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A355751
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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).
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6
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.
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FORMULA
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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)
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EXAMPLE
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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.
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MAPLE
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restart; for n to 20 do (1/2)*(9*(2*n - 1)^3-1); end do;
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CROSSREFS
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Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352222, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A352759, A355752, A355753.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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