A139275 -id:A139275 - cp's OEIS frontend

A139591 A139275(n) followed by 18-gonal number A051870(n+1).

0, 1, 9, 18, 34, 51, 75, 100, 132, 165, 205, 246, 294, 343, 399, 456, 520, 585, 657, 730, 810, 891, 979, 1068, 1164, 1261, 1365, 1470, 1582, 1695, 1815, 1936, 2064, 2193, 2329, 2466, 2610, 2755, 2907, 3060, 3220, 3381, 3549, 3718, 3894, 4071, 4255, 4440, 4632

Offset: 0

Omar E. Pol, May 03 2008

Sequence found by reading the line from 0, in the direction 0, 9, ... and the line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the triangular numbers A000217.

			Array begins:
   0,   1;
   9,  18;
  34,  51;
  75, 100;
  ...

Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.

Cf. A000217, A046092, A051870, A139275, A077221, A139592, A139593, A139595, A139596, A139597, A139598.

Partial sums of A047393.

Array read by rows: row n gives 8*n^2 + n, 8*(n+1)^2 - 7*(n+1).
G.f.: -x*(7*x+1)/((x-1)^3*(x+1)). - Colin Barker, Oct 16 2012
a(n) = 2*n^2 + (7/2)*n + (3/4)*((-1)^n-1). - Sean A. Irvine, Jul 14 2022

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

N. J. A. Sloane

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

			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.

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.

Kade Robertson, Table of n, a(n) for n = 0..600 [This replaces an earlier b-file computed by T. D. Noe]
Nikola Adžaga, Andrej Dujella, Dijana Kreso, and Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Tom Beldon and Tony Gardiner, Triangular numbers and perfect squares, The Mathematical Gazette, 2002, pp. 423-431, esp pp. 424-426.
K. S. Brown, Square Triangular Numbers.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
Yong-Gao Chen and Jin-Hui Fang, Triangular numbers in geometric progression, INTEGERS 7 (2007), #A19.
F. Javier de Vega, On the parabolic partitions of a number, J. Alg., Num. Theor., and Appl. (2023) Vol. 61, No. 2, 135-169.
Colin Dickson et al., ratio of integers = sqrt(2), thread in newsgroup alt.math.recreational, March 7, 2004.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
Bill Gosper, The Triangular Squares, 2014.
Jon Grantham and Hester Graves, The abc Conjecture Implies That Only Finitely Many Cullen Numbers Are Repunits, arXiv:2009.04052 [math.NT], 2020. Mentions this sequence.
Gillian Hatch, Pythagorean Triples and Triangular Square Numbers, Mathematical Gazette 79 (1995), 51-55.
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Roger B. Nelsen, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
A. Nowicki, The numbers a^2+b^2-dc^2, J. Int. Seq. 18 (2015) # 15.2.3.
Matthew Parker, The first 5000 square triangular numbers (7-Zip compressed file).
J. L. Pietenpol, A. V. Sylwester, E. Just, and R. M. Warten, Problem E1473, Amer. Math. Monthly, Vol. 69, No. 2 (Feb. 1962), pp. 168-169. (From the editorial note on p. 169 of this source, we learn that the question about the existence of perfect squares in the sequence of triangular numbers cropped up in the Euler-Goldbach Briefwechsel of 1730; the translation into English of the relevant letters can be found at Correspondence of Leonhard Euler with Christian Goldbach (part II), pp. 614-615.) - _José Hernández_, May 24 2022
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
D. A. Q., Triangular square numbers: a postscript, Math. Gaz., 56 (1972), 311-314.
K. Ramsey, Re_Generalized_Proof_re_Square_Triangular_Numbers. [Broken link]
K. Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
Fabio Roman, On certain ratios regarding integer numbers which are both triangulars and squares, arXiv:1703.06701 [math.NT], 2017.
Jon E. Schoenfield, Prime factorization of a(n) for n = 1..300.
Jaap Spies, A Bit of Math, The Art of Problem Solving, Jaap Spies Publishers (2019).
Ralf Stephan, Boring proof of a nonlinearity.
UWC, Problem A, Nieuw Archief voor Wiskunde, Dec 2004; Jaap Spies, Solution.
Michel Waldschmidt, Continued fractions, École de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Wikipedia, Square triangular number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

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.

a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
   zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
-- Reinhard Zumkeller, Oct 12 2011

[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

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

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 *)

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

;; 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

;; 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

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

A139271 a(n) = 2n(4*n-3).

