A156638 -id:A156638 - cp's OEIS frontend

A060464 Numbers that are not congruent to 4 or 5 mod 9.

0, 1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 15, 16, 17, 18, 19, 20, 21, 24, 25, 26, 27, 28, 29, 30, 33, 34, 35, 36, 37, 38, 39, 42, 43, 44, 45, 46, 47, 48, 51, 52, 53, 54, 55, 56, 57, 60, 61, 62, 63, 64, 65, 66, 69, 70, 71, 72, 73, 74, 75, 78, 79, 80, 81, 82, 83, 84, 87, 88, 89, 90, 91

Offset: 1

N. J. A. Sloane, Apr 10 2001

Conjecture: n is a sum of three cubes iff n is in this sequence.
As of their 2009 paper, Elsenhans and Jahnel did not know of a sum of three cubes that gives 33 or 42.
The problem with 33 is cracked, see links below: 8866128975287528^3 + (-8778405442862239)^3 + (-2736111468807040)^3 = 33. - Alois P. Heinz, Mar 11 2019
Numbers that are congruent to {0, 1, 2, 3, 6, 7, 8} mod 9. - Wesley Ivan Hurt, Jul 21 2016
Heath-Brown conjectures that n is a sum of three cubes in infinitely many ways iff n is in this sequence (and not at all otherwise). See his paper for a conjectural asymptotic. - Charles R Greathouse IV, Mar 12 2019
The problem with 42 is cracked by Andrew Booker from University of Bristol and Andrew Sutherland from Massachusetts Institute of Technology, see the link below: 42 = (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3. - Jianing Song, Sep 07 2019
A third solution to writing 3 as a sum of three third powers was found by the same team using 4 million computer-hours. 3 = 569936821221962380720^3 + (-569936821113563493509)^3 + (-472715493453327032)^3. - Peter Luschny, Sep 20 2019

			30 belongs to this sequence because it has the partition as sum of 3 cubes 30 = (-283059965)^3 + (-2218888517)^3 + (2220422932)^3. - _Artur Jasinski_, Apr 30 2010, edited by _M. F. Hasler_, Nov 10 2015

R. K. Guy, Unsolved Problems in Number Theory, Section D5.
Cohen H. 2007. Number Theory Volume I: Tools and Diophantine Equations. Springer Verlag p. 380. - Artur Jasinski, Apr 30 2010

Harry J. Smith, Table of n, a(n) for n = 1..2000
Nikos Bagis, On the numbers that are sums of three cubes, arXiv:2009.11972 [math.GM], 2020.
Andrew R. Booker, Cracking the problem with 33, March 2019
Andrew R. Booker and Brady Haran, 42 is the new 33, Numberphile video (2019)
Andrew R. Booker and Brady Haran, NEWS: The Mystery of 42 is Solved, Numberphile video (2019)
Tim Browning and Brady Haran, The Uncracked Problem with 33, Numberphile video (2015)
Tim Browning and Brady Haran, 74 is cracked, Numberphile video (2016)
Jean-Louis Colliot-Thélène and Olivier Wittenberg, Groupe de Brauer et points entiers de deux familles de surfaces cubiques affines, Amer. J. Math. 134:5 (2012), pp. 1303-1327.
Andreas-Stephan Elsenhans and Jörg Jahnel, List of solutions of x^3 + y^3 + z^3 = n for n < 1000 neither a cube nor twice a cube
A.-S. Elsenhans, J. Jahnel, New sums of three cubes, Math. Comp. 78 (2009) 1227-1230.
Brady Haran, 569936821221962380720, Numberphile video (2020)
D. R. Heath-Brown, The density of zeros of forms for which weak approximation fails, Mathematics of Computation 59 (1992), pp. 613-623.
Sander G. Huisman, Newer sums of three cubes, arXiv:1604.07746 [math.NT], 2016.
H. Mishima, About n=x^3+y^3+z^3
Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.
University of Bristol, Sum of three cubes for 42 finally solved - using real life planetary computer
Wikipedia, Manin obstruction
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,-1).

