A085787 -id:A085787 - cp's OEIS frontend

A111710 Consider the triangle shown below in which the n-th row contains the n smallest numbers greater than those in the previous row such that the arithmetic mean is an integer. Sequence contains the leading diagonal.

1, 4, 7, 13, 18, 27, 34, 46, 55, 70, 81, 99, 112, 133, 148, 172, 189, 216, 235, 265, 286, 319, 342, 378, 403, 442, 469, 511, 540, 585, 616, 664, 697, 748, 783, 837, 874, 931, 970, 1030, 1071, 1134, 1177, 1243, 1288, 1357, 1404, 1476, 1525, 1600, 1651, 1729

Offset: 1

Amarnath Murthy, Aug 24 2005

			The fourth row is 8,9,10 and 13,(8+9+10 +13)/4 = 10.
Triangle begins:
1
2 4
5 6 7
8 9 10 13
14 15 16 17 18
19 20 21 22 23 27
28 29 30 31 32 33 34

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

Cf. A111711, A111712.

Cf. A085787. - R. J. Mathar, Aug 15 2008

LinearRecurrence[{1, 2, -2, -1, 1}, {1, 4, 7, 13, 18}, 100] (* Paolo Xausa, Feb 09 2024 *)

Vec(x*(1+3*x+x^2)/((1-x)^3*(1+x)^2) + O(x^100)) \\ Colin Barker, Jan 26 2016

a(1)=1, a(2n) = a(2n-1)+3n, a(2n+1)=a(2n)+2n+1. - Franklin T. Adams-Watters, May 01 2006
G.f.: -x*(1+3*x+x^2) / ( (1+x)^2*(x-1)^3 ). a(n+1)-a(n) = A080512(n+1). - R. J. Mathar, May 02 2013
From Colin Barker, Jan 26 2016: (Start)
a(n) = (10*n^2+2*(-1)^n*n+10*n+(-1)^n-1)/16.
a(n) = (5*n^2+6*n)/8 for n even.
a(n) = (5*n^2+4*n-1)/8 for n odd. (End)

More terms from Franklin T. Adams-Watters, May 01 2006

A220082 Numbers k such that 10*k-1 is a square.

Original entry on oeis.org

1, 5, 17, 29, 53, 73, 109, 137, 185, 221, 281, 325, 397, 449, 533, 593, 689, 757, 865, 941, 1061, 1145, 1277, 1369, 1513, 1613, 1769, 1877, 2045, 2161, 2341, 2465, 2657, 2789, 2993, 3133, 3349, 3497, 3725, 3881, 4121, 4285, 4537, 4709, 4973, 5153, 5429, 5617, 5905

Offset: 1

Bruno Berselli, Dec 05 2012

Equivalently, numbers of the form m*(10*m+6)+1, where m=0,-1,1,-2,2,-3,3,...

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

Cf. A085787, A132356 (numbers n such that 10*n+1 is a square).

Cf. numbers n such that k*n-1 is a square: A002522 (k=1), A001844 (k=2), A062317 (k=5).

[n: n in [1..6000] | IsSquare(10*n-1)]; /* or (see the first comment): */ [1] cat [m*(10*m+6)+1: m in [-n,n], n in [1..24]];

I:=[1,5,17,29,53]; [n le 5 select I[n] else Self(n-1) +2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..60]]; // Vincenzo Librandi, Aug 18 2013

A220082:=proc(q)
local n;
for n from 1 to q do if type(sqrt(10*n-1), integer) then print(n);
fi; od; end:
A220082(1000); # Paolo P. Lava, Feb 19 2013

Select[Range[0, 6000], IntegerQ[Sqrt[10 # - 1]] &]
CoefficientList[Series[(1 + 4 x + 10 x^2 + 4 x^3 + x^4) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{1,5,17,29,53},50] (* Harvey P. Dale, Nov 19 2023 *)

G.f.: x*(1+4*x+10*x^2+4*x^3+x^4)/((1+x)^2*(1-x)^3).
a(n) = a(-n+1) = (10*n*(n-1)-(2*n-1)*(-1)^n+3)/4.
For the definition: 10*a(n)-1 = ((10*n-(-1)^n-5)/2)^2.
a(n) = A212570(n)-A212570(n-1) = 4*A085787(n-1)+1 = A132356(n-1)-(2*n-1)*(-1)^n.

