A000290 -id:A000290 - cp's OEIS frontend

A240088 The number of ways of writing n as an ordered sum of a triangular number (A000217), a square (A000290) and a pentagonal number (A000326).

1, 3, 3, 2, 3, 4, 4, 4, 3, 3, 5, 5, 5, 3, 3, 7, 7, 5, 2, 6, 5, 4, 8, 5, 6, 4, 8, 7, 5, 7, 4, 9, 6, 5, 4, 3, 9, 12, 9, 4, 7, 9, 8, 4, 6, 8, 7, 8, 4, 8, 9, 10, 9, 6, 10, 6, 7, 10, 9, 8, 7, 11, 7, 4, 10, 8, 10, 10, 7, 5, 10, 14, 11, 7, 6, 11, 10, 10, 4, 11, 10, 10, 13, 8, 7, 7, 13, 12, 8, 8, 6, 10, 17, 8, 10, 7, 16, 10, 3, 12, 9

Offset: 0

Robert G. Wilson v, Mar 31 2014

nonn

0 and 1 are triangular numbers, square numbers and pentagonal numbers.
It is conjectured that a(n) is always positive - this is one of the conjectures in Conjecture 1.1 of Sun (2009). - N. J. A. Sloane, Apr 01 2014
Note that both the conjecture in A160325 and the conjecture in A160324 imply that a(n) is always positive. - Zhi-Wei Sun, Apr 01 2014
a(n) > 0 for all n < 10^10. - Robert G. Wilson v, Aug 20 2016
Least number to be represented k ways, k >= 1: 0, 3, 1, 5, 10, 19, 15, 22, 31, 51, 61, 37, 82, 71, 126, 96, 92, 136, 162, 187, 206, 276, 191, 261, 236, 247, 317, 302, 401, 292, 422, 547, 456, 544, 551, 612, 591, 577, 521, 666, 742, 726, 682, 877, 796, 1052, 961, 1046, 1171, 1027, ..., . A275999.
Greatest number (conjectured) to be represented k ways, k >= 1: 0, 18, 168, 78, 243, 130, 553, 455, 515, 658, 865, 945, 633, 1918, 2258, 1385, 1583, 2828, 2135, 2335, 2785, 4533, 3168, 3478, 2790, 3868, 4193, 7328, 4953, 5278, 6390, 8148, 8015, 4585, 9160, 10485, 7613, 12333, 12025, 10178, 9923, 9720, 12558, 11340, 17420, 11753, 14893, 16155, 16415, 14343, ..., .
Conjectured lists of numbers that are represented in k >= 1 ways:
1: 0;
2: 3, 18;
3: 1, 2, 4, 8, 9, 13, 14, 35, 98, 168;
4: 5, 6, 7, 21, 25, 30, 34, 39, 43, 48, 63, 78;
5: 10, 11, 12, 17, 20, 23, 28, 33, 69, 193, 203, 230, 243;
6: 19, 24, 32, 44, 53, 55, 74, 90, 111, 130;
7: 15, 16, 27, 29, 40, 46, 56, 60, 62, 68, 73, 84, 85, 95, 108, 113, 123, 135, 139, 163, 165, 273, 553;
8: 22, 26, 42, 45, 47, 49, 59, 65, 83, 88, 89, 93, 112, 119, 125, 134, 140, 144, 186, 205, 233, 244, 320, 405, 455;
9: 31, 36, 38, 41, 50, 52, 58, 100, 109, 124, 160, 214, 249, 308, 358, 515; ..., .

Robert G. Wilson v and Robert Israel, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Science China Mathematics, Vol. 58, No. 7 (2015), 1367-1396; arXiv:0905.0635 [math.NT], 2009-2015 [Edited, _Felix Fröhlich_, Aug 24 2016].
Zhi-Wei Sun, Various new conjectures involving polygonal numbers and primes, a message to the Number Theory List, May 8 2009.

Cf. A000925, A008443, A101428, A115171, A115172, A115173, A115174, A115175, A115176, A115177, A144642.

