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 1-7 of 7 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

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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

A027568 Numbers that are both triangular and tetrahedral.

Original entry on oeis.org

0, 1, 10, 120, 1540, 7140
Offset: 1

Views

Author

Patrick De Geest

Keywords

Comments

From Anthony C Robin, Oct 27 2022: (Start)
For numbers to be triangular and tetrahedral, we look for solutions r*(r+1)*(r+2)/6 = t*(t+1)/2 = a(n). The corresponding r and t are r = A224421(n-1) and t = A102349(n).
Writing m=r+1 and s=2t+1, this problem is equivalent to solving the Diophantine equation 3 + 4*(m^3 - m) = 3*s^2. The integer solutions for this equation are m = 0, 1, 2, 4, 9, 21, 35 and the corresponding values of s are 1, 1, 3, 9, 31, 111, 239. (End)

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Ellipses (Paris), 2008 (entry 10, page 3; entry 120, page 41).
  • L. J. Mordell, Diophantine Equations, Ac. Press, page 258.
  • P. Odifreddi, Il museo dei numeri, Rizzoli, 2014, page 224.
  • J. Roberts, The Lure of the Integers, page 53.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 21.

Links

Crossrefs

Intersection of A000217 and A000292.
Cf. A102349, A102461, A102772, A224421.

Programs

  • Maple
    {seq(binomial(i,3),i=0..100000) } intersect {seq(binomial(k,2), k= 0..100000)}; # Zerinvary Lajos, Apr 26 2008
  • Mathematica
    With[{trno=Accumulate[Range[0,1000]]},Intersection[trno,Accumulate[ trno]]] (* Harvey P. Dale, May 25 2014 *)
  • PARI
    for(n=0,1e3,if(ispolygonal(t=n*(n+1)*(n+2)/6,3),print1(t", "))) \\ Charles R Greathouse IV, Apr 07 2013

A003556 Numbers that are both square and tetrahedral.

Original entry on oeis.org

0, 1, 4, 19600
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

A. J. J. Meyl proved in 1878 that only 1, 4 and 19600 are both square and tetrahedral. See link. [Bernard Schott, Dec 23 2012]

Examples

			From _Bernard Schott_, Dec 23 2012: (Start)
If S(n) = n^2 and T(m) = m*(m+1)*(m+2)/6, then
-> S(1)= T(1) = 1;
-> S(2)= T(2) = 4;
-> S(140) = T(48) = 19600. (End)

References

  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 600.
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, p. 165 (Rev. ed. 1997).

Links

Crossrefs

Intersection of A000290 and A000292.

Programs

  • Mathematica
    Select[Rest[FoldList[Plus, 0, Rest[FoldList[Plus, 0, Range[50000]]]]], IntegerQ[Sqrt[ # ]]&]
Intersection[Binomial[# + 2, 3]&/@Range[0, 10000], Range[0,409000]^2] (* Harvey P. Dale, Feb 01 2011 *)

A053612 Numbers such that 1+2+3...+a(n) = 1+4+9+...+r^2, for some r.

Original entry on oeis.org

1, 10, 13, 645
Offset: 1

Views

Author

Jud McCranie, Mar 19 2000

Keywords

Examples

			1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 2 + 3 + ... + 10, so 10 is in the sequence.

References

  • Joe Roberts, Lure of the Integers, page 245 (entry for 645).

Links

Crossrefs

Cf. A039596, A053611.

A053611 Numbers k such that 1 + 4 + 9 + ... + k^2 = 1 + 2 + 3 + ... + m for some m.

Original entry on oeis.org

1, 5, 6, 85
Offset: 1

Views

Author

Jud McCranie, Mar 19 2000

Keywords

Comments

These are the only possibilities for a sum of the first n squares to equal a triangular number.
From Seiichi Manyama, Aug 25 2019: (Start)
The complete list of solutions to k*(k+1)*(2*k+1)/6 = m*(m+1)/2 is as follows.
(k,m) = (-1, 0), (0, 0), (1, 1), (5, 10), (6, 13), (85, 645),
(-1,-1), (0,-1), (1,-2), (5,-11), (6,-14), (85,-646). (End)

Examples

			1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 2 + 3 + ... + 10, so 5 is in the sequence.

