A276914 -id:A276914 - cp's OEIS frontend

A276915 Indices of triangular numbers in A276914 which are also pentagonal.

0, 1, 10, 143, 1988, 27693, 385710, 5372251, 74825800, 1042188953, 14515819538, 202179284583, 2815994164620, 39221739020101, 546288352116790, 7608815190614963, 105977124316492688, 1476070925240282673, 20559015829047464730, 286350150681424223551

Offset: 0

Daniel Poveda Parrilla, Sep 22 2016

A276914(a(n)) = A014979(n + 1). All numbers which are both triangular and pentagonal can be found in sequence A276914.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13,13,-1).

Cf. A014979, A046175, A276914.

RecurrenceTable[{a[n] == 14 a[n - 1] - a[n - 2] - 4 (-1)^n, a[0] == 0, a[1] == 1}, a, {n, 19}] (* Michael De Vlieger, Sep 23 2016 *)

concat(0, Vec(x*(1-3*x)/((1+x)*(1-14*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 23 2016

a(n) = 14*a(n-1) - a(n-2) - 4*(-1)^n for n>1, a(0)=0, a(1)=1.
a(n) = (A046175(n) + (A046175(n) mod 2))/2.
From Colin Barker, Sep 23 2016: (Start)
G.f.: x*(1 - 3*x) / ((1 + x)*(1 - 14*x + x^2)).
a(n) = 13*a(n-1) + 13*a(n-2) - a(n-3) for n>2.
a(n) = ( -6*(-1)^n + (3+sqrt(3))*(7-4*sqrt(3))^n - (-3+sqrt(3))*(7+4*sqrt(3))^n )/24. (End)

A276916 Subsequence of centered square numbers obtained by adding four triangles from A276914 and a central element, a(n) = 4*A276914(n) + 1.

Original entry on oeis.org

1, 5, 41, 61, 145, 181, 313, 365, 545, 613, 841, 925, 1201, 1301, 1625, 1741, 2113, 2245, 2665, 2813, 3281, 3445, 3961, 4141, 4705, 4901, 5513, 5725, 6385, 6613, 7321, 7565, 8321, 8581, 9385, 9661, 10513, 10805, 11705, 12013, 12961, 13285, 14281, 14621, 15665

Offset: 0

Daniel Poveda Parrilla, Sep 27 2016

All terms of this sequence are centered square numbers. Graphically, each term of the sequence is made of four squares, eight triangles and a central element.

a(A220185(n+1)) = A008844(2n) = A079291(4n+1), which is a square of a Pell number.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008844, A079291, A220185, A276914.

[4*n*(2*n+(-1)^n)+1 : n in [0..60]]; // Wesley Ivan Hurt, Sep 27 2016

A276916:=n->4*n*(2*n+(-1)^n)+1: seq(A276916(n), n=0..60); # Wesley Ivan Hurt, Sep 27 2016

Table[4 n (2 n + (-1)^n) + 1, {n, 0, 44}] (* or *)
CoefficientList[Series[(1 +4x +34x^2 +12x^3 +13x^4)/((1-x)^3*(1+x)^2), {x, 0, 44}], x] (* Michael De Vlieger, Sep 28 2016 *)

Vec((1+4*x+34*x^2+12*x^3+13*x^4)/((1-x)^3*(1+x)^2) + O(x^50)) \\ Colin Barker, Sep 27 2016

[4*n*(2*n+(-1)^n) +1 for n in (0..60)] # G. C. Greubel, Aug 19 2022

a(n) = 4*n*(2*n + (-1)^n) + 1.
a(n) = 4*n*(2*n + 1) + 1 for n even.
a(n) = 4*n*(2*n - 1) + 1 for n odd.
a(n) is sum of two squares; a(n) = k^2 + (k+1)^2 where k = 2n-(n mod 2). - David A. Corneth, Sep 27 2016
From Colin Barker, Sep 27 2016: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4.
G.f.: (1+4*x+34*x^2+12*x^3+13*x^4) / ((1-x)^3*(1+x)^2). (End)
E.g.f.: (1+8*x+8*x^2)*exp(x) - 4*x*exp(-x). - G. C. Greubel, Aug 19 2022

cp's OEIS Frontend

A276915 Indices of triangular numbers in A276914 which are also pentagonal.

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