)
2) from left to right: a(n).
a(n) and e(n) are companions.
The third row is of period 3.
The last digit of a(n) is of period 15; the same is true of e(n).

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

Cf. A000290, A008586, A016742, A016754, A016789, A016921, A016945, A022267, A042965, A240438, A262397, A262523.

a[0] = 0; a[1] = 5; a[2] = 19; a[n_] := a[n] = a[n - 3] + 24 (n - 3) + 40; Table[a@ n, {n, 0, 46}] (* Michael De Vlieger, Oct 09 2015 *)
LinearRecurrence[{2,-1,1,-2,1},{0,5,19,40,69},60] (* Harvey P. Dale, Dec 16 2024 *)

vector(100, n, n--; 4*n^2 + (4*(n+1)-3)\3) \\ Altug Alkan, Oct 07 2015

concat(0, Vec(-x*(x+1)*(3*x^2+4*x+5)/((x-1)^3*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Oct 08 2015

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) - a(n-5), n> 4.
a(n) = A016742(n) + A042965(n).
a(-n) = e(n).
a(-n) + a(n) = 8*n^2.
a(n+2) - 2*a(n+1) + a(n) = period 3:repeat 9, 7, 8.
a(n+3) - a(n-3) =  8*(1 + 6*n).
a(n+7) - a(n-7) = 40*(2 + 3*n).
a(2n+1) = -a(2n) + 6*n + 3.
a(2n+2) = -a(2n+1) + 4*(n+1).
a(3n) = 4*n*(9*n+1) = 8*A022267(n), a(3n+1) = 36*n^2 +28*n +5, a(3n+2) = 36*n^2 +52*n +19.
G.f.: -x*(x+1)*(3*x^2+4*x+5) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Oct 08 2015

A267797 Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

Original entry on oeis.org

76, 1364, 24476, 439204, 7881196, 141422324, 2537720636, 45537549124, 817138163596, 14662949395604, 263115950957276, 4721424167835364, 84722519070079276, 1520283919093591604, 27280388024614569596, 489526700523968661124, 8784200221406821330636

Offset: 1

Altug Alkan, Jan 24 2016

Lucas numbers that are the averages of 2 distinct positive cubes.
Inspired by relation between sequence A024851 and A188378.
Corresponding indices are 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, ...
6*n + 3 is the corresponding form of indices.
Corresponding y values are listed by A188378, for n > 0. Note that corresponding x values are A188378(n) - 2, for n > 0.

			Lucas number 76 is a term because 76 = (3^3 + 5^3) / 2.
Lucas number 1364 is a term because 1364 = (10^3 + 12^3) / 2.
Lucas number 24476 is a term because 24476 = (28^3 + 30^3) / 2.
Lucas number 439204 is a term because 439204 = (75^3 + 77^3) / 2.
Lucas number 7881196 is a term because 7881196 = (198^3 + 200^3) / 2.
Lucas number 141422324 is a term because 141422324 = (520^3 + 522^3) / 2.

Colin Barker, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000032, A016945, A024670, A024851, A049629, A188378.

[Fibonacci(6*n+4)+Fibonacci(6*n+2): n in [1..20]]; // Vincenzo Librandi, Jan 24 2016

Table[Fibonacci[6 n + 4] + Fibonacci[6 n + 2], {n, 1, 20}] (* Vincenzo Librandi, Jan 24 2016 *)
LinearRecurrence[{18,-1},{76,1364},20] (* Harvey P. Dale, Jul 23 2024 *)

l(n) = fibonacci(n+1) + fibonacci(n-1);
is(n) = for(i=ceil(sqrtn(n\2+1, 3)), sqrtn(n-.5, 3), ispower(n-i^3, 3) && return(1));
for(n=1, 120, if(is(2*l(n)), print1(l(n), ", ")));

a(n) = ((5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n)^3 + (5*fibonacci(n)*fibonacci(n+1) - 1 + (-1)^n)^3) / 2;

a(n) = (fibonacci(6*n+4) + fibonacci(6*n+2));

