A190816 -id:A190816 - cp's OEIS frontend

A062786 Centered 10-gonal numbers.

1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051, 1201, 1361, 1531, 1711, 1901, 2101, 2311, 2531, 2761, 3001, 3251, 3511, 3781, 4061, 4351, 4651, 4961, 5281, 5611, 5951, 6301, 6661, 7031, 7411, 7801, 8201, 8611, 9031, 9461, 9901, 10351, 10811

Offset: 1

Jason Earls, Jul 19 2001

Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 10*A000217(n-1) + 1. - Amarnath Murthy, Dec 11 2003
All divisors of a(n) are congruent to 1 or -1, modulo 10; that is, they end in the decimal digit 1 or 9. Proof: If p is an odd prime different from 5 then 5n^2 - 5n + 1 == 0 (mod p) implies 25(2n - 1)^2 == 5 (mod p), whence p == 1 or -1 (mod 10). - Nick Hobson, Nov 13 2006
Centered decagonal numbers. - Omar E. Pol, Oct 03 2011
The partial sums of this sequence give A004466. - Leo Tavares, Oct 04 2021
The continued fraction expansion of sqrt(5*a(n)) is [5n-3; {2, 2n-2, 2, 10n-6}]. For n=1, this collapses to [2; {4}]. - Magus K. Chu, Sep 12 2022
Numbers m such that 20*m + 5 is a square. Also values of the Fibonacci polynomial y^2 - x*y - x^2 for x = n and y = 3*n - 1. This is a subsequence of A089270. - Klaus Purath, Oct 30 2022
All terms can be written as a difference of two consecutive squares a(n) = A005891(n-1)^2 - A028895(n-1)^2, and they can be represented by the forms (x^2 + 2mxy + (m^2-1)y^2) and (3x^2 + (6m-2)xy + (3m^2-2m)y^2), both of discriminant 4. - Klaus Purath, Oct 17 2023

T. D. Noe, Table of n, a(n) for n = 1..1000
Leo Tavares, Illustration: Pentagonal Stars.
Leo Tavares, Illustration: Mid-section Stars.
Leo Tavares, Illustration: Mid-section Star Pillars.
Leo Tavares, Illustration: Trapezoidal Rays.
R. Yin, J. Mu, and T. Komatsu, The p-Frobenius Number for the Triple of the Generalized Star Numbers, Preprints 2024, 2024072280. See p. 2.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001263, A124080, A101321, A028387, A016861, A003154, A005891, A000217, A004466, A144390, A000326, A060544.

Cf. also A016754, A002378, A069099, A045943, A003215, A046092, A001844, A028896, A005448, A024966, A082970.

List([1..50], n-> 1+5*n*(n-1)); # G. C. Greubel, Mar 30 2019

[1+5*n*(n-1): n in [1..50]]; // G. C. Greubel, Mar 30 2019

FoldList[#1+#2 &, 1, 10Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
1+5*Pochhammer[Range[50]-1, 2] (* G. C. Greubel, Mar 30 2019 *)

j=[]; for(n=1,75,j=concat(j,(5*n*(n-1)+1))); j

for (n=1, 1000, write("b062786.txt", n, " ", 5*n*(n - 1) + 1) ) \\ Harry J. Smith, Aug 11 2009