A270594 Number of ordered ways to write n as the sum of a triangular number, a positive square and the square of a generalized pentagonal number (A001318).

Original entry on oeis.org

1, 2, 1, 2, 4, 2, 2, 4, 2, 3, 5, 2, 2, 3, 3, 4, 3, 2, 4, 5, 1, 2, 5, 1, 3, 7, 3, 2, 6, 5, 3, 6, 2, 2, 5, 4, 6, 4, 3, 5, 8, 2, 2, 6, 2, 5, 5, 1, 4, 9, 5, 3, 8, 5, 4, 8, 4, 3, 5, 5, 5, 6, 3, 6, 11, 2, 3, 9, 2, 5, 12, 2, 2, 9, 6, 3, 4, 4, 5, 6, 6, 6, 5, 5, 6, 11, 2, 4, 8, 1

Offset: 1

Zhi-Wei Sun, Mar 19 2016

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 21, 24, 48, 90, 138, 213, 283, 462, 468, 567, 573, 1998, 2068, 2488, 2687, 5208, 5547, 5638, 6093, 6492, 6548, 6717, 7538, 7731, 8522, 14763, 16222, 17143, 24958, 26148.
(ii) Let T(x) = x(x+1)/2, pen(x) = x(3x+1)/2 and hep(x) = x(5x+3)/2. Then any natural number can be written as P(x,y,z) with x, y and z integers, where P(x,y,z) is either of the following polynomials: T(x)^2+T(y)+z(5z+1)/2, T(x)^2+T(y)+z(3z+j) (j = 1,2), T(x)^2+y^2+pen(z), T(x)^2+pen(y)+hep(z), T(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), T(x)^2+pen(y)+z(4z+j) (j = 1,3), T(x)^2+pen(y)+z(5z+j) (j = 1,3,4), T(x)^2+pen(y)+z(11z+7)/2, T(x)^2+y(5y+1)/2+z(3z+2), T(x)^2+hep(y)+z(3z+2), pen(x)^2+T(y)+pen(z), pen(x)^2+T(y)+2*pen(z), pen(x)^2+T(y)+z(9z+7)/2, pen(x)^2+y^2+pen(z), pen(x)^2+2*T(y)+pen(z), pen(x)^2+pen(y)+3*T(z), pen(x)^2+pen(y)+2z^2, pen(x)^2+pen(y)+2*pen(z), pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+pen(y)+z(4z+3), pen(x)^2+pen(y)+z(9z+1)/2, pen(x)^2+pen(y)+3*pen(z), pen(x)^2+pen(y)+z(5z+j) (j = 1,2,3,4), pen(x)^2+pen(y)+z(11z+j)/2 (j = 7,9), pen(x)^2+pen(y)+z(7z+1), pen(x)^2+pen(y)+3*hep(z), pen(x)^2+y(5y+j)/2+z(3z+k) (j = 1,3; k = 1,2), pen(x)^2+hep(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+hep(y)+z(9z+5)/2, pen(y)^2+2pen(y)+z(3z+2), pen(x)^2+2*pen(y)+3*pen(z), (x(5x+1)/2)^2+2*T(y)+pen(z), (x(5x+1)/2)^2+pen(y)+z(7z+3)/2, (x(5x+1)/2)^2+pen(y)+z(4z+1), (x(5x+1)/2)^2+hep(y)+2*pen(z), hep(x)^2+T(y)+2*pen(z), hep(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), hep(x)^2+pen(y)+z(4z+1), hep(x)^2+pen(y)+z(5z+4), 4*pen(x)^2+T(y)+hep(z), 4*pen(x)^2+T(y)+2*pen(z), 4*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), (x(3x+2))^2+y^2+pen(z), (x(3x+2))^2+pen(y)+z(7z+j)/2 (j = 3,5), 2*T(x)^2+T(y)+z(3z+j) (j = 1,2), 2*T(x)^2+y^2+pen(z), 2*T(x)^2+2*T(y)+pen(z), 2*T(x)^2+pen(y)+z(7z+j)/2 (j = 1,5), 2*T(x)^2+pen(y)+z(5z+1), 2*pen(y)^2+T(y)+z(3z+2), 2*pen(x)^2+y^2+pen(z), 2*pen(x)^2+pen(y)+z(7z+3)/2, 2*pen(x)^2+pen(y)+z(4z+j) (j = 1,3), 2*pen(x)^2+pen(y)+z(5z+4), 2*pen(x)^2+pen(y)+z(7z+1), 2*pen(x)^2+hep(y)+2*pen(z), 2*hep(x)^2+pen(y)+z(7z+5)/2, 3*pen(x)^2+T(y)+z(3z+2), 3*pen(x)^2+y^2+pen(z), 3*pen(x)^2+2*T(y)+pen(z), 3*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), 3*pen(x)^2+pen(y)+z(4z+1), 6*pen(x)^2+pen(y)+z(7z+3)/2.
See also A270566 for a similar conjecture involving four powers.
It is known that any positive integer can be written as the sum of a triangular number, a square and an odd square.