Cf. A160324, A160325, A160326. - Zhi-Wei Sun, Apr 01 2014

# requires Maple 17 and up
with(SignalProcessing):
N:= 10000;  # to get terms up to a(N)
A:= Array(0..N,datatype=float);
B:= Array(0..N,datatype=float);
C:= Array(0..N,datatype=float);
for i from 0 to floor(sqrt(N)) do A[i^2]:= 1 od:
for i from 0 to floor((1+sqrt(1+8*N))/2) do B[i*(i-1)/2]:= 1 od:
for i from 0 to floor((1+sqrt(1+24*N))/6) do C[i*(3*i-1)/2]:= 1 od:
R:= Convolution(Convolution(A,B),C);
R:= evalhf(map(round,R));
# Note that a(i) = R[i+1] for i from 0 to N
# Robert Israel, Apr 01 2014

p = Table[n (3n - 1)/2, {n, 0, 26}]; s = Table[n^2, {n, 0, 32}]; t = Table[n (n + 1)/2, {n, 0, 45}]; a = Sort@ Flatten@ Table[ p[[i]] + s[[j]] + t[[k]], {i, 26}, {j, 32}, {k, 45}]; Table[ Count[a, n], {n, 0, 105}]

A014134 Numbers that are not the sum of a square (A000290) and a triangular number (A000217).

Original entry on oeis.org

8, 13, 18, 20, 23, 27, 33, 34, 38, 41, 43, 47, 48, 58, 60, 62, 63, 68, 69, 73, 76, 83, 86, 88, 89, 90, 93, 97, 98, 99, 108, 111, 112, 113, 118, 123, 125, 132, 133, 134, 135, 138, 139, 143, 146, 148, 151, 158, 160, 163, 164, 167, 168, 173, 174

Offset: 1

N. J. A. Sloane

n is in the sequence if for some prime p == 5 or 7 (mod 8), 8*n+1 is divisible by p but not by p^2. The first member of the sequence that does not have this property is 16078. - Robert Israel, Mar 17 2015

R. J. Mathar, Table of n, a(n) for n = 1..4646

Cf. A140867.

N:= 1000: # to generate all terms <= N
{$1 .. N} minus {seq(seq(x*(x+1)/2 + y^2, y = 0 .. floor(sqrt(N - x*(x+1)/2))),
x = 0 .. floor((sqrt(8*N+1)-1)/2))};
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(%,list)); # Robert Israel, Mar 17 2015

fQ[n_] := Block[{k = 0, lmt = 1 + Floor@ Sqrt@ n}, While[k < lmt && !IntegerQ@ Sqrt[ 8(n - k^2) + 1], k++]; k == lmt]; Select[ Range@ 175, fQ@# &] (* Robert G. Wilson v, Nov 29 2015 *)

is_A014134(n)=for(k=0,sqrtint(n*2),issquare(n-k*(k+1)/2)&return);1 /* M. F. Hasler, Jan 05 2009 */

A033452 "STIRLING" transform of squares A000290.

Original entry on oeis.org

0, 1, 5, 22, 99, 471, 2386, 12867, 73681, 446620, 2856457, 19217243, 135610448, 1001159901, 7714225057, 61904585510, 516347066551, 4468588592739, 40058673825258, 371421499686007, 3556976106133821, 35138574378189700, 357654857584636597, 3746672593640388775

Offset: 0

N. J. A. Sloane

nonn

If an integer N is squarefree and has n+2 distinct prime factors then a(n) is the number of product signs needed to write the factorizations of N, so a(n)=A076277(N). - Floor van Lamoen, Oct 17 2002

Convolved with powers of 2 = A058681: (1, 7, 36, 171, 813, ...). Cf. triangle A180338. - Gary W. Adamson, Aug 28 2010

			G.f. = x + 5*x^2 + 22*x^3 + 99*x^4 + 471*x^5 + 2386*x^6 + 12867*x^7 + 73681*x^8 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..574

Partial sums of A005494.

