A129556 -id:A129556 - cp's OEIS frontend

A001652 a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.

0, 3, 20, 119, 696, 4059, 23660, 137903, 803760, 4684659, 27304196, 159140519, 927538920, 5406093003, 31509019100, 183648021599, 1070379110496, 6238626641379, 36361380737780, 211929657785303, 1235216565974040, 7199369738058939, 41961001862379596, 244566641436218639

Offset: 0

N. J. A. Sloane

Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values.
Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571).
Members of Diophantine pairs. Solution to a*(a+1) = 2*b*(b+1) in natural numbers including 0; a = a(n), b = b(n) = A053141(n); The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square.
The three sequences x (A001652), y (A046090) and z (A001653) may be obtained by setting u and v equal to the Pell numbers (A000129) in the formulas x = 2uv, y = u^2 - v^2, z = u^2 + v^2 [Joseph Wiener and Donald Skow]. - Antonio Alberto Olivares, Dec 22 2003
All Pythagorean triples {X(n), Y(n)=X(n)+1, Z(n)} with X M*W(n), where W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given by [2 1 2 / 1 2 2 / 2 2 3]. - Lekraj Beedassy, Aug 14 2006
Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n). - Kenneth J Ramsey, Sep 22 2007
In general, if b(n) = A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); e.g., 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870 = 5741*14+84. - Charlie Marion, Nov 19 2007
Limit_{n -> oo} a(n)/a(n-1) = 3+2*sqrt(2) = A156035. - Klaus Brockhaus, Feb 17 2009
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with pMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = y^2 with pMohamed Bouhamida, Sep 02 2009
a(n+k) = A001541(k)*a(n) + A001542(k)*A001653(n+1) + A001108(k). - Charlie Marion, Dec 10 2010
The numbers 3*A001652 = (0, 9, 60, 357, 2088, 12177, 70980, ...) are all the nonnegative values of X such that X^2 + (X+3)^2 = Z^2 (Z is in A075841). - Bruno Berselli, Aug 26 2010
Let T(n) = n*(n+1)/2 (the n-th triangular number). For n > 0,
  T(a(n) + 2*k*A001653(n+1)) = 2*T(A053141(n-1) + k*A002315(n)) + k^2 and
  T(a(n) + (2*k+1)*A001653(n+1)) = (A001109(n+1) + k*A002315(n))^2 + k*(k+1).
  Also (a(n) + k*A001653(n))^2 + (a(n) + k*A001653(n) + 1)^2 = (A001653(n+1) + k*A002315(n))^2 + k^2. - Charlie Marion, Dec 09 2010
For n>0, A143608(n) divides a(n). - Kenneth J Ramsey, Jun 28 2012
Set a(n)=p; a(n)+1=q; the generated triple x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). - Carmine Suriano, Dec 17 2013
The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1) = 1 and c(1) = 2, then b(n) = c(n-1) and c(n) = 2*c(n-1) + b(n-1). Alternatively, b(n) = c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). - J. M. Bergot, Oct 09 2014
Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see A274307). - Chai Wah Wu, Jun 21 2016
Numbers m such that m^2+(m+1)^2 is a square. - César Aguilera, Aug 14 2017
For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-2). Further, let p(n) = Sum_{i=1..2n} P(a,d,i). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) = A000129(n), the n-th Pell number. - Charlie Marion, Dec 08 2018
The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in A002315 and A001653. The relation k = (x-1)/2 explains Lekraj Beedassy's Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in A001653. - Bernard Schott, Mar 10 2019
Members of Diophantine pairs. Related to solutions of p^2 = 2q^2 + 2 in natural numbers; p = p(n) = 2*sqrt(4T(a(n))+1), q = q(n) = sqrt(8*T(a(n))+1). Note that this implies that 4*T(a(n))+1 is a perfect square (numbers of the form 8*T(n)+1 are perfect squares for all n); these T(a(n))'s are the only solutions to the given Diophantine equation. - Steven Blasberg, Mar 04 2021

			The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
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).