Original entry on oeis.org

0, 2, 20, 54, 104, 170, 252, 350, 464, 594, 740, 902, 1080, 1274, 1484, 1710, 1952, 2210, 2484, 2774, 3080, 3402, 3740, 4094, 4464, 4850, 5252, 5670, 6104, 6554, 7020, 7502, 8000, 8514, 9044, 9590, 10152, 10730, 11324, 11934, 12560, 13202, 13860, 14534, 15224

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 2, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A033585 in the same spiral.
Twice decagonal numbers (or twice 10-gonal numbers). - Omar E. Pol, May 15 2008
a(n) is the number of walks in a cubic lattice of n dimensions that reach the point of origin for the first time after 4 steps. - Shel Kaphan, Mar 20 2023

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.
Cf. A001107.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=16). - Bruno Berselli, Jun 10 2013
Row n=2 of A361397.

Table[8n^2-6n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,2,20},50] (* Harvey P. Dale, Sep 26 2016 *)

a(n)=2*n*(4*n-3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 - 6*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = A001107(n)*2. - Omar E. Pol, May 15 2008
a(n) = 16*n + a(n-1) - 14 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: (2*x)*(7*x+1)/(1-x)^3.
E.g.f.: (8*x^2 + 2*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = Pi/12 + log(2)/2. - Amiram Eldar, Mar 28 2023

Corrected by Harvey P. Dale, Sep 26 2016

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A139278 a(n) = n(8n+7).

Original entry on oeis.org

0, 15, 46, 93, 156, 235, 330, 441, 568, 711, 870, 1045, 1236, 1443, 1666, 1905, 2160, 2431, 2718, 3021, 3340, 3675, 4026, 4393, 4776, 5175, 5590, 6021, 6468, 6931, 7410, 7905, 8416, 8943, 9486, 10045, 10620, 11211, 11818, 12441, 13080

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the segment (0, 15) together with the line from 15, in the direction 15, 46, ..., in the square spiral whose vertices are the triangular numbers A000217.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139276, A139277, A139279, A139280, A139281, A139282.

Table[n (8 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 15, 46}, 50] (* Harvey P. Dale, Oct 07 2015 *)

a(n)=n*(8*n+7) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 7*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-1 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(x+15)/(1-x)^3.
E.g.f.: (8*x^2 + 15*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/49 + (sqrt(2)+1)*Pi/14 - 4*log(2)/7 - sqrt(2)*log(sqrt(2)+1)/7. - Amiram Eldar, Mar 17 2022

A139272 a(n) = n(8n-5).

Original entry on oeis.org

0, 3, 22, 57, 108, 175, 258, 357, 472, 603, 750, 913, 1092, 1287, 1498, 1725, 1968, 2227, 2502, 2793, 3100, 3423, 3762, 4117, 4488, 4875, 5278, 5697, 6132, 6583, 7050, 7533, 8032, 8547, 9078, 9625, 10188, 10767, 11362, 11973, 12600

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 3, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139276 in the same spiral.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(n*k-k+6)/2, this sequence is the case k=16: see Comments lines of A226492.

s=0;lst={s};Do[s+=n++ +3;AppendTo[lst, s], {n, 0, 7!, 16}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
Table[n*(8*n -5), {n,0,50}] (* G. C. Greubel, Jul 18 2017 *)
LinearRecurrence[{3,-3,1},{0,3,22},50] (* Harvey P. Dale, Jan 13 2024 *)

a(n)=n*(8*n-5) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 8*n^2 - 5*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 13 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(13*x + 3)/(1-x)^3.
E.g.f.: (8*x^2 + 3*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = ((sqrt(2)-1)*Pi + 8*log(2) - 2*sqrt(2)*log(sqrt(2)+1))/10. - Amiram Eldar, Mar 17 2022

A139274 a(n) = n(8n-1).