def a(n): return(5*n**2-5*n+1) # Torlach Rush, May 10 2024

[1+5*rising_factorial(n-1, 2) for n in (1..50)] # G. C. Greubel, Mar 30 2019

a(n) = 5*n*(n-1) + 1.
From  Gary W. Adamson, Dec 29 2007: (Start)
Binomial transform of [1, 10, 10, 0, 0, 0, ...];
Narayana transform (A001263) of [1, 10, 0, 0, 0, ...]. (End)
G.f.: x*(1+8*x+x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011
a(n) = A124080(n-1) + 1. - Omar E. Pol, Oct 03 2011
a(n) = A101321(10,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A028387(A016861(n-1))/5 for n > 0. - Art Baker, Mar 28 2019
E.g.f.: (1+5*x^2)*exp(x) - 1. - G. C. Greubel, Mar 30 2019
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(5))) / sqrt(5). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=1} a(n)/n! = 6*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 6/e - 1. (End)
a(n) = A005891(n-1) + 5*A000217(n-1). - Leo Tavares, Jul 14 2021
a(n) = A003154(n) - 2*A000217(n-1). See Mid-section Stars illustration. - Leo Tavares, Sep 06 2021
From Leo Tavares, Oct 06 2021: (Start)
a(n) = A144390(n-1) + 2*A028387(n-1). See Mid-section Star Pillars illustration.
a(n) = A000326(n) + A000217(n) + 3*A000217(n-1). See Trapezoidal Rays illustration.
a(n) = A060544(n) + A000217(n-1). (End)
From Leo Tavares, Oct 31 2021: (Start)
a(n) = A016754(n-1) + 2*A000217(n-1).
a(n) = A016754(n-1) + A002378(n-1).
a(n) = A069099(n) + 3*A000217(n-1).
a(n) = A069099(n) + A045943(n-1).
a(n) = A003215(n-1) + 4*A000217(n-1).
a(n) = A003215(n-1) + A046092(n-1).
a(n) = A001844(n-1) + 6*A000217(n-1).
a(n) = A001844(n-1) + A028896(n-1).
a(n) = A005448(n) + 7*A000217(n).
a(n) = A005448(n) + A024966(n). (End)
From Klaus Purath, Oct 30 2022: (Start)
a(n) = a(n-2) + 10*(2*n-3).
a(n) = 2*a(n-1) - a(n-2) + 10.
a(n) = A135705(n-1) + n.
a(n) = A190816(n) - n.
a(n) = 2*A005891(n-1) - 1. (End)

Better description from Terrel Trotter, Jr., Apr 06 2002

A055096 Triangle read by rows, sums of 2 distinct nonzero squares: T(n,k) = k^2+n^2, (n>=2, 1 <= k <= n-1).

Original entry on oeis.org

5, 10, 13, 17, 20, 25, 26, 29, 34, 41, 37, 40, 45, 52, 61, 50, 53, 58, 65, 74, 85, 65, 68, 73, 80, 89, 100, 113, 82, 85, 90, 97, 106, 117, 130, 145, 101, 104, 109, 116, 125, 136, 149, 164, 181, 122, 125, 130, 137, 146, 157, 170, 185, 202, 221, 145, 148, 153, 160

Offset: 2

Antti Karttunen, Apr 04 2000

Discovered by Bernard Frénicle de Bessy (1605?-1675). - Paul Curtz, Aug 18 2008
Terms that are not hypotenuses in primitive Pythagorean triangles, are replaced by 0 in A222946. - Reinhard Zumkeller, Mar 23 2013
This triangle T(n,k) gives the circumdiameters for the Pythagorean triangles with a = (n+1)^2 - k^2, b = 2*(n+1)*k and c = (n+1)^2 + k^2 (see the Floor van Lamoen entries or comments A063929, A063930, A002283, A003991). See also the formula section. Note that not all Pythagorean triangles are covered, e.g., (9,12,15) does not appear. - Wolfdieter Lang, Dec 03 2014

			The triangle T(n, k) begins:
n\k   1   2   3   4   5   6   7   8   9  10  11 ...
2:    5
3:   10  13
4:   17  20  25
5:   26  29  34  41
6:   37  40  45  52  61
7:   50  53  58  65  74  85
8:   65  68  73  80  89 100 113
9:   82  85  90  97 106 117 130 145
10: 101 104 109 116 125 136 149 164 181
11: 122 125 130 137 146 157 170 185 202 221
12: 145 148 153 160 169 180 193 208 225 244 265
...
13: 170 173 178 185 194 205 218 233 250 269 290 313,
14: 197 200 205 212 221 232 245 260 277 296 317 340 365,
15: 226 229 234 241 250 261 274 289 306 325 346 369 394 421,
16: 257 260 265 272 281 292 305 320 337 356 377 400 425 452 481,
...
Formatted and extended by _Wolfdieter Lang_, Dec 02 2014 (reformatted Jun 11 2015)
The successive terms are (1^2+2^2), (1^2+3^2), (2^2+3^2), (1^2+4^2), (2^2+4^2), (3^2+4^2), ...