T. D. Noe, Table of n, a(n) for n = 0..200
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Christian Aebi and Grant Cairns, Lattice equable quadrilaterals III: tangential and extangential cases, Integers (2023) Vol. 23, #A48.
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
Henk Bruin and Robbert Fokkink, On the records and zeros of a deterministic random walk, arXiv:2503.11734 [math.DS], 2025. See p. 5.
T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.
Thibault Gauthier, Deep Reinforcement Learning for Synthesizing Functions in Higher-Order Logic, arXiv:1910.11797 [cs.AI], 2019. See also EPiC Series in Computing, (2020) Vol. 73, 230-248.
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
Ron Knott, Pythagorean Triples and Online Calculators
A. Martin, Table of prime rational right-angled triangles, The Mathematical Magazine, 2 (1910), 297-324.
A. Martin, Table of prime rational right-angled triangles (annotated scans of a few pages)
A. Martin, On rational right-angled triangles, Proceedings of the Fifth International Congress of Mathematicians (Cambridge, 22-28 August 1912).
S. P. Mohanty, Which triangular numbers are products of three consecutive integers, Acta Math. Hungar., 58 (1991), 31-36.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.
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.
Burkard Polster, Nice merging together, Mathologer video (2015).
B. Polster and M. Ross, Marching in squares, arXiv:1503.04658 [math.HO], 2015.
Zhang Zaiming, Problem #502, Pell's Equation - Once Again, The College Mathematics Journal, 25 (1994), 241-243.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A046090(n) = -a(-1-n).
Cf. A001108, A143608, A089950 (partial sums), A156035.
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; this sequence for k=4; A129556 for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013
Cf. A053141, A001541, A001109, A274307.
Cf. A002315, A001653 (solutions of x^2 - 2*y^2 = -1).
Cf. A000129, A001333, A001542, A002024, A002378, A029549, A048739, A053142, A055997, A075841, A084159, A084703, A097571, A123737.

a:=[0,3];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]+2; od; a; # Muniru A Asiru, Dec 08 2018

a001652 n = a001652_list !! n
a001652_list = 0 : 3 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001652_list)) a001652_list)
-- Reinhard Zumkeller, Jan 10 2012

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ (-2+(r2+1)*(3+2*r2)^n-(r2-1)*(3-2*r2)^n)/4: n in [1..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Feb 17 2009

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3-x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018

A001652 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,3]) ;
    else
        6*procname(n-1)-procname(n-2)+2 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

LinearRecurrence[{7,-7,1}, {0,3,20}, 30] (* Harvey P. Dale, Aug 19 2011 *)
With[{c=3+2*Sqrt[2]},NestList[Floor[c*#]+3&,3,30]] (* Harvey P. Dale, Oct 22 2012 *)
CoefficientList[Series[x (3 - x)/((1 - 6 x + x^2) (1 - x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 21 2014 *)
Table[(LucasL[2*n + 1, 2] - 2)/4, {n, 0, 30}] (* G. C. Greubel, Jul 15 2018 *)

{a(n) = subst( poltchebi(n+1) - poltchebi(n) - 2, x, 3) / 4}; /* Michael Somos, Aug 11 2006 */

concat(0, Vec(x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^50))) \\ Altug Alkan, Nov 08 2015

{a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(2*n+1) - 2)/4), ", ")) \\ G. C. Greubel, Jul 15 2018

(x*(3-x)/((1-6*x+x^2)*(1-x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 08 2019

