cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A133216 Integers that are simultaneously triangular (A000217) and decagonal (A001107).

Original entry on oeis.org

0, 1, 10, 1540, 11935, 1777555, 13773376, 2051297326, 15894464365, 2367195337045, 18342198104230, 2731741367653000, 21166880717817451, 3152427171076225351, 24426562006163234620, 3637898223680596402450, 28188231388231654934425, 4198131397700237172202345
Offset: 1

Views

Author

Richard Choulet, Oct 11 2007; Ant King, Nov 04 2011

Keywords

Comments

Positive terms are of the form (m^2-9)/16 where m runs over the elements of A077443 that are congruent to 5 modulo 8. Correspondingly, for n>1, sqrt(16*a(n)+9) form a subsequence of A077443, while sqrt(8*a(n)+1) form a subsequence of A077442 with indices congruent to 2,3 modulo 4. [Max Alekseyev]

Examples

			The initial terms of the sequences of triangular (A000217) and decagonal (A001107) numbers are 0, 1, 3, 6, 10, 15, ... and 0, 1, 10, 27, ... respectively. As the third number which is common to both sequences is 10, we have a(3) = 10.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{1, 1154, -1154, -1, 1} , {0, 1, 10, 1540, 11935, 1777555}, 17] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)

Formula

a(n) = A000217(A133218(n)) = A001107(A133217(n)).
For n>5, a(n) = 1154*a(n-2) - a(n-4) + 396.
For n>6, a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5).
For n>1, a(n) = 1/64 * ( (9 + 4* sqrt(2)*(-1)^n)*(1+sqrt(2))^(4*n-6) + (9 - 4* sqrt(2)*(-1)^n)*(1-sqrt(2))^(4*n-6) - 22).
a(n) = floor ( 1/64 * (9 + 4*sqrt(2)*(-1)^n) * (1+sqrt(2))^(4*n-6) ).
G.f.: (x^5 + 9*x^4 + 376*x^3 + 9*x^2 + x)/((1 - x)*(x^2 - 34*x + 1)*(x^2 + 34*x + 1)). [corrected by Peter Luschny, Apr 04 2019]
Lim (n -> Infinity, a(2n+1)/a(2n)) = (1/49)*(3649+2580*sqrt(2)).
Lim (n -> Infinity, a(2n)/a(2n-1)) = (1/49)*(193+132*sqrt(2)).

Extensions

Entry revised by N. J. A. Sloane, Nov 06 2011
Term 0 prepended and entry revised accordingly by Max Alekseyev, Nov 06 2011

A133217 Indices of decagonal numbers (A001107) that are also triangular (A000217).

Original entry on oeis.org

0, 1, 2, 20, 55, 667, 1856, 22646, 63037, 769285, 2141390, 26133032, 72744211, 887753791, 2471161772, 30157495850, 83946756025, 1024467105097, 2851718543066, 34801724077436, 96874483708207, 1182234151527715, 3290880727535960, 40161159427864862
Offset: 1

Views

Author

Richard Choulet, Oct 11 2007; Ant King, Nov 04 2011

Keywords

Comments

For n>0, a(n) = (A055979(n) - A056161(n))/2, with those two sequences related through the Diophantine equation 2x^2 + 3x + 2 = r^2. - Richard R. Forberg, Nov 24 2013

Examples

			The third number which is both decagonal (A001107) and triangular (A000217) is A133216(3)=10. As this is the second decagonal number, we have a(3) = 2.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{1, 34, -34, -1, 1} , {0, 1, 2, 20, 55, 667}, 24] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)

Formula

For n>5, a(n) = 34*a(n-2) - a(n-4) - 12.
For n>6, a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).
For n>1, a(n) = 1/16 * ((2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3) - (2*sqrt(2) - (-1)^n)*(1 - sqrt(2))^(2*n - 3) + 6).
For n>1, a(n) = ceiling (1/16*(2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3)).
G.f.: ( 1 - 33*x^2 + 18*x^3 + 2*x^4 ) / ((1 - x ) * (1 - 6*x + x^2 ) * (1 + 6*x + x^2)).
lim (n -> Infinity, a(2n+1)/a(2n)) = 1/7*(43 + 30*sqrt(2)).
lim (n -> Infinity, a(2n)/a(2n-1)) = 1/7*(11 + 6*sqrt(2)).

Extensions

Entry revised by Max Alekseyev, Nov 06 2011
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