Cf. A060465, A060466, A060467, A334521, A334522.

A156638 is the complement of this sequence.

A060464:=Filtered([0..100],n->n mod 9 <>4 and n mod 9 <>5); # Muniru A Asiru, Feb 17 2018

[n : n in [0..150] | n mod 9 in [0, 1, 2, 3, 6, 7, 8]]; // Wesley Ivan Hurt, Jul 21 2016

for n from 0 to 100 do if n mod 9 <> 4 and n mod 9 <> 5 then printf(`%d,`, n) fi:od:

a = {}; Do[If[(Mod[n, 9] == 4) || (Mod[n, 9] == 5), , AppendTo[a, n]], {n, 1, 300}]; a (* Artur Jasinski, Apr 30 2010 *)
Which[Mod[#,9]==4,Nothing,Mod[#,9]==5,Nothing,True,#]&/@Range[0,100] (* Harvey P. Dale, Jul 31 2023 *)

n=-1; for (m=0, 4000, if (m%9!=4 && m%9!=5, write("b060464.txt", n++, " ", m)); if (n==2000, break)) \\ Harry J. Smith, Jul 05 2009

concat(0, Vec(x^2*(x^3+x^2+1)*(x^3+x+1)/((1+x+x^2+x^3+x^4+x^5+x^6)*(x-1)^2) + O(x^100))) \\ Altug Alkan, Nov 06 2015

a(n)=n\7*9+[0, 1, 2, 3, 6, 7, 8][n%7+1] \\ Charles R Greathouse IV, Nov 06 2015

G.f.: x^2*(x^3+x^2+1)*(x^3+x+1) / ( (1+x+x^2+x^3+x^4+x^5+x^6)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jul 21 2016: (Start)
a(n) = a(n-1) + a(n-7) - a(n-8) for n>8; a(n) = a(n-7) + 9 for n>7.
a(n) = (63*n - 63 + 2*(n mod 7) + 2*((n+1) mod 7) - 12*((n+2) mod 7) + 2*((n+3) mod 7) + 2*((n+4) mod 7) + 2*((n+5) mod 7) + 2*((n+6) mod 7))/49.
a(7k) = 9k-1, a(7k-1) = 9k-2, a(7k-2) = 9k-3, a(7k-3) = 9k-6, a(7k-4) = 9k-7, a(7k-5) = 9k-8, a(7k-6) = 9k-9. (End)

More terms from James Sellers, Apr 11 2001

A332437 Decimal expansion of 2*cos(Pi/9).

Original entry on oeis.org

1, 8, 7, 9, 3, 8, 5, 2, 4, 1, 5, 7, 1, 8, 1, 6, 7, 6, 8, 1, 0, 8, 2, 1, 8, 5, 5, 4, 6, 4, 9, 4, 6, 2, 9, 3, 9, 8, 7, 2, 4, 1, 6, 2, 6, 8, 5, 2, 8, 9, 2, 9, 2, 6, 6, 1, 8, 0, 5, 7, 3, 3, 2, 5, 5, 4, 8, 4, 4, 2, 4, 2, 1, 9, 9, 1, 7, 7, 8, 9, 1, 7, 8, 9, 9, 4, 9, 1, 7, 7, 9, 6, 7, 5, 8, 9, 6, 1, 3, 4, 9

Offset: 1

Wolfdieter Lang, Mar 27 2020

This algebraic number called rho(9) of degree 3 = A055034(9) has minimal polynomial C(9, x) = x^3 - 3*x - 1 (see A187360).
rho(9) gives the length ratio diagonal/side of the smallest diagonal in the regular 9-gon.
The length ratio diagonal/side of the second smallest and the third smallest (or the largest) diagonal in the regular 9-gon are rho(9)^2 - 1 = A332438 - 1 and rho(9) + 1, respectively. - Mohammed Yaseen, Oct 31 2020

			rho(9) = 1.87938524157181676810821855464946293987241626852892926618...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 207.

Index entries for algebraic numbers, degree 3.