G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - Simon Plouffe in his 1992 dissertation
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). a_{n} = -1/2 + ((1-2^{1/2})/4)*(3 - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/2})^n. - Antonio Alberto Olivares, Oct 13 2003
a(n) = a(n-2) + 4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). - Pierre CAMI, Mar 30 2005
a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8*A029549)-1)/2. - Bill Gosper, Feb 07 2010
Binomial(a(n)+1,2) = 2*binomial(A053141(n)+1,2) = A029549(n). See A053141. - Bill Gosper, Feb 07 2010
Let b(n) = A046090(n) and c(n) = A001653(n). Then for k>j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n<0, a(n) = -b(-n-1). Also a(n)*a(n+2*k+1) + b(n)*b(n+2*k+1) + c(n)*c(n+2*k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2*k) + b(n)*b(n+2*k) + c(n)*c(n+2*k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003
a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). - Charlie Marion, Jul 01 2003
For n and j >= 1, Sum_{k=0..j} A001653(k)*a(n) - Sum_{k=0...j-1} A001653(k)*a(n-1) + A053141(j) = A001109(j+1)*a(n) - A001109(j)*a(n-1) + A053141(j) = a(n+j). - Charlie Marion, Jul 07 2003
Sum_{k=0...n} (2*k+1)*a(n-k) = A001109(n+1) - A000217(n+1). - Charlie Marion, Jul 18 2003
a(n) = A055997(n) - 1 + sqrt(2*A055997(n)*A001108(n)). - Charlie Marion, Jul 21 2003
a(n) = {A002315(n) - 1}/2. - Lekraj Beedassy, Nov 25 2003
a(2*n+k) + a(k) + 1 = A001541(n)*A002315(n+k). For k>0, a(2*n+k) - a(k-1) = A001541(n+k)*A002315(n); e.g., 803760-119 = 19601*41. - Charlie Marion, Mar 17 2003
a(n) = (A001653(n+1) - 3*A001653(n) - 2)/4. - Lekraj Beedassy, Jul 13 2004
a(n) = {2*A084159(n) - 1 + (-1)^(n+1)}/2. - Lekraj Beedassy, Jul 21 2004
a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8*a(n) +4) + 1, a(1)=0. - Richard Choulet, Sep 18 2007
As noted (Sep 20 2006), a(n) = 5*(a(n-1) + a(n-2)) - a(n-3) + 4. In general, for n > 2*k, a(n) = A001653(k)*(a(n-k) + a(n-k-1) + 1) - a(n-2*k-1) - 1. Also a(n) = 7*(a(n-1) - a(n-2)) + a(n-3). In general, for n > 2*k, A002378(k)*(a(n-k)-a(n-k-1)) + a(n-2*k-1). - Charlie Marion, Dec 26 2007
In general, for n >= k >0, a(n) = (A001653(n+k) - A001541(k) * A001653(n) - 2*A001109(k-1))/(4*A001109(k-1)); e.g., 4059 = (33461-3*5741-2*1)/(4*1); 4059 = (195025-17*5741-2*6)/(4*6). - Charlie Marion, Jan 21 2008
From Charlie Marion, Jan 04 2010: (Start)
a(n) = ( (1 + sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 2)/4 = (A001333(2n+1) - 1)/2.
a(2*n+k-1) = Pell(2*n-1)*Pell(2*n+2*k) + Pell(2*n-2)*Pell(2*n+2*k+1) + A001108(k+1);
a(2*n+k) = Pell(2*n)*Pell(2*n+2*k+1) + Pell(2*n-1)*Pell(2*n+2*k+2) - A055997(k+2). (End)
a(n) = A048739(2*n-1) for n > 0. - Richard R. Forberg, Aug 31 2013
a(n+1) = 3*a(n) + 2*A001653(n) + 1 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014
a(n)^2 + (a(n)+1)^2 = A001653(n+1)^2. - Pierre CAMI, Mar 30 2005; clarified by Hermann Stamm-Wilbrandt, Aug 31 2014
a(n+1) = 3*A001541(n) + 10*A001109(n) + A001108(n). - Hermann Stamm-Wilbrandt, Sep 09 2014
For n>0, a(n) = Sum_{k=1..2*n} A000129(k). - Charlie Marion, Nov 07 2015
a(n) = 3*A053142(n) - A053142(n-1). - R. J. Mathar, Feb 05 2016
E.g.f.: (1/4)*(-2*exp(x) - (sqrt(2) - 1)*exp((3-2*sqrt(2))*x) + (1 + sqrt(2))*exp((3+2*sqrt(2))*x)). - Ilya Gutkovskiy, Apr 11 2016
a(n) = A001108(n) + 2*sqrt(A000217(A001108(n))). - Dimitri Papadopoulos, Jul 06 2017
a(A000217(n-1)) = ((A001653(n)+1)/2) * ((A001653(n)-1)/2), n > 1. - Ezhilarasu Velayutham, Mar 10 2019
a(n) = ((a(n-1)+1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
In general, for each k >= 0, a(n) = ((a(n-k)+a(k-1)+1)*(a(n-k)-a(k)))/a(n-2*k) for n > 2*k. - Charlie Marion, Dec 27 2020
A generalization of the identity a(n)^2 + A046090(n)^2 = A001653(n+1)^2 follows. Let P(k,n) be the n-th k-gonal number. Then P(k,a(n)) + P(k,A046090(n)) = P(k,A001653(n+1)) + (4-k)*A001109(n). - Charlie Marion, Dec 07 2021
a(n) = A046090(n)-1 = A002024(A029549(n)). - Pontus von Brömssen, Sep 11 2024

