A137694 -id:A137694 - cp's OEIS frontend

A137693 Numbers n such that 3n^2-n = 6k^2-2k for some integer k>0.

7, 7887, 9101399, 10503006367, 12120460245927, 13987000620793199, 16140986595935105527, 18626684544708490984767, 21495177823607002661315399, 24805416581757936362666985487, 28625429240170834955515039936407, 33033720537740561780727993419627999

Offset: 1

M. F. Hasler, Feb 08 2008

Also indices of pentagonal numbers which are twice some other pentagonal number.
Note that A000326(n) = 2 A000326(k) <=> n(3n-1)=2k(3k-1), which is easily solved by standard Pell-type techniques (cf. link to D. Alpern's quadratic solver). Here we consider only positive solutions.
Inspired by a recent comment on A000326 by R. J. Mathar.

Colin Barker, Table of n, a(n) for n = 1..250
Dario Alpern, Quadratic two integer variable equation solver
Index entries for linear recurrences with constant coefficients, signature (1155,-1155,1).

Cf. A000326, A136112-A136118, A135768-A135769, A135771, A137694.

CoefficientList[Series[x (-7+198x+x^2)/((x-1)(x^2-1154x+1)),{x,0,20}],x] (* or *) Join[{0},LinearRecurrence[{1155,-1155,1},{7,7887,9101399}, 20]] (* Harvey P. Dale, Jun 21 2011 *)

vector(20,i, (v=if(i>1,[577,408; 816,577]*v-[164;232], [5;7]))[2,1])

a(n) = f^{2n-2}(5,7)[2], where f(x,y) = (577x + 408y - 164, 816x + 577y - 232)
a(n) = (7,7,9,7,7,9,...) mod 10
G.f. x*(-7+198*x+x^2) / ( (x-1)*(x^2-1154*x+1) ). - R. J. Mathar, Apr 17 2011
a(0)=0, a(1)=7, a(2)=7887, a(3)=9101399, a(n)=1155*a(n-1)-1155*a(n-2)+ a(n-3). - Harvey P. Dale, Jun 21 2011

More terms from Harvey P. Dale, Jun 21 2011

A330657 Number of ways the n-th pentagonal number A000326(n) can be written as the difference of two positive pentagonal numbers.

Original entry on oeis.org

0, 0, 0, 1, 1, 0, 2, 0, 0, 1, 0, 1, 1, 1, 0, 1, 2, 0, 2, 1, 1, 3, 1, 0, 2, 3, 0, 3, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 2, 1, 1, 1, 1, 3, 2, 2, 2, 1, 4, 1, 0, 2, 1, 2, 1, 1, 1, 3, 2, 2, 1, 3, 1, 3, 2, 1, 6, 1, 1, 1, 3, 2

Offset: 1

Bradley Klee, Mar 01 2020

nonn

Equivalently, a(n) is the number of triples [n,k,m] with k>0 satisfying the Diophantine equation n*(3*n-1) + k*(3*k-1) - m*(3*m-1) = 0. Any such triple satisfies a triangle inequality, n+k > m. The n for which there is a triple [n,n,m] are listed in A137694. Solutions of the form [n,m-1,m] appear only when n=3*z+1, z > 0. The n for which a(n)=0 are listed in A135768.

			Isosceles case, n=5: 2*5*(3*5-1) - 7*(3*7-1) = 0.

N. J. A. Sloane et al., "sum of 2 triangular numbers is a triangular number", math-fun mailing list, Feb. 19-29, 2020.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
B. Klee, Pentathagorean triples: two easy proofs, seqfan mailing list, Mar. 20, 2020.
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A046079, A309507, A000326.

PentaTriples[PNn_] := Sort[Select[{PNn,
      (-PNn + 3 PNn^2 + # - 3 #^2)/(6 #),
      (-PNn + 3 PNn^2 + # + 3 #^2)/(6 #)
      } & /@ Divisors[PNn*(3*PNn - 1)],
   And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]]
Length[PentaTriples[#]] & /@ Range[100]
a[n_] := Length@FindInstance[n > 0 && y > 0 && z > 0 &&
     n (3 n - 1) + y (3 y - 1) == z (3 z - 1), {y, z}, Integers, 10^9];
a /@ Range[100]

cp's OEIS Frontend

A137693 Numbers n such that 3n^2-n = 6k^2-2k for some integer k>0.

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