Cf. A019879, A055034, A056020, A156638, A187360, A214778, A215885, A332438.

RealDigits[2 * Cos[Pi/9], 10, 100][[1]] (* Amiram Eldar, Mar 27 2020 *)

2*cos(Pi/9) \\ Michel Marcus, Mar 28 2020

rho(9) = 2*cos(Pi/9).
Equals (-1)^(-1/9)*((-1)^(1/9) - i)*((-1)^(1/9) + i). - Peter Luschny, Mar 27 2020
Equals 2*A019879. - Michel Marcus, Mar 28 2020
Equals sqrt(A332438). - Mohammed Yaseen, Oct 31 2020
From Peter Bala, Oct 20 2021: (Start)
The zeros of x^3 - 3*x - 1 are r_1 = -2*cos(2*Pi/9), r_2 = -2*cos(4*Pi/9) and r_3 = -2*cos(8*Pi/9) = 2*cos(Pi/9).
The polynomial x^3 - 3*x - 1 is irreducible over Q (since it is irreducible mod 2) with discriminant equal to 3^4, a square. It follows that the Galois group of the number field Q(2*cos(Pi/9)) over Q is cyclic of order 3.
The mapping r -> 2 - r^2 cyclically permutes the zeros r_1, r_2 and r_3. The inverse cyclic permutation is given by r -> r^2 - r - 2.
The first differences r_1 - r_2, r_2 - r_3 and r_3 - r_1 are the zeros of the cyclic cubic polynomial x^3 - 9*x - 9 of discriminant 3^6.
First quotient relations:
r_1/r_2 = 1 + (r_3 - r_1); r_2/r_3 = 1 + (r_1 - r_2); r_3/r_1 = 1 + (r_2 - r_3);
r_2/r_1 = (r_3 - r_2) - 2; r_3/r_2 = (r_1 - r_3) - 2; r_1/r_3 = (r_2 - r_1) - 2;
r_1/r_2 + r_2/r_3 + r_3/r_1 = 3; r_2/r_1 + r_3/r_2 + r_1/r_3 = -6.
Thus the first quotients r_1/r_2, r_2/r_3 and r_3/r_1 are the zeros of the cyclic cubic polynomial x^3 - 3*x^2 - 6*x - 1 of discriminant 3^6. See A214778.
Second quotient relations:
(r_1*r_2)/(r_3^2) = 3*r_2 - 6*r_1 - 8, with two other similar relations by cyclically permuting the 3 zeros. The three second quotients are the zeros of the cyclic cubic polynomial x^3 + 24*x^2 + 3*x - 1 of discriminant 3^10.
(r_1^2)/(r_2*r_3) = 1 - 3*(r_2 + r_3), with two other similar relations by cyclically permuting the 3 zeros. (End)
Equals i^(2/9) + i^(-2/9). - Gary W. Adamson, Jun 25 2022
Equals Re((4+4*sqrt(3)*i)^(1/3)). - Gerry Martens, Mar 19 2024
From Amiram Eldar, Nov 22 2024: (Start)
Equals Product_{k>=1} (1 - (-1)^k/A056020(k)).
Equals 1 + Product_{k>=1} (1 + (-1)^k/A156638(k)). (End)

A185039 Numbers of the form 9m^2 + 4m, m an integer.

Original entry on oeis.org

0, 5, 13, 28, 44, 69, 93, 128, 160, 205, 245, 300, 348, 413, 469, 544, 608, 693, 765, 860, 940, 1045, 1133, 1248, 1344, 1469, 1573, 1708, 1820, 1965, 2085, 2240, 2368, 2533, 2669, 2844, 2988, 3173, 3325, 3520, 3680, 3885, 4053, 4268, 4444, 4669, 4853, 5088

Offset: 1

N. J. A. Sloane, Feb 04 2012

Also, numbers m such that 9*m+4 is a square. After 0, therefore, there are no squares in this sequence. - Bruno Berselli, Jan 07 2016

Bruno Berselli, Table of n, a(n) for n = 1..1000
S. Cooper and M. D. Hirschhorn, Results of Hurwitz type for three squares. Discrete Math. 274 (2004), no. 1-3, 9-24. See B(q).
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Characteristic function is A205809.
Numbers of the form 9*n^2+k*n, for integer n: A016766 (k=0), A132355 (k=2), this sequence (k=4), A057780 (k=6), A218864 (k=8). [Jason Kimberley, Nov 08 2012]
For similar sequences of numbers m such that 9*m+k is a square, see list in A266956.

