A082043 -id:A082043 - cp's OEIS frontend

A082044 Main diagonal of A082043: a(n) = n^4 + 2*n^2 + 1.

1, 4, 25, 100, 289, 676, 1369, 2500, 4225, 6724, 10201, 14884, 21025, 28900, 38809, 51076, 66049, 84100, 105625, 131044, 160801, 195364, 235225, 280900, 332929, 391876, 458329, 532900, 616225, 708964, 811801, 925444, 1050625, 1188100

Offset: 0

Paul Barry, Apr 03 2003

a(n) = longest side b of all integer-sided triangles with sides a <= b <= c and inradius n >= 1. Triangle has sides (n^2+2, n^4+2*n^2+1, n^4+3*n^2+1).

			G.f. = 1 + 4*x + 25*x^2 + 100*x^3 + 289*x^4 + 676*x^5 + 1369*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A002061 A002522, A059826, A082043, A082047.

See A120062 for sequences related to integer-sided triangles with integer inradius n.

[(n^2+1)^2: n in [0..40]]; // G. C. Greubel, Dec 24 2022

seq(fibonacci(3,n)^2,n=0..33); # Zerinvary Lajos, Apr 09 2008

Fibonacci[3,Range[0,40]]^2 (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,4,25,100,289},40] (* Harvey P. Dale, Feb 27 2015 *)

a(n) = n^4+2*n^2+1; \\ Michel Marcus, Jan 22 2016

[(n^2+1)^2 for n in range(41)] # G. C. Greubel, Dec 24 2022

a(n) = n^4 + 2*n^2 + 1.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Feb 27 2015
a(n) = (4*A000217(n-1)^2 + 2*A002061(n))^2 / a(n-1). - Bruce J. Nicholson, Apr 17 2017
a(n) = A002522(n)^2 = (n^2 + 1)^2 = a(-n) for all n in Z. - Michael Somos, Apr 17 2017
G.f.: (1 - x + 15*x^2 + 5*x^3 + 4*x^4) / (1 - x)^5. - Michael Somos, Apr 17 2017
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=0} 1/a(n) = Pi^2*csch(Pi)^2/4 + Pi*coth(Pi)/4 + 1/2.
Sum_{n>=0} (-1)^n/a(n) = Pi^2*csch(Pi)*coth(Pi)/4 + Pi*csch(Pi)/4 + 1/2. (End)
E.g.f.: (1 + 3*x + 9*x^2 + 6*x^3 + x^4)*exp(x). - G. C. Greubel, Dec 24 2022

A082045 Diagonal sums of number array A082043.

Original entry on oeis.org

1, 2, 6, 20, 59, 150, 336, 680, 1269, 2218, 3674, 5820, 8879, 13118, 18852, 26448, 36329, 48978, 64942, 84836, 109347, 139238, 175352, 218616, 270045, 330746, 401922, 484876, 581015, 691854, 819020, 964256, 1129425, 1316514, 1527638

Offset: 0

Paul Barry, Apr 03 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A082043, A082047.

[(n^5+10*n^3+19*n+30)/30: n in [0..40]]; // G. C. Greubel, Dec 24 2022

LinearRecurrence[{6,-15,20,-15,6,-1},{1,2,6,20,59,150},41] (* Harvey P. Dale, Mar 18 2018 *)

[(n^5+10*n^3+19*n+30)/30 for n in range(41)] # G. C. Greubel, Dec 24 2022

a(n) = (n^5+10*n^3+19*n+30)/30.
G.f.: (1-4*x+9*x^2-6*x^3+4*x^4)/(1-x)^6. - R. J. Mathar, Oct 29 2014
E.g.f.: (1/30)*(30 +30*x +45*x^2 +35*x^3 +10*x^4 +x^5)*exp(x). - G. C. Greubel, Dec 24 2022

A062938 a(n) = n(n+1)(n+2)(n+3)+1 = (n^2 + 3n + 1)^2.