Reinhard Zumkeller, Rows n = 2..121 of triangle, flattened
M. de Frénicle, Méthode pour trouver la solutions des problèmes par les exclusions, in: "Divers ouvrages de mathématiques et de physique, par Messieurs de l'Académie royale des sciences", Paris, 1693, pp 1-44.
Antti Karttunen, Larger table, showing also locations of 4k+1 primes and squares
Eric Weisstein's World of Mathematics, Congruum Problem.
Index entries for sequences related to sums of squares

Sorting gives A024507. Count of divisors: A055097, Möbius: A055132. For trinv, follow A055088.
Cf. A001844 (right edge), A002522 (left edge), A033429 (central column).
Cf. A000290, A033583, A079273, A190816, A226141, A331987.

a055096 n k = a055096_tabl !! (n-1) !! (k-1)
a055096_row n = a055096_tabl !! (n-1)
a055096_tabl = zipWith (zipWith (+)) a133819_tabl a140978_tabl
-- Reinhard Zumkeller, Mar 23 2013

[n^2+k^2: k in [1..n-1], n in [2..15]]; // G. C. Greubel, Apr 19 2023

sum2distinct_squares_array := (n) -> (((n-((trinv(n-1)*(trinv(n-1)-1))/2))^2)+((trinv(n-1)+1)^2));

T[n_, k_]:= (n+1)^2 + k^2; Table[T[n, k], {n,15}, {k,n}]//Flatten (* Jean-François Alcover, Mar 16 2015, after Reinhard Zumkeller *)

def A055096(n,k): return n^2 + k^2
flatten([[A055096(n,k) for k in range(1,n)] for n in range(2,16)]) # G. C. Greubel, Apr 19 2023

a(n) = sum2distinct_squares_array(n).
T(n, 1) = A002522(n).
T(n, n-1) = A001844(n-1).
T(2*n-2, n-1) = A033429(n-1).
T(n,k) = A133819(n,k) + A140978(n,k) = (n+1)^2 + k^2, 1 <= k <= n. - Reinhard Zumkeller, Mar 23 2013
T(n, k) = a*b*c/(2*sqrt(s*(s-1)*(s-b)*(s-c))) with s =(a + b + c)/2 and the substitution a = (n+1)^2 - k^2, b = 2*(n+1)*k and c = (n+1)^2 + k^2 (the circumdiameter for the considered Pythagorean triangles). - Wolfdieter Lang, Dec 03 2014
From Bob Selcoe, Mar 21 2015:  (Start)
T(n,k) = 1 + (n-k+1)^2 + Sum_{j=0..k-2} (4*j + 2*(n-k+3)).
T(n,k) = 1 + (n+k-1)^2 - Sum_{j=0..k-2} (2*(n+k-3) - 4*j).
Therefore: 4*(n-k+1) + Sum_{j=0..k-2} (2*(n-k+3) + 4*j) = 4*n(k-1) - Sum_{j=0..k-2} (2*(n+k-3) - 4*j). (End)
From G. C. Greubel, Apr 19 2023: (Start)
T(2*n-3, n-1) = A033429(n-1).
T(2*n-4, n-2) = A079273(n-1).
T(2*n-2, n) = A190816(n).
T(3*n-4, n-1) = 10*A000290(n-1) = A033583(n-1).
Sum_{k=1..n-1} T(n, k) = A331987(n-1).
Sum_{k=1..floor(n/2)} T(n-k, k) = A226141(n-1). (End)

Edited: in T(n, k) formula by Reinhard Zumkeller k < n replaced by k <= n. - Wolfdieter Lang, Dec 02 2014

Made definition more precise, changed offset to 2. - N. J. A. Sloane, Mar 30 2015

A172117 a(n) = n(n+1)(20*n-17)/6.

Original entry on oeis.org

0, 1, 23, 86, 210, 415, 721, 1148, 1716, 2445, 3355, 4466, 5798, 7371, 9205, 11320, 13736, 16473, 19551, 22990, 26810, 31031, 35673, 40756, 46300, 52325, 58851, 65898, 73486, 81635, 90365, 99696, 109648, 120241, 131495, 143430, 156066

Offset: 0

Vincenzo Librandi, Jan 26 2010

Generated by the formula n*(n+1)*(2*d*n-2*d+3)/6 for d=10.
This sequence is related to A051624 by a(n) = n*A051624(n) - Sum_{i=0..n-1} A051624(i) = n*(n+1)*(20*n-17)/2; in fact, this is the case d=10 in the identity n*(n*(d*n-d+2)/2) - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)*(2*d*n-2*d+3)/6. - Bruno Berselli, Aug 26 2010
Also, a(n) = n*A190816(n) - Sum_{i=0..n-1} A190816(i) for n>0. - Bruno Berselli, Dec 18 2013
Starting with offset 1, the sequence is the binomial transform of (1, 22, 41, 20, 0, 0, 0, ...). - Gary W. Adamson, Jul 31 2015

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93. [Bruno Berselli, Feb 13 2014]

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Bruno Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian), 2008.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A051624.