Vec(4*x*(19-x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

a(n) = A000032(A016945(n)), for n > 0.
a(n) = A188378(n)^3 - 3*A188378(n)^2 + 6*A188378(n) - 4, for n > 0.
From Colin Barker, Jan 24 2016: (Start)
a(n) = (9+4*sqrt(5))^(-n)*(2-sqrt(5)+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)).
a(n) = 18*a(n-1)-a(n-2) for n>2.
G.f.: 4*x*(19-x) / (1-18*x+x^2).
(End)

A364414 Numbers k with the property that the second part of the symmetric representation of sigma(k) is an octagon of width 1 and one of its vertices is also the central vertex of the first valley of the largest Dyck path of the diagram.

Original entry on oeis.org

21, 27, 33, 39, 51, 57, 63, 69, 81, 87, 93, 99, 111, 117, 123, 129, 141, 147, 153, 159, 171, 177, 183, 189, 201, 207, 213, 219, 231, 237, 243, 249, 261, 267, 273, 279, 291, 297, 303, 309, 321, 327, 333, 339, 351, 357

Offset: 1

Omar E. Pol, Jul 23 2023

nonn

Conjecture 1: These are the numbers > 9 that are congruent to {3, 9, 21, 27} mod 30.
Conjecture 2: These are the terms > 9 of A016945 except the terms ending in 5.
Conjecture 3: The polygon mentioned in the definition is an "S"-shaped concave octagon.
Conjecture 4: Every term of this sequence has as nearest neighbor a term of A091999.
Conjecture 5: The terms of A091999 greater than 2 are the numbers k with the property that the first part of the symmetric representation of sigma(k) is an octagon.
Conjecture 6: The octagon mentioned in the definition shares at least an edge with the octagon mentioned in conjecture 5.
Also the row numbers of the triangle A364639 where the rows start with [0, 0, 1, 0, -1]. - Omar E. Pol, Aug 23 2023

			The symmetric representation of sigma(21) in the first quadrant looks like this:
   _ _ _ _ _ _ _ _ _ _ _
  |_ _ _ _ _ _ _ _ _ _ _|
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.
Its second part is an octagon of width 1 and one of its vertices is also the central vertex of the first valley of the largest Dyck path of the structure, so 21 is in the sequence.
Note that 10 is not in the sequence because the second part of the symmetric representation of sigma(10) is an octagon of width 1 in accordance with the definition but none of its vertices is the central vertex of the first valley of the largest Dyck path of the diagram.

Subsequence of A016945.
Cf. A000203, A001068, A045572, A091999, A196020, A235791, A236104, A237270 (parts), A237271, A237591, A237593, A244145, A245092, A249223, A249351
(widths), A262626, A362866, A364639, A365081.

A365081 Numbers k with the property that the symmetric representation of sigma(k) has four parts and its second part is an octagon of width 1 and one of the vertices of the octagon is also the central vertex of the first valley of the largest Dyck path of the diagram.

Original entry on oeis.org

21, 27, 33, 39, 51, 57, 69, 87, 93, 111, 123, 129, 141, 159, 177, 183, 201, 213, 219, 237, 249, 267, 291, 303, 309, 321, 327, 339, 381

Offset: 1

Omar E. Pol, Aug 20 2023

Also the row numbers of the triangle A364639 where the rows are [0, 0, 1, 0, -1, 1] or where the rows start with [0, 0, 1, 0, -1, 1] and the remaining terms are zeros.
Observation: the first 29 terms coincide with the first 29 terms of A161345 that are >= 21.
Apparently a(n)=A127329(n) for n>2. - R. J. Mathar, Sep 05 2023

			The symmetric representation of sigma(21) in the first quadrant looks like this:
   _ _ _ _ _ _ _ _ _ _ _
  |_ _ _ _ _ _ _ _ _ _ _|
                        |
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                        |_ _ _
                        |_ _  |_
                            |_ _|_
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.
There are four parts (or polygons) and its second part is an octagon of width 1 and one of the vertices of the octagon is also the central vertex of the first valley of the largest Dyck path of the structure so 21 is in the sequence.