Additional comments from Wolfdieter Lang, Feb 10 2000

A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

Original entry on oeis.org

0, 2, 5, 15, 32, 90, 189, 527, 1104, 3074, 6437, 17919, 37520, 104442, 218685, 608735, 1274592, 3547970, 7428869, 20679087, 43298624, 120526554, 252362877, 702480239, 1470878640, 4094354882, 8572908965, 23863649055, 49966575152

Offset: 0

N. J. A. Sloane and Jeffrey Shallit

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Problem J/A2, Jul/August 1973 pp. 73-74; solution Jan 1974 (see link).
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
A. J. Gottlieb, How four dogs meet in a field, etc. (scanned copy)
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
J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000124, A006452, A006454, A098586, A098790.

Cf. numbers m such that k*A000217(m)+1 is a square: this sequence for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

a006451 n = a006451_list !! n
a006451_list = 0 : 2 : 5 : 15 : map (+ 2)
   (zipWith (-) (map (* 6) (drop 2 a006451_list)) a006451_list)
-- Reinhard Zumkeller, Jan 10 2012

N:= 100: # to get a(0) to a(N)
A[0]:= 0: A[1]:= 2: A[2]:= 5: A[3]:= 15:
for n from 4 to N do A[n]:= 6*A[n-2] - A[n-4] + 2 od:
seq(A[n],n=0..N); # Robert Israel, Aug 26 2014

LinearRecurrence[{1,6,-6,-1,1},{0,2,5,15,32},30] (* Harvey P. Dale, Jul 17 2013 *)
Select[Range[10^6], IntegerQ@ Sqrt[# (# + 1)/2 + 1] &] (* Michael De Vlieger, Apr 25 2017 *)

for(n=1,10000,t=n*(n+1)/2+1;if(issquare(t), print1(n,", "))) \\ Joerg Arndt, Oct 10 2009

G.f.: x*(-2-3*x+2*x^2+x^3)/(x-1)/(x^2+2*x-1)/(x^2-2*x-1). Conjectured (correctly) by Simon Plouffe in his 1992 dissertation.
a(n) = 6*a(n-2) - a(n-4) + 2 with a(0)=0, a(1)=2, a(2)=5, a(3)=15. - Zak Seidov, Apr 15 2008
a(n) = 3*a(n-2) + 4*sqrt((a(n-2)^2 + a(n-2))/2 + 1) + 1 with a(0) = 0, a(1) = 2. - Raphie Frank, Feb 02 2013
a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5); a(0)=0, a(1)=2, a(2)=5, a(3)=15, a(4)=32. - Harvey P. Dale, Jul 17 2013
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n+1) = A098790(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n) = A098586(2*n-1), for n>0. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(n) = 8*sqrt(T(a(n-2)) + 1) + a(n-4) where T(a(n)) = A000217(a(n)), and a(-1) = -1, a(0)=0, a(1)=2, a(2)=5. - Vladimir Pletser, Apr 29 2017

