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 1-6 of 6 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

A029547 Expansion of g.f. 1/(1 - 34*x + x^2).

Original entry on oeis.org

1, 34, 1155, 39236, 1332869, 45278310, 1538129671, 52251130504, 1775000307465, 60297759323306, 2048348816684939, 69583562007964620, 2363792759454112141, 80299370259431848174, 2727814796061228725775, 92665403695822344828176, 3147895910861898495432209
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Chebyshev sequence U(n,17)=S(n,34) with Diophantine property.
b(n)^2 - 2*(12*a(n))^2 = 1 with the companion sequence b(n)=A056771(n+1). - Wolfdieter Lang, Dec 11 2002
More generally, for t(m) = m + sqrt(m^2-1) and u(n) = (t(m)^(n+1) - 1/t(m)^(n+1))/(t(m) - 1/t(m)), we can verify that ((u(n+1) - u(n-1))/2)^2 - (m^2-1)*u(n)^2 = 1. - Bruno Berselli, Nov 21 2011
a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,33}. - Milan Janjic, Jan 26 2015

Links

Crossrefs

A091761 is an essentially identical sequence.
Cf. A200441, A200724.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).

Programs

  • GAP
    m:=17;; a:=[1,2*m];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 22 2019
  • Magma
    I:=[1,34]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011
  • Maple
    with (combinat):seq(fibonacci(4*n+4,2)/12, n=0..15); # Zerinvary Lajos, Apr 21 2008
  • Mathematica
    Table[GegenbauerC[n, 1, 17], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
LinearRecurrence[{34,-1},{1,34},20] (* Vincenzo Librandi, Nov 22 2011 *)
ChebyshevU[Range[21] -1, 17] (* G. C. Greubel, Dec 22 2019 *)
  • PARI
    A029547(n, x=[0,1],A=[17,72*4;1,17]) = vector(n,i,(x*=A)[1]) \\ M. F. Hasler, May 26 2007
  • PARI
    vector( 21, n, polchebyshev(n-1, 2, 17) ) \\ G. C. Greubel, Dec 22 2019
  • Sage
    [lucas_number1(n,34,1) for n in range(1, 16)] # Zerinvary Lajos, Nov 07 2009
  • Sage
    [chebyshev_U(n,17) for n in (0..20)] # G. C. Greubel, Dec 22 2019

Formula

a(n) = 34*a(n-1) - a(n-2), a(-1)=0, a(0)=1.
a(n) = S(n, 34) with S(n, x):= U(n, x/2) Chebyshev's polynomials of the 2nd kind. See A049310. - Wolfdieter Lang, Dec 11 2002
a(n) = (ap^(n+1) - am^(n+1))/(ap - am) with ap = 17+12*sqrt(2) and am = 17-12*sqrt(2).
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*binomial(n-k, k)*34^(n-2*k).
a(n) = sqrt((A056771(n+1)^2 -1)/2)/12.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) with a(-1)=0, a(0)=1, a(1)=34. Also a(n) = (sqrt(2)/48)*((17+12*sqrt(2))^n-(17-12*sqrt(2))^n) = (sqrt(2)/48)*((3+2*sqrt(2))^(2n+2)-(3-2*sqrt(2))^(2n+2)) = (sqrt(2)/48)*((1+sqrt(2))^(4n+4)-(1-sqrt(2))^(4n+4)). - Antonio Alberto Olivares, Mar 19 2008
a(n) = Sum_{k=0..n} A101950(n,k)*33^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 0} (1 + 1/a(n)) = 1/4*(4 + 3*sqrt(2)).
Product {n >= 1} (1 - 1/a(n)) = 2/17*(4 + 3*sqrt(2)). (End)
E.g.f.: exp(17*x)*(24*cosh(12*sqrt(2)*x) + 17*sqrt(2)*sinh(12*sqrt(2)*x))/24. - Stefano Spezia, Apr 16 2023