Original entry on oeis.org

0, 7, 30, 69, 124, 195, 282, 385, 504, 639, 790, 957, 1140, 1339, 1554, 1785, 2032, 2295, 2574, 2869, 3180, 3507, 3850, 4209, 4584, 4975, 5382, 5805, 6244, 6699, 7170, 7657, 8160, 8679, 9214, 9765, 10332, 10915, 11514, 12129, 12760, 13407, 14070, 14749

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the triangular numbers A000217.

Polygonal number connection: 2*P_n + 5*S_n where P_n is the n-th pentagonal number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010

			a(1) = 16*1 + 0 - 9 = 7; a(2) = 16*2 + 7 - 9 = 30; a(3) = 16*3 + 30 - 9 = 69. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

[n*(8*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

A139274:=n->n*(8*n-1): seq(A139274(n), n=0..100); # Wesley Ivan Hurt, Dec 04 2016

CoefficientList[Series[x (9 x + 7)/(1 - x)^3, {x, 0, 43}], x] (* Michael De Vlieger, Jan 11 2020 *)
Table[n(8n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,30},50] (* Harvey P. Dale, Apr 01 2024 *)

a(n)=n*(8*n-1) \\ Charles R Greathouse IV, Jun 17 2017

Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = (1/3) * Sum_{i=n..(7*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(9*x+7)/(1-x)^3.
E.g.f.: (8*x^2 + 7*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 4*log(2) + sqrt(2)*log(sqrt(2)+1) - (sqrt(2)+1)*Pi/2. - Amiram Eldar, Mar 18 2022

A139276 a(n) = n(8n+3).

Original entry on oeis.org

0, 11, 38, 81, 140, 215, 306, 413, 536, 675, 830, 1001, 1188, 1391, 1610, 1845, 2096, 2363, 2646, 2945, 3260, 3591, 3938, 4301, 4680, 5075, 5486, 5913, 6356, 6815, 7290, 7781, 8288, 8811, 9350, 9905, 10476, 11063, 11666, 12285, 12920

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 11,..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139272 in the same spiral.

			a(1)=16*1+0-5=11; a(2)=16*2+11-5=38; a(3)=16*3+38-5=81. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139277, A139278, A139279, A139280, A139281, A139282.

a[n_]:=n*(8*n+3); a[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3,-3,1},{0,11,38},50] (* Harvey P. Dale, Jul 02 2022 *)

a(n)=n*(8*n+3) \\ Charles R Greathouse IV, Jun 16 2017

a(n) = 8*n^2 + 3*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-5 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(5*x + 11)/(1-x)^3.
E.g.f.: (8*x^2 + 11*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/9 - (sqrt(2)-1)*Pi/6 - 4*log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3. - Amiram Eldar, Mar 17 2022

A154260 Numbers of the form m(4m +- 1)/2.

Original entry on oeis.org

0, 7, 9, 30, 34, 69, 75, 124, 132, 195, 205, 282, 294, 385, 399, 504, 520, 639, 657, 790, 810, 957, 979, 1140, 1164, 1339, 1365, 1554, 1582, 1785, 1815, 2032, 2064, 2295, 2329, 2574, 2610, 2869, 2907, 3180, 3220, 3507, 3549, 3850, 3894, 4209, 4255, 4584

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Also integers of the form Sum_{k = 1..j} k/4 = j*(j + 1)/8. - Alonso del Arte, Jan 20 2012
Numbers h such that 32*h + 1 is a square. - Bruno Berselli, Mar 30 2014
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(16*n))*(1 - q^(16*n-7))*(1 - q^(16*n-9)) = 1 - q^7 - q^9 + q^30 + q^34 - q^69 - q^75 + + - - .... - Peter Bala, Dec 24 2024

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A001318, A074378, A057569, A057570.