More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001

Edited by N. J. A. Sloane, Oct 24 2009, following discussions by several correspondents in the Sequence Fans Mailing List, Oct 10 2009

A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

Original entry on oeis.org

0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, 5570039304932024, 77580639439715775, 1080558912851088832, 15050244140475527879, 209622859053806301480

Offset: 0

N. J. A. Sloane

(a(n)+1)^3 - a(n)^3 is a square (that of A001570(n)).
The ratio A001570(n)/a(n) tends to sqrt(3) = 1.73205... as n increases. - Pierre CAMI, Apr 21 2005
Define a(1)=0 a(2)=7 such that 3*(a(1)^2) + 3*a(1) + 1 = j(1)^2 = 1^2 and 3*(a(2)^2) + 3*a(2) + 1 = j(2)^2 = 13^2. Then a(n) = a(n-2) + 8*sqrt(3*(a(n-1)^2) + 3*a(n-1) + 1). Another definition : a(n) such that 3*(a(n)^2) + 3*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
a(n) = A001353(n)*A001075(n+1). For n>0, the triple {a(n), a(n)+1=A001922(n), A001570(n)} forms a near-isosceles triangle with angle 2*Pi/3 bounded by the consecutive sides. - Lekraj Beedassy, Jul 21 2006
Numbers n such that A003215(n) is a square, cf. A006051. - Joerg Arndt, Jan 02 2017

			G.f. = 7*x + 104*x^2 + 1455*x^3 + 20272*x^4 + 282359*x^5 + 3932760*x^6 + ... - _Michael Somos_, Aug 17 2018

J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
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).

T. D. Noe, Table of n, a(n) for n = 0..200
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
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
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001570, A001922, A006051.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; A233450 for k=3; A001652 for k=4; A129556 for k=5; this sequence for k=6. - Bruno Berselli, Dec 16 2013

[Round(-1/2 - (1/6)*Sqrt(3)*(7-4*Sqrt(3))^n + (1/6)*Sqrt(3)*(7+4*Sqrt(3))^n + (1/4)*(7+4*Sqrt(3))^n + (1/4)*(7-4*Sqrt(3))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017

A001921:=z*(-7+z)/(z-1)/(z**2-14*z+1); # Conjectured by Simon Plouffe in his 1992 dissertation.

t = {0, 7}; Do[AppendTo[t, 14*t[[-1]] - t[[-2]] + 6], {20}]; t (* T. D. Noe, Aug 17 2012 *)
LinearRecurrence[{15, -15, 1}, {0, 7, 104}, 19] (* Michael De Vlieger, Jan 02 2017 *)
a[ n_] := -1/2 + (ChebyshevT[n + 1, 7] - ChebyshevT[n, 7]) / 12; (* Michael Somos, Aug 17 2018 *)

concat(0, Vec(x*(x-7)/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, Jan 06 2015

{a(n) = -1/2 + (polchebyshev(n + 1, 1, 7) - polchebyshev(n, 1, 7)) / 12}; /* Michael Somos, Aug 17 2018 */

G.f.: x*(-7 + x)/(x - 1)/(x^2 - 14*x + 1) (see Simon Plouffe in Maple section).
a(n) = (A028230(n+1)-1)/2. - R. J. Mathar, Mar 19 2009
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). - Colin Barker, Jan 06 2015
a(n) = -1 - a(-1-n) for all n in Z. - Michael Somos, Aug 17 2018

More terms from James Sellers, Jul 04 2000

A221874 Numbers m such that 10*m^2 + 6 is a square.