Subsequence of A016945, A264102, A280107 and A364414.

Cf. A033676, A161345, A196020, A235791, A236104, A237270 (parts), A237271, A237591, A237593, A240062, A245092, A249351 (widths), A262626, A364639.

A380346 Number of corona for a hexagon of edge n with diamonds of side 1.

Original entry on oeis.org

18, 198, 1298, 5778, 19602, 54758, 132498, 287298, 571538, 1060902

Offset: 0

Craig Knecht, Jan 22 2025

The number of diamonds that can surround a hexagon(n) fall into four categories: A016945(n), A016945(n) + 1, A016945(n) + 2, and A016945(n) + 3.
The number of coronal tilings for A016945(n) is 2.
The number of coronal tilings for A016945(n) + 1 is 9,36,81,144,225, see A016766.
The number of coronal tilings for A016945(n) + 2 is 6,96,486,1536,3750,7776,14406 = 6*A000583.
The number of coronal tilings for A016945(n) + 3 is 1,64,729,4096,15625, see A001014.
A008793 looks at the enumeration of diamonds inside the hexagon. In contrast this looks at the enumeration of diamond corona of the hexagon.

Craig Knecht, 198 diamond corona of the H1 hexagon.
Craig Knecht, Coronal corners.
Craig Knecht, Example for the sequence.
Craig Knecht, H1 corona sorted by the number of coronal tiles.
Craig Knecht, Hexagon diamond corona inside a hexagon diamond tiling.
Craig Knecht, Feihu Liu, and Guoce Xin, Enumeration of Corona for Lozenge Tilings, arXiv:2504.19269 [math.CO], 2025.
Walter Trump, Lozenge-Coronae-Around-Hexagon.

Cf. A001014, A008793, A016766, A016945, A380416.

a(n) = n^6 + 6*n^5 + 21*n^4 + 44*n^3 + 60*n^2 + 48*n + 18 (conjectured).

A383403 Partial sums of the sum of the divisors of the numbers of the form 6*k + 3, k >= 0.

Original entry on oeis.org

4, 17, 41, 73, 113, 161, 217, 295, 367, 447, 551, 647, 771, 892, 1012, 1140, 1296, 1488, 1640, 1822, 1990, 2166, 2406, 2598, 2826, 3060, 3276, 3564, 3824, 4064, 4312, 4632, 4968, 5240, 5552, 5840, 6136, 6539, 6923, 7243, 7607, 7943, 8375, 8765, 9125, 9573, 9989, 10469, 10861

Offset: 0

Omar E. Pol, Apr 27 2025

Partial sums of the sum of the divisors of A016945.
See the illustration of a(3) and a(10) as the total area (or total number of cells) in the diagram of the symmetric representation of sigma in the Links section.
Also consider a spiral similar to the spiral described in A239660 but with six wedges on the triangular grid. A "diamond" formed by two adjacent triangles has area 1. a(n) is the total number of diamonds (or the total area) in the third wedge after n + 1 turns. The spiral can be visualized from the top view of the stepped pyramid described in A274536. The graph is named W3 in the Plot 6 of the Links section.

			For n = 3 the first four terms of the numbers of the form 6*k + 3, k >= 0, are [3, 9, 15, 21]. The divisors of them are [1, 3], [1, 3, 9], [1, 3, 5, 15], [1, 3, 7, 21]. The sum of the divisors of them are [4, 13, 24, 32] respectively, and the sum of all divisors of them are 4 + 13 + 24 + 32 = 73, so a(3) = 73.

Omar E. Pol, Illustration of a(3) = 73
Omar E. Pol, Illustration of a(10) = 551
Omar E. Pol, Plot 6. Area of the spiral in the six wedges

Sequences of the same family are A363161, A365442, this sequence, A365444, A383405, A365446.