Cf. A180338.

a := n -> add(Stirling2(n, j)*j^2, j=0..n): seq(a(n), n=0..20); # Zerinvary Lajos, Apr 18 2007
# second Maple program:
b:= proc(n, m) option remember;
     `if`(n=0, m^2, m*b(n-1, m)+b(n-1, m+1))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..23);  # Alois P. Heinz, Aug 04 2021

max = 20; Clear[g]; g[max + 2] = 1; g[k_] := g[k] = 2 - 1/(1 - k*x)/(1 - x/(x - 1/g[k + 1])); gf = 1/x + 1/x^2 - g[0]/x^2; CoefficientList[ Series[gf, {x, 0, max}], x] (* Jean-François Alcover, Jan 24 2013, after Sergei N. Gladkovskii *)

{a(n) = if( n<0, 0, n! * polcoeff( (exp(x + x * O(x^n)) - 1) * exp( exp(x + x * O(x^n)) - 1 + x), n))}; /* Michael Somos, Mar 28 2012 */

Representation as an infinite series: a(n) = (Sum_{k>=1} k^n*k*(k-2)/k!)/exp(1), n >= 1. This is a Dobinski-type summation formula. - Karol A. Penson, Mar 21 2002
a(n) = A005493(n) - A000110(n+1). - Floor van Lamoen and Christian Bower, Oct 16 2002. (n^2 has e.g.f.: e^x * (x^2+x), a(n) thus has e.g.f: e^(e^x-1) * ( (e^x-1)^2 + (e^x-1) ) which simplifies to e^(e^x-1) * (e^2x - e^x). A005493 has e.g.f.: e^(e^x+2x-1), A000110 has e.g.f.: e^(e^x-1), A000110(n+1) has as e.g.f.: derivative of A000110 which is e^(e^x+x-1).) [corrected by Georg Fischer, Jun 17 2020]
a(n) = Bell(n+2) - 2*Bell(n+1). - Vladeta Jovovic, Jul 28 2003
G.f.: sum{k>=0, k^2*x^k/prod[l=1..k, 1-lx]}. - Ralf Stephan, Apr 18 2004
E.g.f.: exp( exp(x) - 1 + x) * (exp(x) - 1). - Michael Somos, Mar 28 2012
a(n) = A123158(n,3). - Philippe Deléham, Oct 06 2006
G.f.: G(0)/x -1/x, where G(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (2*x+x*k-1)*(3*x+x*k-1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Feb 25 2014

A046172 Indices of pentagonal numbers (A000326) that are also squares (A000290).

Original entry on oeis.org

1, 81, 7921, 776161, 76055841, 7452696241, 730288175761, 71560788528321, 7012226987599681, 687126683996240401, 67331402804643959601, 6597790348171111800481, 646516122717964312487521, 63351982236012331511976561, 6207847743006490523861215441

Offset: 1

Eric W. Weisstein

if P_x = y^2 is a pentagonal number that is also a square, the least both pentagonal and square number that is greater as P_x, is P_(49*x + 40*y - 8) = (60*x + 49*y - 10)^2 (in fact, P_(49*x + 40*y - 8) - (60*x + 49*y - 10)^2 = (3/2)*x^2 - (1/2)*x - y^2). - Richard Choulet, Apr 28 2009
a(n)*(3*a(n)-1)/2 = m^2 is equivalent to the Pell equation (6*a(n)-1)^2 - 6*(2*m)^2 = 1 or x(n)^2 - 6*y(n)^2 = 1. - Paul Weisenhorn, May 15 2009
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^4 = 49 + 20*sqrt(6). - Ant King, Nov 07 2011
Numbers k such that the k-th pentagonal number is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014
Indices of pentagonal numbers (A000326) that are also centered octagonal numbers (A016754). - Colin Barker, Jan 11 2015

Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7; http://dx.doi.org/10.1080/0020739X.2016.1164346

Colin Barker, Table of n, a(n) for n = 1..503
Leonhard Euler, De solutione problematum diophanteorum per numeros integros, section 21.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
W. Sierpiński, Sur les nombres pentagonaux, Bull. Soc. Roy. Sci. Liege 33 (1964) 513-517.
Eric Weisstein's World of Mathematics, Pentagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A036353, A046173.

Cf. A000217, A000290, A000326, A251914, A248205.

