A001082 -id:A001082 - cp's OEIS frontend

A154105 a(n) = 12n^2 + 18n + 7.

7, 37, 91, 169, 271, 397, 547, 721, 919, 1141, 1387, 1657, 1951, 2269, 2611, 2977, 3367, 3781, 4219, 4681, 5167, 5677, 6211, 6769, 7351, 7957, 8587, 9241, 9919, 10621, 11347, 12097, 12871, 13669, 14491, 15337, 16207, 17101, 18019, 18961, 19927, 20917, 21931

Offset: 0

Klaus Brockhaus, Jan 04 2009

a(n) is the number of partitions with three integral dissimilar components of the number 12(n+1), e.g for n=0, 12 may be partitioned in the 7 ways (1,2,9), (1,3,8), (1,4,7), (1,5,6), (2,3,7), (2,4,6) and (3,4,5). - Ian Duff, Jan 31 2010

Sequence found by reading the line from 7, in the direction 7, 37, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, May 08 2018

			a(2) = 12*2^2 + 18*2 + 7 = 91 = 6*14 + 7 = 6*A014106(2) + 7.
a(3) = a(2) + 24*3 + 6 = 91 + 72 + 6 = 169.
a(-4) = 12*4^2 - 18*4 + 7 = 127 = 2*64 - 1 = 2*A085473(3) - 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
John Elias, Animated Illustration: Starburst Hexagrams
Leo Tavares, Illustration: Hexagonal Halos
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001082, A014106, A152746, A153286, A085473.

Cf. A003215, A000290, A000217.

Magma
```
[ 12*n^2+18*n+7: n in [0..40] ];
```

Table[12*n^2 + 18*n + 7, {n, 0, 42}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)
LinearRecurrence[{3,-3,1}, {7,37,91}, 25] (* G. C. Greubel, Sep 02 2016 *)

a(n)=12*n^2+18*n+7 \\ Charles R Greathouse IV, Sep 02 2016

G.f.: (7 + 16*x + x^2)/(1-x)^3.
a(n) = 6*A014106(n) + 7.
a(0) = 7; for n > 0, a(n) = a(n-1) + 24*n + 6.
a(-n-1) = 2*A085473(n) - 1. - Bruno Berselli, Sep 05 2011
E.g.f.: (7 + 30*x + 12*x^2)*exp(x). - G. C. Greubel, Sep 02 2016
a(n) = 1 + A152746(n+1). - Omar E. Pol, May 08 2018
a(n) = A003215(n) + 6*A000290(n+1) + 6*A000217(n). - Leo Tavares, Sep 12 2022

A271713 Numbers n such that 3*n - 5 is a square.

Original entry on oeis.org

2, 3, 7, 10, 18, 23, 35, 42, 58, 67, 87, 98, 122, 135, 163, 178, 210, 227, 263, 282, 322, 343, 387, 410, 458, 483, 535, 562, 618, 647, 707, 738, 802, 835, 903, 938, 1010, 1047, 1123, 1162, 1242, 1283, 1367, 1410, 1498, 1543, 1635, 1682, 1778, 1827, 1927, 1978, 2082, 2135, 2243, 2298

Offset: 1

Juri-Stepan Gerasimov, Apr 12 2016

Quasipolynomial of order 2 and degree 2. - Charles R Greathouse IV, Apr 12 2016
From Ray Chandler, Apr 13 2016: (Start)
Square roots of resulting squares gives A001651.
Sequence is the union of A141631 and A271740. (End)

			a(3) = 7 because 3*7 - 5 = 16 = 4^2.

Index entries for linear recurrences with constant coefficients, signature (1, 2, -2, -1, 1).

Cf. numbers n such that 3*n + k is a square: A120328 (k=-6), this sequence (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), A271675 (k=4), A100536 (k=6).

Cf. A001651, A141631, A271740.

[ n: n in [0..2500] | IsSquare(3*n - 5)];

A271713:=n->`if`(issqr(3*n-5), n, NULL): seq(A271713(n), n=1..5000); # Wesley Ivan Hurt, Apr 13 2016

Select[Range@ 2300, IntegerQ@ Sqrt[3 # - 5] &] (* Michael De Vlieger, Apr 12 2016 *)

is(n)=issquare(3*n-5) \\ Charles R Greathouse IV, Apr 12 2016

a(n)=((n\2*3-(-1)^n)^2+5)/3 \\ Charles R Greathouse IV, Apr 12 2016

from _future_ import division
A271713_list = [(n**2+5)//3 for n in range(10**6) if not (n**2+5) % 3] # Chai Wah Wu, Apr 13 2016

G.f.: x*(2 + x + x^3 + 2*x^4)/((1 - x)^3*(1 + x)^2). - Ilya Gutkovskiy, Apr 12 2016
a(n) = (3/2)*n^2 + O(n). - Charles R Greathouse IV, Apr 12 2016
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5. - Wesley Ivan Hurt, Apr 13 2016

A154106 a(n) = 12n^2 + 22n + 11.

