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a(n) is the number of partitions of n into 2 positive cubes.
+10
9
0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
COMMENTS
In other words, number of solutions to the equation x^3 + y^3 = n with x >= y > 0. - Antti Karttunen, Aug 28 2017
FORMULA
a(n) = [x^n y^2] Product_{k>=1} 1/(1 - y*x^(k^3)). - Ilya Gutkovskiy, Apr 23 2019
MATHEMATICA
Table[Count[IntegerPartitions[n, {2}], _?(AllTrue[Surd[#, 3], IntegerQ]&)], {n, 0, 110}] (* Harvey P. Dale, Nov 23 2022 *)
Product of n-th prime and n-th prime written backwards.
+10
8
4, 9, 25, 49, 121, 403, 1207, 1729, 736, 2668, 403, 2701, 574, 1462, 3478, 1855, 5605, 976, 5092, 1207, 2701, 7663, 3154, 8722, 7663, 10201, 31003, 75007, 98209, 35143, 91567, 17161, 100147, 129409, 140209, 22801, 117907, 58843, 127087
COMMENTS
a(8) = 1729 is the second taxicab number, also called the Hardy-Ramanujan number (see A001235, A011541 and A133029).
EXAMPLE
a(8) = 1729 because the 8th prime is 19 and 19 written backwards is 91 and 19*91 = 1729.
MATHEMATICA
#*FromDigits[Reverse[IntegerDigits[#]]] & /@ Prime[Range[1, 50]] (* G. C. Greubel, Oct 02 2017 *) #*IntegerReverse[#]&/@Prime[Range[40]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 29 2021 *)
PROG
(PARI) vector(60, n, prime(n)*subst(Polrev(digits(prime(n))), x, 10)) \\ Michel Marcus, Dec 17 2014
a(n) is the smallest number that is the sum of n positive cubes in two ways.
+10
8
1729, 251, 219, 157, 158, 131, 132, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126
COMMENTS
This is r(n,3,2) in Alter's notation.
EXAMPLE
a(2) = 1729 = 12^3 + 1^3 = 10^3 + 9^3 (the famous Hardy-Ramanujan number).
a(3) = 251 = 5^3 + 5^3 + 1^3 = 6^3 + 3^3 + 2^3.
Positive centered cube numbers that can be written as the difference of two positive cubes: a(n) = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1 and n > 0.
+10
8
91, 201159, 15407765, 295233841, 2746367559, 16448122691, 73287987409, 264133278045, 811598515091, 2202365761759, 5410166901741, 12249942682409, 25914353312575, 51755729480091, 98389720844009, 179211321358741, 314429627203659, 533744613620855, 879807401606341, 1412624924155809
COMMENTS
Numbers A > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 2n - 1, with C > D > B > 0, and A = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1, and where A = a(n) (this sequence), B = A352756(n), C = A352757(n) and D = A352758(n). There are infinitely many such numbers a(n) = A in this sequence.
LINKS
A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998. Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).
FORMULA
a(n) = (2*n - 1)*(3*(2*n - 1)^2 + 4)*((2*n - 1)^2*(3*(2*n - 1)^2 + 4)^2 + 3)/4.
a(n) can be extended for negative n such that a(-n) = -a(n+1).
EXAMPLE
a(1) = 91 belongs to the sequence because 91 = 3^3 + 4^3 = 6^3 - 5^3 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 201159 belongs to the sequence because 201159 = 46^3 + 47^3 = 151^3 - 148^3 and 151 - 148 = 3 = 2*2 - 1.
a(3) = (2*3 - 1)*(3*(2*3 - 1)^2 + 4)*((2*3 - 1)^2*(3*(2*3 - 1)^2 + 4)^2 + 3)/4 = 15407765.
MAPLE
restart; for n to 20 do (1/4)*(2*n-1)*(3*(2*n-1)^2+4)*((2*n-1)^2*(3*(2*n-1)^2+4)^2+3) end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352221, A352222, A352223, A352224, A352225, A352756, A352757, A352758, A352759, A355751, A355752, A355753.
Positive numbers k such that the centered cube number k^3 + (k+1)^3 is equal to the difference of two positive cubes and to A352755(n).
+10
8
3, 46, 197, 528, 1111, 2018, 3321, 5092, 7403, 10326, 13933, 18296, 23487, 29578, 36641, 44748, 53971, 64382, 76053, 89056, 103463, 119346, 136777, 155828, 176571, 199078, 223421, 249672, 277903, 308186, 340593, 375196, 412067, 451278, 492901, 537008, 583671, 632962, 684953, 739716, 797323, 857846, 921357
COMMENTS
Numbers B > 0 such that the centered cube number B^3 + (B+1)^3 is equal to the difference of two positive cubes, i.e., A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 2n - 1, with C > D > B > 0, and A > 0, A = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1, and where A = A352755(n), B = a(n) (this sequence), C = A352757(n) and D = A352758(n). There are infinitely many such numbers a(n) = B in this sequence.
