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 1-4 of 4 results.

A001110 Square triangular numbers: numbers that are both triangular and square.

Original entry on oeis.org

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004
For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007
The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012
Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012
For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013
For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014
The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014
The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016
Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

Examples

			a(2) = ((17 + 12*sqrt(2))^2 + (17 - 12*sqrt(2))^2 - 2)/32 = (289 + 24*sqrt(2) + 288 + 289 - 24*sqrt(2) + 288 - 2)/32 = (578 + 576 - 2)/32 = 1152/32 = 36 and 6^2 = 36 = 8*9/2 => a(2) is both the 6th square and the 8th triangular number.

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 38, 204.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923; see Vol. 2, p. 10.
  • Martin Gardner, Time Travel and other Mathematical Bewilderments, Freeman & Co., 1988, pp. 16-17.
  • Miodrag S. Petković, Famous Puzzles of Great Mathematicians, Amer. Math. Soc. (AMS), 2009, p. 64.
  • J. H. Silverman, A Friendly Introduction to Number Theory, Prentice Hall, 2001, p. 196.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-259.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 93.

Links

Crossrefs

Cf. A001108, A001109.
Other S_r type sequences are S_4=A000290, S_5=A004146, S_7=A054493, S_8=A001108, S_9=A049684, S_20=A049683, S_36=this sequence, S_49=A049682, S_144=A004191^2.
Cf. A001014; intersection of A000217 and A000290; A010052(a(n))*A010054(a(n)) = 1.
Cf. A005214, A054686, A232847 and also A233267 (reveals an interesting divisibility pattern for this sequence).
Cf. A240129 (triangular numbers that are squares of triangular numbers), A100047.
See A229131, A182334, A299921 for near-misses.

Programs

  • Haskell
    a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
   zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
-- Reinhard Zumkeller, Oct 12 2011
  • Magma
    [n le 2 select n-1 else Floor((6*Sqrt(Self(n-1)) - Sqrt(Self(n-2)))^2): n in [1..20]]; // Vincenzo Librandi, Jul 22 2015
  • Maple
    a:=17+12*sqrt(2); b:=17-12*sqrt(2); A001110:=n -> expand((a^n + b^n - 2)/32); seq(A001110(n), n=0..20); # Jaap Spies, Dec 12 2004
A001110:=-(1+z)/((z-1)*(z**2-34*z+1)); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    f[n_]:=n*(n+1)/2; lst={}; Do[If[IntegerQ[Sqrt[f[n]]],AppendTo[lst,f[n]]],{n,0,10!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Table[(1/8) Round[N[Sinh[2 n ArcSinh[1]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
Transpose[NestList[Flatten[{Rest[#],34Last[#]-First[#]+2}]&, {0,1},20]][[1]]  (* Harvey P. Dale, Mar 25 2011 *)
LinearRecurrence[{35, -35, 1}, {0, 1, 36}, 20] (* T. D. Noe, Mar 25 2011 *)
LinearRecurrence[{6,-1},{0,1},20]^2 (* Harvey P. Dale, Oct 22 2012 *)
(* Square = Triangular = Triangular = A001110 *)
ChebyshevU[#-1,3]^2==Binomial[ChebyshevT[#/2,3]^2,2]==Binomial[(1+ChebyshevT[#,3])/2,2]=={1,36,1225,41616,1413721}[[#]]&@Range[5]
True (* Bill Gosper, Jul 20 2015 *)
L=0;r={};Do[AppendTo[r,L];L=1+17*L+6*Sqrt[L+8*L^2],{i,1,19}];r (* Kebbaj Mohamed Reda, Aug 02 2023 *)
  • PARI
    a=vector(100);a[1]=1;a[2]=36;for(n=3,#a,a[n]=34*a[n-1]-a[n-2]+2);a \\ Charles R Greathouse IV, Jul 25 2011
  • Scheme
    ;; With memoizing definec-macro from Antti Karttunen's IntSeq-library.
(definec (A001110 n) (if (< n 2) n (+ 2 (- (* 34 (A001110 (- n 1))) (A001110 (- n 2))))))
;; Antti Karttunen, Dec 06 2013
  • Scheme
    ;; For testing whether n is in this sequence:
(define (inA001110? n) (and (zero? (A068527 n)) (inA001109? (floor->exact (sqrt n)))))
(define (inA001109? n) (= (* 8 n n) (floor->exact (* (sqrt 8) n (ceiling->exact (* (sqrt 8) n))))))
;; Antti Karttunen, Dec 06 2013