			a(21) = 1 since 21 = 1*2/2 + 4^2 + (1*(3*1+1)/2)^2.
a(24) = 1 since 24 = 5*6/2 + 3^2 + (0*(3*0-1)/2)^2.
a(468) = 1 since 468 = 0*1/2 + 18^2 + (3*(3*3-1)/2)^2.
a(7538) = 1 since 7538 = 64*65/2 + 47^2 + (6*(3*6+1)/2)^2.
a(7731) = 1 since 7731 = 82*83/2 + 62^2 + (4*(3*4-1)/2)^2.
a(8522) = 1 since 8522 = 127*128/2 + 13^2 + (3*(3*3+1)/2)^2.
a(14763) = 1 since 14763 = 164*165/2 + 33^2 + (3*(3*3-1)/2)^2.
a(16222) = 1 since 16222 = 168*169/2 + 45^2 + (1*(3*1-1)/2)^2.
a(17143) = 1 since 17143 = 182*183/2 + 21^2 + (2*(3*2+1)/2)^2.
a(24958) = 1 since 24958 = 216*217/2 + 39^2 + (1*(3*1-1)/2)^2.
a(26148) = 1 since 26148 = 10*11/2 + 142^2 + (7*(3*7+1)/2)^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
B. K. Oh and Z.-W. Sun, Mixed sums of squares and triangular numbers (III), J. Number Theory 129(2009), 964-969.
Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Z.-W. Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015.

Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566.

pQ[n_]:=pQ[n]=IntegerQ[n]&&IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[pQ[Sqrt[n-x^2-y(y+1)/2]],r=r+1],{x,1,Sqrt[n]},{y,0,(Sqrt[8(n-x^2)+1]-1)/2}];Print[n," ",r];Continue,{n,1,90}]

A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12x, 5x, 28x, 9x, 44x, 13x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).

Original entry on oeis.org

0, 1, 12, 5, 28, 9, 44, 13, 60, 17, 76, 21, 92, 25, 108, 29, 124, 33, 140, 37, 156, 41, 172, 45, 188, 49, 204, 53, 220, 57, 236, 61, 252, 65, 268, 69, 284, 73, 300, 77, 316, 81, 332, 85, 348, 89, 364, 93, 380, 97, 396, 101, 412, 105, 428, 109, 444, 113, 460, 117, 476

Offset: 0

Paul D. Hanna, Jul 14 2005

Simple continued fraction expansion of 2*(e - 1)/(e + 1) = 2*tanh(1/2) = 1/(1 + 1/(12 + 1/(5 + 1/(28 + ...)))). - Peter Bala, Oct 01 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. continued fraction expansions: A004273 ( tanh(1) ), A204877 ( 3*tanh(1/3) ), A130824 ( tanh(1/2) ).

a(n)=polcoeff(x*(1+12*x+3*x^2+4*x^3)/(1-x^2)^2+x*O(x^n),n)

G.f.: x*((1+3*x^2) + 4*x*(3+x^2))/(1-x^2)^2 = sum_{n>=0} a(n)*x^n.
From Carl R. White, Feb 11 2010: (Start)
a(n) = sign(n) * (2*n+1) * (3*cos(Pi*n)+5)/2.
a(2n+1) = a(2n-1) + 4, a(2n+2) = a(2n) + 16, with a(0)=0, a(1)=1, a(2)=12. (End)
a(n) = (5+3*(-1)^n)*(2*n-1)/2, with a(0)=0. Sum_{i=0..n} a(i) = A085787(A042948(n)). - Bruno Berselli, Jan 20 2012

A118282 Conjectured largest number that is not the sum of three generalized n-gonal numbers, or -1 if there is no largest number.