[0] cat &cat[[9*n^2-4*n,9*n^2+4*n]: n in [1..32]]; // Bruno Berselli, Feb 04 2011

CoefficientList[Series[x*(5+8*x+5*x^2)/((x+1)^2*(1-x)^3), {x,0,50}], x] (* G. C. Greubel, Jun 20 2017 *)
LinearRecurrence[{1,2,-2,-1,1},{0,5,13,28,44},50] (* Harvey P. Dale, Jan 23 2018 *)

x='x+O('x^50); Vec(x*(5+8*x+5*x^2)/((x+1)^2*(1-x)^3)) \\ G. C. Greubel, Jun 20 2017

From Bruno Berselli, Feb 04 2012: (Start)
G.f.: x*(5+8*x+5*x^2)/((x+1)^2*(1-x)^3).
a(n) = a(-n+1) = (18*n*(n-1)+(2*n-1)*(-1)^n+1)/8 = A004526(n)*A156638(n). (End).

A266956 Numbers m such that 9*m+7 is a square.

Original entry on oeis.org

1, 2, 18, 21, 53, 58, 106, 113, 177, 186, 266, 277, 373, 386, 498, 513, 641, 658, 802, 821, 981, 1002, 1178, 1201, 1393, 1418, 1626, 1653, 1877, 1906, 2146, 2177, 2433, 2466, 2738, 2773, 3061, 3098, 3402, 3441, 3761, 3802, 4138, 4181, 4533, 4578, 4946, 4993, 5377, 5426

Offset: 1

Bruno Berselli, Jan 07 2016

Equivalently, numbers of the form h*(9*h+8)+1, where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ...
Also, integer values of k*(k+8)/9 plus 1.
It is easy to see that the Diophantine equation 9*x+3*j+1 = y^2 has infinitely many solutions in integers (x,y) for any j in Z. It follows a table with j = -5..5:
...
j = -5, x:  2, 7, 15, 30, 46, 71,  95, 130, 162, 207, 247, ...
j = -4, x:  3, 4, 20, 23, 55, 60, 108, 115, 179, 188, 268, ...
j = -3, x:  1, 8, 12, 33, 41, 76,  88, 137, 153, 216, 236, ...
j = -2, x:  1, 6, 14, 29, 45, 70,  94, 129, 161, 206, 246, ...
j = -1, x:  2, 3, 19, 22, 54, 59, 107, 114, 178, 187, 267, ...
j =  0, x:  0, 7, 11, 32, 40, 75,  87, 136, 152, 215, 235, ... (A132355)
j =  1, x:  0, 5, 13, 28, 44, 69,  93, 128, 160, 205, 245, ... (A185039)
j =  2, x:  1, 2, 18, 21, 53, 58, 106, 113, 177, 186, 266, ... (A266956)
j =  3, x: -1, 6, 10, 31, 39, 74,  86, 135, 151, 214, 234, ... (A266957)
j =  4, x: -1, 4, 12, 27, 43, 68,  92, 127, 159, 204, 244, ... (A266958)
j =  5, x:  0, 1, 17, 20, 52, 57, 105, 112, 176, 185, 265, ... (A218864)
...
The general closed form of these sequences is:
b(n,j) = (18*(n-1)*n + s(j)*(2*n-1)*(-1)^n + t(j))/8, where s(j) = 6*(-j) + 18*floor(j/3) - (-1)^floor(2*(j+1)/3) + 6 and t(j) = 4*(-j) + 4*floor((j+1)/3) + 5.
a(2m) - a(2m-1) gives the odd numbers (A005408); a(2m+1) - a(2m) gives the multiples of 16 (A008598).