LinearRecurrence[{99, -99, 1}, {1, 81, 7921}, 13] (* Ant King, Nov 07 2011 *)
Table[Round[(1 + x^(2*n+1))^2 / (12*x^(2*n+1)) /. x->5+2*Sqrt@6],{n,0,99}] (* Federico Provvedi, Apr 24 2023 *)

a(n) = 98*a(n-1) - a(n-2) - 16; g.f.: x*(1 - 18*x + x^2)/((1-x)*(1 - 98*x + x^2)). - Warut Roonguthai Jan 05 2001 - Corrected by Colin Barker, Jan 11 2015
a(n+1) = 49*a(n) - 8 + 10*sqrt(8*(3*a(n)^2 - a(n))) with a(1) = 1. - Richard Choulet, Apr 28 2009
a(n) = 1/6+((5 + 2*sqrt(6))^(2*n+1)/12) + ((5 - 2*sqrt(6))^(2*n+1)/12) for n >= 0. - Richard Choulet, Apr 29 2009
From Paul Weisenhorn, May 15 2009: (Start)
x(n+2) = 98*x(n+1) - x(n) with x(1)=5, x(2)=485;
y(n+2) = 98*y(n+1) - y(n) with y(n)=A046173(n)*2;
m(n+2) = 98*m(n+1) - m(n) with m(n)=A046173(n);
a(n) = A072256(n)^2.
(End)
a(n) = b(n)*b(n), b(n) = 10*b(n-1)- b(n-2), b(1)=1, b(2)=9, b(n)=((5 + sqrt(24))^n - (5 - sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, Sep 21 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).
a(n) = ceiling((1/12)*(sqrt(3) + sqrt(2))^(4*n-2)).
(End)
a(n) = (1 + x^(2n+1))^2 / (12*x^(2*n+1)), with x = 5 + 2*sqrt(6). - Federico Provvedi, Apr 24 2023

A113257 Ascending descending base exponent transform of squares (A000290).

Original entry on oeis.org

1, 5, 266, 268722, 4682453347, 2978988815561863, 722638800922610642480852, 22529984108212742763058965679103268, 57286470055793196612331429228839529219232484069

Offset: 1

Jonathan Vos Post, Jan 07 2006

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. The smallest prime in this sequence is a(2) = 5. What is the next prime? What is the first square value after 1?

			a(1) = 1 because (1^2)^(1^2) = 1^1 = 1.
a(2) = 5 because (1^2)^(4^1) + (4^1)^(1^4) = 1^4 + 4^1 = 5.
a(3) = 266 = 1^9 + 4^4 + 9^1.
a(4) = 268722 = 1^16 + 4^9 + 9^4 + 16^1.
a(5) = 4682453347 = 1^25 + 4^16 + 9^9 + 16^4 + 25^1.
a(6) = 2978988815561863 = 1^36 + 4^25 + 9^16 + 16^9 + 25^4 + 36^1.
a(7) = 722638800922610642480852 = 1^49 + 4^36 + 9^25 + 16^16 + 25^9 + 36^4 + 49^1.
a(8) = 22529984108212742763058965679103268 = 1^64 + 4^49 + 9^36 + 16^25 + 25^16 + 36^9 + 49^4 + 64^1.
a(9) = 57286470055793196612331429228839529219232484069 = 1^81 + 4^64 + 9^49 + 16^36 + 25^25 + 36^16 + 49^9 + 64^4 + 81^1.

G. C. Greubel, Table of n, a(n) for n = 1..30

Cf. A000290, A005408, A113122, A113153, A113154.

Table[Sum[(k^2)^((n - k + 1)^2), {k, 1, n}], {n, 1, 10}] (* G. C. Greubel, May 18 2017 *)

for(n=1,10, print1(sum(k=1,n, (k^2)^((n-k+1)^2) ), ", ")) \\ G. C. Greubel, May 18 2017

a(n) = Sum_{i=1..n} (i^2)^((n-i+1)^2).
a(n) = Sum_{i=1..n} (A000290(i))^(A000290(n-i+1)).
log(a(n)) ~ n^2 * (-1 + 2*LambertW(exp(1/2)*n/2))^3 / (4*LambertW(exp(1/2)*n/2)^2). - Vaclav Kotesovec, Jun 07 2025

a(4) and a(5) corrected by Giovanni Resta, Jun 13 2016

A165517 Indices of the least triangular numbers (A000217) for which three consecutive triangular numbers sum to a perfect square (A000290).