Cf. A000203, A016945, A237593, A239660, A274536.

Accumulate@ Array[DivisorSigma[1, 6 # + 3] &, 55, 0]

PARI
```
a(n) = sum(k=0, n, sigma(6*k+3));
```

a(n) = Sum_{k=0..n} sigma(6*k+3).

a(n) = (11*Pi^2/24) * n^2 + O(n*log(n)). - Amiram Eldar, Apr 28 2025

A016955 a(n) = (6*n + 3)^11.

Original entry on oeis.org

177147, 31381059609, 8649755859375, 350277500542221, 5559060566555523, 50542106513726817, 317475837322472439, 1532278301220703125, 6071163615208263051, 20635899893042801193, 62050608388552823487, 168787390185178426269, 422351360321044921875, 984770902183611232881

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. A016763, A016945, A016946, A016947, A016948, A016949, A016950, A016951, A016952, A016953, A016954.

Subsequence of A008455.

[(6*n+3)^11: n in [0..20]]; // Vincenzo Librandi, May 06 2011

a[n_] := (6*n + 3)^11; Array[a, 50, 0] (* Amiram Eldar, Mar 30 2022 *)

From Amiram Eldar, Mar 30 2022: (Start)
a(n) = A016945(n)^11.
a(n) = 3^11*A016763(n).
Sum_{n>=0} 1/a(n) = 2047*zeta(11)/362797056.
Sum_{n>=0} (-1)^n/a(n) = 50521*Pi^11/2633035913625600. (End)

A228935 a(n) = (3 - 6n)(-1)^n.

Original entry on oeis.org

0, 3, -9, 15, -21, 27, -33, 39, -45, 51, -57, 63, -69, 75, -81, 87, -93, 99, -105, 111, -117, 123, -129, 135, -141, 147, -153, 159, -165, 171, -177, 183, -189, 195, -201, 207, -213, 219, -225, 231, -237, 243, -249, 255, -261, 267, -273, 279, -285, 291

Offset: 0

Giovanni Artico, Oct 25 2013

Optimal simple continued fraction (with signed denominators) of tan(1/3).
This expansion is a simple continued fraction a(0) + 1/(a(1) + 1/(a(2) + ...)) with all numerators 1, and the numbers a(n) are signed integers computed with an algorithm that guarantees the fastest convergence for this type of continued fraction.
The algorithm is the same described in A133593. It is as usually based on the Euclidean division algorithm, but instead of taking the floor of the quotients, as it happens for standard continued fractions, the nearest integer is taken, so the remainders can be also negative and the quotients also.
Generally the convergents of this expansion are a subset of the convergents of the standard continued fraction expansion, and they are no more alternatively lesser and greater than the given number: in this special case of tan(1/3) they are all lesser.
The terms of this sequence can be generated by this formula: a(n) = -3*(-1)^n*(2n-1) for n > 0 (the first term is a(0)=0).
Repeating the expansion for other numbers of type 1/k a common pattern seems to emerge. Examples:
tan(1/4) gives 0, 4, -12, 20, -28, 36, -44, 52, -60, 68, -76, 84, ...
tan(1/5) gives 0, 5, -15, 25, -35, 45, -55, 65, -75, 85, -95, 105, ...
so it seems that in general the terms of tan(1/k) are generated by the formula a(n) = (-1)^(n+1)*k*(2n-1) for n > 0. This formula gives this expansion for tan(1/k):
tan(1/k) = 1/(k+1/(-3k+1/(5k-1/(-7k+1/(9k+1/...))))).
This expansion can be rewritten in an equivalent form:
tan(1/k) = 1/(k-1/(3k-1/(5k-1/(7k-1/(9k-1/...))))).
The rule for converting the first form to the second is this: if two consecutive terms have different sign put a minus before the fraction line, otherwise put plus, and take the absolute value of the terms in the denominators.
This general expansion seems to be valid for any real value of k and resembles Lambert's expansion of tan(k).
The unsigned version of this sequence is A016945. - Colin Barker, Oct 26 2013

			tan(1/3) = 0 + 1/(3 + 1/(-9 + 1/(15 + 1/(-21 + 1/(27 + ...))))) or
tan(1/3) = 0 + 1/(3 - 1/(9 - 1/(15 - 1/(21 - 1/(27 - ...))))).