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

Cf. numbers m such that 9*m+i: A132355 (i=1), A185039 (i=4), this sequence (i=7), A005563 (i=9), A266957 (i=10), A266958 (i=13), A218864 (i=16), A008865 (i=18, without -2).
Cf. A156638: square roots of 9*a(n)+7.
Cf. A005408, A008598.

Magma
```
[n: n in [0..6000] | IsSquare(9*n+7)];
```

[(18*(n-1)*n-7*(2*n-1)*(-1)^n+1)/8: n in [1..50]];

Select[Range[0, 6000], IntegerQ[Sqrt[9 # + 7]] &]
Table[(18 (n - 1) n - 7 (2 n - 1) (-1)^n + 1)/8, {n, 1, 50}]

for(n=0, 6000, if(issquare(9*n+7), print1(n, ", ")))

vector(50, n, n; (18*(n-1)*n-7*(2*n-1)*(-1)^n+1)/8)

from gmpy2 import is_square
[n for n in range(6000) if is_square(9*n+7)]

[(18*(n-1)*n-7*(2*n-1)*(-1)**n+1)/8 for n in range(1, 60)]

[n for n in (0..6000) if is_square(9*n+7)]

[(18*(n-1)*n-7*(2*n-1)*(-1)^n+1)/8 for n in (1..50)]

G.f.: x*(1 + x + 14*x^2 + x^3 + x^4)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (18*(n-1)*n - 7*(2*n-1)*(-1)^n + 1)/8.
a(n) = A218864(n) + 1.

A262711 Numbers k such that sum of digits of k^2 is 7.

Original entry on oeis.org

4, 5, 32, 40, 49, 50, 149, 320, 400, 490, 500, 1049, 1490, 3200, 4000, 4900, 5000, 10490, 14900, 32000, 40000, 49000, 50000, 104900, 149000, 320000, 400000, 490000, 500000, 1049000, 1490000, 3200000, 4000000, 4900000, 5000000, 10490000, 14900000

Offset: 1

Vincenzo Librandi, Sep 28 2015

Subsequence of A156638. [Bruno Berselli, Sep 28 2015]

			4 is in sequence because 4^2 = 16 and 1+6 = 7.

Cf. sum of digits of n^2 is k: A052216 (k=4), this sequence (k=7), A262712 (k=9), A262713 (k=10).
Cf. A004159, A061910, A156638.
Cf. A215614.

[n: n in [1..2*10^7] | &+Intseq(n^2) eq 7];

Select[Range[10^7], Total[IntegerDigits[#^2]] == 7 &]

for(n=1, 1e8, if (sumdigits(n^2) == 7, print1(n", "))) \\ Altug Alkan, Sep 28 2015

A334521 Numbers that are not congruent to +- mod 9 and are neither a cube nor twice a cube.

Original entry on oeis.org

3, 6, 7, 9, 10, 11, 12, 15, 17, 18, 19, 20, 21, 24, 25, 26, 28, 29, 30, 33, 34, 35, 36, 37, 38, 39, 42, 43, 44, 45, 46, 47, 48, 51, 52, 53, 55, 56, 57, 60, 61, 62, 63, 65, 66, 69, 70, 71, 72, 73, 74, 75, 78, 79, 80, 81, 82, 83, 84, 87, 88, 89, 90, 91, 92, 93, 96, 97, 98, 99, 100, 101, 102, 105, 106, 107, 108, 109

Offset: 1

N. J. A. Sloane, May 08 2020

nonn

Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.

Cf. A000578, A060464, A156638.

Complement of A334522.

A334522 Numbers that are either congruent to +-4 mod 9, or else are cubes or twice cubes.