Original entry on oeis.org

11, 45, 103, 185, 291, 421, 575, 753, 955, 1181, 1431, 1705, 2003, 2325, 2671, 3041, 3435, 3853, 4295, 4761, 5251, 5765, 6303, 6865, 7451, 8061, 8695, 9353, 10035, 10741, 11471, 12225, 13003, 13805, 14631, 15481, 16355, 17253, 18175, 19121

Offset: 0

Klaus Brockhaus, Jan 04 2009

Sequence found by reading the line from 11, in the direction 11, 45,..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

			a(3) = 12*3^2 + 22*3 + 11 = 185 = 2*3*29 + 11 = 2*3*A016969(4) + 11.
a(4) = a(3) +24*4 +10 = 185 +96 +10 = 291.

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

Cf. A016969 (6n+5), A153286, A185918, A194454.

Magma
```
[ 12*n^2+22*n+11: n in [0..39] ];
```

Table[12n^2+22n+11,{n,0,50}]  (* Harvey P. Dale, Mar 16 2011 *)
LinearRecurrence[{3,-3,1},{11,45,103}, 25] (* G. C. Greubel, Sep 02 2016 *)

a(n)=12*n^2+22*n+11 \\ Charles R Greathouse IV, Oct 16 2015

G.f.: (1 +x)*(11 +x)/(1-x)^3.
a(n) = 2*n*A016969(n+1) + 11.
a(0) = 11; for n > 0, a(n) = a(n-1) + 24*n + 10.
a(n) = 2 + A185918(n+1). - Omar E. Pol, Jul 18 2012
E.g.f.: (11 + 34*x + 12*x^2)*exp(x). - G. C. Greubel, Sep 02 2016

A158480 a(n) = 12*n^2 + 1.

Original entry on oeis.org

1, 13, 49, 109, 193, 301, 433, 589, 769, 973, 1201, 1453, 1729, 2029, 2353, 2701, 3073, 3469, 3889, 4333, 4801, 5293, 5809, 6349, 6913, 7501, 8113, 8749, 9409, 10093, 10801, 11533, 12289, 13069, 13873, 14701, 15553, 16429, 17329, 18253, 19201, 20173, 21169

Offset: 0

Vincenzo Librandi, Mar 20 2009

The identity (12*n^2 + 1)^2 - (36*n^2 + 6)*(2*n)^2 = 1 can be written as a(n)^2 - A158479(n)*A005843(n)^2 = 1.

Sequence found by reading the line from 13, in the direction 13, 49, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

			a(1) = 12*1^2 + 1 = 13.
a(2) = 12*2^2 + 1 = 49.
a(3) = 12*3^2 + 1 = 109.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Vincenzo Librandi, X^2-AY^2=1 [broken link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A158479.

List([1..40], n-> 12*n^2 + 1); # G. C. Greubel, Nov 06 2019

I:=[13,49,109];[n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2) +Self(n-3): n in [1..50]];

seq(12*n^2 +1, n=0..45); # G. C. Greubel, Nov 06 2019

LinearRecurrence[{3,-3,1}, {13,49,109}, 40]
12*Range[40]^2 +1 (* G. C. Greubel, Nov 06 2019 *)

PARI
```
a(n)=12*n^2+1
```

[12*n^2 +1 for n in (1..40)] # G. C. Greubel, Nov 06 2019

a(n) = A010014(n)/2. - Vladimir Joseph Stephan Orlovsky, May 18 2009
G.f: (13*x^2 + 10*x + 1)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 1 + A135453(n). - Omar E. Pol, Jul 18 2012
a(n) = (A016969(n-1)*A016921(n) + 4)/3. - Hilko Koning, Oct 25 2019
E.g.f.: (1 + 12*x + 12*x^2)*exp(x). - G. C. Greubel, Nov 06 2019
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(12))*coth(Pi/sqrt(12)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(12))*csch(Pi/sqrt(12)))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(12))*sinh(Pi/sqrt(6)).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(12))*csch(Pi/sqrt(12)). (End)

a(0)=1 prepended by Alois P. Heinz, Jun 12 2021

A294655 Expansion of Product_{k>=1} 1/((1 - x^(2k-1))^(k(3k-2))(1 - x^(2k))^(k(3*k+2))).