LINKS
A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.
FORMULA
a(n) = ((2*n - 1)*(3*(2*n - 1)^2 + 4) - 1)/2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 72, with a(1) = 3, a(2) = 46 and a(3) = 197.
a(n) can be extended for negative n such that a(-n) = -a(n+1) - 1.
G.f.: x*(3 + 34*x + 31*x^2 + 4*x^3)/(1 - x)^4. - Stefano Spezia, Apr 08 2022
EXAMPLE
a(1) = 3 is a term because 3^3 + 4^3 = 6^3 - 5^3 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 46 is a term because 46^3 + 47^3 = 151^3 - 148^3 and 151 - 148 = 3 = 2*2 - 1.
a(3) = ((2*3 - 1)*(3*(2*3 - 1)^2 + 4) - 1)/2 = 197.
a(4) = 3*197 - 3*46 + 3 + 72 = 528.
MAPLE
restart; for n to 20 do (1/2)* ((2*n - 1)*(3*(2*n - 1)^2 + 4) - 1); end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352222, A352223, A352224, A352225, A352755, A352757, A352758, A352759, A352760, A352761, A352762.
a(n) = (3*(2*n - 1)^2*((2*n - 1)^2 + 2) + 2*n + 1)/2 for n > 0.
+10
8
6, 151, 1016, 3753, 10090, 22331, 43356, 76621, 126158, 196575, 293056, 421361, 587826, 799363, 1063460, 1388181, 1782166, 2254631, 2815368, 3474745, 4243706, 5133771, 6157036, 7326173, 8654430, 10155631, 11844176, 13735041, 15843778, 18186515, 20779956, 23641381, 26788646, 30240183, 34015000, 38132681
COMMENTS
Numbers C > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 such that the difference C - D is odd, C - D = 2*n - 1, and the difference between the positive cubes C^3 - D^3 is equal to a centered cube number, C^3 - D^3 = B^3 + (B+1)^3, with C > D > B > 0, and A > 0, A = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1, and where A = A352755(n), B = A352756(n), C = a(n) (this sequence), and D = A352758(n). There are infinitely many such numbers a(n) = C in this sequence.
LINKS
A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.
FORMULA
a(n) = (3*(2*n - 1)^2*((2*n - 1)^2 + 2) + 2*n + 1)/2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 576*(n - 2), with a(1) = 6, a(2) = 151 and a(3) = 1016.
a(n) can be extended for negative n such that a(-n) = a(n+1) - (2*n + 1).
G.f.: x*(6 + 121*x + 321*x^2 + 123*x^3 + 5*x^4)/(1 - x)^5. - Stefano Spezia, Apr 08 2022
EXAMPLE
a(1) = 6 belongs to the sequence as 6^3 - 5^3 = 3^3 + 4^3 = 91 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 151 belongs to the sequence as 151^3 - 148^3 = 46^3 + 47^3 = 201159 and 151 - 148 = 3 = 2*2 - 1.
a(3) = (3(2*3 - 1)^2*((2*3 - 1)^2 + 2) + 2*3 + 1)/2 = 1016.
a(4) = 3*a(3) - 3*a(2) + a(1) + 576*2 = 3*1016 - 3*151 + 6 + 576*2 = 3753.
MAPLE
restart; for n to 20 do (1/2)*(3*(2*n - 1)^2*((2*n - 1)^2 + 2) + 2*n + 1); end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352223, A352224, A352225, A352755, A352756, A352758, A352759, A352760, A352761, A352762.
Centered cube numbers that are the difference of two positive cubes; a(n) = 27*t^3*(27*t^6 + 1)/4 with t = 2*n-1.
+10
8
189, 3587409, 355957875, 7354447191, 70607389041, 429735975669, 1932670025559, 7006302268875, 21612640524741, 58809832966521, 144757538551899, 328260072633759, 695228576765625, 1389765141771741, 2643927354266751, 4818621138983379, 8458493032498509, 14364150148238625, 23685527077994691, 38040743821584231
COMMENTS
Numbers A > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 3*(2*n - 1) == 3 (mod 6), with (for n > 1) C > D > B > 0, and A = as 27*t^3*(27*t^6 + 1)/4 with t = 2*n-1, and where A = a(n) (this sequence), B = A355751(n), C = A355752(n) and D = A355753(n). There are infinitely many such numbers a(n) = A in this sequence.
LINKS
A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998. Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).
FORMULA
a(n) = 27*(2*n - 1)^3*(27*(2*n - 1)^6 + 1)/4.
a(n) can be extended for negative n such that a(-n) = -a(n+1).