Original entry on oeis.org

0, -1, 0, 0, 307, -1, 2027, 5200, 18180, -1, 10795, -1, 87740, -1, 75150, 212048, 122818, -1, 146970, 199153, 585513

Offset: 3

T. D. Noe, Apr 21 2006

sign

Extensive calculations show that if a(n)>=0, then every number greater than a(n) can be represented as the sum of three generalized n-gonal numbers. a(n)=0 for n=3 and 6 because generalized triangular and generalized hexagonal numbers are the same a triangular numbers and every number can be written as the sum of three triangular numbers. When n is a multiple of 4, there is an infinite set of numbers not representable. For n=14, there appears to be a sparse, but infinite, set of numbers not representable. See A118283 for the number of numbers not representable.

Cf. A001318 (generalized pentagonal numbers), A085787 (generalized heptagonal numbers), A001082 (generalized octagonal numbers), A118277 (generalized 9-gonal numbers), A118278-A118285.

A153784 4 times heptagonal numbers: a(n) = 2n(5*n-3).

Original entry on oeis.org

0, 4, 28, 72, 136, 220, 324, 448, 592, 756, 940, 1144, 1368, 1612, 1876, 2160, 2464, 2788, 3132, 3496, 3880, 4284, 4708, 5152, 5616, 6100, 6604, 7128, 7672, 8236, 8820, 9424, 10048, 10692, 11356, 12040, 12744, 13468, 14212, 14976, 15760, 16564, 17388, 18232, 19096

Offset: 0

Omar E. Pol, Jan 02 2009

Sequence found by reading the line from 0, in the direction 0, 4, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000566, A085787, A087348, A135706, A152745, A152773.

s=0;lst={s};Do[s+=n;AppendTo[lst,s],{n,4,6!,20}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
Table[2n(5n-3),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,4,28},50] (* Harvey P. Dale, Mar 19 2015 *)

a(n)=2*n*(5*n-3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 10*n^2 - 6*n = 4*A000566(n) = 2*A135706(n).
a(n) = 20*n + a(n-1) - 16 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = A087348(n) - 1, n >= 1. - Omar E. Pol, Jul 18 2012
a(0)=0, a(1)=4, a(2)=28, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Mar 19 2015
From Elmo R. Oliveira, Dec 15 2024: (Start)
G.f.: 4*x*(1 + 4*x)/(1 - x)^3.
E.g.f.: 2*exp(x)*x*(2 + 5*x).
a(n) = A152745(n) - n. (End)

A158186 a(n) = 10n^2 - 7n + 1.

Original entry on oeis.org

1, 4, 27, 70, 133, 216, 319, 442, 585, 748, 931, 1134, 1357, 1600, 1863, 2146, 2449, 2772, 3115, 3478, 3861, 4264, 4687, 5130, 5593, 6076, 6579, 7102, 7645, 8208, 8791, 9394, 10017, 10660, 11323, 12006, 12709, 13432, 14175, 14938, 15721, 16524, 17347

Offset: 0

Reinhard Zumkeller, Mar 13 2009

Sequence found by reading the segment (1, 4) together with the line (one of the diagonal axes) from 4, in the direction 4, 27, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Sep 10 2011

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001622, A008596, A010010, A033571, A085787.

Table[10n^2-7n+1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{1,4,27},50] (* Harvey P. Dale, Apr 06 2020 *)

a(n)=10*n^2-7*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = (2*n-1)*(5*n-1).
a(n) = A033571(n) - A008596(n) = A010010(n) - A033571(n).
G.f.: (1+x+18*x^2)/(1-x)^3. - Jaume Oliver Lafont, Mar 27 2009
a(n) = a(n-1) + 20*n - 17 (with a(0)=1). - Vincenzo Librandi, Dec 03 2010
Sum_{n>=0} 1/a(n) = 1 + (2*sqrt(1+2/sqrt(5))*Pi - 2*sqrt(5)*log(phi) - 5*log(5) + 8*log(2))/12, where phi is the golden ratio (A001622). - Amiram Eldar, Sep 22 2022