Original entry on oeis.org

0, 5, 14, 63, 152, 637, 1518, 6319, 15040, 62565, 148894, 619343, 1473912, 6130877, 14590238, 60689439, 144428480, 600763525, 1429694574, 5946945823, 14152517272, 58868694717, 140095478158, 582740001359, 1386802264320

Offset: 1

Ant King, Sep 25 2009, Oct 01 2009

Those perfect squares that can be expressed as the sum of three consecutive triangular numbers correspond to integer solutions of the equation T(k)+T(k+1)+T(k+2)=s^2, or equivalently to 3k^2 + 9k + 8 = 2s^2. Hence solutions occur whenever (3k^2 + 9k + 8)/2 is a perfect square, or equivalently when s>=2 and sqrt(24s^2 - 15) is congruent to 3 mod 6. This sequence returns the index of the smallest of the 3 triangular numbers, the values of s^2 are given in A165516 and, with the exception of the first term, the values of s are in A129445.

			The fourth perfect square that can be expressed as the sum of three consecutive triangular numbers is 6241 = T(63) + T(64) + T(65). Hence a(4)=63.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Tom Beldon and Tony Gardiner, Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431.
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Cf. A000290, A129445, A000217, A165516.

I:=[0, 5, 14, 63, 152]; [n le 5 select I[n] else Self(n-1) + 10*Self(n-2) - 10*Self(n-3) - Self(n-4) + Self(n-5): n in [1..50]]; // G. C. Greubel, Oct 21 2018

TriangularNumber[ n_ ]:=1/2 n (n+1);Select[ Range[ 0,10^7 ], IntegerQ[ Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1 ]+TriangularNumber[ #+2 ] ] ] & ]
CoefficientList[Series[x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)), {x,0,50}], x] (* or *) LinearRecurrence[{1,10,-10,-1,1}, {0, 5, 14, 63, 152}, 50] (* G. C. Greubel, Feb 17 2017 *)

x='x+O('x^50); concat([0], Vec(x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)))) \\ G. C. Greubel, Feb 17 2017

a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5).
G.f.: x(x^3 + x^2 - 9x - 5)/((x-1)(x^4 - 10x^2 + 1)).
a(n) = 10*a(n-2) - a(n-4) + 12. - Zak Seidov, Sep 25 2009

a(1) = 0 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander R. Povolotsky

More terms from Zak Seidov, Sep 25 2009

A046177 Squares (A000290) which are also hexagonal numbers (A000384).

Original entry on oeis.org

1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481

Offset: 1

Eric W. Weisstein

Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007
Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011
Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - Colin Barker, Jan 16 2015
Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Colin Barker, Table of n, a(n) for n = 1..327
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Eric Weisstein's World of Mathematics, Square Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (1155,-1155,1).

Cf. A008844, A046176, A253826.
Cf. A001110 (Numbers that are both triangular and square).
Cf. A000290, A000384, A016754, A253826.

LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)

Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015

a(n) = A001110(2n-1). - Alexander Adamchuk, Nov 06 2007
a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009
a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - Richard Choulet, May 01 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
(End)
a(n) = A096979(4*n - 3). - Ivan N. Ianakiev, Sep 05 2016
a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - Ivan N. Ianakiev, Sep 05 2016

A176744 The squares A000290 and the integers which cannot be represented as a sum of two earlier terms of the sequence.

Original entry on oeis.org

0, 1, 3, 4, 9, 11, 16, 21, 23, 25, 31, 33, 36, 38, 43, 49, 51, 64, 77, 81, 83, 91, 96, 100, 118, 121, 135, 144, 150, 163, 169, 176, 189, 196, 203, 211, 213, 223, 225, 230, 237, 243, 256, 278, 283, 289, 291, 315, 324, 350, 361, 390, 395, 400, 408, 430, 437, 441, 484, 497, 510

Offset: 0

Vladimir Shevelev, Apr 25 2010

			3 is the smallest number which is not a sum of 2 numbers of {0,1}. Therefore 3 in the sequence.
4 is a square, and included as such.
5 can be represented by 1+4 (both already in the sequence) and is not included; 6=3+3, 7=3+4, 8=4+4 are also sums of earlier terms: not included.
11 is the smallest number which is not a sum of 2 numbers of {0, 1, 3, 4, 9}. Therefore 11 in the sequence.

Cf. A000290.

