A325173 Perfect squares of the form a + b^2 + c^3, where a,b,c are consecutive numbers.
9, 144, 1089, 5184, 18225, 51984, 127449, 278784, 558009, 1040400, 1830609, 3069504, 4941729, 7683984, 11594025, 17040384, 24472809, 34433424, 47568609, 64641600, 86545809, 114318864, 149157369, 192432384, 245705625, 310746384, 389549169, 484352064, 597655809, 732243600
Offset: 1
Examples
9 = 0 + 1^2 + 2^3. 0,1,2 are consecutive numbers and 9 is a perfect square. Hence, 9 is a member of the sequence. 18225 = 24 + 25^2 + 26^3. 24,25,26 are consecutive numbers and 18225 is a perfect square. Hence 18225 is a member of the sequence.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).
Crossrefs
Programs
-
PARI
a(n) = n^2*(2 + n^2)^2 \\ David A. Corneth, Sep 11 2019
-
PARI
Vec(9*x*(1 + x)*(1 + 8*x + 22*x^2 + 8*x^3 + x^4) / (1 - x)^7 + O(x^40)) \\ Colin Barker, Sep 11 2019
Formula
From Colin Barker, Sep 05 2019: (Start)
G.f.: 9*x*(1 + x)*(1 + 8*x + 22*x^2 + 8*x^3 + x^4) / (1 - x)^7.
a(n) = n^2*(2 + n^2)^2.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>7.
(End)
E.g.f.: exp(x)*x*(9 + 63*x + 114*x^2 + 69*x^3 + 15*x^4 + x^5). - Conjectured by Stefano Spezia, Sep 05 2019 after Colin Barker
From Chai Wah Wu, Sep 10 2019: (Start)
Above conjectures are true. Proof: k + (k+1)^2 + (k+2)^3 = (k+1)*(k+3)^2 and thus is a perfect square if and only if k+1 = n^2 is a perfect square. This implies that (k+1)*(k+3)^2 = n^2*(n^2+2)^2.
(End)