Typo in definition corrected by Reinhard Zumkeller, Dec 03 2009

A194801 Square array read by antidiagonals: T(n,k) = k((n+1)k-n+1)/2, k = 0, +- 1, +- 2,..., n >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 3, 0, 1, 2, 4, 1, 0, 1, 3, 5, 4, 6, 0, 1, 4, 6, 7, 9, 3, 0, 1, 5, 7, 10, 12, 9, 10, 0, 1, 6, 8, 13, 15, 15, 16, 6, 0, 1, 7, 9, 16, 18, 21, 22, 16, 15, 0, 1, 8, 10, 19, 21, 27, 28, 26, 25, 10, 0, 1, 9, 11, 22, 24, 33, 34, 36, 35

Offset: 0

Omar E. Pol, Feb 05 2012

Note that a single formula gives several types of numbers. Row 0 lists 0 together the Molien series for 3-dimensional group [2,k]+ = 22k. Row 1 lists, except first zero, the squares repeated. If n >= 2, row n lists the generalized (n+3)-gonal numbers, for example: row 2 lists the generalized pentagonal numbers A001318. See some other examples in the cross-references section.

			Array begins:
(A008795): 0, 1,  0,  3,  1,  6,  3, 10,   6,  15,  10...
(A008794): 0, 1,  1,  4,  4,  9,  9, 16,  16,  25,  25...
A001318:   0, 1,  2,  5,  7, 12, 15, 22,  26,  35,  40...
A000217:   0, 1,  3,  6, 10, 15, 21, 28,  36,  45,  55...
A085787:   0, 1,  4,  7, 13, 18, 27, 34,  46,  55,  70...
A001082:   0, 1,  5,  8, 16, 21, 33, 40,  56,  65,  85...
A118277:   0, 1,  6,  9, 19, 24, 39, 46,  66,  75, 100...
A074377:   0, 1,  7, 10, 22, 27, 45, 52,  76,  85, 115...
A195160:   0, 1,  8, 11, 25, 30, 51, 58,  86,  95, 130...
A195162:   0, 1,  9, 12, 28, 33, 57, 64,  96, 105, 145...
A195313:   0, 1, 10, 13, 31, 36, 63, 70, 106, 115, 160...
A195818:   0, 1, 11, 14, 34, 39, 69, 76, 116, 125, 175...

Rows (0-11): 0 together with A008795, (truncated A008794), A001318, A000217, A085787, A001082, A118277, A074377, A195160, A195162, A195313, A195818
Columns (0-9): A000004, A000012, A001477, (truncated A000027), A016777, (truncated A008585), A016945, (truncated A016957), A017341, (truncated A017329).
Cf. A139600.

A195016 a(n) = (n(5n+7)-(-1)^n+1)/2.

Original entry on oeis.org

0, 7, 17, 34, 54, 81, 111, 148, 188, 235, 285, 342, 402, 469, 539, 616, 696, 783, 873, 970, 1070, 1177, 1287, 1404, 1524, 1651, 1781, 1918, 2058, 2205, 2355, 2512, 2672, 2839, 3009, 3186, 3366, 3553, 3743, 3940, 4140, 4347, 4557, 4774, 4994

Offset: 0

Omar E. Pol, Sep 26 2011

Sequence found by reading the line from 0, in the direction 0, 7,..., and the same line from 0, in the direction 0, 17,..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. Axis perpendicular to the main axis A195015 in the same spiral.

Also sequence found by reading the line from 0, in the direction 0, 7,..., and the same line from 0, in the direction 0, 17,..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. This line is parallel to A153126 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000566, A005476, A028895, A033571, A085787, A152965, A153126, A195013, A195014, A195015.

&cat[[n*t,(n+1)*t] where t is 10*n+7: n in [0..22]]; // Bruno Berselli, Oct 14 2011

LinearRecurrence[{2, 0, -2, 1}, {0, 7, 17, 34}, 50] (* Paolo Xausa, Feb 09 2024 *)

n*(10*n-3), if n >= 1, and (2*n+1)*(5*n+1)-1, if n >= 0, interleaved.