References

  • E. T. Avanesov, The Diophantine equation 3y(y+1) = x(x+1)(2x+1), Volz. Mat. Sb. Vyp., 8 (1971), 3-6.
  • R. K. Guy, Unsolved Problems in Number Theory, Section D3.
  • Joe Roberts, Lure of the Integers, page 245 (entry for 645).

Links

Crossrefs

Cf. A039596, A053612 (values of m).
Cf. A000217, A000330.

Programs

  • Maple
    istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then RETURN(true) else RETURN(false); fi; end;
M:=1000; for n from 1 to M do if istriangular(n*(n+1)*(2*n+1)/6) then lprint(n,n*(n+1)*(2*n+1)/6); fi; od: # N. J. A. Sloane
# second Maple program:
q:= n-> issqr(8*sum(j^2, j=1..n)+1):
select(q, [$1..100])[];  # Alois P. Heinz, Oct 10 2024
  • Mathematica
    Select[Range[90], IntegerQ[(Sqrt[(4/3) * (# + 3 * #^2 + 2 * #^3) + 1] - 1)/2] &] (* Harvey P. Dale, Sep 22 2014 *)

Extensions

Edited by N. J. A. Sloane, May 25 2008

A136276 Consider pairs of nonnegative integers (m,k) such that 2^2 + 4^2 + 6^2 + ... + (2m)^2 = k(k+1); sequence gives k values.

Original entry on oeis.org

0, 4, 7, 84
Offset: 1

Views

Author

Ken Knowlton (www.KnowltonMosaics.com), Mar 29 2008

Keywords

Comments

The problem arises when trying to build a square pyramid out of dominoes. The solution (m,k) = (3,7) for example corresponds to building a pyramid with layers of sizes 2 X 2, 4 X 4 and 6 X 6 from one set of double-6 dominoes.
The three nonzero solutions use one double-3 set, one double-6 set and one double-83 set. (The sequence 3,6,83 is too short to warrant a separate entry.)
The problem is equivalent to finding integers (m,k) such that 2m(m+1)(m+2)/3 = k*(k+1). This is a nonsingular cubic, so by Siegel's theorem, there are only finitely many solutions. - N. J. A. Sloane, May 25 2008. See also the articles by Stroeker and Tzanakis and Stroeker and de Weger.

Examples

			The known solutions are (m,k) = (0,0), (2,4), (3,7) and (17,84). There are no other solutions.

Links

Crossrefs

Cf. A039596, A053611, A053612.

Programs

  • Maple
    Simple-minded Maple program from N. J. A. Sloane:
f1:=m-> 1+8*m*(m+1)*(2*m+1)/3;
for m from 0 to 10^8 do if issqr(f1(m)) then lprint( m, (-1+sqrt(f1(m)))/2); fi; od;

Extensions

Edited by N. J. A. Sloane, May 25 2008, Aug 17 2008
May 26 2008: John Cannon used MAGMA to show there are no further solutions (see link)

A307492 Numbers that are both centered square and square pyramidal.

Original entry on oeis.org

1, 5, 42925, 1026745
Offset: 1

Views

Author

Ilya Gutkovskiy, Apr 10 2019

Keywords

Comments

If it exists, a(5) > 10^29. - Bert Dobbelaere, Apr 12 2019

Links

Crossrefs

Intersection of A000330 and A001844.
Cf. A008844, A039596, A131750, A307491.

Programs

  • Mathematica
    csQ[n_] := IntegerQ[Sqrt[2*n-1]]; Select[Table[n(n+1)(2n+1)/6, {n, 0, 1000}], csQ] (* Amiram Eldar, Apr 11 2019 *)
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