Original entry on oeis.org

1, 1, 6, 14, 45, 106, 290, 683, 1698, 3918, 9179, 20640, 46444, 101819, 222092, 475886, 1012270, 2124725, 4425195, 9118705, 18648048, 37797126, 76062443, 151889787, 301296200, 593593192, 1162276735, 2261819285, 4376578818, 8421295585, 16118902083, 30694325652, 58164428059

Offset: 0

Ilya Gutkovskiy, Nov 06 2017

nonn

Euler transform of the generalized octagonal numbers (A001082).

Vaclav Kotesovec, Table of n, a(n) for n = 0..5000
N. J. A. Sloane, Transforms
Eric Weisstein's World of Mathematics, Octagonal Number

Cf. A000294, A001082, A274998, A294591, A294622, A294654, A294691, A294692.

nmax = 32; CoefficientList[Series[Product[1/((1 - x^(2 k - 1))^(k (3 k - 2)) (1 - x^(2 k))^(k (3 k + 2))), {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d (d^2 + d - Ceiling[d/2]^2), {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 32}]

G.f.: Product_{k>=1} 1/(1 - x^k)^A001082(k+1).

a(n) ~ exp(Pi * 2^(3/2) * n^(3/4) / (3*5^(1/4)) + 3*Zeta(3) * sqrt(5*n) / (2*Pi^2) - (45*Zeta(3)^2 / Pi^5 + Pi/6) * 5^(1/4) * (n^(1/4) / 2^(5/2)) + 225 * Zeta(3)^3 / (4*Pi^8) - Zeta(3) / (32*Pi^2) + 1/8) * Pi^(1/8) / (A^(3/2) * 2^(77/48) * 5^(5/32) * n^(21/32)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Nov 07 2017

A303301 Square array T(n,k) read by antidiagonals upwards in which row n is obtained by taking the general formula for generalized n-gonal numbers: m((n - 2)m - n + 4)/2, where m = 0, +1, -1, +2, -2, +3, -3, ... and n >= 5. Here n >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, -3, 0, 1, -2, 0, 0, 1, -1, 1, -8, 0, 1, 0, 2, -5, -3, 0, 1, 1, 3, -2, 0, -15, 0, 1, 2, 4, 1, 3, -9, -8, 0, 1, 3, 5, 4, 6, -3, -2, -24, 0, 1, 4, 6, 7, 9, 3, 4, -14, -15, 0, 1, 5, 7, 10, 12, 9, 10, -4, -5, -35, 0, 1, 6, 8, 13, 15, 15, 16, 6, 5, -20, -24, 0, 1, 7, 9, 16, 18, 21, 22, 16, 15, -5, -9, -48

Offset: 0

Omar E. Pol, Jun 08 2018

Note that the formula mentioned in the definition gives several kinds of numbers, for example:
Row 0 and row 1 give A317300 and A317301 respectively.
Row 2 gives A001057 (canonical enumeration of integers).
Row 3 gives 0 together with A008795 (Molien series for 3-dimensional representation of dihedral group D_6 of order 6).
Row 4 gives A008794 (squares repeated) except the initial zero.
Finally, for n >= 5 row n gives the generalized k-gonal numbers (see Crossrefs section).

			Array begins:
------------------------------------------------------------------
n\m  Seq. No.    0   1  -1   2  -2   3   -3    4   -4    5   -5
------------------------------------------------------------------
0    A317300:    0,  1, -3,  0, -8, -3, -15,  -8, -24, -15, -35...
1    A317301:    0,  1, -2,  1, -5,  0,  -9,  -2, -14,  -5, -20...
2    A001057:    0,  1, -1,  2, -2,  3,  -3,   4,  -4,   5,  -5...
3   (A008795):   0,  1,  0,  3,  1,  6,   3,  10,   6,  15,  10...
4   (A008794):   0,  1,  1,  4,  4,  9,   9,  16,  16,  25,  25...
5    A001318:    0,  1,  2,  5,  7, 12,  15,  22,  26,  35,  40...
6    A000217:    0,  1,  3,  6, 10, 15,  21,  28,  36,  45,  55...
7    A085787:    0,  1,  4,  7, 13, 18,  27,  34,  46,  55,  70...
8    A001082:    0,  1,  5,  8, 16, 21,  33,  40,  56,  65,  85...
9    A118277:    0,  1,  6,  9, 19, 24,  39,  46,  66,  75, 100...
10   A074377:    0,  1,  7, 10, 22, 27,  45,  52,  76,  85, 115...
11   A195160:    0,  1,  8, 11, 25, 30,  51,  58,  86,  95, 130...
12   A195162:    0,  1,  9, 12, 28, 33,  57,  64,  96, 105, 145...
13   A195313:    0,  1, 10, 13, 31, 36,  63,  70, 106, 115, 160...
14   A195818:    0,  1, 11, 14, 34, 39,  69,  76, 116, 125, 175...
15   A277082:    0,  1, 12, 15, 37, 42,  75,  82, 126, 135, 190...
...