G.f.: x*(7+3*x)/((1+x)*(1-x)^3). - Bruno Berselli, Oct 14 2011

Concise definition by Bruno Berselli, Oct 14 2011

A255934 Number of ways to write n as the sum of four unordered generalized octagonal numbers.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 3, 2, 3, 2, 1, 3, 3, 3, 3, 2, 4, 2, 1, 3, 2, 3, 4, 3, 4, 2, 2, 4, 4, 3, 4, 3, 6, 5, 2, 4, 3, 4, 5, 4, 6, 4, 1, 4, 5, 4, 5, 5, 7, 4, 1, 5, 5, 5, 6, 5, 8, 5, 3, 4, 6, 6, 6, 6, 7, 6, 3, 6, 6, 5, 6, 6, 10, 7, 1, 5, 8, 7, 7, 7, 8, 5, 3, 6, 7, 6, 8, 7, 10, 8, 3

Offset: 0

Zhi-Wei Sun, Mar 11 2015

nonn

I have proved that a(n) > 0 for all n, i.e., any nonnegative integer can be expressed as the sum of four generalized octagonal numbers. I can show that a(n) = 1 if 3*n+4 is among 7, 13, 19, 31, 43, 2^(2k), 5*2^(2k+1), 11*2^(2k+1), 23*2^(2k+1) (k = 0,1,2,...), and conjecture the converse.

I also conjecture that each nonnegative integer can be written as the sum of two heptagonal numbers, a second heptagonal number and a generalized heptagonal number.

			a(60) = 1 since 60 = 1*(3*1-2) + (-1)*(3*(-1)-2) + 3*(3*3-2) + (-3)*(3*(-3)-2).
a(1876) = 1 since 1876 = (-5)*(3*(-5)-2) + (-5)*(3*(-5)-2) + 11*(3*11-2) + (-21)*(3*(-21)-2).
a(15700) = 1 since 15700 = 11*(3*11-2) + (-21)*(3*(-21)-2) + 43*(3*43-2) + (-53)*(3*(-53)-2).
a(21844) = 1 since 21844 = 43*(3*43-2) + 43*(3*43-2) + 43*(3*43-2) + 43*(3*43-2).
a(30036) = 1 since 30036 = (-21)*(3*(-21)-2) + (-21)*(3*(-21)-2) + 43*(3*43-2) + (-85)*(3*(-85)-2).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.
Zhi-Wei Sun, A result similar to Lagrange's theorem, arXiv:1503.03743 [math.NT], 2015.

Cf. A000566, A001082, A085787, A147875, A255350.

T[n_]:=Union[Table[x(3x-2),{x,-Floor[(Sqrt[3n+1]-1)/3],Floor[(Sqrt[3n+1]+1)/3]}]]
Do[r=0;Do[If[n-Part[T[n],x]-Part[T[n],y]-Part[T[n],z]

cp's OEIS Frontend

A111710 Consider the triangle shown below in which the n-th row contains the n smallest numbers greater than those in the previous row such that the arithmetic mean is an integer. Sequence contains the leading diagonal.

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A220082 Numbers k such that 10*k-1 is a square.

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A270594 Number of ordered ways to write n as the sum of a triangular number, a positive square and the square of a generalized pentagonal number (A001318).

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A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12*x, 5*x, 28*x, 9*x, 44*x, 13*x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).

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A118282 Conjectured largest number that is not the sum of three generalized n-gonal numbers, or -1 if there is no largest number.

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A153784 4 times heptagonal numbers: a(n) = 2*n*(5*n-3).

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A158186 a(n) = 10*n^2 - 7*n + 1.

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A194801 Square array read by antidiagonals: T(n,k) = k*((n+1)*k-n+1)/2, k = 0, +- 1, +- 2,..., n >= 0.

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A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12x, 5x, 28x, 9x, 44x, 13x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).

A153784 4 times heptagonal numbers: a(n) = 2n(5*n-3).

A158186 a(n) = 10n^2 - 7n + 1.

A194801 Square array read by antidiagonals: T(n,k) = k((n+1)k-n+1)/2, k = 0, +- 1, +- 2,..., n >= 0.

A195016 a(n) = (n(5n+7)-(-1)^n+1)/2.