Index entries for linear recurrences with constant coefficients, signature (-2,-1).

Cf. A016945, A133593, A228936.

SCF := proc (n, q::posint)::list; local L, i, z; Digits := 10000; L := [round(n)]; z := n; for i from 2 to q do if z = op(-1, L) then break end if; z := 1/(z-op(-1, L)); L := [op(L), round(z)] end do; return L end proc
SCF(tan(1/3), 50)  # Giovanni Artico, Oct 26 2013

Join[{0},LinearRecurrence[{-2,-1},{3,-9},50]] (* Harvey P. Dale, Mar 10 2015 *)

Vec(3*x*(1-x)/(x+1)^2+O(x^100)) \\ Colin Barker, Oct 26 2013

a(n) = A016945(n-1)*(-1)^(n-1).
Formulae for the general case tan(1/k):
a(0)=0; for n > 0, a(n) = (-1)^(n+1)*k*(2n-1).
tan(1/k) = 1/(k + 1/(-3k + 1/(5k - 1/(-7k + 1/(9k + 1/...))))) or equivalently
tan(1/k) = 1/(k - 1/(3k - 1/(5k - 1/(7k - 1/(9k - 1/...))))).
From Colin Barker, Oct 26 2013: (Start)
a(n) = -2*a(n-1) - a(n-2) for n > 2.
G.f.: 3*x*(1-x) / (x+1)^2. (End)

A251758 Let n>=2 be a positive integer with divisors 1 = d_1 < d_2 < ... < d_k = n, and s = d_1d_2 + d_2d_3 + ... + d_(k-1)*d_k. The sequence lists the values a(n) = floor(n^2/s).

Original entry on oeis.org

2, 3, 1, 5, 1, 7, 1, 2, 1, 11, 1, 13, 1, 2, 1, 17, 1, 19, 1, 2, 1, 23, 1, 4, 1, 2, 1, 29, 1, 31, 1, 2, 1, 4, 1, 37, 1, 2, 1, 41, 1, 43, 1, 2, 1, 47, 1, 6, 1, 2, 1, 53, 1, 4, 1, 2, 1, 59, 1, 61, 1, 2, 1, 4, 1, 67, 1, 2, 1, 71, 1, 73, 1, 2, 1, 6, 1, 79, 1, 2, 1

Offset: 2

Michel Lagneau, Dec 08 2014

nonn

s is always less than n^2 and if n is a prime number then s divides n^2.
For n >= 2, the sequence has the following properties:
a(n) = n if n is prime.
a(n) = 1 if n is in A005843 and > 2;
a(n) <= 2 if n is in A016945 and > 3;
a(n) <= 4 if n is in A084967 and > 5;
a(n) <= 6 if n is in A084968 and > 7;
a(n) = 8: <= 35336848261, ...;
a(n) <= 10 if n is in A084969 and > 11;
a(n) <= 12 if n is in A084970 and > 13;
a(n) = 14: 6678671, ...;
This is different from A250480 (a(n) = n for all prime n, and a(n) = A020639(n) - 1 for all composite n), which thus satisfies the above conditions exactly, while with this sequence A020639(n)-1 gives only the guaranteed upper limit for a(n) at composite n. Note that the first different term does not occur until at n = 2431 = 11*13*17, for which a(n) = 9. (See the example below.)
Conjecture: Terms x, where a(x)=n, x=p#k/p#j, p#i is the i-th primorial, k>j is suitable large k and j is the number of primes less than n. As an example, n=9, x = p#7/p#4 = 2431. For n=10, x = p#6/p#4 = 143 although 121 = 11^2 is the least x where a(x)=10 (see formula section). For n=8, x = p#12/p#4, p#13/p#4, p#14/p#4, p#15/p#4, p#16/p#4, etc. But is p#12/p#4 the least such x? - Robert G. Wilson v, Dec 18 2014
n^2/s is only an integer iff n is prime. - Robert G. Wilson v, Dec 18 2014
First occurrence of n >= 1: 4, 2, 3, 25, 5, 49, 7, ??? <= 35336848261, 2431, 121, 11, 169, 13, 6678671, 7429, 289, 17, 361, 19, 31367009, 20677, 529, 23, ..., . - Robert G. Wilson v, Dec 18 2014