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

Original entry on oeis.org

1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001
Offset: 0

Views

Author

Henry Bottomley, Aug 16 2000

Keywords

Comments

The sequence satisfies the Pell equation a(n)^2 - 18 * A202299(n+1)^2 = 1. - Vincenzo Librandi, Dec 19 2011
Also numbers n such that n - 1 and 2*n + 2 are squares. - Arkadiusz Wesolowski, Mar 15 2015
And they, n - 1 and 2*n + 2, are the squares of A005319 and A003499. - Michel Marcus, Mar 15 2015
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3*A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019

Examples

			G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...

Links

Crossrefs

Cf. A001075, A001541, A001091, A001079, A023038, A011943, A001081, A023039, A001085 and note relationship with square triangular number sequences A001110 and A001109. A091761.
Row 3 of array A188644.

Programs

  • Magma
    I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011
  • Mathematica
    LinearRecurrence[{34,-1},{1,17},30] (* Vincenzo Librandi, Dec 18 2011 *)
a[ n_] := ChebyshevT[ 2 n, 3]; (* Michael Somos, May 28 2014 *)
  • Maxima
    makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); /* Vincenzo Librandi, Dec 18 2011 */
  • PARI
    {a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */
  • Sage
    [lucas_number2(n,34,1)/2 for n in range(0,15)] # Zerinvary Lajos, Jun 27 2008

Formula

a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = 16*A001110(n) + 1 = A001541(2n) = (4*A001109(n))^2 + 1 = 3*A001109(2n-1) - A001109(2n-2) = A001109(2n) - 3*A001109(2n-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n).
G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019
a(n) = cosh(2n*arcsinh(sqrt(8))). - Herbert Kociemba, Apr 24 2008
a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011
a(-n) = a(n). - Michael Somos, May 28 2014
a(n) = sqrt(1 + 32*9*A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019

Extensions

More terms from James Sellers, Sep 07 2000
Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A350984 a(0)=0, a(1)=18, a(2)=612; a(n) = 35*(a(n-1)-a(n-2))+a(n-3).

Original entry on oeis.org

0, 18, 612, 20790, 706248, 23991642, 815009580, 27686334078, 940520349072, 31950005534370, 1085359667819508, 36870278700328902, 1252504116143363160, 42548269670174018538, 1445388664669773267132, 49100666329102117063950, 1667977266524802206907168, 56662126395514172917779762, 1924844320180957076997604740
Offset: 0

Views

Author

N. J. A. Sloane, Mar 08 2022

Keywords

References

  • P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139.

Links

Crossrefs

Cf. A001541, A001542, A350983.
Equals 18*A029547(n-1) for n >= 1.
Equals 18*A091761.

Programs

  • Mathematica
    CoefficientList[Series[18*x/(x^2 - 34*x + 1), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 03 2022 *)

Formula

From Chai Wah Wu, Mar 08 2022: (Start)
a(n) = 34*a(n-1) - a(n-2) for n > 1.
G.f.: 18*x/(x^2 - 34*x + 1). (End)

A383742 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where column k is the expansion of g.f. x/(1 - A002203(k)*x + (-1)^k*x^2).

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 6, 5, 4, 0, 1, 14, 35, 12, 5, 0, 1, 34, 197, 204, 29, 6, 0, 1, 82, 1155, 2772, 1189, 70, 7, 0, 1, 198, 6725, 39236, 39005, 6930, 169, 8, 0, 1, 478, 39203, 551532, 1332869, 548842, 40391, 408, 9, 0, 1, 1154, 228485, 7761996, 45232349, 45278310, 7722793, 235416, 985, 10
Offset: 0

Views

Author

Seiichi Manyama, May 07 2025

Keywords

Examples

			Square array begins:
  0,  0,    0,     0,       0,        0, ...
  1,  1,    1,     1,       1,        1, ...
  2,  2,    6,    14,      34,       82, ...
  3,  5,   35,   197,    1155,     6725, ...
  4, 12,  204,  2772,   39236,   551532, ...
  5, 29, 1189, 39005, 1332869, 45232349, ...