Cf. similar sequences listed in A237616.

[n*(n+1)*(20*n-17)/6: n in [0..50]]; // Vincenzo Librandi, Aug 01 2015

Table[(20n^3+3n^2-17n)/6,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,1,23,86},40] (* Harvey P. Dale, May 15 2011 *)

a(n)=n*(20*n^2+3*n-17)/6 \\ Charles R Greathouse IV, Jan 11 2012

[sum( (-1)^j*(20-j)*binomial(n+2-j, 3-j) for j in (0..1)) for n in (0..50)] # G. C. Greubel, Apr 15 2022

G.f.: x*(1+19*x)/(1-x)^4. - Bruno Berselli, Aug 26 2010
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Harvey P. Dale, May 15 2011
a(n) = Sum_{i=0..n-1} (n-i)*(20*i+1), with a(0)=0. - Bruno Berselli, Feb 11 2014
E.g.f.: (1/6)*x*(6 + 63*x + 20*x^2)*exp(x). - G. C. Greubel, Apr 15 2022

A202141 a(n) = 13n^2 - 16n + 5.

Original entry on oeis.org

5, 2, 25, 74, 149, 250, 377, 530, 709, 914, 1145, 1402, 1685, 1994, 2329, 2690, 3077, 3490, 3929, 4394, 4885, 5402, 5945, 6514, 7109, 7730, 8377, 9050, 9749, 10474, 11225, 12002, 12805, 13634, 14489, 15370, 16277, 17210, 18169, 19154, 20165, 21202, 22265

Offset: 0

Bruno Berselli, Dec 12 2011

Numbers of the form (r*n - r + 1)^2 + ((r+1)*n - r)^2; in this case, r=2.

Inverse binomial transform of this sequence: 5,-3, 26, 0, 0 (0 continued).

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

Cf. A190816 (r=1), A154355 (r=3), A161587.

Magma
```
[13*n^2-16*n+5: n in [0..42]];
```

A202141:=n->13*n^2-16*n+5: seq(A202141(n), n=0..100); # Wesley Ivan Hurt, Oct 09 2017

Table[13 n^2 - 16 n + 5, {n, 0, 42}]
LinearRecurrence[{3,-3,1},{5,2,25},50] (* Harvey P. Dale, Aug 23 2025 *)

for(n=0, 42, print1(13*n^2-16*n+5", "));

G.f.: (5 - 13*x + 34*x^2)/(1-x)^3.
a(n) = A161587(n-1) + 1 with A161587(-1) = 4.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. - Wesley Ivan Hurt, Oct 09 2017
E.g.f.: (5 - 3*x + 13*x^2)*exp(x). - Elmo R. Oliveira, Oct 20 2024

A226292 (10n^2+4n+(1-(-1)^n))/8.

Original entry on oeis.org

2, 6, 13, 22, 34, 48, 65, 84, 106, 130, 157, 186, 218, 252, 289, 328, 370, 414, 461, 510, 562, 616, 673, 732, 794, 858, 925, 994, 1066, 1140, 1217, 1296, 1378, 1462, 1549, 1638, 1730, 1824, 1921, 2020, 2122, 2226, 2333, 2442, 2554, 2668, 2785, 2904, 3026, 3150

Offset: 1

Yosu Yurramendi, Jun 02 2013

The number of binary pattern classes in the (3,n)-rectangular grid with 2 '1's and (n-2) '0's: two patterns are in same class if one of them can be obtained by a reflection or 180-degree rotation of the other, n<10.
Column k=2 of A226290.
For n even, a(n) is A202803; for n odd, a(n) is A190816.
Number of lattice points (x,y) in the region bounded by y < 3x, y > x/2 and x <= n. - Wesley Ivan Hurt, Oct 31 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A190816, A202803, A226290.