Alois P. Heinz, Antidiagonals n = 0..200 (first 45 antidiagonals from Robert G. Wilson v)

Columns 0..2 are A000004, A000012, A023445.
Column 3 gives A001477 which coincides with the row numbers.
Main diagonal gives A292551.
Row 0-2 gives A317300, A317301, A001057.
Row 3 gives 0 together with A008795.
Row 4 gives A008794.
For n >= 5, rows n gives the generalized n-gonal numbers: A001318 (n=5), A000217 (n=6), A085787 (n=7), A001082 (n=8), A118277 (n=9), A074377 (n=10), A195160 (n=11), A195162 (n=12), A195313 (n=13), A195818 (n=14), A277082 (n=15), A274978 (n=16), A303305 (n=17), A274979 (n=18), A303813 (n=19), A218864 (n=20), A303298 (n=21), A303299 (n=22), A303303 (n=23), A303814 (n=24), A303304 (n=25), A316724 (n=26), A316725 (n=27), A303812 (n=28), A303815 (n=29), A316729 (n=30).
Cf. A194801, A195152.
Cf. A317302 (a similar table but with polygonal numbers).

t[n_, r_] := PolygonalNumber[n, If[OddQ@ r, Floor[(r + 1)/2], -r/2]]; Table[ t[n - r, r], {n, 0, 11}, {r, 0, n}] // Flatten (* also *)
(* to view the square array *)  Table[ t[n, r], {n, 0, 15}, {r, 0, 10}] // TableForm (* Robert G. Wilson v, Aug 08 2018 *)

T(n,k) = A194801(n-3,k) if n >= 3.

A153794 4 times octagonal numbers: a(n) = 4n(3*n-2).

Original entry on oeis.org

0, 4, 32, 84, 160, 260, 384, 532, 704, 900, 1120, 1364, 1632, 1924, 2240, 2580, 2944, 3332, 3744, 4180, 4640, 5124, 5632, 6164, 6720, 7300, 7904, 8532, 9184, 9860, 10560, 11284, 12032, 12804, 13600, 14420, 15264, 16132, 17024

Offset: 0

Omar E. Pol, Jan 19 2009

Sequence found by reading the segment (0, 4) together with the line from 4, in the direction 4, 32, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

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

Cf. A000567, A139267, A152751, A153808.

s=0;lst={s};Do[s+=n;AppendTo[lst,s],{n,4,7!,24}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
Table[4n(3n-2),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,4,32},41] (* Harvey P. Dale, Jul 14 2011 *)
4*PolygonalNumber[8,Range[0,40]] (* Harvey P. Dale, Dec 04 2022 *)

a(n) = 12*n^2 - 8*n; \\ Altug Alkan, Aug 29 2016

a(n) = 12*n^2 - 8*n = 4*A000567(n) = 2*A139267(n).
a(n) = 24*n + a(n-1) - 20 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(0)=0, a(1)=4, a(2)=32, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jul 14 2011
G.f.: 4*(x + 5*x^2)/(1-x)^3. - Harvey P. Dale, Jul 14 2011
E.g.f.: 4*x*(1 + 3*x)*exp(x). - G. C. Greubel, Aug 29 2016

A185212 a(n) = 12n^2 - 8n + 1.

Original entry on oeis.org

1, 5, 33, 85, 161, 261, 385, 533, 705, 901, 1121, 1365, 1633, 1925, 2241, 2581, 2945, 3333, 3745, 4181, 4641, 5125, 5633, 6165, 6721, 7301, 7905, 8533, 9185, 9861, 10561, 11285, 12033, 12805, 13601, 14421, 15265, 16133, 17025, 17941, 18881, 19845, 20833

Offset: 0

Reinhard Zumkeller, Dec 20 2012

Sequence found by reading the line from 1, in the direction 1, 5, and the same line from 5, in the direction 5, 33, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, May 08 2018

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Square Block Rays
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

For n > 0: odd terms in A001859.
Cf. A001082.
Cf. A016754, A000384.