EXAMPLE
a(1) = 189 belongs to the sequence because 189 = 4^3 + 5^3 = 6^3 - 3^3 and 6 - 3 = 3 = 3*(2*1 - 1).
a(2) = 3587409 belongs to the sequence because 3587409 = 121^3 + 122^3 = 369^3 - 360^3 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 27*(2*3 - 1)^3*(27*(2*3 - 1)^6 + 1)/4 = 355957875.
MAPLE
restart; for n from 1 to 20 do 27*(2*n-1)^3*(27*(2*n-1)^6+1)*(1/4); end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352221, A352222, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A355751, A355752, A355753.
Sum of two coprime cubes in at least three ways.
+10
7
15170835645, 208438080643, 320465258659, 1658465000647, 3290217425101, 3938530307257, 7169838686017, 13112542594333, 24641518275703, 36592635038993, 36848138663889, 41332017729268, 74051580874005
Products of three distinct primes of the form 6*k + 1.
+10
7
1729, 2821, 3367, 3913, 4123, 4921, 5551, 5719, 6097, 6643, 7189, 7657, 8029, 8113, 8827, 8911, 9139, 9331, 9373, 9709, 9919, 10507, 10621, 11137, 11557, 12649, 12901, 13237, 13699, 13741, 14287, 14497, 14539, 14833, 14911, 15067, 15799, 15841
COMMENTS
a(1) = 1729 is the Hardy-Ramanujan number (see taxicab numbers in A001235, A011541). Equivalently, products of three distinct primes of the form 3*k + 1. - Omar E. Pol, Feb 17 2018
LINKS
G. L. Honaker, Jr. and Chris Caldwell, Prime Curios!, Number 1729.
EXAMPLE
The first three primes of the form 6*k + 1 are 7, 13 and 19, so a(1) = 7*13*19 = 1729. - Omar E. Pol, Feb 17 2018
MATHEMATICA
Module[{nn=40, prs}, prs=Select[6*Range[nn]+1, PrimeQ]; Take[Times@@@ Subsets[ prs, {3}]//Union, nn]] (* Harvey P. Dale, Feb 17 2018 *)
PROG
(PARI) fct(n, o=[1])=if(n>1, concat(apply(t->vector(t[2], i, t[1]), Vec(factor(n)~))), o) \\ after M. F. Hasler in A027746is(n) = my(f=fct(n)); if(#f!=3 || f!=vecsort(f, , 8), return(0), for(k=1, #f, if((f[k]-1)/6!=ceil((f[k]-1)/6), return(0)))); 1 \\ Felix Fröhlich, Jul 07 2021
Numbers that are sums of two coprime positive cubes.
+10
7
2, 9, 28, 35, 65, 91, 126, 133, 152, 189, 217, 341, 344, 351, 370, 407, 468, 513, 539, 559, 637, 730, 737, 793, 854, 855, 1001, 1027, 1072, 1241, 1332, 1339, 1343, 1358, 1395, 1456, 1547, 1674, 1729, 1843, 1853, 2060, 2071, 2198, 2205, 2224, 2261, 2322, 2331, 2413
COMMENTS
Not a subsequence of A020898: non-cubefree members of this sequence include 152, 189, 344, 351, 513, 1072. - Robert Israel, Mar 16 2016
FORMULA
Erdős & Mahler shows that a(n) < kn^(3/2) for some k. Erdős later gives an elementary proof. - Charles R Greathouse IV, Dec 05 2012
EXAMPLE
28 is in the sequence since 1^3 + 3^3 = 28 and (1, 3) = 1.
MAPLE
N:= 10000: # to get all terms <= N
S:= {2, seq(seq(x^3 + y^3, y = select(t -> igcd(t, x)=1, [$x+1 .. floor((N - x^3)^(1/3))])), x = 1 .. floor((N/2)^(1/3)))}:
MATHEMATICA
nn = 2500; Union[Flatten[Table[If[CoprimeQ[x, y] == True, x^3 + y^3, {}], {x, nn^(1/3)}, {y, x, (nn - x^3)^(1/3)}]]]
Select[Range@ 2500, Length[PowersRepresentations[#, 2, 3] /. {{0, _} -> Nothing, {a_, b_} /; ! CoprimeQ[a, b] -> Nothing}] > 0 &] (* Michael De Vlieger, Mar 15 2016 *)
PROG
(PARI) is(n)=for(k=1, (n\2+.5)^(1/3), if(gcd(k, n)==1&&ispower(n-k^3, 3), return(1))); 0 \\ Charles R Greathouse IV, Apr 13 2012 (PARI) list(lim)=my(v=List()); forstep(x=1, lim^(1/3), 2, forstep(y=2, (lim-x^3+.5)^(1/3), 2, if(gcd(x, y)==1, listput(v, x^3+y^3))); forstep(y=1, min((lim-x^3+.5)^(1/3), x), 2, if(gcd(x, y)==1, listput(v, x^3+y^3)))); vecsort(Vec(v), , 8) \\ Charles R Greathouse IV, Dec 05 2012
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