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URL: https://oeis.org/A325173

⇱ A325173 - OEIS

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A325173
Perfect squares of the form a + b^2 + c^3, where a,b,c are consecutive numbers.
1
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
1,1
FORMULA
a(n) = A000290(A054602(n)). - Michel Marcus, Sep 05 2019
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)
EXAMPLE
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.
PROG
(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
CROSSREFS
Intersection of A000290 and A027620.
Cf. A005563 (the indices n that give these squares), A054602.
Sequence in context: A017198 A281789 A134176 * A067415 A393407 A069134
KEYWORD
nonn,easy
AUTHOR
Philip Mizzi, Sep 05 2019
STATUS
approved