Original entry on oeis.org

1, 25, 121, 361, 841, 1681, 3025, 5041, 7921, 11881, 17161, 24025, 32761, 43681, 57121, 73441, 93025, 116281, 143641, 175561, 212521, 255025, 303601, 358801, 421201, 491401, 570025, 657721, 755161, 863041, 982081, 1113025, 1256641

Offset: 0

Amarnath Murthy, Jul 05 2001

a(n) = product of first four terms of an arithmetic progression + n^4, where the first term is 1 and the common difference is n. E.g. a(1) = 1*2*3*4 +1^4 =25, a(4) = 1*5*9*13 + 4^4= 841 etc. - Amarnath Murthy, Sep 19 2003
Is it possible for one of the squares to be the sum of two or more lesser squares each used only once? - J. M. Bergot, Feb 17 2011
Yes, in fact a(1)-a(11) are examples. - Charles R Greathouse IV, Jun 28 2011
This sequence demonstrates that the product of any 4 consecutive integers plus 1 is a square.  The square roots are in A028387. - Harvey P. Dale, Oct 19 2011
The sum of three consecutive terms of the sequence is divisible by 3. The quotient is a square number: [a(n)+a(n+1)+a(n+2)]/3=(n^2+5*n+7)^2. - Carmine Suriano, Jan 23 2012
All terms end with 1 or 5. - Uri Geva, Jan 06 2024

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 19.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 85.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A028387, A052762, A082043.

[(n^2+3*n+1)^2: n in [0..50]]; // G. C. Greubel, Dec 24 2022

Table[(n^2+3*n+1)^2, {n,0,50}]
Times@@#+1&/@Partition[Range[0,50],4,1]  (* Harvey P. Dale, Apr 02 2011 *)

j=[]; for(n=0,70,j=concat(j,(n^2+3*n+1)^2)); j

{ for (n=0, 1000, write("b062938.txt", n, " ", (n^2 + 3*n + 1)^2) ) } \\ Harry J. Smith, Aug 14 2009

[(n^2+3*n+1)^2 for n in range(51)] # G. C. Greubel, Dec 24 2022

a(n+1) = numerator( ((n+2)! + (n-2)!)/n! ), for n>=2. - Artur Jasinski, Jan 09 2007; corrected by Michel Marcus, Dec 25 2022
a(n) = A028387(n)^2. - Jaroslav Krizek, Oct 31 2010
a(n) = n*(n+1)*(n+2)*(n+3)+1^4 = 1*(1+n)*(1+2*n)*(1+3*n)+n^4 =(n^2+3*n+1)^2; in general, n*(n+k)*(n+2*k)*(n+3*k)+k^4 = k*(k+n)*(k+2*n)*(k+3*n)+n^4 = (n^2+3*k*n+k^2)^2. - Charlie Marion, Jan 13 2011
G.f.: (1+20*x+6*x^2-4*x^3+x^4)/(1-x)^5. - Colin Barker, Jun 30 2012
a(n) = A052762(n+3) + 1. - Bruce J. Nicholson, Apr 22 2017
Sum_{n>=0} 1/a(n) = (Pi^2/5)*(1+t^2) - 2*sqrt(5)*Pi*t/25 - 1, where t = tan(Pi*sqrt(5)/2). - Amiram Eldar, Apr 03 2022
E.g.f.: (1 +24*x +36*x^2 +12*x^3 +x^4)*exp(x). - G. C. Greubel, Dec 24 2022

More terms from Jason Earls, Harvey P. Dale and Dean Hickerson, Jul 06 2001

A017402 a(n) = (11*n+1)^2.

Original entry on oeis.org

1, 144, 529, 1156, 2025, 3136, 4489, 6084, 7921, 10000, 12321, 14884, 17689, 20736, 24025, 27556, 31329, 35344, 39601, 44100, 48841, 53824, 59049, 64516, 70225, 76176, 82369, 88804, 95481, 102400

Offset: 0

N. J. A. Sloane

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

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m=3), A016814 (m=4), A016862 (m=5), A016922 (m=6), A016994 (m=7), A017078 (m=8), A017174 (m=9), A017282 (m=10), this sequence (m=11), A017534 (m=12), A134934 (m=14).

Cf. A017534, A082043.