A176744 := proc(n) option remember; if n <=1 then n; else for a from procname(n-1)+1 do
if issqr(a) then return a; end if; isrep := false; for i from 1 to n-1 do for j from i to n-1 do if procname(i)+procname(j) = a then isrep := true; end if; end do: end do: if not isrep then return a; end if; end do:
end if; end proc: seq(A176744(n),n=0..60) ; # R. J. Mathar, Oct 29 2010

a[n_] := a[n] = Module[{tt, k}, If[n == 0, 0, tt = Total /@ Tuples[Array[a, n-1], {2}]; For[k = a[n-1]+1, True, k++, If[IntegerQ@Sqrt@k, Return[k], If[FreeQ[tt, k], Return[k]]]]]];
Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Aug 02 2022 *)

Definition rephrased, more examples added, and sequence extended beyond 51 by R. J. Mathar, Oct 29 2010

A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).

Original entry on oeis.org

1, 8, 105, 1456, 20273, 282360, 3932761, 54776288, 762935265, 10626317416, 148005508553, 2061450802320, 28712305723921, 399910829332568, 5570039304932025, 77580639439715776, 1080558912851088833, 15050244140475527880, 209622859053806301481

Offset: 0

N. J. A. Sloane

Also larger of two consecutive integers whose cubes differ by a square. Defined by a(n)^3 - (a(n) - 1)^3 = square.
Let m be the n-th ratio 2/1, 7/4, 26/15, 97/56, 362/209, ... Then a(n) = m*(2-m)/(m^2-3). The numerators 2, 7, 26, ... of m are A001075. The denominators 1, 4, 15, ... of m are A001353.
From Colin Barker, Jan 06 2015: (Start)
Also indices of centered triangular numbers (A005448) which are also centered square numbers (A001844).
Also indices of centered hexagonal numbers (A003215) which are also centered octagonal numbers (A016754).
Also positive integers x in the solutions to 3*x^2 - 4*y^2 - 3*x + 4*y = 0, the corresponding values of y being A156712.
(End)

			8 is in the sequence because 3*8^2 - 3*8 + 1 = 169 is a square and also a centered hexagonal number. - _Colin Barker_, Jan 07 2015

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..800
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
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A001570, A001844, A001921, A003215.

Cf. A005448, A006051, A016754, A076139, A156712.

I:=[1, 8, 105]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Apr 16 2012

seq(simplify((1 +ChebyshevU(n,7) +ChebyshevU(n-1,7))/2), n=0..30); # G. C. Greubel, Oct 07 2022

With[{s1=3+2Sqrt[3],s2=3-2Sqrt[3],t1=7+4Sqrt[3],t2=7-4Sqrt[3]}, Simplify[ Table[(s1 t1^n+s2 t2^n+6)/12,{n,0,20}]]] (* or *) LinearRecurrence[ {15,-15,1},{1,8,105},21] (* Harvey P. Dale, Aug 14 2011 *)
CoefficientList[Series[(1-7*x)/(1-15*x+15*x^2-x^3),{x,0,30}],x] (* Vincenzo Librandi, Apr 16 2012 *)

Vec((1-7*x)/(1-15*x+15*x^2-x^3) + O(x^100)) \\ Colin Barker, Jan 06 2015

[(1+chebyshev_U(n,7) +chebyshev_U(n-1,7))/2 for n in range(30)] # G. C. Greubel, Oct 07 2022

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (s1*t1^n + s2*t2^n + 6)/12 where s1 = 3 + 2*sqrt(3), s2 = 3 - 2*sqrt(3), t1 = 7 + 4*sqrt(3), t2 = 7 - 4*sqrt(3).
a(n) = A001075(n)*A001353(n+1).
G.f.: (1-7*x)/((1-x)*(1-14*x+x^2)). - Simon Plouffe (in his 1992 dissertation) and Colin Barker, Jan 01 2012
a(n) = A076139(n+1) - 7*A076139(n). - R. J. Mathar, Jul 14 2015
a(n) = (1/2)*(1 + ChebyshevU(n, 7) + ChebyshevU(n-1, 7)). G. C. Greubel, Oct 07 2022
a(n) = 1 - a(-1-n) = 1 + A001921(n) for all integers n. - Michael Somos, Jul 10 2025

Additional comments from James R. Buddenhagen, Mar 04 2001
Name improved by Colin Barker, Jan 07 2015
Edited by Robert Israel, Feb 20 2017

A046195 Indices of heptagonal numbers (A000566) which are also squares (A000290).