Original entry on oeis.org

1, 5, 43, 191, 1633, 7253, 62011, 275423, 2354785, 10458821, 89419819, 397159775, 3395598337, 15081612629, 128943316987, 572704120127, 4896450447169, 21747674952197, 185936173675435, 825838944063359, 7060678149219361, 31360132199455445

Offset: 1

Bruno Berselli, Jan 28 2013

The Diophantine equation 10*x^2 + k = y^2, for |k| < 10, has integer solutions with the following k values:
k =  1, the nonnegative x values are in A084070;
k = -1,               "                 A097315;
k =  4,               "                 2*A084070;
k = -4,               "                 2*A097315;
k =  6,               "                 this sequence;
k = -6,               "                 A221875;
k =  9,               "                 A075836;
k = -9,               "                 A052454.
a(n+1)/a(n) tends alternately to (sqrt(2)+sqrt(5))^2/3 and (2*sqrt(2)+sqrt(5))^2/3; a(n+2)/a(n) tends to A176398^2.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).

Cf. A097315, A084070, A075836, A052454, A129556, A221875.

Subsequence of A031150.

m:=22; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+x)*(1+4*x+x^2)/((1-6*x-x^2)*(1+6*x-x^2))));

A221874:=proc(q)
local n;
for n from 1 to q do if type(sqrt(10*n^2+6),integer) then print(n);
fi; od; end:
A221874(100000000000000000); # Paolo P. Lava, Feb 11 2013

LinearRecurrence[{0, 38, 0, -1}, {1, 5, 43, 191}, 22]

makelist(expand(((-5*(-1)^n+2*sqrt(10))*(3+sqrt(10))^(2*floor(n/2))-(5*(-1)^n+2*sqrt(10))*(3-sqrt(10))^(2*floor(n/2)))/10), n, 1, 22);

G.f.: x*(1+x)*(1+4*x+x^2)/((1-6*x-x^2)*(1+6*x-x^2)).
a(n) = ((-5*(-1)^n+2*t)*(3+t)^(2*floor(n/2)) - (5*(-1)^n+2*t)*(3-t)^(2*floor(n/2)))/10, where t=sqrt(10).
a(n) = 2*A129556(n) + 1.
a(n)*a(n-3) - a(n-1)*a(n-2) = -36 + 12(-1)^n.

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

Original entry on oeis.org

0, 1, 6, 15, 64, 153, 638, 1519, 6320, 15041, 62566, 148895, 619344, 1473913, 6130878, 14590239, 60689440, 144428481, 600763526, 1429694575, 5946945824, 14152517273, 58868694718, 140095478159, 582740001360, 1386802264321, 5768531318886, 13727927165055

Offset: 1

Bruno Berselli, Dec 10 2013

For n>1, partial sums of A080872 starting from A080872(1).

			153 is in the sequence because 3*153*154/2+1 = 188^2.

Bruno Berselli, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Sequence A129444 gives n+1.
Cf. A000217, A080872, A129445 (square roots of 3*A000217(a(n))+1), A132596 (numbers m such that 3*A000217(m) is a square).
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; this sequence for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6.

LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 6, 15, 64}, 30]

G.f.: x^2*(1 + 5*x - x^2 - x^3) / ((1 - x)*(1 - 10*x^2 + x^4)).
a(n) = a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=0, a(2)=1, a(3)=6, a(4)=15, a(5)=64.
a(n) = -1/2 + ( (-3*(-1)^n + 2*sqrt(6))*(5 + 2*sqrt(6))^floor(n/2) - (3*(-1)^n + 2*sqrt(6))*(5 - 2*sqrt(6))^floor(n/2) )/12.

A129557 Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).