[(11*n+1)^2: n in [0..50]]; // G. C. Greubel, Dec 24 2022

(11*Range[0,30]+1)^2 (* or *) LinearRecurrence[{3,-3,1},{1,144,529},30] (* Harvey P. Dale, May 05 2014 *)

a(n)=(11*n+1)^2 \\ Charles R Greathouse IV, Jun 17 2017

[(11*n+1)^2 for n in range(51)] # G. C. Greubel, Dec 24 2022

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 05 2014
From G. C. Greubel, Dec 24 2022: (Start)
G.f.: (1 + 141*x + 100*x^2)/(1-x)^3.
E.g.f.: (1 + 143*x + 121*x^2)*exp(x). (End)

A017534 a(n) = (12*n + 1)^2.

Original entry on oeis.org

1, 169, 625, 1369, 2401, 3721, 5329, 7225, 9409, 11881, 14641, 17689, 21025, 24649, 28561, 32761, 37249, 42025, 47089, 52441, 58081, 64009, 70225, 76729, 83521, 90601, 97969, 105625, 113569, 121801

Offset: 0

N. J. A. Sloane

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

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m=3), A016814 (m=4), A016862 (m=5), A016922 (m=6), A016994 (m=7), A017078 (m=8), A017174 (m=9), A017282 (m=10), A017402 (m=11), this sequence (m=12), A134934 (m=14).

Cf. A082043.

I:=[1, 169, 625]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Jul 07 2012

CoefficientList[Series[(1+166*x+121*x^2)/(1-x)^3,{x,0,50}],x] (* Vincenzo Librandi, Jul 07 2012 *)
LinearRecurrence[{3,-3,1},{1,169,625},30] (* Harvey P. Dale, Feb 27 2023 *)

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

[(12*n+1)^2 for n in range(51)] # G. C. Greubel, Dec 24 2022

G.f.: (1 + 166*x + 121*x^2 )/(1-x)^3. - R. J. Mathar, Mar 10 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 07 2012
E.g.f.: (1 + 168*x + 144*x^2)*exp(x). - G. C. Greubel, Dec 24 2022

A082105 Array A(n, k) = (kn)^2 + 4(k*n) + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 13, 13, 1, 1, 22, 33, 22, 1, 1, 33, 61, 61, 33, 1, 1, 46, 97, 118, 97, 46, 1, 1, 61, 141, 193, 193, 141, 61, 1, 1, 78, 193, 286, 321, 286, 193, 78, 1, 1, 97, 253, 397, 481, 481, 397, 253, 97, 1, 1, 118, 321, 526, 673, 726, 673, 526, 321, 118, 1

Offset: 0

Paul Barry, Apr 03 2003

			Array, A(n, k), begins as:
  1,  1,   1,   1,   1,   1, ... A000012;
  1,  6,  13,  22,  33,  46, ... A028872;
  1, 13,  33,  61,  97, 141, ... A082109;
  1, 22,  61, 118, 193, 286, ... ;
  1, 33,  97, 193, 321, 481, ... ;
  1, 46, 141, 286, 481, 726, ... ;
Triangle, T(n, k), begins as:
  1;
  1,  1;
  1,  6,   1;
  1, 13,  13,   1;
  1, 22,  33,  22,   1;
  1, 33,  61,  61,  33,   1;
  1, 46,  97, 118,  97,  46,   1;
  1, 61, 141, 193, 193, 141,  61,  1;
  1, 78, 193, 286, 321, 286, 193, 78,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A016897, A028872, A047673, A082043, A082046, A082106, A082107, A082109.

[(k*(n-k))^2 + 4*(k*(n-k)) + 1: k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 22 2022

T[n_, k_]:= (k*(n-k))^2 + 4*(k*(n-k)) + 1;
Table[T[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2022 *)

def A082105(n,k): return (k*(n-k))^2 + 4*(k*(n-k)) + 1
flatten([[A082105(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 22 2022

A(n, k) = (k*n)^2 + 4*(k*n) + 1 (square array).
A(n, n) = T(2*n, n) = A082106(n) (main diagonal).
T(n, k) = A(n-k, k) (number triangle).
Sum_{k=0..n} T(n, k) = A082107(n) (diagonal sums).
T(n, n-1) = A028872(n-1), n >= 1.
T(n, n-2) = A082109(n-2), n >= 2.
From G. C. Greubel, Dec 22 2022: (Start)
Sum_{k=0..n} (-1)^k * T(n, k) = ((1+(-1)^n)/2)*A016897(n-1).
T(2*n+1, n+1) = A047673(n+1), n >= 0.
T(n, n-k) = T(n, k). (End)