Formula

a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
G.f.: x*(1 + x) / (( 1 - x )*( 1 - 34*x + x^2 )).
a(n-1) * a(n+1) = (a(n)-1)^2. - Colin Dickson, posting to alt.math.recreational, Mar 07 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2). - Lekraj Beedassy, Jun 27 2001
a(n) - a(n-1) = A046176(n). - Sophie Kuo (ejiqj_6(AT)yahoo.com.tw), May 27 2006
a(n) = A001109(n)^2 = A001108(n)*(A001108(n)+1)/2 = (A000129(n)*A001333(n))^2 = (A000129(n)*(A000129(n) + A000129(n-1)))^2. - Henry Bottomley, Apr 19 2000
a(n) = (((17+12*sqrt(2))^n) + ((17-12*sqrt(2))^n)-2)/32. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2). See UWC problem link and solution. - Jaap Spies, Dec 12 2004
From Antonio Alberto Olivares, Nov 07 2003: (Start)
a(n) = 35*(a(n-1) - a(n-2)) + a(n-3);
a(n) = -1/16 + ((-24 + 17*sqrt(2))/2^(11/2))*(17 - 12*sqrt(2))^(n-1) + ((24 + 17*sqrt(2))/2^(11/2))*(17 + 12*sqrt(2))^(n-1). (End)
a(n+1) = (17*A029547(n) - A091761(n) - 1)/16. - R. J. Mathar, Nov 16 2007
a(n) = A001333^2 * A000129^2 = A000129(2*n)^2/4 = binomial(A001108,2). - Bill Gosper, Jul 28 2008
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper, Jul 25 2008
a(n) = (1/8)*(sinh(2*n*arcsinh(1)))^2. - Artur Jasinski, Feb 10 2010
a(n) = floor((17 + 12*sqrt(2))*a(n-1)) + 3 = floor(3*sqrt(2)/4 + (17 + 12*sqrt(2))*a(n-1) + 1). - Manuel Valdivia, Aug 15 2011
a(n) = (A011900(n) + A001652(n))^2; see the link about the generalized proof of square triangular numbers. - Kenneth J Ramsey, Oct 10 2011
a(2*n+1) = A002315(n)^2*(A002315(n)^2 + 1)/2. - Ivan N. Ianakiev, Oct 10 2012
a(2*n+1) = ((sqrt(t^2 + (t+1)^2))*(2*t+1))^2, where t = (A002315(n) - 1)/2. - Ivan N. Ianakiev, Nov 01 2012
a(2*n) = A001333(2*n)^2 * (A001333(2*n)^2 - 1)/2, and a(2*n+1) = A001333(2*n+1)^2 * (A001333(2*n+1)^2 + 1)/2. The latter is equivalent to the comment above from Ivan using A002315, which is a bisection of A001333. Using A001333 shows symmetry and helps show that a(n) are both "squares of triangular" and "triangular of squares". - Richard R. Forberg, Aug 30 2013
a(n) = (A001542(n)/2)^2.
From Peter Bala, Apr 03 2014: (Start)
a(n) = (T(n,17) - 1)/16, where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = U(n-1,3)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -17; 1, 18].
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = A096979(2*n-1) for n > 0. - Ivan N. Ianakiev, Jun 21 2014
a(n) = (6*sqrt(a(n-1)) - sqrt(a(n-2)))^2. - Arkadiusz Wesolowski, Apr 06 2015
From Daniel Poveda Parrilla, Jul 16 2016 and Sep 21 2016: (Start)
a(n) = A000290(A002965(2*n)*A002965(2*n + 1)) (after Hugh Darwen).
a(n) = A000217(2*(A000129(n))^2 - (A000129(n) mod 2)).
a(n) = A000129(n)^4 + Sum_{k=0..(A000129(n)^2 - (A000129(n) mod 2))} 2*k. This formula can be proved graphically by taking the corresponding triangle of a square triangular number and cutting both acute angles, one level at a time (sum of consecutive even numbers), resulting in a square of squares (4th powers).
a(n) = A002965(2*n)^4 + Sum_{k=A002965(2*n)^2..A002965(2*n)*A002965(2*n + 1) - 1} 2*k + 1. This formula takes an equivalent sum of consecutives, but odd numbers. (End)
E.g.f.: (exp((17-12*sqrt(2))*x) + exp((17+12*sqrt(2))*x) - 2*exp(x))/32. - Ilya Gutkovskiy, Jul 16 2016