Cf. similar sequences listed in A219257 and A299645.

k:=8; f:=func; [0] cat [f(n*m): m in [-1, 1], n in [1..25]]; // Bruno Berselli, Nov 14 2012

Select[Union[Flatten[Table[{n (4n - 1)/2, n (4n + 1)/2}, {n, 0, 199}]]], IntegerQ] (* Alonso del Arte, Jan 20 2012 *)

print1(0);forstep(n=2,1e2,2,print1(", "n*(4*n-1)/2", "n*(4*n+1)/2)) \\ Charles R Greathouse IV, Jan 20 2012

print1(s=0);for(n=1,1e3,s+=n/4;if(denominator(s)==1,print1(s", "))) \\ Charles R Greathouse IV, Jan 20 2012

def A154260(n): return (n>>1)*((n<<2)+(-3 if n&1 else -1)) # Chai Wah Wu, Mar 11 2025

From R. J. Mathar, Jan 07 2009: (Start)
A139274 UNION A139275.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
G.f.: x^2*(7 + 2x + 7x^2)/((1+x)^2*(1-x)^3). (End)
From G. C. Greubel, Sep 08 2016: (Start)
a(n) = (1/4)*(8*n^2 + 6*(-1)^n*n - 8*n - 3*(-1)^n + 3).
E.g.f.: (1/4)*( (3 + 8*x^2)*exp(x) - 3*(1 + 2*x)*exp(-x) ). (End)
From Amiram Eldar, Mar 17 2022: (Start)
Sum_{n>=2} 1/a(n) = 8 - (sqrt(2)+1)*Pi.
Sum_{n>=2} (-1)^n/a(n) = 2*sqrt(2)*log(sqrt(2)+1) - 8*(1-log(2)). (End)
a(n) = (n-1)*(4*n-3)/2 if n is odd and a(n) = n*(4*n-1)/2 if n is even. - Chai Wah Wu, Mar 11 2025

A139277 a(n) = n(8n+5).

Original entry on oeis.org

0, 13, 42, 87, 148, 225, 318, 427, 552, 693, 850, 1023, 1212, 1417, 1638, 1875, 2128, 2397, 2682, 2983, 3300, 3633, 3982, 4347, 4728, 5125, 5538, 5967, 6412, 6873, 7350, 7843, 8352, 8877, 9418, 9975, 10548, 11137, 11742, 12363, 13000

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139273 in the same spiral.

Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250.

Cf. A139271, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Table[n (8 n + 5), {n, 0, 50}] (* Bruno Berselli, Aug 22 2018 *)
LinearRecurrence[{3,-3,1},{0,13,42},50] (* Harvey P. Dale, Dec 04 2018 *)

a(n)=n*(8*n+5) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 5*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x*{c+8 + (8-c)*x}/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 3 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
Sum_{n>=1} 1/a(n) = (sqrt(2)-1)*Pi/10 - 4*log(2)/5 + sqrt(2)*log(sqrt(2)+1)/5 + 8/25. - Amiram Eldar, Mar 18 2022
E.g.f.: exp(x)*x*(13 + 8*x). - Elmo R. Oliveira, Dec 15 2024

cp's OEIS Frontend

A139591 A139275(n) followed by 18-gonal number A051870(n+1).

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

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A139271 a(n) = 2*n*(4*n-3).

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A139273 a(n) = n*(8*n - 3).

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A139278 a(n) = n*(8*n+7).

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A139272 a(n) = n*(8*n-5).

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A139274 a(n) = n*(8*n-1).

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A139271 a(n) = 2n(4*n-3).

A139273 a(n) = n(8n - 3).

A139278 a(n) = n(8n+7).

A139272 a(n) = n(8n-5).

A139274 a(n) = n(8n-1).

A139276 a(n) = n(8n+3).

A154260 Numbers of the form m(4m +- 1)/2.

A139277 a(n) = n(8n+5).