[(10*n^2+4*n+(1-(-1)^n))/8: n in [1..50]]; // Vincenzo Librandi, Sep 04 2013

A226292:=n->(10*n^2+4*n+(1-(-1)^n))/8: seq(A226292(n), n=1..50); # Wesley Ivan Hurt, Oct 31 2014

CoefficientList[Series[(2 + 2 x + x^2) / ((1 + x) (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 04 2013 *)
LinearRecurrence[{2,0,-2,1},{2,6,13,22},60] (* Harvey P. Dale, Feb 01 2019 *)

a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n>4, a(1)=2, a(2)=6, a(3)=13, a(4)=22.
a(n) = 2*a(n-2)-a(n-4)+10 for n>4, a(1)=2, a(2)=6, a(3)=13, a(4)=22.
a(n) = a(n-1)+a(n-2)-a(n-3)+5 for n>3, a(1)=2, a(2)=6, a(3)=13.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3)+(-1)^n for n>3, a(1)=2, a(2)=6, a(3)=13.
a(n) = 2*a(n-1)-a(n-2)+2+(1-(-1)^n)/2 for n>2, a(1)=2, a(2)=6.
G.f.: x*(2+2*x+x^2)/((1+x)*(1-x)^3). - Bruno Berselli, Jun 03 2013

More terms from Vincenzo Librandi, Sep 04 2013

A088307 Triangle, read by rows, T(n,k) = n^2 + k^2 if gcd(n,k)=1, otherwise 0.

Original entry on oeis.org

2, 5, 0, 10, 13, 0, 17, 0, 25, 0, 26, 29, 34, 41, 0, 37, 0, 0, 0, 61, 0, 50, 53, 58, 65, 74, 85, 0, 65, 0, 73, 0, 89, 0, 113, 0, 82, 85, 0, 97, 106, 0, 130, 145, 0, 101, 0, 109, 0, 0, 0, 149, 0, 181, 0, 122, 125, 130, 137, 146, 157, 170, 185, 202, 221, 0

Offset: 1

Reinhard Zumkeller, Nov 05 2003

(n^2-k^2, 2*k*n, T(n,k)) is a primitive Pythagorean triple iff T(n,k) > 0.

			Triangle begins:
   2;
   5,  0;
  10, 13,  0;
  17,  0, 25,  0;
  26, 29, 34, 41,  0;
  37,  0,  0,  0, 61, 0;
  ...

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A000007, A000010, A001844, A002522, A008527, A053818.

Cf. A057427, A070216, A078370, A079273, A190816.

function A088307(n,k)
  if GCD(k,n) eq 1 then return n^2+k^2;
  else return 0;
  end if; return A088307;
end function;
[A088307(n,k): k in [1..n], n in [1..13]]; // G. C. Greubel, Dec 16 2022

Table[If[CoprimeQ[n,k],n^2+k^2,0],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Jul 13 2018 *)

def A088307(n,k):
    if (gcd(n,k)==1): return n^2 + k^2
    else: return 0
flatten([[A088307(n,k) for k in range(1,n+1)] for n in range(1,14)]) # G. C. Greubel, Dec 16 2022

T(n, n) = 2*A000007(n-1).
T(n, 1) = A002522(n).
T(2*n+1, 2) = A078370(n).
Sum_{k=1..n} A057427(T(n,m)) = A000010(n).
From G. C. Greubel, Dec 15 2022: (Start)
T(n, n-1) = A001844(n).
T(n, n-2) = ((1-(-1)^n)/2) * A008527((n+1)/2).
T(2*n, n) = 5*A000007(n-1).
T(2*n+1, n) = A079273(n+1).
T(2*n-1, n) = A190816(n).
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = A053818(n+1) + [n=1]. (End)

cp's OEIS Frontend

A062786 Centered 10-gonal numbers.

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A055096 Triangle read by rows, sums of 2 distinct nonzero squares: T(n,k) = k^2+n^2, (n>=2, 1 <= k <= n-1).

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A172117 a(n) = n*(n+1)*(20*n-17)/6.

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A202141 a(n) = 13*n^2 - 16*n + 5.

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A226292 (10*n^2+4*n+(1-(-1)^n))/8.

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A088307 Triangle, read by rows, T(n,k) = n^2 + k^2 if gcd(n,k)=1, otherwise 0.

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A172117 a(n) = n(n+1)(20*n-17)/6.

A202141 a(n) = 13n^2 - 16n + 5.

A226292 (10n^2+4n+(1-(-1)^n))/8.