Original entry on oeis.org

1, 2, 4, 5, 8, 13, 14, 16, 22, 23, 27, 31, 32, 40, 41, 49, 50, 54, 58, 59, 64, 67, 68, 76, 77, 85, 86, 94, 95, 103, 104, 112, 113, 121, 122, 125, 128, 130, 131, 139, 140, 148, 149, 157, 158, 166, 167, 175, 176, 184, 185, 193, 194, 202, 203, 211, 212, 216, 220, 221, 229, 230, 238, 239, 247, 248, 250, 256, 257, 265

Offset: 1

N. J. A. Sloane, May 08 2020

nonn

Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.

Cf. A000578, A060464, A156638.

Complement of A334521.

A299647 Positive solutions to x^2 == -2 (mod 11).

Original entry on oeis.org

3, 8, 14, 19, 25, 30, 36, 41, 47, 52, 58, 63, 69, 74, 80, 85, 91, 96, 102, 107, 113, 118, 124, 129, 135, 140, 146, 151, 157, 162, 168, 173, 179, 184, 190, 195, 201, 206, 212, 217, 223, 228, 234, 239, 245, 250, 256, 261, 267, 272, 278, 283, 289, 294, 300, 305, 311, 316

Offset: 1

Bruno Berselli, Mar 06 2018

Positive numbers congruent to {3, 8} mod 11.

Equivalently, interleaving of A017425 and A017485.

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

Subsequence of A106252, A279000.
Cf. A017425, A017485.
Cf. A017497: positive solutions to x == -2 (mod 11).
Cf. A017437: positive solutions to x^3 == -2 (mod 11).
Nonnegative solutions to x^2 == -2 (mod j): A005843 (j=2), A001651 (j=3), A047235 (j=6), A156638 (j=9), this sequence (j=11).

List([1..60], n -> 5*n-2+(2*n-(-1)^n-3)/4);

[(11(2n-1)-(-1)^n)>>2 for n in 1:60] # Peter Luschny, Mar 07 2018

[5*n-2+(2*n-(-1)^n-3)/4: n in [1..60]];

Table[5 n - 2 + (2 n - (-1)^n - 3)/4, {n, 1, 60}]
CoefficientList[ Series[(3 + 5x + 3x^2)/((x - 1)^2 (x + 1)), {x, 0, 57}], x] (* or *)
LinearRecurrence[{1, 1, -1}, {3, 8, 14}, 58] (* Robert G. Wilson v, Mar 08 2018 *)

makelist(5*n-2+(2*n-(-1)^n-3)/4, n, 1, 60);

vector(60, n, nn; 5*n-2+(2*n-(-1)^n-3)/4)

[5*n-2+(2*n-(-1)**n-3)/4 for n in range(1, 60)]

[5*n-2+(2*n-(-1)^n-3)/4 for n in (1..60)]

O.g.f.: x*(3 + 5*x + 3*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (-1 + 12*exp(x) - 11*exp(2*x) + 22*x*exp(2*x))*exp(-x)/4.
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 5*n - 2 + (2*n - (-1)^n - 3)/4.
a(n) = 4*n - 1 + floor((n - 1)/2) + floor((3*n - 1)/3).
a(n+k) - a(n) = 11*k/2 + (1 - (-1)^k)*(-1)^n/4.
a(n+k) + a(n) = 11*(2*n + k - 1)/2 - (1 + (-1)^k)*(-1)^n/4.
E.g.f.: 3 + ((22*x - 11)*exp(x) - exp(-x))/4. - David Lovler, Aug 08 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(5*Pi/22)*Pi/11. - Amiram Eldar, Feb 27 2023
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = cosec(3*Pi/22)/2.
Product_{n>=1} (1 + (-1)^n/a(n)) = sec(5*Pi/22)*sin(2*Pi/11). (End)

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A060464 Numbers that are not congruent to 4 or 5 mod 9.

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A332437 Decimal expansion of 2*cos(Pi/9).

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A185039 Numbers of the form 9*m^2 + 4*m, m an integer.

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A266956 Numbers m such that 9*m+7 is a square.

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A262711 Numbers k such that sum of digits of k^2 is 7.

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A334521 Numbers that are not congruent to +- mod 9 and are neither a cube nor twice a cube.

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A185039 Numbers of the form 9m^2 + 4m, m an integer.