Haskell
```
a185212 = (+ 1) . (* 4) . a000567
```

Table[12n^2-8n+1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{1,5,33},50] (* Harvey P. Dale, Jul 07 2015 *)

a(n)=12*n^2-8*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 4*A000567(n) + 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0)=1, a(1)=5, a(2)=33. - Harvey P. Dale, Jul 07 2015
G.f.: (-1 - 2*x - 21*x^2)/(-1+x)^3. - Harvey P. Dale, Jul 07 2015
E.g.f.: (12*x^2 + 4*x + 1)*exp(x). - G. C. Greubel, Jun 25 2017
a(n) = A016754(n-1) + 4*A000384(n). - Leo Tavares, May 21 2022
From Amiram Eldar, May 28 2022: (Start)
Sum_{n>=0} 1/a(n) = sqrt(3)*Pi/8 - 3*log(3)/8 + 1.
Sum_{n>=0} (-1)^n/a(n) = Pi/8 - sqrt(3)*arccoth(sqrt(3))/2 + 1. (End)

A194454 a(n) = 12n^2 + 2n + 1.

Original entry on oeis.org

1, 15, 53, 115, 201, 311, 445, 603, 785, 991, 1221, 1475, 1753, 2055, 2381, 2731, 3105, 3503, 3925, 4371, 4841, 5335, 5853, 6395, 6961, 7551, 8165, 8803, 9465, 10151, 10861, 11595, 12353, 13135, 13941, 14771, 15625, 16503, 17405, 18331, 19281

Offset: 0

Bruno Berselli, Aug 24 2011

A142241 gives the first differences.
Inverse binomial transform of this sequence: 1, 14, 24, 0, 0 (0 continued).
a(n)*a(n-1)-11 is a square, precisely 4*A051866(n)^2.
Sequence found by reading the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

			Using these numbers we can write:
  1, 15, 53, 115, 201, 311, 445,  603,  785,  991, 1221, ...
  0,  0,  1,  15,  53, 115, 201,  311,  445,  603,  785, ...
  0,  0,  0,   0,   1,  15,  53,  115,  201,  311,  445, ...
  0,  0,  0,   0,   0,   0,   1,   15,   53,  115,  201, ...
  0,  0,  0,   0,   0,   0,   0,    0,    1,   15,   53, ...
  0,  0,  0,   0,   0,   0,   0,    0,    0,    0,    1, ...
  ======================================================
  The sums of the columns give the sequence A172073 (after 0):
  1, 15, 54, 130, 255, 441, 700, 1044, 1485, 2035, 2706, ...

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

Cf. A154106, A172073, A049453.

Magma
```
[12*n^2+2*n+1: n in [0..40]];
```

Table[12 n^2 + 2 n + 1, {n, 0, 50}] (* Vincenzo Librandi, Mar 26 2013 *)

PARI
```
for(n=0, 40, print1(12*n^2+2*n+1", "));
```

G.f.: (1+x)*(1+11*x)/(1-x)^3.
a(n) = A154106(-n-1).
a(n) = 2*A049453(n) + 1.
a(n) = A051866(n) + A051866(n+1). - Charlie Marion, Nov 15 2019
E.g.f.: exp(x)*(1 + 14*x + 12*x^2). - Stefano Spezia, Nov 15 2019

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}]

cp's OEIS Frontend

A154105 a(n) = 12*n^2 + 18*n + 7.

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A271713 Numbers n such that 3*n - 5 is a square.

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A154106 a(n) = 12*n^2 + 22*n + 11.

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A158480 a(n) = 12*n^2 + 1.

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A294655 Expansion of Product_{k>=1} 1/((1 - x^(2*k-1))^(k*(3*k-2))*(1 - x^(2*k))^(k*(3*k+2))).

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A303301 Square array T(n,k) read by antidiagonals upwards in which row n is obtained by taking the general formula for generalized n-gonal numbers: m*((n - 2)*m - n + 4)/2, where m = 0, +1, -1, +2, -2, +3, -3, ... and n >= 5. Here n >= 0.

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A154105 a(n) = 12n^2 + 18n + 7.

A154106 a(n) = 12n^2 + 22n + 11.

A294655 Expansion of Product_{k>=1} 1/((1 - x^(2k-1))^(k(3k-2))(1 - x^(2k))^(k(3*k+2))).

A303301 Square array T(n,k) read by antidiagonals upwards in which row n is obtained by taking the general formula for generalized n-gonal numbers: m((n - 2)m - n + 4)/2, where m = 0, +1, -1, +2, -2, +3, -3, ... and n >= 5. Here n >= 0.

A153794 4 times octagonal numbers: a(n) = 4n(3*n-2).

A185212 a(n) = 12n^2 - 8n + 1.

A194454 a(n) = 12n^2 + 2n + 1.