Original entry on oeis.org

1, 6, 49, 961, 8214, 70225, 1385329, 11844150, 101263969, 1997643025, 17079255654, 146022572641, 2880599856289, 24628274808486, 210564448483921, 4153822995125281, 35513955194580726, 303633788691241009, 5989809878370798481, 51211098762310597974

Offset: 1

Eric W. Weisstein

(10 * a(n) - 3)^2 - 40 * (A046196(n))^2 = 9. - Ant King, Nov 12 2011
Also numbers n such that the n-th heptagonal number is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014
Also indices of heptagonal numbers (A000566) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 19 2015
Also nonnegative integers y in the solutions to 2*x^2-5*y^2+4*x+3*y+2+2 = 0, the corresponding values of x being A251927. - Colin Barker, Dec 11 2014

Colin Barker, Table of n, a(n) for n = 1..950
Eric Weisstein's World of Mathematics, Heptagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (1,0,1442,-1442,0,-1,1).

Cf. A000566, A036254, A046196, A251927, A253920.

for n from 1 to 10000 do m:=sqrt((5*n^2-3*n)/2):
if (trunc(m)=m) then print(n,m): end if: end do: # Paul Weisenhorn, May 01 2009

LinearRecurrence[{1 ,0, 1442, -1442, 0, -1, 1}, {1, 6, 49, 961, 8214, 70225, 1385329}, 17] (* Ant King, Nov 12 2011 *)

From Paul Weisenhorn, May 01 2009: (Start)
Pell equations: r^2-10*s^2=1 with solution (19,6)
(10*n-3)^2-10*(2*m)^2=9; basic solutions: (7,-2); (7,+2)((57,18);
with x=10*n-3; y=2*m; A=(19+6*sqrt(10))^2; B=(19-6*sqrt(10))^2 one get
x(3*k)+sqrt(10)*y(3*k)=(7-2*sqrt(10))*A^k;
x(3*k+1)+sqrt(10)*y(3*k+1)=(7+2*sqrt(10))*A^k;
x(3*k+2)+sqrt(10)*y(3*k+2)=(57+18*sqrt(10))*A^k;
with the eigenvalues A=721+228*sqrt(10); B=721-228*sqrt(10)
one get the recurrences with 1442=4*19*19-2
x(k+6)=1442*x(k+3)-x(k); y(k+6)=1442*y(k+3)-y(k);
m(k+6)=1442*m(k+3)-m(k); n(k+6)=1442*n(k+3)-n(k)-432;
and the explicit formulas
x(3*k+1)=(7*(A^k+B^k)+2*sqrt(10)*(A^k-B^k))/2;
x(3*k+2)=(57*(A^k+B^k)+18*sqrt(10)*(A^k-B^k))/2;
x(3*k)=(7*(A^k+B^k)-2*sqrt(10)*(A^k-B^k))/2;
y(3*k+1)=(7*(A^k-B^k)/sqrt(10)+2*(A^k+B^k)/2;
y(3*k+2)=(57*(A^k-B^k)/sqrt(10)+18*(A^k+B^k))/2;
y(3*k)=(7*(A^k-B^k)/sqrt(10)-2*(A^k+B^k))/2;
n(k)=(x(k)+3)/10; m(k)=y(k)/2;
(End)
a(n) = +a(n-1) +1442*a(n-3) -1442*a(n-4) -a(n-6) +a(n-7). G.f.: -x*(1+5*x+43*x^2-530*x^3+43*x^4+5*x^5+x^6) / ( (x-1)*(x^6-1442*x^3+1) ). - R. J. Mathar, Aug 01 2010
a(n) = 1442*a(n-3) - a(n-6) - 432. - Ant King, Nov 12 2011

More terms from Colin Barker, Dec 11 2014

cp's OEIS Frontend

A240088 The number of ways of writing n as an ordered sum of a triangular number (A000217), a square (A000290) and a pentagonal number (A000326).

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A014134 Numbers that are not the sum of a square (A000290) and a triangular number (A000217).

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A033452 "STIRLING" transform of squares A000290.

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A046172 Indices of pentagonal numbers (A000326) that are also squares (A000290).

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A113257 Ascending descending base exponent transform of squares (A000290).

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A165517 Indices of the least triangular numbers (A000217) for which three consecutive triangular numbers sum to a perfect square (A000290).

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A046177 Squares (A000290) which are also hexagonal numbers (A000384).

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A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).