A082046 Square array, A(n, k) = (kn)^2 + 3k*n + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 11, 11, 1, 1, 19, 29, 19, 1, 1, 29, 55, 55, 29, 1, 1, 41, 89, 109, 89, 41, 1, 1, 55, 131, 181, 181, 131, 55, 1, 1, 71, 181, 271, 305, 271, 181, 71, 1, 1, 89, 239, 379, 461, 461, 379, 239, 89, 1, 1, 109, 305, 505, 649, 701, 649, 505, 305, 109, 1

Offset: 0

Paul Barry, Apr 03 2003

			Array, A(n, k), begins as:
  1,  1,   1,   1,   1,    1,    1,    1, ... A000012;
  1,  5,  11,  19,  29,   41,   55,   71, ... A028387;
  1, 11,  29,  55,  89,  131,  181,  239, ... A082108;
  1, 19,  55, 109, 181,  271,  379,  505, ... A069131;
  1, 29,  89, 181, 305,  461,  649,  869, ... ;
  1, 41, 131, 271, 461,  701,  991, 1331, ... ;
  1, 55, 181, 379, 649,  991, 1405, 1891, ... ;
  1, 71, 239, 505, 869, 1331, 1891, 2549, ... ;
Antidiagonals, T(n, k), begin as:
  1;
  1,  1;
  1,  5,   1;
  1, 11,  11,   1;
  1, 19,  29,  19,   1;
  1, 29,  55,  55,  29,   1;
  1, 41,  89, 109,  89,  41,   1;
  1, 55, 131, 181, 181, 131,  55,  1;
  1, 71, 181, 271, 305, 271, 181, 71,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A028387, A057721, A069131, A072025, A082039, A082043, A082047, A082105, A082108.

[(k*(n-k))^2 + 3*(k*(n-k)) + 1: k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 22 2022

T[n_, k_]:= (k*(n-k))^2 + 3*(k*(n-k)) + 1;
Table[T[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2022 *)

def A082046(n,k): return (k*(n-k))^2 + 3*(k*(n-k)) + 1
flatten([[A082046(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 22 2022

A(n, k) = (k*n)^2 + 3*k*n + 1 (square array).
A(k, n) = A(n, k).
A(n, n) = T(2*n, n) = A057721(n).
A(n, n+1) = A072025(n).
T(n, k) = (k*(n-k))^2 + 3*k*(n-k) + 1 (antidiagonals).
Sum_{k=0..n} T(n, k) = A082047(n) (antidiagonal sums).
From G. C. Greubel, Dec 22 2022: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*(1 + (-1)^n)*(1 - 2*n).
T(2*n+1, n-1) = T(2*n-1, n-1) = A072025(n-1). (End)

A082110 Array A(n,k) = (kn)^2 + 5(k*n) + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 15, 15, 1, 1, 25, 37, 25, 1, 1, 37, 67, 67, 37, 1, 1, 51, 105, 127, 105, 51, 1, 1, 67, 151, 205, 205, 151, 67, 1, 1, 85, 205, 301, 337, 301, 205, 85, 1, 1, 105, 267, 415, 501, 501, 415, 267, 105, 1, 1, 127, 337, 547, 697, 751, 697, 547, 337, 127, 1