Crossrefs

Columns k=0..6 give A001477, A000129, A001109, A041085(n-1), A091761, A292423, A097731(n-1).
Rows n=0..5 give A000004, A000012, A002203, A383720, A383740, A383741.
Main diagonal gives A380083.
Cf. A028412.

Programs

  • Mathematica
    A[n_, k_] := Fibonacci[k*n, 2]/Fibonacci[k, 2]; A[n_, 0] := n; Table[A[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 08 2025 *)
  • PARI
    pell(n) = ([2, 1; 1, 0]^n)[2, 1];
a(n, k) = if(k==0, n, pell(k*n)/pell(k));

Formula

A(0,k) = 0, A(1,k) = 1; A(n,k) = A002203(k) * A(n-1,k) - (-1)^k * A(n-2,k) for n > 1.
A(n,k) = Pell(k*n)/Pell(k) for k > 0.

A011906 If b(n) is A011900(n) and c(n) is A001109(n), then a(n) = b(n)*c(n) = b(n) + (b(n)+1) + (b(n)+2) + ... + c(n).

Original entry on oeis.org

1, 18, 525, 17340, 586177, 19896030, 675781821, 22956120408, 779829016225, 26491211221770, 899921240562957, 30570830315362260, 1038508305678375841, 35278711540581704598, 1198437683944896688125, 40711602541832856049200, 1382996048733983114022337
Offset: 1

Views

Author

Mario Velucchi (mathchess(AT)velucchi.it)

Keywords

Examples

			a(3) = 525 = 15*35 = 15 + 16 + ... + 35.

References

  • Mario Velucchi "From the desk of ... Mario Velucchi" in 'Mathematics and Informatics quarterly' volume 7 - 2/1997, p. 81.

Links

Crossrefs

Cf. A001109, A011900, A029547, A091761.

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-23*x+33*x^2-3*x^3)/((1-x)*(1-34*x+x^2)*(1-6*x+x^2)) )); // G. C. Greubel, Oct 18 2024
  • Maple
    A011900 := proc(n) coeftayl( (1-4*x+x^2)/((1-x)*(1-6*x+x^2)),x=0,n) ; end proc: A001109 := proc(n) coeftayl( x/(1-6*x+x^2),x=0,n) ; end proc: A011906 := proc(n) A001109(n)*A011900(n-1) ; end proc: seq(A011906(n),n=1..30) ; # R. J. Mathar, Apr 15 2010
  • Mathematica
    LinearRecurrence[{41, -246, 246, -41, 1}, {1, 18, 525, 17340, 586177}, 20] (* Paul Cleary, Dec 05 2015 *)
CoefficientList[Series[(-1 + 23*x - 33*x^2 + 3*x^3)/((x - 1)*(x^2 - 34*x + 1)*(1 - 6*x + x^2)), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 16 2017 *)
  • SageMath
    def A011906_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1-23*x+33*x^2-3*x^3)/((1-x)*(1-34*x+x^2)*(1-6*x+x^2)) ).list()
a=A011906_list(30); a[1:] # G. C. Greubel, Oct 18 2024

Formula

From R. J. Mathar, Apr 15 2010: (Start)
G.f.: x*(1-23*x+33*x^2-3*x^3)/((1-x)*(1-34*x+x^2)*(1-6*x+x^2)).
a(n) = 41*a(n-1) -246*a(n-2) +246*a(n-3) -41*a(n-4) +a(n-5). (End)
Lim_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jul 17 2020
a(n) = (1/16)*(1 - A029547(n) + 41*A091761(n) + 8*A001109(n)). - G. C. Greubel, Oct 18 2024

Extensions

More terms from R. J. Mathar, Apr 15 2010
Showing 1-6 of 6 results.

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