			For n = 2431 = 11*13*17, we have (as the eight divisors of 2431 are [1, 11, 13, 17, 143, 187, 221, 2431]) a(n) = floor((2431*2431) / ((1*11)+(11*13)+(13*17)+(17*143)+(143*187)+(187*221)+(221*2431))) = floor(5909761/608125) = floor(9.718) = 9.

Michel Lagneau, Table of n, a(n) for n = 2..10000
International Mathematical Olympiad, Problems, IMO-2002, Problem 4.

Cf. A000040 (prime numbers), A005843 (even numbers), A016945 (6n+3), A084967 (GCD( 5k, 6) =1), A084968 (GCD( 7k, 30) =1), A084969 (GCD( 11k, 30) =1), A084970 (Numbers whose smallest prime factor is 13).
Cf. also A020639 (the smallest prime divisor), A055396 (its index) and arrays A083140 and A083221 (Sieve of Eratosthenes).
Differs from A250480 for the first time at n = 2431, where a(2431) = 9, while A250480(2431) = 10.
Cf. A078730 (sum of products of two successive divisors of n).

with(numtheory):nn:=100:
for n from 2 to nn do:
   x:=divisors(n):n0:=nops(x):s:=sum('x[i]*x[i+1]','i'=1..n0-1):
   z:=floor(n^2/s):printf(`%d, `,z):
od:

f[n_] := Floor[ n^2/Plus @@ Times @@@ Partition[ Divisors@ n, 2, 1]]; Array[f, 81, 2] (* Robert G. Wilson v, Dec 18 2014 *)

a(n) <= A250480(n), and especially, for all composite n, a(n) < A020639(n). [Cf. the Comments section above.] - Antti Karttunen, Dec 09 2014
From Robert G. Wilson v, Dec 18 2014: (Start)
a(n) = floor(n^2/A078730(n));
a(n) = n iff n is prime. (End)

Comments section edited by Antti Karttunen, Dec 09 2014

Instances of n for which a(n) = 8 and 14 found by Robert G. Wilson v, Dec 18 2014

cp's OEIS Frontend

A249827 Row 3 of A246278: replace in 2n each prime factor p(k) with prime p(k+2).

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A262997 a(n+3) = a(n) + 24*n + 40, a(0)=0, a(1)=5, a(2)=19.

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A267797 Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

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A364414 Numbers k with the property that the second part of the symmetric representation of sigma(k) is an octagon of width 1 and one of its vertices is also the central vertex of the first valley of the largest Dyck path of the diagram.

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A365081 Numbers k with the property that the symmetric representation of sigma(k) has four parts and its second part is an octagon of width 1 and one of the vertices of the octagon is also the central vertex of the first valley of the largest Dyck path of the diagram.

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A380346 Number of corona for a hexagon of edge n with diamonds of side 1.

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A383403 Partial sums of the sum of the divisors of the numbers of the form 6*k + 3, k >= 0.

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A016955 a(n) = (6*n + 3)^11.

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A228935 a(n) = (3 - 6n)(-1)^n.

A251758 Let n>=2 be a positive integer with divisors 1 = d_1 < d_2 < ... < d_k = n, and s = d_1d_2 + d_2d_3 + ... + d_(k-1)*d_k. The sequence lists the values a(n) = floor(n^2/s).