| Displaying 1-7 of 7 results found.
page
1
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.
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.
a(n) = (3*(2*n - 1)^2*((2*n - 1)^2 + 2) - 2*n + 3)/2 for n > 0.
+10
7
5, 148, 1011, 3746, 10081, 22320, 43343, 76606, 126141, 196556, 293035, 421338, 587801, 799336, 1063431, 1388150, 1782133, 2254596, 2815331, 3474706, 4243665, 5133728, 6156991, 7326126, 8654381, 10155580, 11844123, 13734986, 15843721, 18186456, 20779895, 23641318, 26788581, 30240116, 34014931, 38132610
COMMENTS
Numbers D > 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 of the positive cubes C^3 - D^3 is equal to centered cube numbers, 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 = A352757(n), and D = a(n) (this sequence). There are infinitely many such numbers a(n) = D 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 + 3)/2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 576*(n - 2), with a(1) = 5, a(2) = 148 and a(3) = 1011.
a(n) can be extended for negative n such that a(-n) = a(n+1) + (2n + 1).
G.f.: x*(5 + 123*x + 321*x^2 + 121*x^3 + 6*x^4)/(1 - x)^5. - Stefano Spezia, Apr 08 2022
EXAMPLE
a(1) = 5 belongs to the sequence as 6^3 - 5^3 = 3^3 + 4^3 = 91 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 148 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 + 3)/2 = 1011.
a(4) = 3*a(3) - 3*a(2) + a(1) + 576*2 = 3*1011 - 3*148 + 5 + 576*2 = 3746.
MAPLE
restart; for n to 20 do (1/2)*(3*(2*n - 1)^2*((2*n - 1)^2 + 2) - 2*n + 3); end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352222, A352224, A352225, A352755, A352756, A352757, 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 A352759(n).
+10
6
4, 121, 562, 1543, 3280, 5989, 9886, 15187, 22108, 30865, 41674, 54751, 70312, 88573, 109750, 134059, 161716, 192937, 227938, 266935, 310144, 357781, 410062, 467203, 529420, 596929, 669946, 748687, 833368, 924205, 1021414, 1125211, 1235812, 1353433
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 = 3 (2n - 1) == 3 (mod 6), with C > D > B > 0, and A > 0, A = 27*t^3 * (27*t^6 + 1) /4 with t = 2*n-1, and where A = A352759(n), B = a(n) (this sequence), C = A355752(n) and D = A355753(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) = (9*(2*n - 1)^3 - 1) / 2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 216, with a(1) = 4, a(2) = 121 and a(3) = 562.
a(n) can be extended for negative n such that a(-n) = - a(n+1) - 1.
G.f.: x*(4+105*x+102*x^2+5*x^3)/(1-x)^4.
E.g.f.: 5 + exp(x)*(-5+9*x+54*x^2+36*x^3). (End)
EXAMPLE
a(1) = 4 is a term because 4^3 + 5^3 = 6^3 - 3^3 and 6 - 3 = 3 = 3*(2*1 - 1).
a(2) = 121 is a term because 121^3 + 122^3 = 369^3 - 360^3 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = (9*(2*3 - 1)^3 - 1) / 2 = 562.
a(4) = 3*562 - 3*121 + 4 + 216 = 1543.
MAPLE
restart; for n to 20 do (1/2)*(9*(2*n - 1)^3-1); end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352222, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A352759, A355752, A355753.
a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 + 1) / 2.
+10
6
6, 369, 2820, 10815, 29538, 65901, 128544, 227835, 375870, 586473, 875196, 1259319, 1757850, 2391525, 3182808, 4155891, 5336694, 6752865, 8433780, 10410543, 12715986, 15384669, 18452880, 21958635, 25941678, 30443481, 35507244, 41177895, 47502090, 54528213, 62306376, 70888419, 80327910, 90680145, 102002148, 114352671, 127792194
COMMENTS
Numbers C > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 such that the difference (C - D) == 3 (mod 6), C - D = 3 (2n - 1) for 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 = 27*t^3 *(27*t^6+1)/4 with t = 2*n-1, and where A = A352759(n), B = A355751(n), C = a(n) (this sequence), and D = A355753(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)*( 3*(2*n - 1)^3 + 1) / 2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 1728*(n - 2), with a(1) = 6, a(2) = 369 and a(3) = 2820.
a(n) can be extended for negative n such that a(-n) = a(n+1) - 3*(2*n + 1).
G.f.: -3*x*(2+113*x+345*x^2+115*x^3+x^4) / (x-1)^5 . - R. J. Mathar, Aug 03 2022
EXAMPLE
a(1) = 6 belongs to the sequence as 6^3 - 3^3 = 4^3 + 5^3 = 189 and 6 - 3 = 3 = 3 (2*1 - 1).
a(2) = 369 belongs to the sequence as 369^3 - 360^3 = 121^3 + 122^3 = 3587409 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 3*(2*3 - 1)*( 3*(2*3 - 1)^3 + 1) / 2 = 2820.
a(4) = 3*a(3) - 3*a(2) + a(1) + 1728*2 = 3*2820 - 3*369 + 6 + 1728*2 = 10815.
MAPLE
restart; for n to 20 do (1/2)* 3*(2*n - 1)*(3*(2*n - 1)^3+1); end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A352759, A355751, A355753.
a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 - 1) / 2 for n > 0.
+10
6
3, 360, 2805, 10794, 29511, 65868, 128505, 227790, 375819, 586416, 875133, 1259250, 1757775, 2391444, 3182721, 4155798, 5336595, 6752760, 8433669, 10410426, 12715863, 15384540, 18452745, 21958494, 25941531, 30443328, 35507085, 41177730, 47501919, 54528036, 62306193, 70888230, 80327715, 90679944, 102001941, 114352458, 127791975
COMMENTS
Numbers D > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 such that the difference C - D == 3 (mod 6), C - D = 3*(2*n - 1) for n > 1, and the difference of the positive cubes C^3 - D^3 is equal to centered cube numbers, with C > D > B > 0, and A > 0, A = 27*t^3 *(27*t^6+1)/4 with t = 2*n-1, and where A = A352759(n), B = A355751(n), C = A355752(n), and D = a(n) (this sequence). There are infinitely many such numbers a(n) = D 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)*( 3*(2*n - 1)^3 - 1) / 2 for n > 0.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 1728*(n - 2), with a(1) = 3, a(2) = 360 and a(3) = 2805.
a(n) can be extended for negative n such that a(-n) = a(n+1) + (2n + 1).
G.f.: -3*x*(1+115*x+345*x^2+113*x^3+2*x^4) / (x-1)^5 . - R. J. Mathar, Aug 03 2022
EXAMPLE
a(1) = 3 belongs to the sequence as 6^3 - 3^3 = 4^3 + 5^3 = 189 and 6 - 3 = 3 = 3*(2*1 - 1).
a(2) = 360 belongs to the sequence as 369^3 - 360^3 = 121^3 + 122^3 = 3587409 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 3*(2*3 - 1)*( 3*(2*3 - 1)^3 - 1) / 2 = 2805.
a(4) = 3*a(3) - 3*a(2) + a(1) + 1728*2 = 3*2805 - 3*360 + 3 + 1728*2 = 10794.
MAPLE
restart; for n to 20 do (1/2)* 3*(2*n - 1)*(3*(2*n - 1)^3-1); end do;
CROSSREFS
Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352222, A352224, A352225, A352755, A352756, A352757, A352759, A352759, A355751, A355752.
Search completed in 0.022 seconds
|