A005214 Triangular numbers together with squares (excluding 0).

Original entry on oeis.org

1, 3, 4, 6, 9, 10, 15, 16, 21, 25, 28, 36, 45, 49, 55, 64, 66, 78, 81, 91, 100, 105, 120, 121, 136, 144, 153, 169, 171, 190, 196, 210, 225, 231, 253, 256, 276, 289, 300, 324, 325, 351, 361, 378, 400, 406, 435, 441, 465, 484, 496, 528, 529, 561, 576, 595, 625, 630, 666, 676
Offset: 1

Views

Author

Russ Cox, Jun 14 1998

Keywords

References

  • Douglas R. Hofstadter, Fluid Concepts and Creative Analogies: Computer Models of the Fundamental Mechanisms of Thought, (together with the Fluid Analogies Research Group), NY: Basic Books, 1995, p. 15.

Links

  • Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
  • Douglas R. Hofstadter, Analogies and Sequences: Intertwined Patterns of Integers and Patterns of Thought Processes, DIMACS Conference on Challenges of Identifying Integer Sequences, Rutgers University, October 10 2014; Part 1, Part 2.
  • Eric Weisstein's World of Mathematics, Square Triangular Number.

Crossrefs

Union of A000290 and A000217.
Cf. A001110, A054686, A157259, A117704 (first differences), A193711 (partial sums), A193748, A193749 (partitions into).
Cf. A010052, A010054, A193714, A193715.
Cf. A241241 (subsequence), A242401 (complement).

Programs

  • Haskell
    import Data.List.Ordered (union)
a005214 n = a005214_list !! (n-1)
a005214_list = tail $ union a000290_list a000217_list
-- Reinhard Zumkeller, Feb 15 2015, Aug 03 2011
  • Maple
    a := proc(n) floor(sqrt(n)): floor(sqrt(n+n)):
`if`(n+n = %*% + % or n = %% * %%, n, NULL) end: # Peter Luschny, May 01 2014
  • Mathematica
    With[{upto=700},Module[{maxs=Floor[Sqrt[upto]], maxt=Floor[(Sqrt[8upto+1]- 1)/2]},Union[Join[Range[maxs]^2, Table[(n(n+1))/2,{n,maxt}]]]]] (* Harvey P. Dale, Sep 17 2011 *)
  • PARI
    upTo(lim)=vecsort(concat(vector(sqrtint(lim\1),n,n^2), vector(floor(sqrt(2+2*lim)-1/2),n,n*(n+1)/2)),,8) \\ Charles R Greathouse IV, Aug 04 2011
  • PARI
    isok(m) = ispolygonal(m,3) || ispolygonal(m,4); \\ Michel Marcus, Mar 13 2021

Formula

From Reinhard Zumkeller, Aug 03 2011: (Start)
A010052(a(n)) + A010054(a(n)) > 0.
A010052(a(A193714(n))) = 1.
A010054(a(A193715(n))) = 1. (End)
a(n) ~ c * n^2, where c = 3 - 2*sqrt(2) = A157259 - 4 = 0.171572... . - Amiram Eldar, Apr 04 2025

A123596 Squares alternating with triangular numbers.