Offset: 0

Paul Barry, Apr 04 2003

			Square array, A(n, k), begins as:
  1,   1,   1,   1,    1,    1,    1,    1,    1, ... A000012;
  1,   7,  15,  25,   37,   51,   67,   85,  105, ... A082111;
  1,  15,  37,  67,  105,  151,  205,  267,  337, ... A082112;
  1,  25,  67, 127,  205,  301,  415,  547,  697, ...
  1,  37, 105, 205,  337,  501,  697,  925, 1185, ...
  1,  51, 151, 301,  501,  751, 1051, 1401, 1801, ...
  1,  67, 205, 415,  697, 1051, 1477, 1975, 2545, ...
  1,  85, 267, 547,  925, 1401, 1975, 2647, 3417, ...
  1, 105, 337, 697, 1185, 1801, 2545, 3417, 4417, ...
Antidiagonals, T(n, k), begins as:
  1;
  1,  1;
  1,  7,   1;
  1, 15,  15,   1;
  1, 25,  37,  25,   1;
  1, 37,  67,  67,  37,   1;
  1, 51, 105, 127, 105,  51,   1;
  1, 67, 151, 205, 205, 151,  67,  1;
  1, 85, 205, 301, 337, 301, 205, 85,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A082039, A082043, A082046, A082105, A082111, A082112, A082113, A082114.

[(k*(n-k))^2 + 5*(k*(n-k)) + 1: k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 22 2022

T[n_, k_]:= (k*(n-k))^2 + 5*(k*(n-k)) + 1;
Table[T[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2022 *)

def A082110(n,k): return (k*(n-k))^2 + 5*(k*(n-k)) + 1
flatten([[A082110(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 22 2022

A(n, k) = (k*n)^2 + 5*(k*n) + 1 (Square array).
A(k, n) = A(n, k).
A(2, k) = A082111(k).
A(3, k) = A082112(k).
A(n, n) = T(2*n, n) = A082113(n) (main diagonal).
T(n, k) = (k*(n-k))^2 + 5*k*(n-k) + 1 (number triangle).
Sum_{k=0..n} T(n, k) = A082114(n) (diagonal sums of the array).
From G. C. Greubel, Dec 22 2022: (Start)
T(n, n-k) = T(n, k).
Sum_{k=0..n} (-1)^k*T(n, k) = (1 - 3*n)*(1 + (-1)^n)/2. (End)

A082107 Diagonal sums of number array A082105.

Original entry on oeis.org

1, 2, 8, 28, 79, 190, 406, 792, 1437, 2458, 4004, 6260, 9451, 13846, 19762, 27568, 37689, 50610, 66880, 87116, 112007, 142318, 178894, 222664, 274645, 335946, 407772, 491428, 588323, 699974, 828010, 974176, 1140337, 1328482, 1540728

Offset: 0

Paul Barry, Apr 03 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A082043, A082045, A082046, A082105, A082106, A082109, A082110.

[(n^5+20*n^3+9*n+30)/30: n in [0..50]]; // G. C. Greubel, Dec 22 2022

LinearRecurrence[{6,-15,20,-15,6,-1}, {1,2,8,28,79,190}, 51] (* G. C. Greubel, Dec 22 2022 *)

[(n^5+20*n^3+9*n+30)/30 for n in range(51)] # G. C. Greubel, Dec 22 2022

a(n) = (n^5 + 20*n^3 + 9*n + 30)/30.
G.f.: (1-4*x+11*x^2-10*x^3+6*x^4)/(1-x)^6 . - R. J. Mathar, Mar 27 2019
E.g.f.: (1/30)*(30 +30*x +75*x^2 +45*x^3 +10*x^4 +x^5)*exp(x). - G. C. Greubel, Dec 22 2022

cp's OEIS Frontend

A082044 Main diagonal of A082043: a(n) = n^4 + 2*n^2 + 1.

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A082045 Diagonal sums of number array A082043.

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A062938 a(n) = n*(n+1)*(n+2)*(n+3)+1 = (n^2 + 3*n + 1)^2.

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A017402 a(n) = (11*n+1)^2.

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A017534 a(n) = (12*n + 1)^2.

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A082105 Array A(n, k) = (k*n)^2 + 4*(k*n) + 1, read by antidiagonals.

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A082046 Square array, A(n, k) = (k*n)^2 + 3*k*n + 1, read by antidiagonals.

A062938 a(n) = n(n+1)(n+2)(n+3)+1 = (n^2 + 3n + 1)^2.

A082105 Array A(n, k) = (kn)^2 + 4(k*n) + 1, read by antidiagonals.

A082046 Square array, A(n, k) = (kn)^2 + 3k*n + 1, read by antidiagonals.

A082110 Array A(n,k) = (kn)^2 + 5(k*n) + 1, read by antidiagonals.