Original entry on oeis.org

1, 4, 34, 151, 1291, 5734, 49024, 217741, 1861621, 8268424, 70692574, 313982371, 2684456191, 11923061674, 101938642684, 452762361241, 3870983965801, 17193046665484, 146995452057754, 652883010927151, 5581956194228851, 24792361368566254

Offset: 1

Alexander Adamchuk, Apr 20 2007

Corresponding numbers m such that centered pentagonal number A005891(m) = (5*m^2 + 5*m + 2)/2 is a perfect square are listed in A129556 = {0, 2, 21, 95, 816, 3626, 31005, ...}.

Also positive integers x in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of y being A254332. - Colin Barker, Jan 28 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Centered Pentagonal Number.
Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).

Cf. A005891 (centered pentagonal numbers).
Cf. A129556 (k such that A005891(k) is a perfect square).
Cf. A000290, A005667, A254332, A254333.

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n,1,40000} ]
CoefficientList[Series[(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)),{x,0,30}],x] (* Vincenzo Librandi, Apr 11 2012 *)

A129557()={ for(n=1,1000000000, f=(5*n^2+5*n+2)/2 ; if(issquare(f), print1(sqrtint(f), ", ") ; ); ) ; } \\ R. J. Mathar, Oct 11 2007

Vec(x*(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)) + O(x^100)) \\ Colin Barker, Jan 28 2015

a(n) = sqrt( (5*A129556(n)^2 + 5*A129556(n) + 2)/2 ).
For n >= 5, a(n) = 38*a(n-2) - a(n-4). - Max Alekseyev, May 08 2009
G.f.: x*(1-x)*(1 + 5*x + x^2)/((1 + 6*x - x^2)*(1 - 6*x - x^2)). - Colin Barker, Apr 11 2012
From Andrea Pinos, Oct 07 2022: (Start)
The ratios of successive terms converge to two different limits:
  lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7 + 2*sqrt(10))/3;
  upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13 + 4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).
a(n) = (A005667(n) - (-1)^n*A005667(n-1))/4. (End)

More terms from R. J. Mathar, Oct 11 2007

More terms from Max Alekseyev, May 08 2009

A120744 Least k>0 such that a centered polygonal number nk(k+1)/2+1 is a perfect square; or -1 if no such number exists.

Original entry on oeis.org

2, -1, 1, 3, 2, 7, 15, 1, 16, 8, 14, 4, 5, 15, 1, 2, 5, -1, 6, 3, 2, 39, 6, 1, 21, 7, 110, 3, 15, 7, 15, -1, 2, 8, 1, 4, 989, 8, 14, 2, 45, 15, 9, 4, 5, 335, 9, 1, 29, -1, 30, 15, 10, 415, 6, 2, 10, 32, 54, 3, 77, 55, 1, 5, 2, 7, 47750, 11, 15, 23, 47, -1, 48, 24, 16, 12, 5, 8, 2639, 1, 6720, 704, 38, 4, 2, 39, 505, 3, 13, 56, 9, 20, 13, 1631, 41

Offset: 1

Alexander Adamchuk, Apr 26 2007

sign

			a(5) = 2 because A129556(2) = 2>1 and A129556(1) = 0<1.

Max Alekseyev, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Centered Polygonal Number

Cf. A001542, A006451, A001921, A129444, A001570, A001652, A129556, A053606, A105038, A105040, A053141, A061278.

a(n) = -1 for n in A166259.

a(n) = 1 for n = k^2-1.

Edited and b-file provided by Max Alekseyev, Jan 20 2010

A166259 Positive integers n such that a centered polygonal number nk(k+1)/2+1 is not a square for any k > 0.

Original entry on oeis.org

2, 18, 32, 50, 72, 98, 128, 162, 200, 242, 338, 392, 450, 512, 578, 648, 722, 882, 968, 1058, 1152, 1250, 1352, 1458, 1682, 1800, 1922, 2048, 2178, 2312, 2401, 2450, 2662, 2738, 2809, 2888, 3042, 3174, 3200, 3362, 3528, 3698, 3750, 4050, 4225, 4232, 4418, 4489, 4608, 4802

Offset: 1

Alexander Adamchuk, Oct 10 2009

nonn

Positive integers n such that A120744(n) = -1.