Original entry on oeis.org

0, 0, 1, 1, 4, 3, 9, 6, 16, 10, 25, 15, 36, 21, 49, 28, 64, 36, 81, 45, 100, 55, 121, 66, 144, 78, 169, 91, 196, 105, 225, 120, 256, 136, 289, 153, 324, 171, 361, 190, 400, 210, 441, 231, 484, 253, 529, 276, 576, 300, 625, 325, 676, 351, 729, 378, 784, 406, 841, 435
Offset: 0

Views

Author

Peter Hansen (babyskbaby(AT)web.de), Nov 14 2006

Keywords

Comments

Rearrangement of A054686.
For n >= 2, a(n) is the number of distinct positive slopes of least squares regression lines fitted to n points (j,y_j), 1 <= j <= n, where all y_j are 0 or 1. The total number of distinct slopes is 2*a(n)+1 (a(n) positive, a(n) negative, and the zero slope). - Pontus von Brömssen, Mar 10 2024

Links

Crossrefs

Cf. A002620, A054686, A124093.

Programs

  • Magma
    [(3*n^2-1+(n^2+1)*(-1)^n)/16: n in [0..10]]; // G. C. Greubel, Oct 26 2017
  • Mathematica
    CoefficientList[Series[x^2*(1+x+x^2)/((1-x)^3*(1+x)^3), {x, 0, 50}], x] (* or *) Table[(3*n^2-1+(n^2+1)*(-1)^n)/16, {n,0,50}] (* G. C. Greubel, Oct 26 2017 *)
With[{nn=30},Riffle[Range[0,nn]^2,Accumulate[Range[0,nn]]]] (* or *) LinearRecurrence[{0,3,0,-3,0,1},{0,0,1,1,4,3},60] (* Harvey P. Dale, Feb 11 2020 *)
  • PARI
    {a(n) = if(n%2,(n^2-1)/8,n^2/4)} \\ Michael Somos, Nov 18 2006

Formula

a(2*n) = n^2, a(2*n+1) = (n^2+n)/2.
From R. J. Mathar, Feb 12 2010: (Start)
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x^2*(1+x+x^2)/((1-x)^3*(1+x)^3). (End)
a(n) = (3*n^2-1+(n^2+1)*(-1)^n)/16. - Luce ETIENNE, May 30 2015

Extensions

Edited by Michael Somos, and several other correspondents, Nov 14 2005

A124093 Triangular numbers alternating with squares.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 6, 9, 10, 16, 15, 25, 21, 36, 28, 49, 36, 64, 45, 81, 55, 100, 66, 121, 78, 144, 91, 169, 105, 196, 120, 225, 136, 256, 153, 289, 171, 324, 190, 361, 210, 400, 231, 441, 253, 484, 276, 529, 300, 576, 325, 625, 351, 676, 378, 729, 406, 784, 435, 841
Offset: 0

Views

Author

N. J. A. Sloane, based on a suggestion from Robert G. Wilson v, Nov 27 2006

Keywords

Links

Crossrefs

Cf. A123596. Rearrangement of A054686.

Programs

  • Maple
    a:=proc(n) if n mod 2 = 0 then n*(n+2)/8 else (n-1)^2/4 fi end: seq(a(n),n=0..70); # Emeric Deutsch, Nov 29 2006
  • Mathematica
    tr=Table[{k(k+1)/2,k^2},{k,0,100}]//Flatten (Seidov)
With[{nn=30},Riffle[Accumulate[Range[0,nn]],Range[0,nn]^2]] (* Harvey P. Dale, Jul 13 2014 *)
  • PARI
    concat([0,0], Vec(-x^2*(x^3+x+1)/((x-1)^3*(x+1)^3) + O(x^100))) \\ Colin Barker, May 28 2015

Formula

a(n) = n(n+2)/8 if n is even; a(n) = (n-1)^2/4 if n is odd (n>=0). - Emeric Deutsch, Nov 29 2006
a(n) = (3*n^2-2*n+2-(n^2-6*n+2)*(-1)^n)/16. - Luce ETIENNE, May 28 2015
a(n) = 3*a(n-2)-3*a(n-4)+a(n-6) for n>5. - Colin Barker, May 28 2015
G.f.: -x^2*(x^3+x+1) / ((x-1)^3*(x+1)^3). - Colin Barker, May 28 2015

Extensions

More terms from Zak Seidov, Nov 28 2006
More terms from Emeric Deutsch, Nov 29 2006
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.