Eric Weisstein's World of Mathematics, Centered Polygonal Number.

Cf. A120744, A001542, A006451, A001921, A129444, A001570, A001652, A129556, A053606, A105038, A105040, A053141, A061278.

Edited and extended by Max Alekseyev, Jan 20 2010

A233474 Numbers m such that 5*T(m)-1 is a square, where T = A000217.

Original entry on oeis.org

1, 4, 52, 169, 1993, 6436, 75700, 244417, 2874625, 9281428, 109160068, 352449865, 4145207977, 13383813460, 157408743076, 508232461633, 5977387028929, 19299449728612, 226983298356244, 732870857225641, 8619387950508361, 27829793124845764

Offset: 1

Bruno Berselli, Dec 11 2013

Sum_{i=1..infinity} 1/a(i) = 1.2758228304947598524736181699610...

			169 is in the sequence because 5*169*170/2-1 = 268^2.

Bruno Berselli, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1)

Cf. A000217, A129556 (numbers m such that 5*A000217(m)+1 is a square), A132593 (numbers m such that 5*A000217(m) is a square).

Cf. numbers m such that k*A000217(m)-1 is a square: A072221 for k=1; m=1 for k=2; this sequence for k=5.

LinearRecurrence[{1, 38, -38, -1, 1}, {1, 4, 52, 169, 1993}, 30]

G.f.: x*(1 + 3*x + 10*x^2 + 3*x^3 + x^4) / ((1 - x)*(1 - 38*x^2 + x^4)).
a(n) = a(n-1) + 38*a(n-2) - 38*a(n-3) - a(n-4) + a(n-5) for n>5, a(1)=1, a(2)=4, a(3)=52, a(4)=169, a(5)=1993.
a(n) = -1/2 + ( (15 + 4*sqrt(10)*(-1)^n)*(19 - 6*sqrt(10))^floor(n/2) + (15 - 4*sqrt(10)*(-1)^n)*(19 + 6*sqrt(10))^floor(n/2) )/20.

A145856 Least number k>1 such that centered n-gonal number n*k(k-1)/2+1 is a perfect square, or 0 if no such k exists.

Original entry on oeis.org

3, 0, 2, 4, 3, 8, 16, 2, 17, 9, 15, 5, 6, 16, 2, 3, 6, 0, 7, 4, 3, 40, 7, 2, 22, 8, 111, 4, 16, 8, 16, 0, 3, 9, 2, 5, 990, 9, 15, 3, 46, 16, 10, 5, 6, 336, 10, 2, 30, 0, 31, 16, 11, 416, 7, 3, 11, 33, 55, 4, 78, 56, 2, 6, 3, 8, 47751, 12, 16, 24, 48, 0, 49, 25, 17, 13, 6, 9, 2640, 2, 6721

Offset: 1

Alexander Adamchuk, Oct 22 2008

nonn

Jonathan Vos Post, When Centered Polygonal Numbers are Perfect Squares, submitted to Mathematics Magazine, 4 May 2004, manuscript no. 04-1165, unpublished, available upon request. - Jonathan Vos Post, Oct 25 2008

Eric Weisstein's World of Mathematics, Centered Polygonal Numbers
Index entries for sequences related to centered polygonal numbers

Cf. A120744, A001542, A166259, A006451, A001921, A129444, A001570, A001652, A129556, A053606, A105038, A105040, A053141, A061278.

a(n) = 0 for n in A166259.

a(n) = A120744(n) + 1. - Alexander Adamchuk, Oct 10 2009

Edited by Max Alekseyev, Jan 23 2010

cp's OEIS Frontend

A001652 a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.

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A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

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A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

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A221874 Numbers m such that 10*m^2 + 6 is a square.

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A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

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A129557 Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).

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A166259 Positive integers n such that a centered polygonal number nk(k+1)/2+1 is not a square for any k > 0.